Instructor's Solution Manual Introduction to Electrodynamics Fourth Edition

Instructor’s Solution Manual Introduction to Electrodynamics Fourth Edition David J. Griffiths 2014

2

Contents 1 Vector Analysis

4

2 Electrostatics

26

3 Potential

53

4 Electric Fields in Matter

92

5 Magnetostatics

110

6 Magnetic Fields in Matter

133

7 Electrodynamics

145

8 Conservation Laws

168

9 Electromagnetic Waves

185

10 Potentials and Fields

210

11 Radiation

231

12 Electrodynamics and Relativity

262

c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

3

Preface Although I wrote these solutions, much of the typesetting was done by Jonah Gollub, Christopher Lee, and James Terwilliger (any mistakes are, of course, entirely their fault). Chris also did many of the figures, and I would like to thank him particularly for all his help. If you find errors, please let me know ([email protected]). David Griffiths


3

CHAPTER 1. VECTOR ANALYSIS

CHAPTER 1. VECTOR ANALYSIS 4

3 CHAPTER 1. VECTOR ANALYSIS 3


Chapter 1 Chapter 1 Chapter 1

1 VectorChapter Analysis

Vector Analysis Vector Analysis Vector Analysis Problem 1.1 ✒ ✣

Problem 1.1

}} }

✒ ✣

|C| sin θ2

C

+

C

C

✒ ✣

B

(a) From the diagram, |B1.1 + C| cos θ3 = |B| cos θ1 + |C| cos θ2 . Multiply by |A|. Problem Problem 1.1 |A||B + C| cos θ(a) |A||B| θ1 + |A||C| cos θ2θ.3 = |B| cos θ1 + |C| cos θ2 . thecos diagram, |B + C| cos 3 =From From the diagram, |B + C| cos ✓= |B| cos ✓1θ ++|C| cos ✓2cos . Multiply by |A|. 3 = |A||B + C| cos θ |A||B| cos θ2θ.2 . So:(a)A·(B + C) = A·B + A·C. (Dot product (a) From the diagram, |B |B| cos|A||C| θ1 + |C| cos 3 + C| cos θ3 = is 1 distributive)

|C| sin

B

B

C

+

C

}

C

+

|A||B + C| cos ✓3 + =A·(B |A||B| ✓=1 A·B + |A||C| cos ✓(Dot 2 . cosproduct |C| sin θ2 |A||B C| cos+ θ3cos = |A||B| cos+θ1A·C. + |A||C| θ2 . So: C) is distributive) θ2 So: A·(B C) = A·B + A·C. (Dot product is Similarly: |B ++C| sin θ = |B| sin θ + |C| sin θ Mulitply by |A| n ˆ. 2 . distributive) So: A·(B3 + C) = A·B 1+ A·C. (Dot product is distributive) θ3 ✯ θ2 θ3 = |B|sin sinθθ1n + |C| sin θ2 . Mulitply by |A| n ˆB . |A||B + C| sin θ3 n ˆSimilarly: = |A||B||B sin+θ1C|n ˆ sin + |A||C| θ3 ✯ |B| sin θ1 2 ˆ. θ2 Similarly: |B + C| sin θ = |B| sin θ + |C| sin θ . Mulitply by |A| n ˆ . Similarly: |B +|A||B C| sin+ ✓C| = |B| sin ✓ + |C| sin ✓ . Mulitply by |A| n ˆ . 3 1 2 3 sin θ3 n ˆ of =1 the |A||B| sin θ2it ˆfollows + |A||C| sin θ2 n ˆ. θ1 θ3 ✯ B 1n |B| sin ✲ If n ˆ is|A||B the unit vector pointing out $|B| sin |A||B C|the sin θunit ˆsin =vector |A||B| sin θpage, ˆ +sin |A||C| n ˆthat . θ1 3n 1n θ1 + C| sin ✓3 n ˆ+is = |A||B| ✓1 n ˆ +pointing |A||C| ✓2ofn ˆ .sin If ˆn out theθ2page, it follows that! "#θ1B $ !! "# A "# θ2 $ !✲ "# $ ✲ A×(B C)the = unit (A×B) (A×C). (Cross product is distributive) |B| cos θ1 $ ! |C|"# cos If n ˆ vector is the+unit vector out pointing out of theit page, it follows that A ! "# $ If + n ˆ is pointing of the page, follows that A×(B + C) = (A×B) + (A×C). (Cross product is distributive) |B| cos θ1 |C| A cos θ2 A×(B + C) = (A×B) + (A×C). (Cross product is distributive) |B| cos θ1 |C| cos θ2 A⇥(B + C)case, = (A⇥B) + E. (A⇥C). product is distributive) (b) For the general see G. Hay’s (Cross Vector and Tensor Analysis, Chapter 1, Section 71,(dot product) and general E. Hay’s Vector and Analysis, Tensor Analysis, Section 7 (dot (b)(b) ForFor the the general case,case, see G.see E. G. Hay’s Vector and Tensor Chapter 1,Chapter Section 7 (dot product) and product) Section (cross product) Section 8 (cross (b) For8 the general case, see G. E. product) Hay’s Vector and Tensor Analysis, Chapter 1, Section 7 (dot product) and Section 8 (cross product) Section 8 (cross product) Problem 1.2 Problem Problem 1.2 1.2 Problem 1.2 C CC ✻ TheThe triple cross-product is not isinassociative. general associative. For example, triple cross-product not in general associative. For example, The triple cross-product is not in general For example, ✻ ✻ The triple cross-product is not in general associative. For example, suppose A = B and C is perpendicular to A, as in the diagram. suppose = B and C istoperpendicular to diagram. A, as in the diagram. suppose A = B and C isAperpendicular A, as in the ✲ out-of-the-page, andasA×(B×C) points down, suppose A Then = Then B (B×C) and(B×C) C points is perpendicular to A, in theA×(B×C) diagram. ✲AA==BB ✲ A = B points out-of-the-page, and points down, Then (B×C) points A×(B×C) down, andpoints hasout-of-the-page, magnitude ABC. and But (A×B) = 0, sopoints (A×B)×C = 0 ̸= ❂ Then (B⇥C) out-of-the-page, and A⇥(B⇥C) points down, and has magnitude ABC. But (A×B) = 0, so=(A×B)×C =B×C 0 ̸= ❂ and has ABC. But (A×B) A×(B×C). ❂ ❄ A×(B×C) and magnitude has magnitude ABC. But (A⇥B)==0,0,so so (A×B)×C (A⇥B)⇥C = 00 6≠= B×C ❄ A×(B×C). A×(B×C) B×C ❄ A×(B×C). A×(B×C) A⇥(B⇥C).

}

}

z

Problem 1.3 z✻✻ Problem 1.3 1.3 Problem 1.3 Problem z✻ √ p A = +1 x ˆ + 1√ y ˆ −p1 ˆ z; A = 3; B√= 1 x ˆ +1y ˆ + 1ˆ z√ ; A =x ˆy ˆ1Aˆ ˆ z;=A 3; B ˆ y ˆ+√ + B =1y 3. ✣B 3; ˆB =1 = 1;x ˆ1Ax +=+1 1y ˆ3; ˆ zˆz;;1B 3.+ 1 ˆz; A = +1 ˆ +1 + 1x ˆ+−1 y z= ; 1A+1 ˆ1= + ˆ z B131√ = x ˆ= ˆcos 3 cos θ+⇒ θ. A·B = +1 +x − 11 y =− 1= AB cospθ = p B √ √ 11 √ √ ✣ θ ✣ ✲By & AB A·B + 11A·B 11−1 == 1%+1 = cos ✓=◦=1 3=3AB 3cos cos ✓θ⇒ ) cos ✓ 3= =cos ..θ ⇒ cos θ. 3 1 − 1 cos = A·B = +1=++1 1− = = AB θ = 3 θ cos θ 1 +cos 3 θ θ θ = cos 3 3 ≈ 70.5288 ✲y ❲ ✲y % & % & 1 θ= A ✰ cos◦−1 31 ≈ 70.5288◦ ✓ = 70.5288 −1cos1 1 3 ⇡ x ❲ θ = cos ❲ 3 ≈ 70.5288 A ✰ A ✰ x Problem 1.4 x Problem 1.4 The cross-product of any two vectors in the plane will give a vector perpendicular to the plane. For example, Problem 1.4 Problem 1.4 we might pick the base (A) and the left side (B): The cross-product of any two vectors in the plane will give a vector perpendicular to the plane. For example,

The cross-product The of any two vectorsofin the plane willingive a vector to the plane. example, cross-product any two vectors the plane willperpendicular give a vector perpendicular to For the plane. For exam we might pick the base (A) and the left side (B): c base we Pearson might pick base (A) and the NJ. leftAllside (B): ⃝2005 Upper Saddle River, rights reserved. This material is we might pick the (A)Education, and the theInc., left side (B): A=

protected under all copyright laws as they currently exist. No portion of this material may be

1x ˆreproduced, + 2y ˆ +in0 any ˆ z; B =or by 1x âny+means, 0y ˆ +without 3ˆ z. permission in writing from the publisher. form

A = −1 x ˆ +2y ˆ⃝2005 + 0ˆ zPearson ; B = Education, −1 ˆ x + 0Inc., y ˆ +Upper 3ˆ z. Saddle River, NJ. All rights reserved. c

This material is protected under all copyright laws as they currently exist. No portion of this material may be c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is reproduced, in any form or by any means, without permission in writing from the publisher. protected under copyright laws as they currently exist. c ⃝2005 Pearson Education, Inc., Upper Saddle River, NJ.all All rights reserved. This material is No portion of this material may be in anyNo form or by any means, without permission in writing from the publisher. protected under all copyright laws as theyreproduced, currently exist. portion of this material may be

reproduced, in any form or by any means, without permission in writing from the publisher.

5


x ˆ y ˆˆ z 1 2 0 = 6x ˆ + 3y ˆ + 2ˆ z. 1 0 3 This has the right direction, but the wrong magnitude. To make a unit vector out of it, simply divide by its length: p A⇥B 6 |A⇥B| = 36 + 9 + 4 = 7. n ˆ = |A ˆ + 37 y ˆ + 27 ˆ z. ⇥B| = 7 x A⇥B =

Problem 1.5

A⇥(B⇥C) =

(By Cz

x ˆ Ax

Bz Cy ) (Bz Cx

y ˆ Ay

Bx Cz ) (Bx Cy

ˆ z Az

By Cx )

=x ˆ[Ay (Bx Cy By Cx ) Az (Bz Cx Bx Cz )] + y ˆ() + ˆ z() (I’ll just check the x-component; the others go the same way) =x ˆ(Ay Bx Cy Ay By Cx Az Bz Cx + Az Bx Cz ) + y ˆ() + ˆ z(). C(A·B) = [Bx (Ax Cx + Ay Cy + Az Cz ) Cx (Ax Bx + Ay By + Az Bz )] x ˆ + () y ˆ + () ˆ z =x ˆ(Ay Bx Cy + Az Bx Cz Ay By Cx Az Bz Cx ) + y ˆ() + ˆ z(). They agree.

B(A·C) Problem 1.6

A⇥(B⇥C)+B⇥(C⇥A)+C⇥(A⇥B) = B(A·C) C(A·B)+C(A·B) A(C·B)+A(B·C) B(C·A) = 0. So: A⇥(B⇥C) (A⇥B)⇥C = B⇥(C⇥A) = A(B·C) C(A·B). If this is zero, then either A is parallel to C (including the case in which they point in opposite directions, or one is zero), or else B·C = B·A = 0, in which case B is perpendicular to A and C (including the case B = 0.) Conclusion: A⇥(B⇥C) = (A⇥B)⇥C () either A is parallel to C, or B is perpendicular to A and C. Problem 1.7

r

r

= (4 x ˆ + 6y ˆ + 8ˆ z) p = 4+4+1= 3

rˆ

=

r r

=

2 ˆ 3x

2 ˆ 3y

(2 x ˆ + 8y ˆ + 7ˆ z) = 2 x ˆ

2y ˆ+ ˆ z

+ 13 ˆ z

Problem 1.8 ¯y + A¯z B ¯z = (cos Ay + sin Az )(cos By + sin Bz ) + ( sin Ay + cos Az )( sin By + cos Bz ) (a) A¯y B = cos2 Ay By + sin cos (Ay Bz + Az By ) + sin2 Az Bz + sin2 Ay By sin cos (Ay Bz + Az By ) + cos2 Az Bz = (cos2 + sin2 )Ay By + (sin2 + cos2 )Az Bz = Ay By + Az Bz . X (b) (Ax )2 + (Ay )2 + (Az )2 = ⌃3i=1 Ai Ai = ⌃3i=1 ⌃3j=1 Rij Aj ⌃3k=1 Rik Ak = ⌃j,k (⌃i Rij Rik ) Aj Ak . ⇢ 1 if j = k This equals A2x + A2y + A2z provided ⌃3i=1 Rij Rik = 0 if j 6= k Moreover, if R is to preserve lengths for all vectors A, then this condition is not only sufficient but also necessary. For suppose A = (1, 0, 0). Then ⌃j,k (⌃i Rij Rik ) Aj Ak = ⌃i Ri1 Ri1 , and this must equal 1 (since we 2 2 2 want Ax +Ay +Az = 1). Likewise, ⌃3i=1 Ri2 Ri2 = ⌃3i=1 Ri3 Ri3 = 1. To check the case j 6= k, choose A = (1, 1, 0). Then we want 2 = ⌃j,k (⌃i Rij Rik ) Aj Ak = ⌃i Ri1 Ri1 + ⌃i Ri2 Ri2 + ⌃i Ri1 Ri2 + ⌃i Ri2 Ri1 . But we already know that the first two sums are both 1; the third and fourth are equal, so ⌃i Ri1 Ri2 = ⌃i Ri2 Ri1 = 0, and so ˜ = 1, where R ˜ is the transpose of R. on for other unequal combinations of j, k. X In matrix notation: RR c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

6


CHAPTER 1. VECTOR ANALYSIS 5 CHAPTER 1. VECTOR ANALYSIS 5 Problem 1.9 y✻ y z′ ✻ ✻ y✻ y ❃ z′ ✻ ✻ ✿ ✿ ❃ down the axis: ✲ x LookingLooking down the axis: ✒ ❄ ✲x Looking down the axis: ✒ ❄ ′ ■ &y ′ ■z ✰ & ✠ y ✰ & z ′ ✰ z x ✠ x &x ✰′ z x

A 120 rotation carries the z axis into the y (= z) axis, y into x (= y), and x into z (= x). So Ax = Az , Ay = Ax , Az = Ay . 0 1 0 01 R = @1 0 0A 0 10 Problem 1.10

(a) No change. (Ax = Ax , Ay = Ay , Az = Az ) (b) A !

A, in the sense (Ax =

Ax , Ay =

Ay , Az =

Az )

(c) (A⇥B) ! ( A)⇥( B) = (A⇥B). That is, if C = A⇥B, C ! C . No minus sign, in contrast to behavior of an “ordinary” vector, as given by (b). If A and B are pseudovectors, then (A⇥B) ! (A)⇥(B) = (A⇥B). So the cross-product of two pseudovectors is again a pseudovector. In the cross-product of a vector and a pseudovector, one changes sign, the other doesn’t, and therefore the cross-product is itself a vector. Angular momentum (L = r⇥p) and torque (N = r⇥F) are pseudovectors. (d) A·(B⇥C) ! ( A)·(( B)⇥( C)) = A·(B⇥C). So, if a = A·(B⇥C), then a ! changes sign under inversion of coordinates. Problem 1.11

a; a pseudoscalar

(a)rf = 2x x ˆ + 3y 2 y ˆ + 4z 3 ˆ z (b)rf = 2xy 3 z 4 x ˆ + 3x2 y 2 z 4 y ˆ + 4x2 y 3 z 3 ˆ z (c)rf = ex sin y ln z x ˆ + ex cos y ln z y ˆ + ex sin y(1/z) ˆ z Problem 1.12 (a) rh = 10[(2y 6x 18) x ˆ + (2x 8y + 28) y ˆ]. rh = 0 at summit, so 2y 6x 18 = 0 2y 18 24y + 84 = 0. 2x 8y + 28 = 0 =) 6x 24y + 84 = 0 22y = 66 =) y = 3 =) 2x 24 + 28 = 0 =) x = 2. Top is 3 miles north, 2 miles west, of South Hadley. (b) Putting in x = 2, y = 3: h = 10( 12 12 36 + 36 + 84 + 12) = 720 ft. (c) Putting in x = 1, y = 1: rh = 10[(2 6 18) x ˆ + (2 p |rh| = 220 2 ⇡ 311 ft/mile; direction: northwest.

8 + 28) y ˆ] = 10( 22 x ˆ + 22 y ˆ) = 220( x ˆ+y ˆ).

c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be in any form or All by rights any means, without c ⃝2005 Pearson Education, Inc.,reproduced, Upper Saddle River, NJ. reserved. Thispermission material isin writing from the publisher. c ⃝2005 Pearson Education, Inc., Upper Saddle NJ. All rights reserved. This material is protected under all copyright laws River, as they currently exist. No portion of this material may be protected under all copyright laws as they exist. No portion of this may the be publisher. reproduced, in any form or bycurrently any means, without permission in material writing from reproduced, in any form or by any means, without permission in writing from the publisher.

7

CHAPTER 1. VECTOR ANALYSIS Problem 1.13

r

= (x

x0 ) x ˆ + (y

y0 ) y ˆ + (z

r

z0) ˆ z;

=

p (x

x0 )2 + (y

y 0 )2 + (z

z 0 )2 .

(a) r( r 2 ) =

@ @x [(x

@ @ x0 )2 +(y y 0 )2 +(z z 0 )2 ] x ˆ + @y () y ˆ + @z () ˆ z = 2(x x0 ) x ˆ +2(y y 0 ) y ˆ +2(z z 0 ) ˆ z=2

(b) r( r1 ) =

@ @x [(x

x0 )2 + (y

(c)

1 2 ()

= =

()

@ @x (

r

n

3 2

3 2

2(x x ) x ˆ 0 [(x x ) x ˆ + (y 0

)=n

r

r

n 1@ @x

1 2 ()

y 0 )2 + (z 3 2

z 0 )2 ] 1 2 ()

2(y y ) y ˆ y )y ˆ + (z z 0 ) ˆ z] = 0

0

=n

r

r 2r

n 1 1 1 (2

x)

1 2

x ˆ+

3 2

@ @y () 0

1 2

y ˆ+

2(z z ) ˆ z 3 (1/ r ) r =

=n

r

n 1

1 2

.

ˆ z

(1/ r 2 ) rˆ .

rˆ x , so

Problem 1.14

@ @z ()

r

r( r n ) = n

r

n 1

rˆ

y = +y cos + z sin ; multiply by sin : y sin = +y sin cos + z sin2 . z = y sin + z cos ; multiply by cos : z cos = y sin cos + z cos2 . Add: y sin + z cos = z(sin2 + cos2 ) = z. Likewise, y cos z sin = y. @y @y @z @z So @y = cos ; @z = sin ; @y = sin ; @z = cos . Therefore ) @f @y @f @z (rf )y = @f @y = @y @y + @z @y = + cos (rf )y + sin (rf )z So rf transforms as a vector. @f @y @f @z (rf )z = @f sin (rf )y + cos (rf )z @z = @y @z + @z @z =

qed

Problem 1.15 (a)r·va =

@ 2 @x (x )

+

@ 2 @y (3xz )

(b)r·vb =

@ @x (xy)

+

@ @y (2yz)

(c)r·vc =

@ 2 @x (y )

+

@ @y (2xy

Problem 1.16

+

+

@ @z (

2xz) = 2x + 0

@ @z (3xz)

+ z2) +

2x = 0.

= y + 2z + 3x.

@ @z (2yz)

= 0 + (2x) + (2y) = 2(x + y)

h @ @ @ @ r·v = @x ( rx3 ) + @y ( ry3 ) + @z ( rz3 ) = @x x(x2 + y 2 + z 2 ) h i h i 3 3 @ @ + @y y(x2 + y 2 + z 2 ) 2 + @z z(x2 + y 2 + z 2 ) 2 3

5

= () 2 + x( 3/2)() 2 2x + () 5 + z( 3/2)() 2 2z = 3r 3 3r

3 2

5

3 2

i

3

+ y( 3/2)() 2 2y + () 2 5 2 (x + y 2 + z 2 ) = 3r 3 3r

3

= 0.

This conclusion is surprising, because, from the diagram, this vector field is obviously diverging away from the origin. How, then, can r·v = 0? The answer is that r·v = 0 everywhere except at the origin, but at the origin our calculation is no good, since r = 0, and the expression for v blows up. In fact, r·v is infinite at that one point, and zero elsewhere, as we shall see in Sect. 1.5. Problem 1.17 v y = cos vy + sin vz ; v z = ⇣sin vy + cos v⌘z . ⇣ ⌘ @v y @vy @vy @y @vy @z @vz @vz @z z @y cos + @v sin . Use result in Prob. 1.14: @y = @y cos + @y sin = @y @y + @z @y @y @y + @z @y ⇣ ⌘ ⇣ ⌘ @vy @vy @vz @vz = @y cos + @z sin cos + @y cos + @z sin sin . ⇣ ⌘ ⇣ ⌘ @vy @vy @y @vy @z @v z @vz @vz @y @vz @z = sin + cos = + sin + + cos @z @z @y @z @z @z ⇣@z ⌘ @y @z ⇣ @z @z ⌘ @vy @vy @vz @vz = sin + cos . So @y sin + @z cos @y sin + @z cos c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

8 @v y @y

+

@v z @z

=

@vy @y

cos2

@vy @z

+


sin cos +

sin cos +

@vz @y

=

@vy @y

cos

+ sin

+

2

2

@vz @y

sin cos + @vz @z

cos

sin2

@vz @z

@vz @z

sin2

+

@vy @y

@vy @z

sin2

sin cos

2

+ cos2

=

@vy @y

+

@vz @z .

X

Problem 1.18 (a) r⇥va =

x ˆ

y ˆ

@ @x 2

@ @y

x ˆ

y ˆ

ˆ z

@ @x

@ @y

@ @z

x 3xz

(b) r⇥vb =

ˆ z @ @z

=x ˆ(0

6xz) + y ˆ(0 + 2z) + ˆ z(3z 2

0) =

6xz x ˆ + 2z y ˆ + 3z 2 ˆ z.

2xz

2

=x ˆ(0

2y) + y ˆ(0

3z) + ˆ z(0

x) =

2y x ˆ

3z y ˆ

xˆ z.

xy 2yz 3xz

(c) r⇥vc =

x ˆ

y ˆ

@ @x 2

@ @y

ˆ z @ @z

=x ˆ(2z

2z) + y ˆ(0

0) + ˆ z(2y

2y) = 0.

y (2xy + z ) 2yz 2

Problem 1.19 As we go from point A to point B (9 o’clock to 10 o’clock), x increases, y increases, vx increases, and vy decreases, so @vx /@y > 0, while @vy /@y < 0. On the circle, vz = 0, and there is no dependence on z, so Eq. 1.41 says ✓ ◆ @vy @vx r⇥v =ˆ z @x @y

y v v

v

B

A

x

points in the negative z direction (into the page), as the right hand rule would suggest. (Pick any other nearby points on the circle and you will come to the same conclusion.) [I’m sorry, but I cannot remember who suggested this cute illustration.]

z v Problem 1.20

v = yx ˆ + xy ˆ; or v = yz x ˆ + xz y ˆ + xy ˆ z; or v = (3x2 z or v = (sin x)(cosh y) x ˆ (cos x)(sinh y) y ˆ; etc. Problem 1.21 (i) r(f g) =

@(f g) @x

=f

⇣

x ˆ+

@g @x

(iv) r·(A⇥B) =

= =

x ˆ+

y ˆ+

(Ay Bz

Az By ) +

Ay

(v) r⇥ (f A) =

⇣

@(f g) @z

@ @ Ax Bz ) + @z (Ax By @y (Az Bx @A @B @Bz @Az @Bx y y z Ay @x + Bz @x Az @x By @x + Az @y + Bx @A @y @By @A x x +Ax @z + By @A Ay @B Bx @zy @z @z ⇣ ⌘ ⇣ ⌘ @A @A @Az @Ax y z x Bx @A + By @A + Bz @xy @y @z @z @x @y @ @x

@(f Az ) @y

3xz 2 ) ˆ z;

⇣ ⌘ ⇣ ⌘ ⇣ ⌘ @g @g @f @g @f ˆ z = f @x + g @f x ˆ + f + g y ˆ + f + g ˆ z @x @y @z @z ⌘ ⇣ ⌘ @y y ˆ + @g z + g @f ˆ + @f ˆ + @f z = f (rg) + g(rf ). qed @z ˆ @x x @y y @z ˆ

@(f g) @y @g @y

z3) x ˆ + 3y ˆ + (x3

@Bx @z

@(f Ay ) @z

@Bz @x

⌘

x ˆ+

⇣

Az

⇣

@(f Ax ) @z

@By @x

⌘

Ay Bx )

z Ax @B @y

Ax

⇣

@Bz @y

x Bz @A @y

@By @z

= B· (r⇥A) A· (r⇥B). ⌘ ⇣ ⌘ @(f Ay ) @(f Az ) @(f Ax ) y ˆ + ˆ z @x @x @y @Bx @y

⌘

qed


9


⇣ ⌘ ⇣ ⌘ @A @f @f z x z = f @A f @zy Ay @f ˆ + f @A f @A Az @f ˆ @y + Az @y @z x @z + Ax @z @x @x y ⇣ ⌘ @Ay @f @f @Ax + f @x + Ay @x f @y Ax @y ˆ z h⇣ ⌘ ⇣ ⌘ i @Ay @Ay @Ax @Az @Ax z = f @A x ˆ + y ˆ + ˆ z @y @z @x h⇣ @z ⌘ @x⇣ ⌘@y ⇣ ⌘ i @f @f @f @f @f Ay @f A x ˆ + A A y ˆ + A A z z x z x y @z @y @x @z @y @x ˆ = f (r⇥A) A⇥ (rf ). qed Problem 1.22

⇣ ⌘ ⇣ ⌘ @By @By @By @Bx @Bx x (a) (A·r) B = Ax @B + A + A x ˆ + A + A + A y ˆ y z x y z @x @y @z @y @z ⇣ ⌘@x @Bz @Bz @Bz + Ax @x + Ay @y + Az @z ˆ z. xx ˆ+y y ˆ+z ˆ z =p . Let’s just do the x component. x2 +y 2 +z 2 ⇣ ⌘ @ @ @ p 2 x2 2 [(ˆ r·r)ˆ r]x = p1 x @x + y @y + z @z x +y +z n h i h i h 1 1 p 3 2y + zx = 1r x p1 + x( 12 ) (p1 )3 2x + yx 2( )

(b) ˆ r=

r r

=

1 r

x r

1 r3

x3 + xy 2 + xz 2

=

1 r

x r

x r3

1 1 p 3 2z 2( )

x2 + y 2 + z 2

=

1 r

x r

io

x r

= 0.

Same goes for the other components. Hence: (ˆ r·r) ˆ r=0. ⇣ ⌘ @ @ @ (c) (va ·r) vb = x2 @x + 3xz 2 @y 2xz @z (xy x ˆ + 2yz y ˆ + 3xz ˆ z)

= x2 (y x ˆ + 0y ˆ + 3z ˆ z) + 3xz 2 (x x ˆ + 2z y ˆ + 0ˆ z) 2xz (0 x ˆ + 2y y ˆ + 3x ˆ z) 2 2 2 3 2 2 = x y + 3x z x ˆ + 6xz 4xyz y ˆ + 3x z 6x z ˆ z = x2 y + 3z 2 x ˆ + 2xz 3z 2

2y y ˆ

3x2 z ˆ z

Problem 1.23 (ii) [r(A·B)]x =

@ @x (Ax Bx

+ Ay By + Az Bz ) =

[A⇥(r⇥B)]x = Ay (r⇥B)z

@Ax @x Bx

Az (r⇥B)y = Ay

x + Ax @B @x +

@By @x

@Bx @y

@Ay @x By

+ Ay

@Bx @z

Az

@By @x

+

@Az @x Bz

z + Az @B @x

@Bz @x

@A

@Ax @Az x [B⇥(r⇥A)]x = By @xy Bz @A @y @z @x @Bx @Bx @ @ @ x [(A·r)B]x = Ax @x + Ay @y + Az @z Bx = Ax @B @x + Ay @y + Az @z @Ax @Ax @Ax [(B·r)A]x = Bx @x + By @y + Bz @z

So [A⇥(r⇥B) + B⇥(r⇥A) + (A·r)B + (B·r)A]x @B @Ay @Bz @Az x x x x = Ay @xy Ay @B Az @B By @A Bz @A @y @z + Az @x + By @x @y @z + Bz @x @Bx @Bx @Ax @Ax @Ax x +Ax @B @x + Ay @y + Az @z + Bx @x + By @y + Bz @z @Bx x = Bx @A @x + Ax @x + By

/

/

/

@Ax @y

+Bz + + Az = [r(A·B)]x (same for y and z) @Ax @z

z + @A @x

@Ay @x

(vi) [r⇥(A⇥B)]x = =

@Ax @z

/ + Ay @B@x @B / + @B@x + @B@z/ @z

x + @A @y x

y

z

/

@Bx @y

/

x + @B @y

x

@ @ @ Ay Bx ) @y (A⇥B)z @z (A⇥B)y = @y (Ax By @By @Ay @Ax @Bx @Az Ay @y @y By + Ax @y @y Bx @z Bx

[(B·r)A (A·r)B + A(r·B) B(r·A)]x @Ax @Ax x x x = Bx @A Ax @B Ay @B @x + By @y + Bz @z @x @y

x Az @B @z + Ax

@Bx @x

@ Ax Bz ) @z (Az Bx @Bx @Ax z Az @z + @z Bz + Ax @B @z

+

@By @y


z + @B @z

Bx

@Ax @x

+

@Ay @y

z + @A @z

10 x = By @A @y + Ax

/

@Bx @x

/

x + @B @x +

CHAPTER 1. VECTOR ANALYSIS @By @y

+

@Bz @z x Bz @A @z

+ Bx

/

/

@Ax @x

@Ay @y

@Ax @x

@Az @z

x x + Ay @B + Az @B + @y @z = [r⇥(A⇥B)]x (same for y and z)

Problem 1.24 r(f /g) =

@ @ @ ˆ + @y (f /g) y ˆ + @z (f /g) ˆ z @x (f /g) x @f @g @f @g @g g @y f @y g @x f @x g @f f x ˆ+ y ˆ + @z g2 @z ˆ z g2 hg2 ⇣ ⌘ ⇣ @f @f @f @g @g 1 ˆ + @y y ˆ + @z ˆ z f @x x ˆ + @y y ˆ g 2 g @x x

= =

r·(A/g) = = =

+

@ @ @ @x (Ax /g) + @y (Ay /g) + @z (Az /g) @A y @g @g x g Ay g @A Ax @x g @Az A @g @x + @y g2 @y + @z g2 z @x h g⇣2 ⌘ ⇣ @Ay @g @g @Ax @Az 1 Ax @x + Ay @y g2 g @x + @y + @z

[r⇥(A/g)]x = = = =

@ @ @y (Az /g) @z (Ay /g) @Az @A @g g @y Az @y g @zy Ay @g @z h g⇣2 ⌘g2 ⇣ @Ay @g @Az 1 Az @y g2 g @y @z g(r⇥A)x +(A⇥rg)x (same for g2

Ay @g @z

⌘i

@g z @z ˆ

=

+ Az @g @z

grf f rg . g2

⌘i

=

qed

gr·A A·rg . g2

qed

⌘i

y and z). qed

Problem 1.25 (a) A⇥B =

x ˆ y ˆ ˆ z x 2y 3z = x ˆ(6xz) + y ˆ(9zy) + ˆ z( 2x2 3y 2x 0

@ @ r·(A⇥B) = @x (6xz) + @y (9zy) + ⇣ ⌘ @ @ r⇥A = x ˆ @y (3z) @z (2y) + y ˆ ⇣ ⌘ @ @ r⇥B = x ˆ @y (0) @z ( 2x) + y ˆ ?

r·(A⇥B) = B·(r⇥A)

(b) A·B = 3xy

4xy =

6y 2 )

@ 2 6y 2 ) @z ( 2x @ @ @z (x) @x (3z) @ @z (3y)

A·(r⇥B) = 0

= 6z + 9z + 0 = 15z ⇣ ⌘ @ @ +ˆ z @x (2y) @y (x) = 0; B·(r⇥A) = 0 ⇣ ⌘ @ @ @ (0) + ˆ z ( 2x) (3y) = 5ˆ z; A·(r⇥B) = @x @x @y

15z

( 15z) = 15z. X

@ @ xy ; r(A·B) = r( xy) = x ˆ @x ( xy) + y ˆ @y ( xy) =

yx ˆ

xy ˆ

x ˆ y ˆ ˆ z A⇥(r⇥B) = x 2y 3z = x ˆ( 10y) + y ˆ(5x); B⇥(r⇥A) = 0 0 0 5 ⇣ ⌘ @ @ @ (A·r)B = x @x + 2y @y + 3z @z (3y x ˆ 2x y ˆ) = x ˆ(6y) + y ˆ( 2x) ⇣ ⌘ @ @ (B·r)A = 3y @x 2x @y (x x ˆ + 2y y ˆ + 3z ˆ z) = x ˆ(3y) + y ˆ( 4x)

A⇥(r⇥B) + B⇥(r⇥A) + (A·r)B + (B·r)A = 10y x ˆ + 5x y ˆ + 6y x ˆ 2x y ˆ + 3y x ˆ 4x y ˆ = yx ˆ xy ˆ = r·(A·B). X ⇣ ⌘ ⇣ @ @ @ @ @ (c) r⇥(A⇥B) = x ˆ @y ( 2x2 6y 2 ) @z (9zy) + y ˆ @z (6xz) @x ( 2x2 6y 2 ) + ˆ z @x (9zy) =x ˆ( 12y 9y) + y ˆ(6x + 4x) + ˆ z(0) = 21y x ˆ + 10x y ˆ r·A =

@ @x (x)

+

@ @y (2y)

+

@ @z (3z)

= 1 + 2 + 3 = 6; r·B =

@ @x (3y)

+

@ @y (

⌘

@ @y (6xz)

2x) = 0


11

CHAPTER 1. VECTOR ANALYSIS (B·r)A (A·r)B + A(r·B) = r⇥(A⇥B). X

B(r·A) = 3y x ˆ

4x y ˆ

6y x ˆ + 2x y ˆ

18y x ˆ + 12x y ˆ=

21y x ˆ + 10x y ˆ

Problem 1.26 (a)

@ 2 Ta @x2

= 2;

(b)

@ 2 Tb @x2

=

(c)

@ 2 Tc @x2

= 25Tc ; @@yT2c =

(d)

@ 2 vx @x2 @ 2 vy @x2 @ 2 vz @x2

@ 2 Ta @y 2

@ 2 Tb @y 2

=

=

@ 2 Ta @z 2

@ 2 Tb @z 2

= 0 ) r2 Ta = 2.

=

Tb ) r2 Tb =

2

3Tb =

3 sin x sin y sin z.

2

9Tc ) r2 Tc = 0. 9 2 2 = 2 ; @@yv2x = @@zv2x = 0 ) r2 vx = 2 > = @2v @2v r2 v = 2 x ˆ + 6x y ˆ. = @y2y = 0 ; @z2y = 6x ) r2 vy = 6x > ; @ 2 vz @ 2 vz 2 = @y2 = @z2 = 0 ) r vz = 0

Problem 1.27

r·(r⇥v) = ⇣ 2 @ vz = @x @y

@ @x

⇣

16Tc ; @@zT2c =

@vy @z

@vz @y

@ 2 vz @y @x

⌘

+

From Prob. 1.18: r⇥va = Problem 1.28

⇣

⌘

@ 2 vx @y @z

@ @ @ @x @y @z @t @t @t @x @y @z

@vx @z

@ @y

@ 2 vx @z @y

⌘

⇣

@vy @vz @ @x + @z @x ⇣ 2 ⌘ @ vy @ 2 vy + @z @x @x @z

@vx @y

=x ˆ

@2t @y @z

@2t @z @y

+y ˆ

@2t @z @x

⌘

= 0, by equality of cross-derivatives.

6xz x ˆ+2z y ˆ+3z 2 ˆ z ) r·(r⇥va ) =

x ˆ y ˆ ˆ z r⇥(rt) =

+

@2t @x @z

@ @x (

@ @ 6xz)+ @y (2z)+ @z (3z 2 ) =

+ˆ z

@2t @x @y

6z+6z = 0.

@2t @y @x

= 0, by equality of cross-derivatives. In Prob. 1.11(b), rf = 2xy 3 z 4 x ˆ + 3x2 y 2 z 4 y ˆ + 4x2 y 3 z 3 ˆ z, so x ˆ y ˆ ˆ z @ @ @ r⇥(rf ) = @x @y @z 3 4 2 2 4 2 3 3 2xy z 3x y z 4x y z =x ˆ(3 · 4x2 y 2 z 3 4 · 3x2 y 2 z 3 ) + y ˆ(4 · 2xy 3 z 3 2 · 4xy 3 z 3 ) + ˆ z(2 · 3xy 2 z 4

Problem 1.29 (a) (0, 0, 0) (1, 0, 0) (1, 1, 0) R Total: (b) (0, 0, 0) (0, 0, 1) (0, 1, 1) R Total:

3 · 2xy 2 z 4 ) = 0. X

R R1 ! (1, 0, 0). x : 0 ! 1, y = z = 0; dl = dx x ˆ; v · dl = x2 dx; v · dl R= 0 x2 dx = (x3 /3)|10 = 1/3. ! (1, 1, 0). x = 1, y : 0 ! 1, z = 0; dl = dy y ˆ; v · dl = 2yz dy = 0; v · dl = 0. R R1 ! (1, 1, 1). x = y = 1, z : 0 ! 1; dl = dz ˆ z; v · dl = y 2 dz = dz; v · dl = 0 dz = z|10 = 1.

v · dl = (1/3) + 0 + 1 = 4/3.

R ! (0, 0, 1). x = y = 0, z : 0 ! 1; dl = dz ˆ z; v · dl = y 2 dz = 0; v · dl = 0. R R1 ! (0, 1, 1). x = 0, y : 0 ! 1, z = 1; dl = dy y ˆ; v · dl = 2yz dy = 2y dy; v · dl = 0 2y dy = y 2 |10 = 1. R R1 ! (1, 1, 1). x : 0 ! 1, y = z = 1; dl = dx x ˆ; v · dl = x2 dx; v · dl = 0 x2 dx = (x3 /3)|10 = 1/3.

v · dl = 0 + 1 + (1/3) = 4/3.

(c) x = y = z : 0 ! 1; dx = dy = dz; v · dl = x2 dx + 2yz dy + y 2 dz = x2 dx + 2x2 dx + x2 dx = 4x2 dx; R R1 v · dl = 0 4x2 dx = (4x3 /3)|10 = 4/3. H (d) v · dl = (4/3) (4/3) = 0. c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

12


Problem 1.30 x, y : 0 ! 1, z = 0; da = dx dy ˆ z; v · da = y(z 2

3) dx dy =

2

R 3y dx dy; v · da =

3

R2 0

dx

R2 0

y dy =

3(x|20 )( y2 |20 ) = 3(2)(2) = 12. In Ex. 1.7 we got 20, for the same boundary line (the square in the xy-plane), so the answer is no: the surface integral does not depend only on the boundary line. The total flux for the cube is 20 + 12 = 32. Problem 1.31 R R T d⌧ = z 2 dx dy dz. You can do the integrals in any order—here it is simplest to save z for last: Z ✓Z ◆ Z z2 dx dy dz.

R (1 y z) The sloping surface is x+y +z = 1, so the x integral is 0 dx = 1 y z. For a given z, y ranges from 0 to R (1 z) (1 z) 1 z, so the y integral is 0 (1 y z) dy = [(1 z)y (y 2 /2)]|0 = (1 z)2 [(1 z)2 /2] = (1 z)2 /2 = R R 1 z2 2 4 3 1 2 1 z z4 z5 1 2 (1/2) z + (z /2). Finally, the z integral is 0 z ( 2 z + 2 ) dz = 0 ( 2 z 3 + z2 ) dz = ( z6 4 + 10 )|0 = 1 6

1 4

+

1 10

= 1/60.

Problem 1.32 T (b) = 1 + 4 + 2 = 7; T (a) = 0. ) T (b)

T (a) = 7.

rT = (2x + 4y)ˆ x + (4x + 2z 3 )ˆ y + (6yz 2 )ˆ z; rT ·dl = (2x + 4y)dx + (4x + 2z 3 )dy + (6yz 2 )dz 9 R R1 1 > (a) Segment 1: x : 0 ! 1, y = z = dy = dz = 0. rT ·dl = 0 (2x) dx = x2 0 = 1. =R R R1 b 1 rT ·dl = 7. X Segment 2: y : 0 ! 1, x = 1, z = 0, dx = dz = 0. rT ·dl = 0 (4) dy = 4y|0 = 4. a > R R1 2 1 ; 3 Segment 3: z : 0 ! 1, x = y = 1, dx = dy = 0. rT ·dl = 0 (6z ) dz = 2z 0 = 2. 9 R R1 > (b) Segment 1: z : 0 ! 1, x = y = dx = dy = 0. rT ·dl = 0 (0) dz = 0. > R R1 > 1 Segment 2: y : 0 ! 1, x = 0, z = 1, dx = dz = 0. rT ·dl = 0 (2) dy = 2y|0 = 2. = R b R R1 rT ·dl = 7. X a > Segment 3: x : 0 ! 1, y = z = 1, dy = dz = 0. rT ·dl = 0 (2x + 4) dx > > 1 ; = (x2 + 4x) 0 = 1 + 4 = 5. (c) x : 0 ! 1, y = x, z = x2 , dy = dx, dz = 2x dx.

rT ·dl = (2x + 4x)dx + (4x + 2x6 )dx + (6xx4 )2x dx = (10x + 14x6 )dx. Rb R1 1 rT ·dl = 0 (10x + 14x6 )dx = (5x2 + 2x7 ) 0 = 5 + 2 = 7. X a

Problem 1.33

r·v = y + 2z + 3x o R R RR nR 2 (r·v)d⌧ = (y + 2z + 3x) dx dy dz = (y + 2z + 3x) dx dy dz 0 ⇥ ⇤2 (y + 2z)x + 32 x2 0 = 2(y + 2z) + 6 n o R R2 = (2y + 4z + 6)dy dz 0 ⇥ 2 ⇤2 y + (4z + 6)y 0 = 4 + 2(4z + 6) = 8z + 16

,!

,!

=

R2 0

(8z + 16)dz = (4z 2 + 16z)

2 0

= 16 + 32 = 48.

Numbering the surfaces as in Fig. 1.29: c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

13

CHAPTER 1. VECTOR ANALYSIS R RR 2 (i) da = dy dz x ˆ, x = 2. v·da = 2y dyRdz. v·da = 2y dy dz = 2y 2 0 = 8. (ii) da = dy dz x ˆ, x = 0. v·da = 0. v·daR= 0. RR (iii) da = dx dz y ˆ, y = 2. v·da = 4z dx R dz. v·da = 4z dx dz = 16. (iv) da = dx dz y ˆ, y = 0. v·da = 0. v·da R = 0. (v) da = dx dy ˆ z, z = 2. v·da = 6x dxRdy. v·da = 24. (vi)R da = dx dy ˆ z, z = 0. v·da = 0. v·da = 0. ) v·da = 8 + 16 + 24 = 48 X

Problem 1.34

r⇥v = x ˆ(0 2y) + y ˆ(0 3z) + ˆ z(0 x) = 2y x ˆ 3z y ˆ xˆ z. da = dy dz x ˆ, if we agree that the path integral shall run counterclockwise. So (r⇥v)·da = 2y dy dz. o R R nR 2 z (r⇥v)·da = ( 2y)dy dz 0 z6 2 2 z 2 @ y 0 = (2 z) ⇣ ⌘ @ 2 R2 3 @ = (4 4z + z 2 )dz = 4z 2z 2 + z3 0 0 @ 8 = 8 8 + 83 = @ 3 @ y Meanwhile, v·dl = (xy)dx + (2yz)dy + (3zx)dz. There are three segments. y

,!

=

2

z

z6 @ @@ I (2) (3) @@ @ ? @ @ - y (1) R (1) x = z = 0; dx = dz = 0. y : 0 ! 2. v·dl = 0. (2) x = 0; z = 2 y; dx = 0, dz = dy, y : 2 ! 0. v·dl = 2yz dy. R R0 R2 2 v·dl = 2 2y(2 y)dy = (4y 2y 2 )dy = 2y 2 32 y 3 0 = 8 23 · 8 = 0 R H (3) x = y = 0; dx = dy = 0; z : 2 ! 0. v·dl = 0. v·dl = 0. So v·dl = 83 . X

8 3.

Problem 1.35

R By Corollary 1, (r⇥v)·da should equal 43 . r⇥v = (4z 2 (i) da = dy dz x ˆ, x = 1; y, z : 0 ! 1. (r⇥v)·da = (4z 2 ( 43 z 3

1 0

2x)ˆ x + 2z ˆ z. R R1 2)dy dz; (r⇥v)·da = 0 (4z 2

= 2z) = 2= R (ii) da = dx dy ˆ z, z = 0; x, y : 0 ! 1. (r⇥v)·da = 0; R (r⇥v)·da = 0. (iii) da = dx dz y ˆ, y = 1; x, z : 0 ! 1. (r⇥v)·da = 0; (r⇥v)·da = 0. R (iv) da = dx dz y ˆ, y = 0; x, z : 0 ! 1. (r⇥v)·da = 0; (r⇥v)·da = 0. R (v) da = dx dy ˆ z, z = 1; x, y : 0 ! 1. (r⇥v)·da = 2 dx dy; (r⇥v)·da = 2. R ) (r⇥v)·da = 23 + 2 = 43 . X 4 3

2 3.


2)dz

14


Problem 1.36 (a) Use the product rule r⇥(f A) = f (r⇥A) A ⇥ (rf ) : Z Z Z I Z f (r⇥A) · da = r⇥(f A) · da + [A ⇥ (rf )] · da = f A · dl + [A ⇥ (rf )] · da. qed S

S

S

P

S

(I used Stokes’ theorem in the last step.) (b) Use the product rule r·(A ⇥ B) = B · (r⇥A) A · (r⇥B) : Z Z Z I Z B · (r⇥A)d⌧ = r·(A ⇥ B) d⌧ + A · (r⇥B) d⌧ = (A ⇥ B) · da + A · (r⇥B) d⌧. qed V

V

V

S

V

(I used the divergence theorem in the last step.) Problem 1.37 r = Problem 1.38

p x2 + y 2 + z 2 ;

✓ = cos

1

✓

p

z x2 +y 2 +z 2

◆

;

= tan

1

y x

.

There are many ways to do this one—probably the most illuminating way is to work it out by trigonometry from Fig. 1.36. The most systematic approach is to study the expression: r = xx ˆ+yy ˆ +zˆ z = r sin ✓ cos x ˆ + r sin ✓ sin y ˆ + r cos ✓ ˆ z. @ If I only vary r slightly, then dr = @r (r)dr is a short vector pointing in the direction of increase in r. To make it a unit vector, I must divide by its length. Thus:

ˆ r= @r @r @r @✓ @r @

= sin ✓ cos x ˆ + sin ✓ sin y ˆ + cos ✓ ˆ z; = r cos ✓ cos x ˆ + r cos ✓ sin y ˆ =

@r @r @r @r

; ✓ˆ =

@r 2 @r

r sin ✓ ˆ z;

r sin ✓ sin x ˆ + r sin ✓ cos y ˆ;

@r 2 @

@r @✓ @r @✓

; ˆ=

= sin2 ✓ cos2 @r 2 = @✓ 2 2

@r @ @r @

.

+ sin2 ✓ sin2

r2 cos2 ✓ cos2

= r sin ✓ sin

2

+ cos2 ✓ = 1.

+ r2 cos2 ✓ sin2

+ r2 sin ✓ cos2 2

+ r2 sin2 ✓ = r2 .

= r2 sin ✓. 2

ˆ r = sin ✓ cos x ˆ + sin ✓ sin y ˆ + cos ✓ ˆ z. ) ✓ˆ = cos ✓ cos x ˆ + cos ✓ sin y ˆ sin ✓ ˆ z. ˆ = sin x ˆ + cos y ˆ. Check: ˆ r·ˆ r = sin2 ✓(cos2 + sin2 ) + cos2 ✓ = sin2 ✓ + cos2 ✓ = 1, X ˆ ✓· ˆ = cos ✓ sin cos + cos ✓ sin cos = 0, X etc. sin ✓ ˆ r = sin2 ✓ cos x ˆ + sin2 ✓ sin y ˆ + sin ✓ cos ✓ ˆ z. 2 ˆ cos ✓ ✓ = cos ✓ cos x ˆ + cos2 ✓ sin y ˆ sin ✓ cos ✓ ˆ z. Add these: (1) sin ✓ ˆ r + cos ✓ ✓ˆ = + cos x ˆ + sin y ˆ; ˆ = sin x (2) ˆ + cos y ˆ. Multiply (1) by cos , (2) by sin , and subtract: x ˆ = sin ✓ cos ˆ r + cos ✓ cos ✓ˆ

sin

ˆ.


15

CHAPTER 1. VECTOR ANALYSIS Multiply (1) by sin , (2) by cos , and add: y ˆ = sin ✓ sin ˆ r + cos ✓ sin ✓ˆ + cos

ˆ.

cos ✓ ˆ r = sin ✓ cos ✓ cos x ˆ + sin ✓ cos ✓ sin y ˆ + cos2 ✓ ˆ z. ˆ sin ✓ ✓ = sin ✓ cos ✓ cos x ˆ + sin ✓ cos ✓ sin y ˆ sin2 ✓ ˆ z. Subtract these: ˆ sin ✓ ✓.

ˆ z = cos ✓ ˆ r Problem 1.39

@ (a) r·v1 = r12 @r (r2 r2 ) = r12 4r3 = 4r ⇣ 4⌘ R R R RR R⇡ R (r·v1 )d⌧ = (4r)(r2 sin ✓ dr d✓ d ) = (4) 0 r3 dr 0 sin ✓ d✓ 2⇡ (2)(2⇡) = 4⇡R4 0 d = (4) 4 R R 2 R R ⇡ 2⇡ v1 ·da = (r ˆ r)·(r2 sin ✓ d✓ d ˆ r) = r4 0 sin ✓ d✓ 0 d = 4⇡R4 X (Note: at surface of sphere r = R.) R @ (b) r·v2 = r12 @r r2 r12 = 0 ) (r·v2 )d⌧ = 0 R R 1 R v2 ·da = r (r2 sin ✓ d✓ d ˆ r) = sin ✓ d✓ d = 4⇡. r2 ˆ They don’t agree! The point is that this divergence is zero except at the origin, where it blows up, so our R calculation of (r·v2 ) is incorrect. The right answer is 4⇡.

Problem 1.40 @ 1 @ 1 @ r·v = r12 @r (r2 r cos ✓) + r sin ✓ @✓ (sin ✓ r sin ✓) + r sin ✓ @ (r sin ✓ cos ) 1 1 1 = r2 3r2 cos ✓ + r sin ✓ r 2 sin ✓ cos ✓ + r sin ✓ r sin ✓( sin ) = 3 cos ✓ + 2 cos ✓ sin = 5 cos ✓ sin R R RR R ✓ hR 2⇡ (r·v)d⌧ = (5 cos ✓ sin ) r2 sin ✓ dr d✓ d = 0 r2 dr 02 0 (5 cos ✓ =

=

⇣

R3 3

⌘

(10⇡)

R

⇡ 2

0

sin ) d

i

,!2⇡(5 cos ✓)

d✓ sin ✓

sin ✓ cos ✓ d✓

⇡ 2

,! sin2 ✓ 0 2

=

1 2

5⇡ 3 3 R .

Two surfaces—one the hemisphere: da = R2 sin ✓ d✓ d ˆ r; r = R; : 0 ! 2⇡, ✓ : 0 ! ⇡2 . ⇡ R R R R 2⇡ v·da = (r cos ✓)R2 sin ✓ d✓ d = R3 02 sin ✓ cos ✓ d✓ 0 d = R3 12 (2⇡) = ⇡R3 . ˆ = r dr d ✓ˆ (here ✓ = ⇡ ). r : 0 ! R, : 0 ! 2⇡. other the flat bottom: da = (dr)(r sin ✓ d )(+✓) 2 R R R R 2 R 2⇡ R3 v·da = (r sin ✓)(r dr d ) = r dr d = 2⇡ . 3 0 0 R Total: v·da = ⇡R3 + 23 ⇡R3 = 53 ⇡R3 . X

Problem 1.41 rt = (cos ✓ + sin ✓ cos )ˆ r + ( sin ✓ + cos ✓ cos )✓ˆ +

1 sin /✓

( sin / ✓ sin ) ˆ

r2 t = r·(rt) @ 1 @ 1 @ = r12 @r r2 (cos ✓ + sin ✓ cos ) + r sin ✓ @✓ (sin ✓( sin ✓ + cos ✓ cos )) + r sin ✓ @ ( sin ) 2 1 1 1 2 = r2 2r(cos ✓ + sin ✓ cos ) + r sin ✓ ( 2 sin ✓ cos ✓ + cos ✓ cos sin ✓ cos ) r sin ✓ cos 2 2 1 2 = r sin ✓ [2 sin ✓ cos ✓ + 2 sin ✓ cos 2 sin ✓ cos ✓ + cos ✓ cos sin ✓ cos cos ] ⇥ ⇤ 2 1 2 = r sin cos = 0. ✓ (sin ✓ + cos ✓) cos


16


) r2 t = 0

Check: r cos ✓ = z, r sin ✓ cos = x ) in Cartesian coordinates t = x + z. Obviously Laplacian is zero. Rb Gradient Theorem: a rt·dl = t(b) t(a) Segment 1: ✓ = ⇡2 , = 0, r : 0 ! 2. dl = dr ˆ r; rt·dl = (cos ✓ + sin ✓ cos )dr = (0 + 1)dr = dr. R R2 rt·dl = 0 dr = 2. Segment 2: ✓ =

⇡ 2,

r = 2,

:0!

⇡ 2.

rt·dl = ( sin )(2 d ) =

dl = r sin ✓ d ˆ = 2 d ˆ. ⇡ R R ⇡2 2 sin d . rt·dl = 2 sin d = 2 cos |02 = 0

2.

Segment 3: r = 2, = ⇡2 ; ✓ : ⇡2 ! 0. ˆ rt·dl = ( sin ✓ + cos ✓ cos )(2 d✓) = 2 sin ✓ d✓. dl = r d✓ ✓ˆ = 2 d✓ ✓; R R0 0 rt·dl = ⇡ 2 sin ✓ d✓ = 2 cos ✓| ⇡ = 2. 2 2 Rb Total: a rt·dl = 2 2 + 2 = 2 . Meanwhile, t(b) t(a) = [2(1 + 0)] [0( )] = 2. X

Problem 1.42 From Fig. 1.42, ˆ s = cos x ˆ + sin y ˆ; ˆ = Multiply first by cos , second by sin , and subtract: ˆ sin = cos2 x ˆ s cos ˆ + cos sin y ˆ + sin2 x ˆ So x ˆ = cos ˆ s sin ˆ.

sin x ˆ + cos y ˆ; ˆ z=ˆ z

sin cos y ˆ=x ˆ(sin2

+ cos2 ) = x ˆ.

Multiply first by sin , second by cos , and add: ˆ s sin + ˆ cos = sin cos x ˆ + sin2 y ˆ sin cos x ˆ + cos2 y ˆ=y ˆ(sin2 So y ˆ = sin ˆ s + cos ˆ. ˆ z=ˆ z.

+ cos2 ) = y ˆ.

Problem 1.43 @ @ (a) r·v = 1s @s s s(2 + sin2 ) + 1s @@ (s sin cos ) + @z (3z) 2 2 1 1 2 = s 2s(2 + sin ) + s s(cos sin ) + 3 = 4 + 2 sin2 + cos2 sin2 + 3 2 = 4 + sin + cos2 + 3 = 8. R R R2 R⇡ R5 (b) (r·v)d⌧ = (8)s ds d dz = 8 0 s ds 02 d 0 dz = 8(2) ⇡2 (5) = 40⇡.

Meanwhile, the surface integral has five parts: R R2 R⇡ top: z = 5, da = s ds d ˆ z; v·da = 3z s ds d = 15s ds d . R v·da = 15 0 s ds 02 d = 15⇡. bottom: z = 0, da = s ds d ˆ z; v·da = 3z s ds d = 0. v·da = 0. R ⇡ ˆ back: = 2 , da = ds dz ; v·da = s sin cos ds dz = 0. v·da = 0. R left: = 0, da = ds dz ˆ; v·da = s sin cos ds dz = 0. v·da = 0. front: s = 2, da = s d dz ˆ s; v·da = s(2 + sin2 )s d dz = 4(2 + sin2 )d dz. R R ⇡2 R5 v·da = 4 0 (2 + sin2 )d 0 dz = (4)(⇡ + ⇡4 )(5) = 25⇡. H So v·da = 15⇡ + 25⇡ = 40⇡. X ⇣ ⌘ @ @ @ ˆ (c) r⇥v = 1s @@ (3z) @z (s sin cos ) ˆ s + @z s(2 + sin2 ) @s (3z) ⇣ ⌘ @ + 1s @s (s2 sin cos ) @@ s(2 + sin2 ) ˆ z =

1 s (2s sin

cos

s 2 sin cos ) ˆ z = 0.


17

CHAPTER 1. VECTOR ANALYSIS Problem 1.44 (a) 3(32 )

2(3)

1 = 27

6

1 = 20.

(b) cos ⇡ = -1. (c) zero. (d) ln( 2 + 3) = ln 1 = zero. Problem 1.45 R2 (a) 2 (2x + 3) 13 (x) dx = 13 (0 + 3) = 1. (b) By Eq. 1.94, (1 (c)

R1

1

1), so 1 + 3 + 2 = 6.

x) = (x

1 2 1 3 3

9x2 13 (x + 13 ) dx = 9

=

1 3.

(d) 1 (if a > b), 0 (if a < b). Problem 1.46 ⇥ d ⇤ R1 1 (a) 1 f (x) x dx (x) dx = x f (x) (x)| 1

R1

d (x f (x)) (x) dx. 1 dx df d dx dx (x f (x)) = x dx + dx f

The first term is zero, since (x) = 0 at ±1; ⌘ R 1 ⇣ df So the integral is x + f (x) dx = 0 dx 1

(b)

So, R1

d x dx

(x) =

(x).

f (0) =

= x df + f. R 1 dx f (x) (x) dx. 1

f (0) =

qed

1

d✓ f (x) dx dx = f (x)✓(x)| 1 R1 = f (0) = 1 f (x) (x) dx. So 1

R1

d✓ dx

df ✓(x)dx 1 dx

= f (1)

= (x). qed

R1 0

df dx dx

= f (1)

(f (1)

f (0))

Problem 1.47

(a) ⇢(r) = q 3 (r

r0 ). Check:

(b) ⇢(r) = q 3 (r

a)

q 3 (r).

(c) Evidently ⇢(r) = A (r R R Q = ⇢ d⌧ = A (r

R

⇢(r)d⌧ = q

R

3

(r

r0 ) d⌧ = q.

X

R). To determine the constant A, we require R)4⇡r2 dr = A 4⇡R2 .

So A =

Q 4⇡R2 .

⇢(r) =

Q 4⇡R2

(r

R).

Problem 1.48

(a) a2 + a·a + a2 = 3a2 . R (b) (r

b)2 513

3

(r) d⌧ =

1 2 125 b

=

1 2 125 (4

+ 32 ) =

1 5.

(c) c2 = 25 + 9 + 4 = 38 > 36 = 62 , so c is outside V, so the integral is zero. 2

2

(d) (e (2 x ˆ + 2y ˆ + 2ˆ z)) = (1 x ˆ + 0y ˆ + ( 1) ˆ z) = 1 + 1 = 2 < (1.5)2 = 2.25, so e is inside V, and hence the integral is e·(d e) = (3, 2, 1)·( 2, 0, 2) = 6 + 0 + 2 = -4. Problem 1.49 R First method: use Eq. 1.99 to write J = e r 4⇡ 3(r) d⌧ = 4⇡e Second method: integrating by parts (use Eq. 1.59).

0

= 4⇡.


18

Z

J =

=

Z

V

ˆ r · r(e r2

1 e r2

= 4⇡

r

e

r

) d⌧ +

4⇡r dr + 2

R 0

r

Problem 1.50 (a) r·F1 =

I

e

r

S

Z

+ 4⇡e


r

e R

ˆ r · da. But r e r2

ˆ r · r2 sin ✓ d✓ d ˆ r = 4⇡ r2

= 4⇡

@ @x (0)

+

R

e

@ @y (0)

+

x ˆ y ˆ ˆ z r⇥F1 =

@ @ @ @x @y @z 2

0

=

y ˆ

0 x

F2 is a gradient; F1 is a curl U2 = ⇣ ⌘ @A @Az x For A1 , we want @zy = @A @y @z

1 2

(b) r·F3 =

= 0;

A1 = 13 x2 y ˆ

+e

@ @z

0

x2 = 0 ;

R

r

dr + e

= 4⇡.X

r·F2 =

r

◆

ˆ r=

R

Z

e

r

ˆ r.

sin ✓ d✓ d

Here R = 1, so e @x @x

+

@y @y

+

@z @z

R

= 0.

=1+1+1= 3

x ˆ y ˆ ˆ z 2xˆ y;

x3 + y 2 + z 2 = 0;

e

@ e @r

0

+ 4⇡e

@ x2 = @x

@Az @x

ZR

✓

=

r

r⇥F2 =

@ @ @ @x @y @z

x

y

= 0

z

would do (F2 = rU2 ).

@Ay @x

@Ax @y

= x2 . Ay =

x3 3 ,

Ax = Az = 0 would do it.

(F1 = r⇥A1 ) . (But these are not unique.)

x ˆ y ˆ ˆ z

@ @x (yz)

> :

@Az @y @Ax @z @Ay @x

@ @y (xz)

@Ay @z @Az @x @Ax @y

+

@ @z (xy)

@ @ @ @x @y @z

=x ˆ (x x) + y ˆ (y y) + ˆ z (z z) = 0. yz xz xy So F3 can be written as the gradient of a scalar (F3 = rU3 ) and as the curl of a vector (F3 = r⇥A3 ). In fact, U3 = xyz does the job. For the vector potential, we have 8 > <

+

r⇥F3 =

= yz, which suggests Az = 14 y 2 z + f (x, z); Ay = = xz, suggesting Ax = 14 z 2 x + h(x, y); Az = = xy, so Ay = 14 x2 y + k(y, z); Ax =

Putting this all together: A3 =

1 4

x z2

y2 x ˆ + y x2

z2 y ˆ + z y2

x2 ˆ z

9

1 2 > 4 yz + g(x, y) = 1 2 4 zx + j(y, z) > ; 1 2 4 xy + l(x, z)

(again, not unique).

Problem 1.51 (d) ) (a): Hr⇥F = r⇥( rU ) = 0 (Eq. 1.44 – curl of gradient is always zero). R (a) ) (c): F · dl = (r⇥F) · da = 0 (Eq. 1.57–Stokes’ theorem). Rb Rb Rb Ra H (c) ) (b): a I F · dl F · dl = a I F · dl + b II F · dl = F · dl = 0, so a II Z

b

a I

F · dl =

Z

b

a II

F · dl.

(b) ) (c): same as (c) ) (b), only in reverse; (c) ) (a): same as (a)) (c).

Problem 1.52 (d) ) (a): Hr·F = r·(r⇥W) = 0 (Eq 1.46—divergence of curl is always zero). R (a) ) (c): F · da = (r·F) d⌧ = 0 (Eq. 1.56—divergence theorem).


19

CHAPTER 1. VECTOR ANALYSIS (c) ) (b):

R

I

F · da

R

II

F · da =

H

F · da = 0, so Z Z F · da = I

II

F · da.

H (Note: sign change because for F · da, da is outward, whereas for surface II it is inward.) (b) ) (c): same as (c) ) (b), in reverse; (c)) (a): same as (a)) (c) .

Problem 1.53 In Prob. 1.15 we found that r·va = 0; in Prob. 1.18 we found that r⇥vc = 0. So vc can be written as the gradient of a scalar; va can be written as the curl of a vector. (a) To find t: (1)

= y 2 ) t = y 2 x + f (y, z)

(2)

@t @x @t @y

= 2xy + z 2

(3)

@t @z

= 2yz

From (1) & (3) we get 2xy + z (from (2)) ) 2

(b) To find W:

@f 2 2 @z = 2yz ) f = yz + g(y) ) t = y x @g @y = 0. We may as well pick g = 0; then

@Wy @z

@Wz @y

= x2 ;

@Wx @z

@Wz @x

= 3z 2 x;

@Wy @x

+ yz 2 + g(y), so

@t @y

= 2xy + z 2 +

@g @y

=

t = xy + yz .

@Wx @y

2

=

2

2xz.

Pick Wx = 0; then @Wz = @x @Wy = @x @Wz @y

W= Check:

@Wy @z

=

x2 z y ˆ

3xz 2 ) Wz = 2xz ) Wy = @f @y

+ x2

@g @z

3 2 2 x z + f (y, z) 2 x2 z + g(y, z).

= x2 )

@f @y

@g @z

= 0. May as well pick f = g = 0.

3 2 2 z. 2x z ˆ

r⇥W =

x ˆ

y ˆ

@ @x

@ @y 2

0

x z

ˆ z @ @z 3 2 2 2x z

=x ˆ x2 + y ˆ 3xz 2 + ˆ z ( 2xz).X

You can add any gradient (rt) to W without changing its curl, so this answer is far from unique. Some other solutions: W = xz 3 x ˆ

x2 z y ˆ;

W = 2xyz + xz 3 x ˆ + x2 y ˆ z; W = xyz x ˆ

1 2 2x z

y ˆ + 12 x2 y

3z 2 ˆ z.


20


Problem 1.54 1 @ 1 @ r2 r2 cos ✓ + sin ✓ r2 cos 2 r @r r sin ✓ @✓ 1 1 1 = 2 4r3 cos ✓ + cos ✓ r2 cos + r r sin ✓ r sin ✓ r cos ✓ = [4 sin ✓ + cos cos ] = 4r cos ✓. sin ✓

r·v =

Z

(r·v) d⌧ =

Z

(4r cos ✓)r sin ✓ dr d✓ d = 4 2

0

✓ ◆⇣ ⌘ 1 ⇡ = 2 2

= R4

ZR

⇡R 4

4

+

1 @ r sin ✓ @

r2 cos ✓ cos

Z⇡/2 Z⇡/2 r dr cos ✓ sin ✓ d✓ d 3

0

0

.

Surface consists of four parts: (1) Curved: da = R2 sin ✓ d✓ d ˆ r; r = R. v · da = R2 cos ✓ Z

.

✓ ◆⇣ ⌘ Z⇡/2 Z⇡/2 1 ⇡ ⇡R4 v · da = R cos ✓ sin ✓ d✓ d = R4 = . 2 2 4 0

r dr d✓ ˆ; (3) Back: da = r dr d✓ ˆ;

0

= 0. v · da = r2 cos ✓ sin

= ⇡/2. v · da = Z

v · da =

ZR 0

(r dr d✓) = 0.

r cos ✓ sin 2

Z⇡/2 r dr cos ✓ d✓ = 3

0

ˆ ✓ = ⇡/2. v · da = r2 cos (4) Bottom: da = r sin ✓ dr d ✓; Z H

R2 sin ✓ d✓ d

4

(2) Left: da =

Total:

r2 cos ✓ sin

v · da = ⇡R4 /4 + 0

1 4 4R

v · da =

+ 14 R4 =

ZR 0

⇡R4 4 .

(r dr d✓) = ✓

R

v · da = 0.

r3 cos ✓ dr d✓.

◆ 1 4 R (+1) = 4

1 4 R . 4

(r dr d ) .

Z⇡/2 1 r dr cos d = R4 . 4 3

0

X

Problem 1.55 x ˆ y ˆ ˆ z R @ @ @ r⇥v = @x z (b a). So (r⇥v) · da = (b a)⇡R2 . @y @z = ˆ ay bx 0 v · dl = (ay x ˆ + bx y ˆ) · (dx x ˆ + dy y ˆ + dz ˆ z) = ay dx + bx dy; x2 + y 2 = R2 ) 2x dx + 2y dy = 0, so dy = (x/y) dx. So v · dl = ay dx + bx( x/y) dx = y1 ay 2 bx2 dx. c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

21

CHAPTER 1. VECTOR ANALYSIS For the “upper” semicircle, y = Z

v · dl =

ZR

R

p

R2

aR2 (a + b)x2 p dx = R2 x2

⇢ aR2 sin R

1 2 R (a b) sin 2 1 = ⇡R2 (b a). 2 =

1

a(R2 x2 ) bx2 p R2 x2

x2 , so v · dl =

(x/R)

= +R

1

⇣x⌘

(a + b)

R

1 2 R (a 2

dx.

b) sin

1



( 1)

xp 2 R 2 sin

1

x2 +

(+1) =

R2 sin 2

1 2 R (a 2

1

⇣x⌘

R

R

b)

+R

⇣ ⇡ 2

⇡⌘ 2

And the same for the lower semicircle (y changes sign, but the limits on the integral are reversed) so H v · dl = ⇡R2 (b a). X Problem 1.56 R (1) x = z = 0; dx = dz = 0; y : 0 ! 1. v · dl = (yz 2 ) dy = 0; v · dl = 0. (2) x = 0; z = 2 2y; dz = 2 dy; y : 1 ! 0. v · dl = (yz 2 ) dy + (3y + z) dz = y(2 Z

v · dl = 2

Z0

(2y 3

4y 2 + y

2) dy = 2

1

✓

4y 3 y2 + 3 2

y4 2

2y

2y)2 dy ◆

0

= 1

(3y + 2

14 . 3

(3) x = y = 0; dx = dy = 0; z : 2 ! 0. v · dl = (3y + z) dz = z dz; Z Total:

H

v · dl = 0 +

2=

14 3

v · dl =

Z0

z dz =

z2 2

0 2

=

2.

2

8 3.

H R Meanwhile, Stokes’ thereom says v · dl = (r⇥v) · da. Here da = dy dz x ˆ, so all we need is @ @ (r⇥v)x = @y (3y + z) @z (yz 2 ) = 3 2yz. Therefore Z

(r⇥v) · da = =

Z Z Z

0

=

1

(3

⇥ 3(2

2yz) dy dz =

Z

1

0

2y)

y 4 + 83 y 3

2y 12 (2 5y 2 + 6y

Z

2 2y

(3

0

⇤ 2y)2 dy =

1 0

=

1+

8 3

2yz) dz dy Z

1

( 4y 3 + 8y 2

10y + 6) dy

0

5 + 6 = 83 . X

Problem 1.57 Start at the origin. (1) ✓ =

⇡ 2,

= 0; r : 0 ! 1. v · dl = r cos2 ✓ (dr) = 0.

(2) r = 1, ✓ =

⇡ 2;

R

v · dl = 0.

: 0 ! ⇡/2. v · dl = (3r)(r sin ✓ d ) = 3 d .

R

v · dl = 3


⇡/2 R 0

d =

3⇡ 2 .

2y)2 dy;

22 (3)

=

⇡ 2;


r sin ✓ = y = 1, so r =

1 sin ✓ ,

dr =

1 sin2 ✓

cos ✓ d✓, ✓ :

⇡ 2

! ✓0 ⌘ tan

1

(1/2).

✓ ◆ cos2 ✓ cos ✓ cos ✓ sin ✓ v · dl = r cos2 ✓ (dr) (r cos ✓ sin ✓)(r d✓) = d✓ d✓ sin ✓ sin2 ✓ sin2 ✓ ✓ 3 ◆ ✓ ◆ cos ✓ cos ✓ cos ✓ cos2 ✓ + sin2 ✓ cos ✓ = + d✓ = d✓ = d✓. sin ✓ sin ✓ sin3 ✓ sin2 ✓ sin3 ✓ Therefore

(4) ✓ = ✓0 ,

Z =

⇡ 2;

Z✓0

v · dl =

cos ✓ 1 3 d✓ = sin ✓ 2 sin2 ✓

⇡/2

r:

p

✓0

1 2 · (1/5)

= ⇡/2

1 5 = 2 · (1) 2

1 = 2. 2

5 ! 0. v · dl = r cos2 ✓ (dr) = 45 r dr. Z

Z0 4 4 r2 v · dl = r dr = 5p 5 2 5

0 p

4 5 · = 5 2

= 5

2.

Total: I

v · dl = 0 +

3⇡ +2 2

2=

3⇡ 2 .

R Stokes’ theorem says this should equal (r⇥v) · da   1 @ @ 1 1 @ r⇥v = (sin ✓ 3r) ( r sin ✓ cos ✓) ˆ r+ r cos2 ✓ r sin ✓ @✓ @ r sin ✓ @  1 @ @ ˆ + ( rr cos ✓ sin ✓) r cos2 ✓ r @r @✓ 1 1 1 = [3r cos ✓] ˆ r + [ 6r] ✓ˆ + [ 2r cos ✓ sin ✓ + 2r cos ✓ sin ✓] ˆ r sin ✓ r r ˆ = 3 cot ✓ ˆ r 6 ✓. (1) Back face: da = (2) Bottom: da =

r dr d✓ ˆ; (r⇥v) · da = 0.

R

(r⇥v) · da = 0.

ˆ (r⇥v) · da = 6r sin ✓ dr d . ✓ = r sin ✓ dr d ✓; Z

(r⇥v) · da =

Z1 0

(2) Bottom: dz = 0.

x; dz =

⇡ 2,

so (r⇥v) · da = 6r dr d

Z⇡/2 1 ⇡ 3⇡ 6r dr d =6· · = . X 2 2 2 0

Problem 1.58 v · dl = y dz. (1) Left side: z = a

@ (r3r) ✓ˆ @r

dx; y = 0. Therefore R Therefore v · dl = 0.

R

v · dl = 0.


23


(3) Back: z = a

1 2 y;

dz =

R

1/2 dy; y : 2a ! 0.

v · dl =

R0

1 2

y

2a

2 1y 2 2

dy =

0 2a

=

4a2 4

R Meanwhile, r⇥v = x ˆ, so (r⇥v) · da is the projection of this surface on the x y plane =

1 2

Problem 1.59

= a2 . · a · 2a = a2 . X

1 @ 1 @ 1 @ r2 r2 sin ✓ + sin ✓ 4r2 cos ✓ + r2 tan ✓ r2 @r r sin ✓ @✓ r sin ✓ @ 1 1 4r = 2 4r3 sin ✓ + 4r2 cos2 ✓ sin2 ✓ = sin2 ✓ + cos2 ✓ sin2 ✓ r r sin ✓ sin ✓ cos2 ✓ = 4r . sin ✓

r·v =

Z

Z ✓

 Z⇡/6 Z2⇡ ✓ sin 2✓ (r·v) d⌧ = r sin ✓ dr d✓ d = 4r dr cos2 ✓ d✓ d = R4 (2⇡) + 2 4 0 0 0 ! p ✓ ◆ p 4 ⇡ sin 60 ⇡R4 3 4 = 2⇡R + = ⇡+3 = ⇡R 2⇡ + 3 3 . 12 12 4 6 2 cos2 ✓ 4r sin ✓

◆

ZR

2

⇡/6

3

0

Surface coinsists of two parts: (1) The ice cream: r = R; R4 sin2 ✓ d✓ d . Z

: 0 ! 2⇡; ✓ : 0 ! ⇡/6; da = R2 sin ✓ d✓ d ˆ r; v·da = R2 sin ✓

 Z⇡/6 Z2⇡ 1 2 4 4 v·da = R sin ✓ d✓ d = R (2⇡) ✓ 2 0

(2) The cone: ✓ =

0

⇡ 6;

R

v · da =

⇡/6

= 2⇡R4 0

: 0 ! 2⇡; r : 0 ! R; da = r sin ✓ d dr ✓ˆ = Z

Therefore

1 sin 2✓ 4

⇡R4 2

⇣

p

✓

3 2 r dr d

⇡ 12

R2 sin ✓ d✓ d

1 sin 60 4

ˆ v · da = ✓;

◆

p

⇡R4 = 6

⇡

= p ! 3 3 2

3 r3 dr d

R 2⇡ p p Z 3 Z p R4 3 4 v · da = 3 r dr d = 3 · · 2⇡ = ⇡R . 4 2 0

⇡ 3

p

3 2

+

p ⌘ 3 =

0

⇡R4 12

p 2⇡ + 3 3 .

X.

Problem 1.60 H H R R (a) Corollary 2 says (rT )·dl = 0. Stokes’ theorem says (rT )·dl = [r⇥(rT )]·da. So [r⇥(rT )]·da = 0, and since this is true for any surface, the integrand must vanish: r⇥(rT ) = 0, confirming Eq. 1.44. H H R R (b) Corollary 2 says (r⇥v)·da = 0. Divergence theorem says (r⇥v)·da = r·(r⇥v) d⌧. So r·(r⇥v) d⌧ = 0, and since this is true for any volume, the integrand must vanish: r(r⇥v) = 0, confirming Eq. 1.46. Problem 1.61 H R (a) Divergence theorem: v · da = (r·v) d⌧. Let v = cT , where c is a constant vector. Using product rule #5 in front cover: r·v = r·(cT ) = T (r·c) + c · (rT ). But c is constant so Rr·c = 0. Therefore we have: R R R c · (rT ) d⌧ = T c · da. Since c is constant, take it outside the integrals: c · rT d⌧ = c · T da. But c c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

24


is any constant vector—in particular, it could be be x ˆ, or y ˆ, or ˆ z—so each component of the integral on left equals corresponding component on the right, and hence Z Z rT d⌧ = T da. qed

R R (b) Let v ! (v ⇥ c) in divergence theorem. Then r·(v ⇥ c)d⌧ = (v ⇥ c) · da. Product rule #6 ) r·(v ⇥ c) = c · (r⇥v) v · (r⇥c) = c · (r⇥v). (Note: r⇥c = R0, since c is constant.) R Meanwhile vector indentity (1) says da · (v ⇥ c) = c · (da ⇥ v) = c · (v ⇥ da). Thus c · (r⇥v) d⌧ = c · (v ⇥ da). Take c outside, and again let c be x ˆ, y ˆ, ˆ z then: Z Z (r⇥v) d⌧ = v ⇥ da. qed

R R (c) Let v = T rU in divergence theorem: r·(T rU ) d⌧ = T rU · da. Product rule #(5) ) r·(T rU ) = T r·(rU ) + (rU ) · (rT ) = T r2 U + (rU ) · (rT ). Therefore Z Z 2 T r U + (rU ) · (rT ) d⌧ = (T rU ) · da. qed R R (d) Rewrite (c) with T $ U : U r2 T + (rT ) · (rU ) d⌧ = (U rT ) · da. Subtract this from (c), noting that the (rU ) · (rT ) terms cancel: Z Z T r2 U U r2 T d⌧ = (T rU U rT ) · da. qed

R H (e) Stokes’ theorem: (r⇥v) · da = v · dl. Let v = cTR. By Product Rule H#(7): r⇥(cT ) = T (r⇥c) c ⇥ (rT ) = c ⇥ (rT ) (since c is constant). Therefore, (c ⇥ (rT R )) · da = T c ·Hdl. Use vector indentity #1 to rewrite the first term (c ⇥ (rT )) · da = c · (rT ⇥ da). So c · (rT ⇥ da) = c · T dl. Pull c outside, and let c ! x ˆ, y ˆ, and ˆ z to prove: Z I rT ⇥ da = T dl. qed Problem 1.62 (a) da = R2 sin ✓ d✓ d ˆ r. Let the surface be the northern hemisphere. The x ˆ and y ˆ components clearly integrate to zero, and the ˆ z component of ˆ r is cos ✓, so Z Z ⇡/2 sin2 ✓ ⇡/2 2 2 a = R sin ✓ cos ✓ d✓ d ˆ z = 2⇡R ˆ z sin ✓ cos ✓ d✓ = 2⇡R2 ˆ z = ⇡R2 ˆ z. 2 0 0 H (b) Let T = 1 in Prob. 1.61(a). Then rT = 0, so da = 0. qed (c) This follows from (b). For suppose a = 6 a ; then if you put them together to make a closed surface, 1 2 H da = a1 a2 6= 0. (d) For one such triangle, da = 12 (r ⇥ dl) (since r ⇥ dl is the area ofH the parallelogram, and the direction is perpendicular to the surface), so for the entire conical surface, a = 12 r ⇥ dl. (e) Let T = c · r, and use product rule #4: rT = r(c · r) = c ⇥ (r⇥r) + (c · r)r. But r⇥r = 0, and @ @ @ (c · r)r = (cx @x + cy @y + cz @z )(x x ˆ+yy ˆ +zˆ z) = cx x ˆ + cy y ˆ + cz ˆ z = c. So Prob. 1.61(e) says I I Z Z Z T dl = (c · r) dl = (rT ) ⇥ da = c ⇥ da = c ⇥ da = c ⇥ a = a ⇥ c. qed


25

CHAPTER 1. VECTOR ANALYSIS Problem 1.63 (1)

1 @ r·v = 2 r @r

✓

1 r · r 2

◆

=

1 @ (r) = r2 @r

1 r2 .

For a sphere of radius R: 9 R R 1 R v · da = r · R2 sin ✓ d✓ d ˆ r = R sin > Rˆ = ! ✓ d✓ d = 4⇡R. So divergence R R R 1 R R 2 (r·v) d⌧ = r sin ✓ dr d✓ d = dr sin ✓ d✓ d = 4⇡R. > 2 r ; theorem checks. 0

Evidently there is no delta function at the origin. r⇥ (rn ˆ r) = (except for n =

1 @ 1 @ 1 r2 rn = 2 rn+2 = 2 (n + 2)rn+1 = (n + 2)rn r2 @r r @r r

1

2, for which we already know (Eq. 1.99) that the divergence is 4⇡ 3(r)).

(2) Geometrically, it should be zero. Likewise, the curl in the spherical coordinates obviously gives zero. To be certain there is no lurking delta function here, we integrate over a sphere of radius R, using R H ? Prob. 1.61(b): If r⇥(rn ˆ r) = 0, then (r⇥v) d⌧ = 0 = v ⇥ da. But v = rn ˆ r and da = 2 R sin ✓ d✓ d ˆ r are both in the ˆ r directions, so v ⇥ da = 0. X Problem 1.64 (a) Since the argument is not a function of angle, Eq. 1.73 says  ✓ ◆  2r r3 1 1 d 1 1 d 2 D = r = 4⇡ r2 dr 2 (r2 + ✏2 )3/2 4⇡r2 dr (r2 + ✏2 )3/2  2 3 1 3r 3 r 2r 1 3r2 = = r2 + ✏2 4⇡r2 (r2 + ✏2 )3/2 2 (r2 + ✏2 )5/3 4⇡r2 (r2 + ✏2 )5/2 (b) Setting r ! 0:

D(0, ✏) =

r2 =

3✏2 .X 4⇡(r2 + ✏2 )5/2

3✏2 3 = , 5 4⇡✏ 4⇡✏3

which goes to infinity as ✏ ! 0. X (c) From (a) it is clear that D(r, 0) = 0 for r 6= 0. X (d) ✓ ◆ Z Z 1 r2 1 2 2 2 D(r, ✏) 4⇡r dr = 3✏ dr = 3✏ = 1. X 3✏2 (r2 + ✏2 )5/2 0 (I looked up the integral.) Note that (b), (c), and (d) are the defining conditions for


3

(r).

26

CHAPTER 2. ELECTROSTATICS

Chapter 2

Electrostatics Problem 2.1 (a) Zero. 1 qQ , where r is the distance from center to each numeral. F points toward the missing q. 4⇡✏0 r2 Explanation: by superposition, this is equivalent to (a), with an extra q at 6 o’clock—since the force of all twelve is zero, the net force is that of q only. (b) F =

(c) Zero. 1 qQ , pointing toward the missing q. Same reason as (b). Note, however, that if you explained (b) as 4⇡✏0 r2 a cancellation in pairs of opposite charges (1 o’clock against 7 o’clock; 2 against 8, etc.), with one unpaired q doing the job, then you’ll need a di↵erent explanation for (d). (d)

Problem 2.2 This time the “vertical” components cancel, leaving E= E=

q 1 4⇡✏0 2

r

2

r

sin ✓ x ˆ, or

1 qd 4⇡✏0 z 2 + d 2

2 3/2

x ˆ.

s q

-E AAU A ✓ A z A A As- x q

1 qd From far away, (z d), the field goes like E ⇡ 4⇡✏ z, which, as we shall see, is the field of a dipole. (If we 3 ˆ 0 z set d ! 0, we get E = 0, as is appropriate; to the extent that this configuration looks like a single point charge from far away, the net charge is zero, so E ! 0.)


1 Problem 2.3 CHAPTER 2. ELECTROSTATICS

27 1

Problem 2.3 Problem 2.3 z z ✻ ❑✻ ❑ θ η z dq = λdx ✴L !x "# $✲x θ r z dq = λdx

% L λ dx 1% 2 Ez =1 4πϵ =2 z 2 +2 x2 ; cos θ z= ηz ) L 2 cos θ; 2(η R λ ηcos 0 dxdx 0 2 2 θ = η) z EzE = =4πϵ01 0 L θ; (η = z 2 η r =+z 2x+; xcos ; cos ✓ = r ) z 4⇡✏0 0 r 2 cos ✓; ( % L 1 % 1 =1 4πϵ0 λzLR L0 (z12 +x2 )3/2 dx 1λz 1 = =4πϵ dxdx 2 3/2 '( z0 &(z2 +x 0 2 ) 2 )3/2 (L 4⇡✏ x '(L 10 & h0 1(z√+x 1 λ√ L . =1 4πϵ0 λz1 √z2 x z2 +x(2i L ( =1 4πϵ L zL 2 +L2 1λ 0√zp . . = =4πϵ01λz z z%2 12 zp2 +xx2 ( =0 =4πϵ 2 +L2 0 z z% z L 2 2 2 2 4⇡✏0 1% z λ dx 4⇡✏ 0 1 x dx z +x θ = +L %0λR zx dx 0 − E = −1 4πϵ0LR L 3/2 λ0 dxdx 1 4πϵ η 2 sin (x2x+z ✴L 1 1λ 0 dx2 ) ExEx=x =− 4πϵ θ= − 4πϵ 2 sin 2 +z 2 )3/2 & '( & ' ✲ sin ✓ = η 0 0 0 (x 2 L 2 2 )3/2 4⇡✏10 & 0 r 4⇡✏0 1 & (x +z ! "# $ ( '( 1 1 1' i L x iL h z−√ = −1 4πϵ0 λh √− √ = − . λ ( ( 1 1 1 1 2 2 2 2 x 4πϵ x1 +z( = z1 +L. 1 √p = =− 4πϵ01λ − xp2 +z − 4πϵ01λ 0 z − 0= 2 2 +L2 . z 2 2 2 2 4⇡✏0 z z +L )* + * 4⇡✏0 + x, +z 0 0 λ✓ z +◆ *✓ L + ◆, 1 )* −1 + √z z x ˆ + √L L ˆ z . E =1 1 λ √p z 2 + L2x √p z 2 + L2ˆ ˆ+ zˆ . . EE = = 4πϵ 0 z −1 + 1 + x ˆ + z 2 2 2 2 4πϵ z z+ L L2 z z+ L L2 0 z z 2+ 2+ 4⇡✏ 0 1 λL For z ≫ L you expect it to look like a point charge q = λL: E →1 4πϵ z. It checks, for with z ≫ L the x ˆ λL z 2 ˆ 1 20 ˆ For z≫ L Lyou expect it ittotolook like a point charge q = λL: E → zL. ˆ It checks, for with z ≫ L Lthe x ˆx 4πϵ For z you expect look like a point charge q = L: E ! the ˆ 1 λL 0 z 2 z. It checks, for with z 4⇡✏ z ˆ z. term → 0, and the ˆ z term →1 4πϵ 0 zL.zz. term →! 0, 0, and the ˆ zˆ term →!4πϵ01λz 0Lzzˆ term and the z term 4⇡✏0 z z ˆ Problem 2.4 Problem 2.4 -q 2 . /. a /2 Problem 2.4 2.1, with L → a and z → z +a 22 2 (distance from center of edge to P ), field of one edge is: From Ex. 2 z→ from center of of edge toto P ), field of of one is:is: z 2z+ From Ex. 2.1, with LL →!a2 aand 2 +2 a (distance From Ex. 2.2, with and z ! (distance from center edge P ), field oneedge edge 2 2 1 λa λa-a . E =1 1 -0q 2 .a2. E1E=1= 4πϵ a2q 2 a2 + z + z + 2 2 2 1 4πϵ0 a 4 2 a 24 a 24 4⇡✏0 z 2z+ 2 +4 a2 z z+ 2 +4 a+ 4a 4 4 + 4 z There are 4 sides, and we want vertical components only, so multiply by 4 cos θ =q4 q z 2 a2 : There are 4 sides, and we want vertical components only, so multiply by 4 cos θ = 4 z 2+ :4 z q There are 4 sides, and we want vertical components only, so multiply by 4 cos ✓ = 4 z2 + a a2 : 4 z2 + 4 4λaz 1 4λaz E =1 1 . ˆ z . / 4 az 2 EE = = 4πϵ z2.ˆ . 0 z 2 a+2 /aq2 z 2 a+2 aˆ z. 4πϵ 0 z2 + 4 2 4⇡✏ 0 z 2 +4 a2 z z+ 2 +2 a2 4

Problem 2.5 Problem 2.5 Problem 2.5 ❑ ❑ θ z

θ z r

r

2

0% 1 1θ o ˆ 10%nR λdl cos z. “Horizontal” components cancel, leaving: E =1 4πϵ 2 1 0 λdl ηcos dl θ ✓ˆ z. ˆ “Horizontal” components cancel, leaving: = =4πϵ4⇡✏ “Horizontal” components cancel, leaving:E E z. η 2%r 2 cos 0 0 z 2 2 2 % Rdl = 2πr. So Here,2 η2 =2 r2 +2 z2 , cos θ z= zη (both constants), while Here, z z, cos θ ✓==η (both constants), while So So Here, ηr ==r r++ , cos constants), whiledl =dl2πr. = 2⇡r. r (both 1 λ(2πr)z (2⇡r)z E =1 1 λ(2πr)z z. . zˆ 3/2 EE = = 4⇡✏ ˆ z. ˆ 2 3/2 4πϵ00(r(r22++zz3/2 4πϵ0 (r2 + z 2 )2 ))

r

Problem 2.6 Break it into rings of radius r, and thickness dr, and use Prob. 2.5 to express the field of each ring. Total charge of a ring is · 2⇡r · dr = · 2⇡r, so = dr is the “line charge” of each ring. Ering

1 ( dr)2⇡rz 1 = ; Edisk = 2⇡ z 3/2 2 2 4⇡✏0 (r + z ) 4⇡✏0 1



1

Z

0

R

r (r2

1

c ⃝2005 Pearson Education, Inc., Upper Saddle River, p This material Edisk = NJ. All2⇡rights z reserved. ˆ z. is 2 + z 2 may be protected under all copyright laws as they currently exist. 4⇡✏0No portionz of thisRmaterial reproduced, in any form or by any means, without permission in writing from the publisher.

c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is c protected ⃝2005 Pearson Education, Inc., laws Upper River, NJ. All No rights reserved. This material under all copyright as Saddle they currently exist. portion of this material mayis be protected underinall copyright as means, they currently No portion of this material may be reproduced, any form or laws by any withoutexist. permission in writing from the publisher. reproduced, in any form or by any means, without permission in writing from the publisher.

3/2

+ z2)

dr.

1 CHAPTER 2. ELECTROSTATICS

28 For R

z the second term ! 0, so Eplane =

For z

R,

and E

⇣

⌘

2 p 1 = z1 1 + R z2 R2 +z 2 1 2⇡R2 1 Q = 4⇡✏ 2z 2 = 4⇡✏0 z 2 , 0

Contents

1/2

⇡

1 z

⇣

1 4⇡✏0 2⇡

1

1 R2 2 z2

where Q = ⇡R2 .

ˆ z= ⌘

2✏0

ˆ z.

, so [ ] ⇡

1 z

1 z

+

1 R2 2 z3

=

R2 2z 3 ,

X

Problem 2.7 E is clearly in the z direction. From the diagram, dq = da = R2 sin ✓ d✓ d , r 2 = R2 + z 2 2Rz cos ✓, cos = z Rrcos ✓ .

z ✻

z

ψ

r

So θ

R

✲ y

φ Z R 1 R2 sin ✓ d✓ d (z R cos ✓) Ez = . d = 2⇡. 4⇡✏0 (R2 + z 2 2Rz cos ✓)3/2 ❂ ⇢ Z ⇡ x 1 (z R cos ✓) sin ✓ ✓ = 0 ) u = +1 = (2⇡R2 ) d✓. Let u = cos ✓; du = sin ✓ d✓; . 2 + z2 3/2 ✓ =⇡)u= 1 4⇡✏0 (R 2Rz cos ✓) 0 Z 1 1 !2 σR21sin θ dθ dφ(z z Ru " − R cos θ)du. Integral = can be done by partial fractions—or look it up. 3/2 . dφ = 2π. Ez4⇡✏ = 0 (2⇡R ) 21 (R22 + z 2 2Rzu)3/2 4πϵ0 (R + z − 2Rz cos θ)  ! ⇢ 1 # $ 1 1 2 1 π zu(z −RR cos θ) sin θ 1 2⇡R2 (z R) ( z R) θ = 0 ⇒ u = +1 2 p = = (2⇡R ) = . dθ. z 2 Let |zu = R| cos θ; du =R| − sin θ dθ; . 2 22Rzu 3/2 0 4⇡✏04πϵ0 (2πR σ) z 2 0R2(R |z + +2z+ θ = π ⇒ u = −1 z − 2Rz cos 1 θ)4⇡✏ ! 1 1 z − Ru = (2πR2 σ) Integral can 1 be q done by partial fractions—or look it up. q 1 2 4⇡R2 1 du. 2 4πϵ0the sphere),−1E(R For z > R (outside = 3/2 z. z0 −z2Rzu) 2 z =+ 4⇡✏ 4⇡✏0 z 2 , so E = 4⇡✏ z 2 ˆ 0 &1 % # $ 2 1 2πR σ (z − R) (−z − R) 1 1 zu − R 2 For z < R (inside), E(2πR so E = 0. √ = = − . z = 0,σ) 4πϵ0 z 2 R2 + z 2 − 2Rzu −1 4πϵ0 z 2 |z − R| |z + R| Problem 2.8 According to Prob. 2.7, all shells interior to the point (i.e. at smaller 1r) contribute as though their charge q 1 q 1 4πR2 σ ˆ z. = 4πϵ = z > R (outside thecenter, sphere), Ez all = exterior 2 , so E wereFor concentrated at the while contribute nothing. Therefore: 4πϵ0 z 2 shells 2 0 z 4πϵ z 0

For z < R (inside), Ez = 0, so E = 0.

E(r) =

1 Qint ˆ r, 4⇡✏0 r2

Problem 2.8 to Prob. 2.7,interior all shells to Outside the pointthe (i.e. at smaller r) charge contribute as though where QAccording charge to interior the point. sphere, all the is interior, so their charge int is the total were concentrated at the center, while all exterior shells contribute nothing. Therefore: 1 Q E= r.Qint 1ˆ ˆ r, E(r)4⇡✏ =0 r 2 4πϵ0 r2 Inside the Q sphere, only that fraction of the total which is interior to the point counts: where int is the total charge interior to the point. Outside the sphere, all the charge is interior, so

Qint =

4 3 3 ⇡r Q 4 3 3 ⇡R

=

r3 1 r3 1 1 Q Q, so EE== 1 Q ˆ r= r. r3.Q r2 ˆ 3 3 R 4⇡✏ R 4⇡✏ 2 0 0 R 4πϵ r 0

Inside 2.9 the sphere, only that fraction of the total which is interior to the point counts: Problem @ (a) ⇢ = ✏0 r·E = ✏0 r12 @r r2 · kr3 = ✏0 r12 k(5r4 ) = 5✏0 kr2 .

c ⃝2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be 2012 Pearson Upper from Saddle NJ. All rights reserved. This material is reproduced, in any form or by anycmeans, withoutEducation, permissionInc., in writing theRiver, publisher. protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

4 πr43 3 r3 3 1 r3 31 1 Q r so E = 1 Qr ˆ 1 r.Q Qint = 43 33πr Q = 3 Q, r 1= 3 2 3 Qint = Q = Q, so E = Q ˆ r = r. R 4πϵ R r 4πϵ 0 0R 3 3 2 3 3 πR4 πR R r3 4πϵ10 Rr3 r 1 4πϵ10 R3Q 4 3 πr3 Qint = 43 3 Q = 3 Q, so E = Q ˆ r= r. Problem 2.9 R 4πϵ0 R3 r2 4πϵ0 R3 ! Problem 2.9 1 ∂ 2 ! 3 " 3"πR1 2 (a) ρ = ϵ0 ∇·E = ϵ0 r2 ∂r1 r∂ · kr = ϵ k(5r4 ) = 5ϵ kr . (a) ρ = ϵ0 ∇·E = ϵ0 r2 ∂r r2 · kr3 0=r2ϵ0 r12 k(5r4 ) = 0 5ϵ0 kr2 . Problem 2.9 !#2 " 1 ∂ 1 3 4 2 ) = 4πϵ (b) By law: =Qϵenc )(4πR (a)Gauss’s ρ = ϵ0 ∇·E · #kr3 ==ϵ0ϵ(kR ) = 5ϵ kr2 . kR5 . 0 r 2=∂rϵ0 r E·da 0 r 2 k(5r (b) By Gauss’s law: Qenc = ϵ0$ E·da = ϵ0 (kR3 )(4πR2 )0 = 04πϵ0 kR5 .$ $ R CHAPTER 2. ELECTROSTATICS 2 $ = R (5ϵ $ R0 kr2 )(4πr By direct integration: Qenc = ρ#dτ dr) = 20πϵ0 k 5 0 r$4Rdr 429 = 4πϵ kR5 .! 0 3kr2 )(4πr 2 2 dr) = 20πϵ integration: ρ dτ = (5ϵ k 0 r dr = 04πϵ0 kR5 .! 0 0 (b)ByBydirect Gauss’s law: QencQenc = ϵ= E·da = ϵ (kR )(4πR ) = 4πϵ kR . 0 0 0 $ $ $R 4 Problem 2.10 2 2 5 Problem 2.10integration: QencH = ρ dτ = R (5ϵ By direct kr )(4πr dr) = 20πϵ k r dr = 4πϵ .!the surface 5 0 0 0 kRup 3 2 0 0 Think this cube as one of the charge. of0the 24 which make (b)Think ByofGauss’s = ✏80ofsurrounding = ✏0 (kR )(4⇡R )Each = Each 4⇡✏ kR . squares enc one of thislaw: cubeQas 8E·da surrounding the charge. of the 24 squares which make up the surface Rother RR 4 R Problem 2.10 of this larger cube gets the same fluxRas every one, 2 so: 2 5 By larger direct integration: ⇢ d⌧ (5✏ kr )(4⇡r dr) = 20⇡✏ k r dr = 4⇡✏ kR .X of this cube gets theQsame as=every other one, so: enc = flux 0 0 0 0 0 squares which make up the surface Think of this cube as one of 8 surrounding the charge. Each of the 24 Problem 2.10cube gets the same flux as of this larger every other one, so: % % % the 1charge.% Each of the 24 squares which make up the surface Think of this cube as one of 8 surrounding 1 E·da. E·da = =one, so:% E·da. of this larger cube gets the same flux as every other 24 %E·da 24 one whole 1 Z Z face one largewhole face E·da = 1 large E·da. cube E·da = 24cube E·da. one %face 24 whole large % one q whole q .cube The latter is ϵ10 q, 1by Gauss’s law. Therefore faceE·da = large = 0cube . The latter is ϵ0 q, by Gauss’s law. Therefore %E·da24ϵ 24ϵ one q0 face one Z . The latter is ϵ10 q, by Gauss’s law. Therefore face E·da = q 1 The latter is ✏10 q, by Gauss’s law. Therefore oneE·da = 24ϵ0. 24✏0 face one Problem 2.11 face Problem 2.11 1 # # = E(4πr2 ) =2 1 Qenc Gaussian surface: Inside: E·da Problem 2.11 1 = 0 ⇒ E = 0. ϵ 0 Gaussian surface: Inside: E·da = E(4πr ) = ϵ0 Qenc = 0 ⇒ E = 0. Problem✙2.11 ✙ (As in Prob. 2.7.) 2 # (As in Prob. 2.7.) ✲ 2 2= 0. 1 2 1 Qenc = 0σR ⇒ σR Gaussiansurface: surface:Outside: Inside: E(4πr E·da2= E(4πr E )= ˆ r . r ✲ ✲ Gaussian 1 ) = ϵ)0 ⇒ 2 E= 2 ϵ0 (σ4πR ✲ 2 Gaussian surface: Outside: E(4πr ) = ϵ0 (σ4πR ) ⇒ Eϵ= ˆ r . r r ✙ 0 ϵ0σR r2 2 (As in Prob. 2.7.) ✲ 1 2 2 ✲ Gaussian surface: Outside: E(4πr ) = ϵ0 (σ4πR ) ⇒ E = ˆ r . r ϵ0 r 2

Contents Contents Gaussian surface

Problem 2.12 Problem 2.12 Problem 2.12 Problem 2.12

}} }

1 4 1 34 #H = E · 4πr2 = 221 Qenc E·da πr ρ. 3 So E·da = =E E ·· 4πr 4⇡r ϵ0= = ✏110Q Q= encϵ0=3 1 4 ⇡r3 ⇢. So E·da ϵ0 enc = ϵ✏00 33πr ρ. So # Gaussian surface 21 1 E·da = E ·E4πr = ρrˆ 1Qenc = ϵ10 43 πr3 ρ. So r ✙ ϵ1 =E = 0 r. ✒ r✒ ✙ ⇢rˆ E = rr.. 3ϵ0 3✏0ρrˆ Gaussian surface 3ϵ01 r ✙ Q r. Q = 1 ρrˆ ✒ Problem 1 3E= πR42⇡R ρ, 2E r Q(as3 in Since QtotQ=tot43 = 3ˆ 3ϵ Since ⇢, E4πϵ E ˆ r Prob. (asProb. in2.8). Prob. R 2.13 0 R14⇡✏ 0 ρ, == ˆ r (as in 2.8).2.8). Since Qtot = 433πR 0 3R 4πϵ0 R ❄R % ❄ Gaussian surface Q 4 E 2· 2πs · l = 1ϵ100 Q = ϵ1in λl. So 2.8). r (as Prob. Since QE·da tot = = 3ˆ R 3 πR ρ, E = 4πϵ Renc 0 Problem 2.13 Problem 2.13 Problem 2.13 ✠ ❄ Problem 2.13 ✻ s # %H Gaussian surface Gaussian surface λ= E · 2⇡s · l1= 1 1 Qenc1= 1 1 l. So #E·da E·da Gaussian surface Problem 2.13 E·da =E ·E2πs ·l= = So So ˆ s· 2πs E = (same 2.1). == · l =asQEx. = λl.✏10λl. ✏10Qenc enc ! "# $ E·da 2πϵ0 sE · 2πs · lϵ0= ϵ0ϵ0 Qencϵ0= ϵ0ϵ0 λl. So ✠ ✠ ✻ sl ✻ s✻ # ✠ Gaussian surface s 1 E·da E · (same 2πsas· lEx. = Eq. Qenc = ϵ10 λl. So E =λ λλ ˆ s (same as 2.9). ϵ2.1). 0 E= ˆ s=s(same ˆ s E = as Ex. 2.1). 2⇡✏ E = (same as Ex. 2.1). ˆ s 0 ✠ & ! '( "# ) ✻ 2πϵ2πϵ $ 0s 0s & '( s ) 2πϵ0λs l l l E= ˆ s (same as Ex. 2.1). Problem 2.14 Problem 2.14 2πϵ0 s Problem Problem2.14 2.14& '( ) l H R R This2 material is Problem 2.14 c ⃝2005 Pearson Education, Inc., Upper2 Saddle 1 River, NJ. 1 All rights reserved. =E · 4⇡r = currently QencRiver, = ⇢ portion d⌧ = ✏1of (k¯ r)(¯ r This sin ✓material d¯ r d✓ dis ) c ⃝2005 Pearson Education, Inc., All rights reserved. protected under all E·da copyright laws as Upper they exist. No ✏Saddle ✏0 NJ. 0 0 this material may be Rasr they protected all copyright currently Nowriting portion of this may be 4 Problem 2.14 reproduced, in under any form or by1anylaws means, without permission in from the material publisher. 4⇡k rexist. ⇡k 3 4 = or kInc., 4⇡ r¯ d¯ rSaddle = ✏0River, rinrights . writing in anyEducation, form any Upper means, without permission from the publisher. Gaussian reproduced, surface c ⃝2005 Pearson NJ.✏0All reserved. This material is ✏0 by 4 = 0 protected under all copyright laws as they currently exist. No portion of this material may be r ✙ ✒ reproduced, in any form means, without permission in writing from the publisher. 1 or by any 2 ) E = ⇡kr ˆ r . Gaussian surface 4⇡✏0 r ✙ ✒

#

Problem 2.15 c 2012 Education, (i) QPearson so E =Inc., 0. Upper Saddle River, NJ. All rights reserved. This material is enc = 0, protected under all copyright laws as they currently exist. No portion of this material may be Problem % 2.15 & permission in& writing from the publisher. reproduced, in any form or by any means, without (ii) E·da = E(4πr2 ) = ϵ10 Qenc = ϵ10 ρ dτ = ϵ10 r¯k2 r¯2 sin θ d¯ r dθ dφ (i) Qenc = 0, so E = 0. ' ( &r k r−a & &ˆ r.k 2 d¯ r = 4πk =% 4πk 2 (r −1a) ∴ E =1 ϵ ϵ a 2 1 0 0 |E| (ii) E·da = E(4πr ) = ϵ0 Qenc = ϵ0 ϵ0ρ dτ r= ¯ sin θ d¯ r dθ ✻ dφ ϵ0 r¯2 r

'

& k r−a & 4πk 2r 4πk4πkb(r − a)4πk r = ϵ∴ (bE−=a), so 2 (iii)=E(4πr =r = ϵ0 a) d¯ ϵ0 ϵ0a d¯ 0 ϵ0 r '

( k b−a &ˆ E= 2 rb. 2 d¯ r= (iii) E(4πr ϵ0 ) =r4πk ϵ0 a ' (

4πk ϵ0 (b

− a), so

(

ˆ r.

|E| ✻

✲

Gaussian surface r ✙ ✒

111

Contents 30


Problem 2.15 Qenc = 0, so E = 0. Problem(i)2.15 & & (i) Qenc =%0, so E = 0. 2 ) = ϵ10 QencR= ϵ10 ρ dτR= ϵ10 r¯k2 r¯2 sin θ d¯ r dθ dφ H (ii) E·da = E(4πr 1 1 2 2 '1 k2 r¯( Problem 2.13 (ii) E·da = E(4⇡r ) = Q = ⇢ d⌧ = sin ✓ d¯ r d✓ d enc ✏0 ✏0 ✏ r¯ &r 4πk 2.16 ✓ k 0◆r − a ˆ Problem r. r = 4πk = ϵ0 ϵ0 (r − a) ∴ kE =r ϵ a r2 a d¯ R |E| r 4⇡k 0 ✻ = 4⇡k d¯ r = (r a) ) E = ˆ r . ✏0 ✏0 a 2 ✏0 r & 4πk 4πk b d¯ r = (b − a), so (iii) E(4πr2 ) = R ϵ0 a b ϵ0 ' a d¯ (iii) E(4⇡r2 ) = 4⇡k a), so r( = 4⇡k ✏0 ✏0 (b k b−a ! E✓=b a ◆ 2 ˆ r. Problem Problem 2.13 ✛ E·da = E · 2πs · l = ϵ10 Qenc = ϵ10 ρπs2 l; Problem2.13 2.13k(i) Gaussian surface ϵ0 r. E= ˆ r 2 ✲ Problem 2.16 Problem Problem 2.16 2.16 ✏0 r ρs l r b ˆ s. a E= 2ϵ0 Problem 2.16 2.16 Problem H ✛ E·da = E · 2⇡s · l = 1 Q = ✏10 ⇢⇡s2 l; Gaussian c ⃝2005 Pearson Education, Inc., Upper NJ. All! reserved. This material is✏0 enc !rights !surface ✻ Saddle River, 222 under all copyright✛ laws they currently exist. No portion of may (i) Q l; E·da E 2πs s asGaussian (i) Q = ρπs l;l; ✛ E·da = EE·!·⇢s 2πs = (i) Qenc = ϵϵ110ϵ010ρπs ρπs ✛ E·da= =this ·material 2πs···lll= = ϵϵ110ϵ010be (i) protected(ii) 1 1 enc 2 enc= surface Gaussian surface Gaussian surface E·da =E · 2πs ·l = E= ˆ s.publisher. reproduced, in any form or by any means, without permission in writing from the ϵ0 Qenc = ϵ0 ρπa l; ρs ρs ρs 2✏ 0 lll 2 sˆ E sˆ E = s... E= = ˆ 2ϵ 2ϵ 2ϵ000 E = ρa ˆ s. 2ϵ0 s l

Contents Contents

(ii) (ii) (ii) (ii)

✻ ✻ ✻ sss (iii) lll

✛ H ✛ ✛ ✻Gaussian Gaussian surface Gaussiansurface surface E·da = E · 2⇡s ·1l = ✏10 Qenc = ✏10 ⇢⇡a2 l; ✛ Gaussian!!!surface s 222 E·da ·!··2πs l; E·da = EE⇢a 2πs = Q = ρπa l;l; E·da= =E 2πs = ϵϵ110ϵ010Q Qenc = ϵϵ110ϵ010ρπa ρπa 2 ···lll= enc enc= 1 Q E·da = E · 2πs · l = E = 222 ˆ s. ϵ0 enc = 0; ρa ρa ρa 2✏0 s ˆ sˆ E sˆ E = s...E = 0. E= = 2ϵ 2ϵ 2ϵ000sss l ✛ ✛ ✛ Gaussian Gaussian surface Gaussiansurface surface

✻ ✻ ✻ sss

!!!H 1 = E·da ···ll·l= E·da= ·2πs 2⇡s l==ϵϵ110ϵ01✏Q Q = 0; E·da = EEE···2πs 2πs = QQ ==0; 0;0; E·da ==E enc enc enc enc 00 E = 0. E = 0. EE == 0.0.

(iii) (iii) (iii) (iii)

Contents

|E| ✻

lll

Problem 2.17 ✲ |E| |E| |E|a✻ On the x z plane E = 0 by symmetry. Set up Gaussian “pillbox” with and the a one face b in this plane s ✻ ✻ other at y. c2.17 Pearson Upper River, NJ. Allup rights reserved. This material with is Problem⃝2005 On theEducation, x z planeInc., E= 0 bySaddle symmetry. Set a Gaussian “pillbox” one face in this plane protected under all copyright laws as they currently exist. No portion of this material may be and the other at y. reproduced, in any form or by any means, without permission in writing from the publisher. Gaussian pillbox

R

✲ ✲ 1 1 ✲ E·da = aa E a · A = ✏0bbbQenc = ✏0ssAy⇢; s ⇢ E= yy îsisis(for |y| < d). cc⃝2005 ⃝2005 Pearson material ⃝2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material c PearsonEducation, Education,Inc., Inc.,Upper UpperSaddle SaddleRiver, River,NJ. NJ.All Allrights rightsreserved. reserved.This This material 0maybe protected ✲ protected under all copyright laws as they currently exist. No portion of this material may be protectedunder underall allcopyright copyright lawsas asthey theycurrently currentlyexist. exist.No Noportion portionof ofthis thismaterial material✏may be y ✛ laws ❂

reproduced, reproduced, in any form or by any means, without permission in writing from the publisher. reproduced,in inany anyform formor orby byany anymeans, means,without withoutpermission permissionin inwriting writingfrom fromthe thepublisher. publisher.

Qenc =

1 ϵ0 Adρ

⇒ E=

ρ > d). Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is dy ˆ c(for 2012yPearson ϵ0 protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

E✻ ρd ϵ0

−d

d

✲ y

Problem 2.17 On the x z plane E = 0 by symmetry. Set up a Gaussian “pillbox” with one face in this plane and the other at y. !

Gaussian pillbox ❂ CHAPTER 2. ELECTROSTATICS

E·da = E · A = ϵ10 Qenc = ϵ10 Ayρ; ρ E = yy ˆ (for |y| < d). ϵ0 31

✲ y✛ ρ Qenc =1 ϵ10 Adρ ⇒ E =⇢ d y ˆ (for y > d). Qenc = ✏0 Ad⇢ ) E = ϵd0 y ˆ (for y > d). ✏0 E✻ ρd ϵ0

−d

d

✲ y

Problem 2.18 Problem 2.18 From Prob. 2.12, the field inside the positive sphere is E+ = 3✏⇢0ρr+ , where r+ is the vector from the positive From Prob. 2.12, the field inside the positive sphere is E+ = 0 r+ , where r⇢+ is the vector from the positive center to the point in question. Likewise, the field of the negative3ϵsphere is 3✏0ρr . So the total field is center to the point in question. Likewise, the field of the negative sphere is − 3ϵ0 r− . So the total field is ⇢ EE== ρ(r(r r r) ) r + − + − ✛ r−*r 3✏3ϵ 0 ✯− 0 r+r+ ✕ dd ⇢ρ r But But(see (seediagram) diagram)r+r+ −r r−==d.d.SoSoEE==3✏0 d.d. + + 3ϵ0 Problem Problem2.19 2.19

✓ ◆ Z" Z" # $ ˆ %& rˆηˆ 11 11 ηˆr r⇥E = r⇥ ⇢ d⌧ = r⇥ (sinceρ⇢depends dependson onr′r,0 ,not notr)r) ∇× r 22 ρ dτ = ∇× r2 2 ρ ⇢dτd⌧ (since ∇×E = 4⇡✏ 4⇡✏ 0 0 4πϵ η 4πϵ00 η ✓$ % ◆ rˆηˆ ==0 0 (since 0, from from Prob. Prob. 1.62). 1.63). (sincer⇥ ∇× 22 = rη = 0,

Problem Problem2.20 2.20

' ' 'x x ˆ y ˆy zˆ ˆ ˆ z '' '@ ˆ ' ∂ @∂ @∂ = ' (1) ==k k@x x[ˆ (0 ˆ(0 ˆ z(0 (1)r⇥E ∇×E x(0 −2y) 2y)++y y ˆ(0 −3z) 3z)++ ˆ z(0 −x)] x)]6≠=0,0, 1 1 ' ∂x @y∂y @z∂z ' =k k[ˆ ' xy2yz xy 2yz3zx 3zx' sosoEE ananimpossible impossibleelectrostatic electrostaticfield. field. 1 1is is

' ' 'x ˆ z '' x ˆ y ˆy ˆ zˆ '@ ˆ @∂ @∂ ' '∂ x(2z −2z) 2z)++y y ˆ(0 −0)0)++ ˆ z(2y −2y)] 2y)]==0,0, (2)r⇥E ∇×E (2) ==k k@x [ˆ x[ˆ (2z ˆ(0 ˆ z(2y 2 2 ' =k k ' ∂x @y∂y @z∂z = 2 2 ' 2y 2xy + 2z 2yz ' y 2xy + z 2yz

possibleelectrostatic electrostaticfield. field. sosoEE a apossible 2 2is is

z

Let’s go by the indicated path: E·dl = (y 2 dx + (2xy +

6

c ⃝2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be any form or by any means, without permission in writing from the publisher. 2 r(x0 , y0 , z0 ) zreproduced, )dy + 2yzindz)k

6

Step I: y = z = 0; dy = dz = 0. E·dl = ky 2 dx = 0. Step II: x = x0 , y : 0 ! y0 , z = 0. dx = dz = 0. E·dl = k(2xy + Rz 2 )dy = 2kx0 y dy. R y E·dl = 2kx0 0 0 y dy = kx0 y02 . II

Step III : x = x0 , y = y0 , z : 0 ! z0 ; dx = dy = 0. E·dl = 2kyz dz = 2ky0 z dz.

x

I⇢⇢ ⇢ = ⇢⇢ = ⇢ II


III

-

-y

32 R

III

E·dl = 2y0 k

V (x0 , y0 , z0 ) =

(x0 ,y R0 ,z0 )

R z0


z dz = ky0 z02 .

0

E·dl =

k(x0 y02 + y0 z02 ), or V (x, y, z) =

0

k(xy 2 + yz 2 ).

@ @ @ rV =k[ @x (xy 2 +yz 2 ) x ˆ+ @y (xy 2 +yz 2 ) y ˆ+ @z (xy 2 +yz 2 ) ˆ z]=k[y 2 x ˆ+(2xy+z 2 ) y ˆ+2yz ˆ z]=E. X

Check :

Problem 2.21 Rr V (r) = E·dl. 1

8 R) : E = : Inside the sphere (r < R) :

Rr ⇣

So for r > R: V (r) =

1

RR⇣

and for r < R: V (r) = =

1 q 4⇡✏0 r¯2

1

q 1 4⇡✏0 2R

When r > R, rV =

q @ 4⇡✏0 @r

When r < R, rV =

q 1 @ 4⇡✏0 2R @r

⌘

1 q 4⇡✏0 r¯2

✓

V(r)

1 4⇡✏0 q

⌘

Rr ⇣

d¯ r ◆ 2

r R2

3

ˆ r= ⇣ 3

1 r

d¯ r=

R

.

q 1 r, 4⇡✏0 r 2 ˆ

⌘ 2

r R2

ˆ r=

1 r¯

E=

r 1

⌘

so E =

q 1 4⇡✏0 2R

q 1 r. 4⇡✏0 R3 rˆ

q 1 , 4⇡✏0 r

=

q 1 ¯ 4⇡✏0 R3 r

1 q r. 4⇡✏0 r 2 ˆ

d¯ r=

rV = 2r R2

q 4⇡✏0

h

1 R

q 1 r. 4⇡✏0 r 2 ˆ

1 R3

r 2 R2 2

⌘i

X

q r r; 4⇡✏0 R3 ˆ

ˆ r=

⇣

so E =

rV =

q 1 r.X 4⇡✏0 R3 rˆ

1.6 1.4 1.2

(In the figure, r is in units of R, and V (r) is in units of q/4⇡✏0 R.)

1 0.8 0.6 0.4 0.2 0.5

1

1.5

2

2.5

3

r

Problem 2.22 1 2 E = 4⇡✏ ˆ s (Prob. 2.13). In this case we cannot set the reference point at 1, since the charge itself 0 s extends to 1. Let’s set it at s = a. Then ⇣s⌘ Rs⇣ 1 2 ⌘ 1 V (s) = d¯ s = 2 ln . 4⇡✏0 s¯ a 4⇡✏0 a

(In this form it is clear why a = 1 would be no good—likewise the other “natural” point, a = 0.) rV =

1 4⇡✏0 2

@ @s

ln

Problem 2.23 R0 V (0) = E·dl = 1 =

k ✏0

1

a b

ln

a b

s a

ˆ s=

Rb

1 4⇡✏0 2

k (b a) 1 ✏0 r 2

1+

a b

1 s sˆ

dr

=

Ra b

✓ ◆ k b = ln . ✏0 a

E. X

k (r a) ✏0 r 2

Problem 2.24 Using Eq. 2.22 and the fields from Prob. 2.16: Rb Ra Rb V (b) V (0) = E·dl = E·dl E·dl = 0 0 a

⇢ 2✏0

dr

Ra 0

R0 a

s ds

(0)dr =

⇢a2 2✏0

Rb

k (b a) ✏0 b

k ✏0

ln

a b

+a

1 a

1 ds a s


1 b

Contents

33


=

⇣

⇢ 2✏0

Problem Problem2.25 2.25

⌘

s2 2

a 0

+

⇢a2 2✏0

b

ln s|a =

⇢a2 4✏0

✓

1 + 2 ln

✓ ◆◆ b . a

11 2q 2q q .. ! 4⇡✏ 4πϵ00 z 22+ "dd #22 z + 22 p R L L 1 √z 2 + x2 ) %L $ LL p 2λdx (b) V = 4⇡✏ 10 dx2 = 4⇡✏λ0 ln(x + 2 + x2 )% L z +x √ (b) V = 4πϵ z = ln(x + 4πϵ0 −L z 2 +x2 −L 0 " # ! p p ' & √ 2 2 L + z 2+ L 2 z2 + ( L+ ) L2 √ λ L + z + L p = ln = λ ln L+ z2 +L2 . = 4⇡✏0 ln L + √z 2 + L2 . = 2⇡✏ 2πϵ00 ln z z 4πϵ0 −L + z 2 + L2 (a) (a) VV ==

) %R √ $ R σ 2πr dr σ (* ⇣2 Z 1 2p 2 )% = R 2−z . ⌘ R√ p (c) V = r + z R = 2πσ ( + z 2⇡r dr 1 2 2 4πϵ0 0 pr +z (c) V = = 2⇡ ( r2 +0z 2 ) 2ϵ =0 R2 + z 2 z . 4⇡✏0 0 4⇡✏0 2✏0 0 r2 + z 2 ∂V In each case, by symmetry ∂V ∴ E = − ∂V z. ∂y = ∂x = 0. ∂z ˆ @V In each case, by symmetry @V = = 0. ) E = @V z. @y @x @z ˆ

z

r x

1 10 4πϵ

" # 2qz 1 1 2z ”3/2 ˆ z= z (agrees with Prob. 2.2a). (a) E = − 4πϵ 2q ✓− 21 ◆“ " #3/2 ˆ 0 d 2 2 4πϵ0 z12 + " d #22qz 1 1 z +( 2 ) 2z 2 (a) E = 2q z= ˆ z (agrees with Ex. 2.1). ⌘3/2 ˆ 4⇡✏0 + 2 ⇣ 2 4⇡✏0 z 2 + d 2 3/2 , d 2 z + 2 1 √ 21 λ √1 2z − (−L+√1z2 +L2 ) 12 √z21+L2 2z ˆ z (b) E = − 4πϵ 2 2 2 0 z 2 +L2 ⇢ (L+ z +L ) 1+ 1 1 1 1 , √ 1 2Lλ p2 −L−√z2 +L p √ 1 pˆ (b) E = = − λ √ z p −L+ z2 +L 2z 2 ˆ 2z with ˆ z Ex. 2.1). z z = (agrees 2 2 2 2 2 2 2 2 2 2 4⇡✏4πϵ 2 2 +L ( L +4πϵz0 z+ zL2 + ) L2 z + L2 0 0 (L z 2+ +L2 z + L )(z +Lz )−L ( ) p p z L + z 2 +-L2 L z2 + 2L 1 . L2 , + p = σ 1p 1 ˆ z= ˆ z (agrees with Ex. 2.2). σ2 z2 2 2 2 2 4⇡✏0 zProb. z +2.6). L2 0 √ z 2 +2L z = (z +1L−)√ L ˆ z (agrees with (c) E = − 2ϵ4⇡✏ 2 R +z 2z − 1 ˆ 0 2ϵ0 R2 + z 2 ⇢  1 1 z the right-hand in (a)1 is ˆ 0 , which, naively, suggests E = −∇V p charge 2z (c) EIf = z−q, = then V 1 =p ˆ z (agrees with Prob. 2.6). = 0, in contradiction 2 2 2 2✏0 2 toRProb. 2✏0 is that R + z 2.2b. The point z 2 know V on the z axis, and from this we cannot with the answer we + only ∂V hope to compute Ex = − ∂V ∂x or Ey = − ∂y . That was OK in part (a), because we knew from symmetry that If the right-hand charge in (a) is q, then V = 0 ,sowhich, naively, E =is insufficient rV = 0, in Ex = Ey = 0. But now E points in the x direction, knowing V onsuggests the z axis tocontradiction determine E. 2 with the answer to Prob. 2.2. The point is that we only know V on the z axis, and from this we cannot hope @V to compute Ex = @V @x or Ey = @y . That was OK in part (a), because we knew from symmetry that Ex = Ey = 0. But now E points in the x direction, so knowing V on the z axis is insufficient to determine E.

Problem 2.26 ✻ c ⃝2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

b h

1 V (a) = 4⇡✏0

Z

0

p

2h

✓

2⇡r

r

(where r =

r

◆

2⇡ 1 p h p ( 2h) = dr = 4⇡✏0 2 2✏0 p / 2)


h

r¯ r

a

✲

34

1 V (b) = 4⇡✏0

Z

p

0

2⇡ 1 p 4⇡✏0 2

= =

p 2 2✏0

2h

Z

✓

p

2⇡r r¯

2h

0

q h2 +


◆

dr

p h2 +

r

2

r

p

(where r¯ =

r 2

2h r

p

2h r

q

h2 +

r

p

2

2h r )

dr

q h + p ln(2 h2 + 2

r

p

2

2h r + 2 r

p

p

2h

2h) 0

 p p p h h = p h + p ln(2h + 2 2h 2h) h p ln(2h 2h) 2 2✏0 2 2 p ! p p i h h h 2+ 2 h p ln(2h + 2h) ln(2h p = p 2h) = ln = ln 4✏0 4✏0 2 2✏0 2 2 2 p p i h hh = ln(1 + 2). ) V (a) V (b) = 1 ln(1 + 2) . 2✏0 2✏0

(2 +

p

2)2

2

!

3

Problem 2.27 Problem 2.27 Cut the Cut cylinder into slabs, shownasinshown the figure, the cylinder intoasslabs, in theand figure, and use result of result Prob. of 2.25c, z !with x and ⇢ dx: use Prob.with 2.25c, z → x!and σ → ρ dx: V =

=

! "# $

z+L/2 R

pz+L/2 % &√ ' ρ R2 + x2 2x dx2 R + x − x dx 2ϵ0 z L/2 # z−L/2 p ⇥ p ⇤ z+L/2 )*z+L/2 √ ( √ ⇢ 1 21 + 2 +2R2 ln(x ρ 2 R2 + x2 ) 2 x2 2 2 * 2 x R x + x z−L/2 R + x z) − 2✏0 2 = 2ϵ0 2 x R + x + R ln(x + L/2 8 9 3 r 2 3 9 8 r 2 q q L 2 < = z+ L + R2 +(z+ q q L +2 )R2 + z+ L 2 = < 2 2 ⇢ z+ 2 L L 2 L 2 L 2 2 ( 4 2 5 2zL2 ) . 5 r 2 ρ z+ R + z+ z R + z +R ln 2 L L L L ( ) ( ) ( ) ( ) 2 2 4 r2 4✏0 : = 2 2 R +(z+ 2 ) −(z− ) 2 R +(z− ) +R ln −2zL . 4ϵ0 :(z+ 2 ) 2 2 2 z L + R2 + z L 2 ; ( L +2 )R2 +(z− L; 2 z− )

⇢ 2✏0

=

z− L 2

V =

2

(Note:

2 2 z + L2− &z++zL '2L2+ &z=− Lz 2'2 (Note: 2 2

L2 4 =zL −z 2 −

+ z− zL 2

L2 L zL +2 4 = + z − zL 4 2

L "#

!

$

✲" ✛

$! x

dx

2

L +2zL.) 4 = −2zL.) 2

s (s ◆2 ✓ ◆2 2 , , L 2& +✓ ' '2 z + . .z2 L2 & @V ˆ z⇢ L L2 L 2 2 2 2 z+ 2 R + z 2 z − L2 qL E = rV = ˆ z = ∂V Rˆ + qL zρ+ z +2 2 2 / / @z= −ˆ 2 z+ 2 + 2 z− 2 − R + E = −∇V z 4✏0 = − L L '2 − R + '2 ∂z 4ϵ0 2R + z +22 & 2R + z 2 2 & R + z + L2 R + z − L2 L L z+ 2 z 2 " ) 1 +0 q 1+ q z− L # z+ L 2 L 2 1 2 2+ 1 (+z qL2 )2 2 L 2 1 + R2 + z+q R ( ) 2 2 L 2 2 2 R +(z+ 2 ) R +(z− 2 ) 2L q +R 2 q 2 2 − +R 2+ / 2+ / ' − 2L z + L2 + RL z +2L2 & z L '2L2 + RL z 2L2 & L 2 z + 2 + R + z{z+ 2 z − 2 + R + }z − 2 | $! 1 " # 1 q 1 q 1 2 / − / L 2& '22 + '2 R2 + z +2L2 & R z 22 R + z + L2 R + z − L2 8 s 9 s ✓ ◆2 ✓ ◆2 < = ⎧ ⎫ , Lˆ z⇢ .R22 + ,z L - 2L .2 2⎨ ⎬ E= 2 ˆ R + z + 2 zρ 4✏E0 : 2 z+ L 2 z − L ; − 2L =− 2 R2 + − 2 R2 + ⎭ 4ϵ0 ⎩ 2 2 ⎡ ⎤ River, NJ. All rights reserved. This material is , c 2012 , Inc.,-Upper Saddle Education, - Pearson 2 copyright laws 2 protected under.all as they.currently exist. No portion of this material may be ρ ⎣ L L 2 + z + in any + form R or2by without ⎦ˆ +anyzmeans, − = L − Rreproduced, z. permission in writing from the publisher. 2ϵ0 2 2

Problem 2.28

z ✻

Orient axes so P is on z axis. P

V =

1

%

ρ

dτ.

=

2

Here√ρ is constant, dτ = r sin θ dr dθ dφ,

z

θ r

✲

1 Hello

35

CHAPTER 2. ELECTROSTATICS 2 ⇢ 4 = L 2✏0

s

✓

L R2 + z + 2

◆2

+

s

✓

L 2

R2 + z

◆2

3

5ˆ z.

Problem 2.28 Orient axes so P is ⇢ on z axis. R ⇢ Here ⇢pis constant, d⌧ = r2 sin ✓ dr d✓ d , 1 V = 4⇡✏0 r d⌧. r = z 2 + r2 2rz cos ✓. V = R⇡ 0

p

⇢ 4⇡✏0

R

2 pr sin ✓ dr d✓ d z 2 +r 2 2rz cos ✓

d✓ =

sin ✓ z 2 +r 2 2rz cos ✓

)V =

But ⇢ =

· 2⇡ · 2

q 4 3, 3 ⇡R

( Rz 0

so V (z) =

Contents

0

p

1 rz

=

⇢ 4⇡✏0

R 2⇡

;

1 rz

+

⇣

3q 1 2✏0 4⇡R3

=

1 4⇡✏0

1 2 4⇡✏ r R 00

R

⇢ r

d⌧ =

⇢(r )[ 4⇡ 3 (r

1 4⇡✏0 0

R

2

)

z2 3

R2

r )] d⌧ =

✲y

x❂

=

⌘

⇢(r0 ) r2

θ r

φ

p p ⇡ 1 2rz cos ✓ 0 = rz r2 + z 2 + 2rz r2 + z 2 1 ⇢ 2/z , if r < z, |r z|) = 2/r , if r > z.

r r dr

z

Problem 2.29 r2 V =

z

r2 + z 2

RR 1

r

P

d = 2⇡.

(r + z

1 2 z r dr

z ✻

n

⇢ ✏0

1 z3 z 3

⇣

+

R2 z 2 2

3

z2 R2

⌘

o

=

⇢ 2✏0

⇣

R2

z2 3

q 8⇡✏0 R

⌘

.

✓

3

=

q 8⇡✏0 R

1

d⌧ (since ⇢ is a function of r0 , not r)

r

1 ✏0 ⇢(r).

; V (r) =

r2 R2

2rz

◆

. X

X

Problem 2.30. Problem 2.30.

(a) ˆn ˆn nnalways ˆn (a)Ex. Ex.2.5: 2.4: EEabove = 2✏2ϵσ00n ˆ; ;EEbelow = −2✏2ϵσ00n ˆ(ˆ (ˆ alwayspointing pointingup); up);EEabove Ebelow =✏0ϵσ0n ˆ. .X! above= below= above −E below= Ex. E ==✏0ϵσ0. .X! Ex.2.6: 2.5: At Ateach eachsurface, surface,EE==00one oneside sideand andEE==✏0ϵσ0 other otherside, side,so so ∆E 22

R Prob. rˆ rˆ E == ✏0ϵσ0ˆ rˆ Prob.2.11: 2.11: EEout = ✏0ϵσR r== ✏0ϵσ0ˆ r; ;EEinin==00; ;so so ∆E r. .X! out= rr2 2ˆ 0

✻ s

(b) (b)

H! Outside: E·da E·da==E(2⇡s)l E(2πs)l== 1ϵ10QQenc = ϵ10(2⇡R)l (2πR)l) ⇒EE== ϵσ0RRsˆ ˆ s = σsˆ s(at (atsurface). surface). enc= Outside: ✏0 ✏0 ✏0 s s = ✏0ϵ0ˆ

✻ R

c σ ⃝2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is s.this .X!material Inside: soEE== ∆E enc Inside: QQenc ==0,0, so 0.0. )∴ E ==✏0ϵof0ˆ sˆ protected all copyright laws as they currently exist. No portion may be " #$ under % reproduced, in any form or by any means, without permission in writing from the publisher.

l

2

(c) Vout = R✏R0 r2 σ = R✏Rσ (at surface); Vin = R0 ; so Vout = Vin . X (c) Vout = ϵ0 r = 0ϵ0 (at surface); Vin = ✏Rσ ϵ0 ; so Vout = Vin . ! @Vout @rout = ∂V ∂r =

R2 ✏0Rr22σ = − ϵ0 r 2 =

@Vin in out (at surface); @V = 0 ; so @V = ✏0σ. X @rout @r ∂Vin ∂V ∂Vin −✏0ϵσ0 (at surface); @r = 0 ; so − ∂r ∂r ∂r = − ϵ0 . !

c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be Problemin2.31 reproduced, any form or by any means, without permission in writing from the publisher.

1 4πϵ0

&

qi rij

1 4πϵ0

'

−q a

√q 2a

+ , + 2 q 1 . ∴ W4 = qV = −2 + √ 4πϵ0 a 2

(a) V =

=

1

)

−q2

*

+

1

−q a

(

)

=

q2

q 4πϵ0 a

q2

) −2 +

*

√1 2

*

. (1)

(4)

−

+

+

−

36


Problem 2.31 (a) V =

P

qi rij

(b) W1 = 0, W2 = Wtot =

2 1 q 4⇡✏0 a

=

n

q a

1 4⇡✏0

1 4⇡✏0

n

⇣

1+

q2 a p1 2

⌘

+

pq 2a

+ ✓ ◆ 2 q 1 ) W4 = qV = 2+ p . 4⇡✏0 a 2 1 4⇡✏0

; W3 =

q a

o

⇣

⇣

q 4⇡✏0 a

=

2 pq 2a

q2 a

2+

p1 2

⌘

.

⌘

; W4 = (see (a)). ✓ ◆ o 1 2q 2 1 2 + p12 = 2+ p . 4⇡✏0 a 2

1

1 4⇡✏0

(1) r

(2)

r+

+

r(4) r

(3)

Problem 2.32 Conservation of energy (kinetic plus potential): 1 1 1 qA qB 2 2 mA vA + mB vB + = E. 2 2 4⇡✏0 r At release vA = vB = 0, r = a, so E=

1 qA qB . 4⇡✏0 a

When they are very far apart (r ! 1) the potential energy is zero, so 1 1 1 qA qB 2 2 mA vA + mB vB = . 2 2 4⇡✏0 a Meanwhile, conservation of momentum says mA vA = mB vB , or vB = (mA /mB )vA . So 1 1 2 mA vA + mB 2 2 vA =

s

✓

mA mB

◆2

1 qA qB 2⇡✏0 (mA + mB )a

2 vA =

✓

1 2

✓

◆ mA ; mB

mA mB

◆

2 (mA + mB )vA =

vB =

s

1 qA qB . 4⇡✏0 a

1 qA qB 2⇡✏0 (mA + mB )a

✓

◆ mB . mA

Problem 2.33 From Eq. 2.42, the energy of one charge is W =

1 1 X 1 1 1 ( 1)n q 2 q 2 X ( 1)n qV = (2) = . 2 2 4⇡✏0 na 4⇡✏0 a 1 n n=1

(The factor of 2 out front counts the charges to the left as well as to the right of q.) The sum is can get it from the Taylor expansion of ln(1 + x): ln(1 + x) = x

1 2 1 3 x + x 2 3

ln 2 (you

1 4 x + ··· 4

with x = 1. Evidently ↵ = ln 2 . c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

2

37


Problem 2.34 R (a) W = 12 ⇢V d⌧ . From Prob. 2.21 (or Prob. 2.28): V =  3 q⇢ r 3 4⇡r dr = 3 4✏0 R 3 0 ✓ 2◆ 2 q⇢ 2 qR q 1 3q = R = = . 4 3 5✏0 5✏0 3 ⇡R 4⇡✏0 5 R

1 1 q W = ⇢ 2 4⇡✏0 2R

(b) W =

✏0 2

R

Z

R

✓

r2 R2

◆

2

E 2 d⌧ . Outside (r > R) E =

1 q r 4⇡✏0 r 2 ˆ

⇢ 2✏0

⇣

R2

1 r5 R2 5

R 0

Problem 2.32 ! (a) W = 21 ρV dτ . From Prob. 2.21 (or P ⌘ ⇣ ⌘ 2 q r2 1 = 4⇡✏ 3 Rr 2 & 3 $ R% 0 2R 1 1 q r2 W = ρ ◆ 3 − 2 4πr2 dr ✓ 2 R4πϵ R 3 0 2R 0 q⇢ = R3 % 2 4✏0 R qρ5 2 qR2 q 1 3q = = R = 5ϵ0 5ϵ0 43 πR3 4πϵ0 5 R

; Inside (r < R) E =

(b) W = q 1 r. 4⇡✏0 R3 rˆ

ϵ0 2

!

E 2 dτ . Outside (r > R) E =

*$ ∞ 1 1 2 ϵ0 2 q (r 4π dr) + ∴ W = (Z ) 2 4 Z R ⇣ ⌘2 1 2 (4πϵ ) r 0 R ✏0 1 1 r *% )W = q2 (r2 4⇡ dr) + (4⇡r2 dr) &)∞ % 5& 4 2 (4⇡✏0 )2 R3 R r 0 1 q2 1 )) 1 r = − ) + 6 (✓ ) ◆ ✓ ◆ ✓ ◆ R 1 r R R 5 1 q2 1 1 r5 1 q2 1 1 1 3 q 2 4πϵ0 2 = + 6 = + = .X 4⇡✏0 2 r R R 5 0 4⇡✏0 2 R 5R 4⇡✏0 5 R ,! . V E·da + V E 2 dτ , whe (c) W = ϵ20 S H R arbitrary. Let’s use a sphere of radius a > R. (c) W = ✏20 V E·da + V E 2 d⌧ , where V is large enough to enclose all the charge, but otherwise S 1 q *$ % arbitrary. Let’s use a sphere of radius a > R. Here V = 4⇡✏0 r . &% & 1 q 1 q ϵ0 r2 sin W (Z ✓ ) 2 ◆✓ ◆ ◆2= 2 Z R Z a✓ 4πϵ r 4πϵ r 0 0 ✏0 1 q 1 q 1 q 2r=a W = r2 sin ✓ d✓ d + E 2 d⌧ + (4⇡r/ dr) 2 2 4⇡✏0 r 4⇡✏0 r2 4⇡✏ r 0 0 R ϵ0 q2 1 q2 4π r=a = 4π + + 2 2 5R ⇢ ✓ ◆a 2 (4πϵ ) a (4πϵ ) ( 2 2 0 0 ✏0 q 1 q 4⇡ 1 1 2 / 0 = 4⇡ + + 4⇡q 2 1 q 1 1 1 1 2 (4⇡✏0 )2 a (4⇡✏0 )2 5R (4⇡✏0 )2 r R = = + − + ⇢ 4πϵ 2 a 5R a R 4π 0 1 q2 1 1 1 1 1 3 q2 = + + = .X 4⇡✏0 2 a 5R a R 4⇡✏0 5 R As a → ∞, the contribution from the surf ⇣ ⌘ " # 2 1 q 1 q2 while As a ! 1, the contribution from the surface integral 4⇡✏0 2a goes to 4πϵ zero, the picks volume ( 6a − 1) up integral the slack. 0 2a 5R ⇣ ⌘ 2 1 q 6a 1) picks up the slack. Problem 2.33 4⇡✏0 2a ( 5R

Problem 2.35

✓

◆ 1 q¯ dW = d¯ q V = d¯ q , (¯ q = charge on sphere of radius r). 4⇡✏0 r 4 r3 q¯ = ⇡r3 ⇢ = q 3 (q = total charge on sphere). 3 R 4⇡r2 3q d¯ q = 4⇡r2 dr ⇢ = 4 3 q dr = 3 r2 dr. R ⇡R ✓ 3 3◆ ✓ ◆ 1 qr 1 3q 2 1 3q 2 4 dW = r dr = r dr 3 3 4⇡✏0 R r R 4⇡✏0 R6 ✓ 2◆ Z 1 3q 2 R 4 1 3q 2 R5 1 3q W = r dr = = .X 6 6 4⇡✏0 R 0 4⇡✏0 R 5 4⇡✏0 5 R


✯ r

✮

d¯ q

q¯

c ⃝2005 Pearson Ed protected under al reproduced, in any

38

Problem 2.36 R (a) W = ✏20 E 2 d⌧. ⇣ ⌘2 R b q W = ✏20 4⇡✏ a 0

E=

1 r2

2


1 q 4⇡✏0 r 2

(a < r < b), zero elsewhere. ✓ ◆ Rb 1 q2 1 1 2 q2 4⇡r2 dr = 8⇡✏ = . 2 a r 0 8⇡✏0 a b 2

q 1 q 1 q 1 q 1 (b) W1 = 8⇡✏ , W2 = 8⇡✏ , E1 = 4⇡✏ r (r > a), E2 = 4⇡✏ r (r > b). So 2 ˆ 2 ˆ 0 b 0 r 0 r ⇣ ⌘02 a 2 ⇣ ⌘ 2 R R q q2 1 1 2 1 1 E1 · E2 = 4⇡✏0 E1 · E2 d⌧ = q b r4 4⇡r2 dr = 4⇡✏ . r 4 , (r > b), and hence 4⇡✏0 0b R 2 q 1 1 1 Wtot = W1 + W2 + ✏0 E1 · E2 d⌧ = 8⇡✏ q 2 a1 + 1b 2b = 8⇡✏ a b .X 0 0

Problem 2.37

z

r b r

q2 a q

x

E1 =

1 q1 ˆ r; 4⇡✏0 r2

y

q1

E2 =

where (from the figure)

r

=

Therefore

1 q2 rˆ ; 4⇡✏0 r 2 p

r2 + a2

Wi =

Wi = ✏0

2ra cos ✓,

q1 q2 2⇡ (4⇡)2 ✏0

Z

q1 q2 (4⇡✏0 )2

cos

(r

a cos ✓)

r

3

r

It’s simplest to do the r integral first, changing variables to 2 r d r = (2r As r : 0 ! 1, The

r

r

: a ! 1, so

2a cos ✓) dr

Wi =

q1 q2 8⇡✏0

integral is 1/a, so Wi =

Z

⇡

0

q1 q2 8⇡✏0 a

1

a

Z

0

⇡

1

r

2

1 r2

(r

r

2

cos r2 sin ✓ dr d✓ d ,

a cos ✓)

r

.

sin ✓ dr d✓.

:

(r

)

✓Z

=

Z

dr

sin ✓ d✓ =

a cos ✓) dr = ◆

r dr

.

sin ✓ d✓.

q1 q2 . 4⇡✏0 a

Of course, this is precisely the interaction energy of two point charges. Problem 2.38 (a)

R

=

q ; 4⇡R2

a

=

q ; 4⇡a2

b

=

q . 4⇡b2


Problem 2.34 ! 1 q (a) W = ϵ20 E 2 dτ. E = 4πϵ 2 (a < r < b), zero elsewhere. 0 r ' & " #2 ! $ % !b 1 q2 1 1 b 1 2 q q2 ϵ0 2 W = 2 4πϵ0 4πr dr = 8πϵ0 a r2 = . − a r2 8πϵ0 a b 2

2

1 q 1 q 1 q 1 −q , W2 = 8πϵ , E1 = 4πϵ (b) W1 = 8πϵ r (r > a), E2 = 4πϵ r (r > b). So 2 ˆ 2 ˆ 0 b 0 r 0 r " " #02 a 2 # CHAPTER 2. ELECTROSTATICS 39 2 ! ! ∞ q2 −q 1 1 E1 · E2 = 4πϵ E1 · E2 dτ = − 4πϵ . q 2 b r14 4πr2 dr = − 4πϵ r 4 , (r > b), and hence 0 0 0b $ $ % % ! 2 q 1 1 2 1 1 WRtot = W1 + W − 1b .! = R8πϵ R 2b + ϵ10 qE1 · E2 Rdτa = 8πϵ0 qR2 R a +1 b − 1 ⇣q q q⌘ 0 0 0 a q b (b) V (0) = E·dl = dr (0)dr dr (0)dr = + . 2 2 1 b a 4⇡✏0 r R 4⇡✏0 b R a Problem 2.35 1 4⇡✏0 r

q −q q (a) σR = ; σa = ; σ R=a . RR ! 0 (the charge “drains (0)2 = b 1 (0)dr 4πR2 o↵”); V4πa 4πb2 a

q⌘ . a !a ! R$ 1 q % !0 !0 !b $ 1 q % q q# 1 "q Problem 2.39 + − dr − dr − (b) V (0) = − ∞ E·dl = − ∞ 4πϵ (0)dr − (0)dr = 2 2 b a 4πϵ0 r R 0 r 4πϵ0 b R a qa qb qa + qb (a) a = ; ; . b = R = !0 !a ! R$ 1 q % 1 "q q 4⇡a2 4⇡b2 4⇡R2 (c) σb → 0 (the charge “drains off”); V (0) = − ∞ (0)dr − a 4πϵ − 2 dr − R (0)dr = 0 r 4πϵ0 R a 1 qa + qb (b) Eout = Problem2 2.36 ˆ r, where r = vector from center of large sphere. 4⇡✏0 r qa qb qa + qb σa = − ; qσb = − ; σR = . 2 2 2 1 q(a) 1 a b 4πa 4πb r (r ) is 4πR (c) Ea = ˆ r , E = ˆ r , where the vector from center of cavity a (b). a b b a b 4⇡✏0 ra2 4⇡✏0 rb2 1 qa + qb ˆ r, where r = vector from center of large sphere. (b) Eout = 4πϵ r2 0 (d) Zero. (c)

b

(e)

R

1 q 4⇡✏0 r 2

R0

(0)dr = R

dr

1 ⇣q 4⇡✏0 R

1 qb(but not Ea or Eb ); force on qa and qb still zero. changes (but not a 1or qba); Eoutside changes (c) Ea = ˆ ra , Eb = ˆ rb , where ra (rb ) is the vector from center of cavity a (b). 2 4πϵ0 ra 4πϵ0 rb2 Problem 2.40 (a) No. For example, if it is very close to the wall, it will induce charge of the opposite sign on the wall, (d) Zero. and it will be attracted. (b) No. Typically itRwill be attractive, seeσbfootnote 12changes for an extraordinary counterexample. (e) σ changes (but not but σa or ); Eoutside (but not Ea or Eb ); force on qa and qb still zero.

Problem 2.41 Problem 2.37 Between plates, the E =plates 0; outside = So σ/ϵ0 = Q/ϵ0 A. So Between the plates, E = the 0; outside E = the /✏0plates = Q/✏E0 A. 2 2 ✏0 2 ✏0 PQ=2 ϵ0 E 2 = Q2ϵ0 Q = Q . P = E = = .2 2 2ϵ0 A2 2 2 ✏20 A2 2 2✏0 A22 ϵ0 A

Problem 2.42 Problem 2.38 1 Q Inside, E = 0; outside, E= = 0;4⇡✏ r; soE = Inside, E outside, 2ˆ 0 r

Eave =

1 1 Q 1 1 EQ rave ; )fzz; aver= 2 ˆ ; f2z4πϵ =0 R(E 2 4⇡✏0 R2 ˆ

1 Q r; 4πϵ0 r 2 ˆ

Q )z ; σ = == σ(Eave 2. 4⇡R

z✻

so

✒E θ

Q 4πR2 .

! !$ Q % 1 $ 1 Q % R RF Q = f1z da1= Q 4πR cos θ R2 sin θ dθ dφ 2 Fz = fz da = z4⇡R cos2 ✓ 2R24πϵ sin0 ✓Rd✓ d 2 2 4⇡✏ R2 0 $ % $ Q %2 $ 1 %(π/2 $ % ! Q2 2 1 ⇡/2 1 2 Q 2 (2 = Q R 1 Q 2 2π π/2 sin 2θ dθ sin θ θ cos = = . Q 2 = ⇡/2 Q Q 2 1 1 1 1 πϵ0 4R = 2 2πϵ0 4R. 0= = 2✏0 4⇡R 2⇡ 0 2ϵ0sin4πR ✓ cos ✓ d✓0= ⇡✏0 4R 32πR2 ϵ0 2 sin ✓ 0 2⇡✏0 4R 32⇡R2 ✏0

Problem 2.43 ⃝2005 c Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright as they No portion this material may be law: Say the charge on the inner cylinderlaws is Q, for acurrently length exist. L. The field isofgiven by Gauss’s R reproduced, in1any form or1 by any means, without Q 1 permission in writing from the publisher. E·da = E · 2⇡s · L = ✏0 Qenc = ✏0 Q ) E = 2⇡✏0 L s ˆ s. Potential di↵erence between the cylinders is V (b)

V (a) =

Z

a

b

E·dl =

Q 2⇡✏0 L

As set up here, a is at the higher potential, so V = V (a)

Z

a

b

1 ds = s

V (b) =

Q ln 2⇡✏0 L

Q 2⇡✏0 L

ln


b a

.

✓ ◆ b . a

cylinder is Q, for a length L. The field is given by Gauss’s law: !Say the charge on the inner s. Potential difference between the cylinders is E·da = E · 2πs · L = ϵ10 Qenc = ϵ10 Q ⇒ E = 2πϵQ0 L 1s ˆ V (b) − V (a) = −

"

b

a

Q E·dl = − 2πϵ0 L

"

b

a

Q 1 ds = − ln s 2πϵ0 L

As set up here, a is at the higher potential, so V = V (a) − V (b) = C=

Q V

=

2πϵ0 L b , ln( a )

Q 2πϵ0 L

2πϵ0 % &. ln ab

40 so capacitance per unit length is

ln

# $ b . a

%b& a .


Problem 2.400 L 2⇡✏0 C = VQ = 2⇡✏ . b , so capacitance per unit length is ln( a ) ln ab ϵ0 2 E Aϵ. (a) W = (force)×(distance) = (pressure)×(area)×(distance) = 2 Problem 2.44 ( ' 2 (b) W = (energy per unit volume)×(decrease in volume) = ✏ϵ00 E22 (Aϵ). Same as (a), confirming that the (a) W = (force)⇥(distance) = (pressure)⇥(area)⇥(distance) = E A✏. 2 energy lost is equal to the work done. ⇣ ⌘ 2 Problem 2.41 (b) W = (energy per unit volume)⇥(decrease in volume) = ✏0 E2 (A✏). Same as (a), confirming that the energyFrom lost Prob. is equal to the the field workatdone. 2.4, height z above the center of a square loop (side a) is Problem 2.45 1 4λaz ) E= ˆ z. % 2& 4πϵ a2 (side a) is a 0 2 From Prob. 2.4, the field at height z above the center loop z of + a4square z2 + 2 ✲ ✛ da 2 ✲ ✛ da 2 1 4 az da over a from q 0 to a ¯: ˆ Here λ → σ 2 (see figure), and we integrate E= z. 2 2 4⇡✏0 z 2 + a ✛ ✲ z2 + a 4

a

2

" a¯ 1 a2 a da Here ! (see figure), and we&integrate over a from ¯:a da = 2 du. ) E =da . Let u = 0 to 2σz , soa 2 % 2 2 4πϵ0 4 0 z 2 + a4 z 2 + a2 Z " * 2 +√ ,-a¯2 /4 a ¯ 2 1 a da du a2 a ¯ /4 σz 2u + z2 q√ E = = 12 4σz z . Let u = , so a da = 2 du. −1 tan 4⇡✏ a2 a2 2 = 0 2 0 2 z2 + 4πϵ πϵ0 z4 z 2u + 0 0 z +(u4+ z ) 2 z 0 ) " !#a¯2 /4 . Z a¯2 /4 + a¯2 , / p 2 12σ du z 2 2u + z 2 2 +z p − tan−1 == 4 ztan−1 = (1) ; tan 1 4⇡✏ ⇡✏0 z z (u +zz 2 ) 2u + z 2 πϵ0 0 0 0 q ( ! * ) a ¯2 0 2 2 2 2 + z 2σ 1 = tan 1 tan −1 (1) 1;+ a − π ˆ z. ⇡✏0 zE = πϵ0 tan 2z 2 4

✛

a+da

✲

1 2 % & 2σ ! r tan−1 (∞) −# π = 2σ π − π = σ . ! a → ∞ (infinite plane):"E = πϵ 4 πϵ0 2 4 2ϵ0 0 2 2 a ⇡ a2 1 1 p E= tan 1 + −12 √ ˆ z = tan ˆ z. 2 + (a2 series: z ≫ a (point charge): and as azTaylor 1 4+ x − π4 , ⇡✏ ⇡✏0 Let f (x) = tan2z /2) 0 expand4z

1 2 ′′ ⇥ f (x)1= f (0) + ⇤ ′ (0) f (0) + · · · 2 +⇡ x ⇡ tan (1) ⇡4 xf = ⇡✏ 2 2 4 = 2✏0 . X 0 p 1 a (point charge): Let f (x) + x ⇡4 Inc., , andUpper expand asRiver, a Taylor series: c = tan ⃝2005 Pearson 1Education, Saddle NJ. All rights reserved.

a ! 1 (infinite plane): E = z

2 ⇡✏0

This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any without permission in writing from the publisher. 1 means, 0 2 00

f (x) = f (0) + xf (0) + x f (0) + · · · 2

Here f (0) = tan

1

(1)

⇡ 4

=

⇡ 4

⇡ 4

= 0; f 0 (x) = f (x) =

Thus (since

a2 2z 2

= x ⌧ 1), E ⇡

2 ⇡✏0

⇣

1 a2 4 2z 2

⌘

=

1 1p1 1+(1+x) 2 1+x

=

1p , 2(2+x) 1+x

so f 0 (0) = 14 , so

1 x + ( )x2 + ( )x3 + · · · 4 1 a2 4⇡✏0 z 2

=

q 1 4⇡✏0 z 2 .

X


5 Here f (0) tan−1 (1) − π4 = π4 − CHAPTER 2. = ELECTROSTATICS

π 4

= 0; f ′ (x) =

1√1 1 1+(1+x) 2 1+x

=

1√ 411+x , 2(2+x)

so f ′ (0) = 14 , so

1 x + ( )x2 + ( )x3 + · · · 4 ⇢ ✓ ◆ ! ◆ ✓ ◆ "@ ✓ 1 @ 1 @ k sin ✓ cos 2 sin ✓ cos ✓ sin 2 3k 2σ q a2 1 1 a2 1 σa2k 1 ⇢ = ✏(since r·E = ✏ r + sin ✓ + = Thus = x ≪ 1), E ≈ = . ! 0 2z 2 0 2z 2✓ @✓ 4πϵ0 z 2 4πϵ0r z 2 r2 @r r πϵ0 r4sin r sin ✓ @ r  3 2 Problem12.42 1 2k sin (2 sin ✓ cos ✓ sin ✓) 1 ( k sin ✓ sin ) = ✏0 2 3k + $ + r sin ✓ %& r r r sin ✓# 1 ∂ $ A %r 1 ∂ B sin θ cos φ ρ =⇥ϵ0 ∇·E = ϵ0 r2 + ⇤θ ∂φk✏0 ⇥ ⇤ k✏ 22∂r 0 2 r r r sin = 2 3+ 2 sin (2 cos ✓ sin ✓) sin( = 2 3 + rsin (4 cos2 ✓ 2 + 2 cos2 ✓ 1) ' r ϵ0r 1 1 B sin θ = ϵ⇥0 2 A + (− sin φ) = 2 (A − B sin φ). ⇤ 3k✏ 3k✏ 0 0 sin 2θ ✓ r1) = r cos 2✓). = 2 1 +r sin (2r cos (1 + sin r r2

Problem 2.46

f (x) =

Problem 2.43 1 Q From Prob. 2.12, the field inside a uniformly charged sphere is: E = 4πϵ Problem 2.47 3 r. So the force per unit volume 0 R ) * ) Q *) Q * Q 2 1 Q 3 From the field inside a uniformly E =z 4⇡✏ force per unit volume 3 r. So r,charged and thesphere force is: in the direction on the dτ is: is f =Prob. ρE =2.12, 4 3 r = 3 3 0 R

is f = ⇢E =

Q

4πϵ0 R Q 3 4⇡✏0 R3 r = ✏0

3 πR

ϵ0 4πR 2 Q r, 4⇡R3

and the force in the z direction on d⌧ is: $ %2 3 Q dFz = fz dτ = r cos θ(r2 sin θ dr dθ dφ). ✓ ◆2 3 3 ϵ0Q 4πR dFz = fz d⌧ = r cos ✓(r2 sin ✓ dr d✓ d ). 3 ✏0 4⇡R The total force on the “northern” hemisphere is: 4 3 3 ⇡R

%2 + R is: + π/2 $ + 2π + the “northern” The total force on hemisphere Q 3 3 r dr cos θ sin θ dθ dφ Fz = fz dτ = ✓ ϵ0 4πR ◆2 Z3 R 0 Z ⇡/2 0 Z Z 2⇡ 0 3 Q$ -π/2 . %2 $3 4 % , cos2 ✓ sin Fz = fz d⌧ = 3 3 Q 0 r dr R 0 sin θ -- ✓ d✓ 0 d 3Q2 ✏0 = 4⇡R . (2π) = 3 ✓ ϵ0 4πR ◆2 ✓ ◆4 2 !-0 64πϵ0 R2 ⇡/2 3 Q R4 sin2 ✓ 3Q2 = (2⇡) = . ✏0 4⇡R3 4 2 0 64⇡✏0 R2 Problem 2.44 + + 1 σ 1 σ σR 1 σ Problem 2.48 2 Vcenter = Z da = Z R da = 4πϵ R (2πR ) = 2ϵ 4πϵ η 4πϵ 0 0 0 1 1 1 R 0 Vcenter = da = da = (2⇡R2 ) = r 4⇡✏0 4⇡✏0 R 2✏0 (4⇡✏/0 R Z + 2 da = sin2 ✓sin d✓,θ dθ, da2⇡R = 2πR 1 1 σ Vpole V=pole = da ,da with , with 2 2 2 2 2 2 2 2 2 2 4⇡✏0 4πϵr η r η= =R R+ +R R −2R2R coscos ✓= 2R2R (1(1 − coscos ✓).θ). θ= 0 Z + 2 ⇡/2 -π/2 1 ) 2⇡/2 sin ✓ sin d✓ θ dθ R σRp √ σ(2πR ) π/2 1(2⇡R p √ p √ = = = p= √(2 1(2 cos 1 −✓)cos θ)4⇡✏0 4πϵ 0 0 cos 1 −✓ cos θ2 2✏20 2ϵ0 R0 2R 20 0 1 p √ R σR R σR R σR p= √ = p= √ (1 (1 0) − = 0) . Vcenter = = ( 2 ( 1). . ) Vpole ∴ Vpole − Vcenter 2 − 1). 2✏0 2ϵ0 2✏0 2ϵ0 2✏0 2ϵ0

r R

R θ

Problem Problem 2.492.45 determine the electric and outside the sphere, Gauss’s FirstFirst let’s let’s determine the electric field field insideinside and outside the sphere, usingusing Gauss’s law: law: / 0 + Z + 4 I Z Z r + r ( 4 πkr 2 2 3 ⇡kr (r < (r R),< R), 2 r )¯ ϵ0 E·da = ϵ20E 4πr =Q =d⌧ = ρ dτ (k¯ =r)¯ θ d¯ = 4πkr¯3 d¯ r= ✏0 E·da = ✏0 4⇡r =E Qenc =enc ⇢ r(k¯ sinr✓ sin d¯ r d✓ dr dθ=dφ 4⇡k r r¯=d¯ 4 πkR (r 4 0 ⇡kR (r > R).> R). 0 c ⃝2005 Pearson Education, Inc., Upper 4 Saddle River, NJ. All rights reserved. This material is No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

So Eprotected = 4✏k0 r2under ˆ r (r all < R); E laws = 4✏kR ˆ r (r > R). exist. copyright asr 2they currently 0


42


Method I : ◆2 ◆2 Z ✓ Z ✓ ✏0 R kr2 ✏0 1 kR4 E 2 d⌧ (Eq. 2.45) = 4⇡r2 dr + 4⇡r2 dr 2 0 4✏0 2 R 4✏0 r2 ) ✓ ◆2 (Z R ⇢ ✓ ◆1 ✓ ◆ Z 1 ✏0 k 1 ⇡k 2 R7 1 ⇡k 2 R7 6 8 8 7 = 4⇡ r dr + R dr = +R = +R 2 2 4✏0 8✏0 7 r R 8✏0 7 0 R r

W =

=

✏0 2

Z

⇡k 2 R7 . 7✏0

Method II : W = For r < R, V (r) = = )W = =

1 2

Z

⇢V d⌧

(Eq. 2.43).

( ◆ ✓ ◆ Z r ✓ 2◆ kR4 kr k 1 4 E·dl = dr dr = R 4✏0 r2 4✏0 4✏0 r 1 1 R ✓ ◆ ✓ ◆ 3 3 3 k r R k r R3 + = R3 . 4✏0 3 3 3✏0 4  ✓ ◆ ◆ Z Z ✓ 1 R k r3 2⇡k 2 R 1 6 (kr) R3 4⇡r2 dr = R3 r 3 r dr 2 0 3✏0 4 3✏0 0 4 ⇢ ✓ ◆ 2 4 7 2 7 2 7 2⇡k R 1R ⇡k R 6 ⇡k R R3 = = .X 3✏0 4 4 7 2 · 3✏0 7 7✏0 Z

Z

r

R

✓

R

r3 + 3 1

r R

)

Problem 2.50 E=

rV =

@ A @r

✓

r

e r

◆

ˆ r=

A

⇢

)e

r(

r

e

r

ˆ r = Ae

r2

r

(1 + r)

ˆ r . r2

⇢ = ✏0 r·E = ✏0 A e r (1 + r)r· rˆr2 + rˆr2 ·r e r (1 + r) . But r· rˆr2 = 4⇡ 3 (r) (Eq. 1.99), and e r (1 + r) 3 (r) = 3 (r) (Eq. 1.88). Meanwhile, @ r e r (1 + r) = ˆ r @r e r (1 + r) = ˆ r e r (1 + r) + e r =ˆ r( 2 re r ).  2 2 So rˆr2 ·r e r (1 + r) = r e r , and ⇢ = ✏0 A 4⇡ 3 (r) e r . r ⇢ Z ✓ ◆ Z Z Z 1 e r 3 2 2 2 r Q = ⇢ d⌧ = ✏0 A 4⇡ (r) d⌧ 4⇡r dr = ✏0 A 4⇡ 4⇡ re dr . r 0 ⇣ ⌘ R1 2 But 0 re r dr = 12 , so Q = 4⇡✏0 A 1 = zero. 2 Problem 2.51

Let u ⌘ s/R. Then

1 V = 4⇡✏0

Z

r

da =

2 R V = 4⇡✏0

Z

0

4⇡✏0 1

Z

0

Z

R

0

Z

0

2⇡

p R2 + s2

⇡

p

u 1 + u2

2u cos

1

2Rs cos

d

!

s ds d .

du.


43

CHAPTER 2. ELECTROSTATICS The (double) integral is a pure number; Mathematica says it is 2. So V =

R . ⇡✏0

Problem 2.52 (a) Potential of + is V+ = 2⇡✏0 ln sa+ , where s+ is distance from + (Prob. 2.22). 7 z Potential of is V = + 2⇡✏0 ln sa , where s is distance from . ✻ (x, y, z) ! " s− ✓ λ◆ s − ∴ Total V = s ln . ) Total V = ln 2πϵ0 . s+ 7 s+ 2⇡✏0 # s+ # z a a p s+ = (y − a)2 + z 2 , and p s− = (y + a)2 + z 2 , so ✲ Now ✻ (x, y, z) 2 , and s = y Now s+ = (y a)2!+ z" (y + a)2 + z 2 , so −λ λ $ % ! " √ λ s − s 2 2 ◆  λ − (y+a)2 +z 2 ∴ Total V = ln✓ pλ(y+a). 2 +z (y + ln a)2 (y + z+2 a) + z . (x,2πϵ y, z) = s+0 ln √2(y−a) 0ln= 2πϵ p 2 +z 2 2 2 . V (x, y, z)V = = ln s+ 4πϵ 2⇡✏0 (y a)2 +z 2 (y 0 a)2 (y + z−2 a) + z # #4⇡✏0 2 2 2 2 a a ✲ Now s+ = (y − a) + z , and s− = (y +2 a)2 + z , so y (y+a) +z (4πϵ0 V0 /λ) −λ λ % = k = constant. That is: ! √ are given " by (y−a)2 +z$2 = e 2 (b) Equipotentials 2 2 2 λ 2 2 (yV + (y+a)2 +z(y+a)2 +z (4⇡✏ / )a) + z 2 constant. 2That is: 2 √ (b)V (x, Equipotentials are+given = e+ k= y, z) = ln . ln = 2 +z y 2 2πϵ +λ2ay a2(y−a) + zby =(y2k(y − 22ay a20+0 z 2 ) = ⇒ a) 2 +z 2 y (k 2 − 1) + z (k − 1) + a (k − 1) − 2ay(k + 1) = 0, or 0 & ' 4πϵ (y − 0 2 2 2 2 2 2 2 2 2 a) + z 2 2 k+1 2 y + 2ayy+ + a z+ + z a⇣= − k(y + a + z ) ) y (k 1) + z (k 1) + a (k 1) 2ay(k + 1) = 0, or 2ay ⌘ 2ay k−1 = 0. The equation for a circle, with center at (y0 , 0) and radius R, is k+1 2 2 y 2 + z 2 +(ya− 2ay = 0. The a2circle, with center at (y0 , 0) and radius R, is 2 2 2equation (4πϵ 0 /λ) + zk2 1=byR2(y+a) , or 2y+z + z2 + (y002Vfor − R )− 0 ) given 0 =&0. 'That is: (b) Equipotentialsyare = k 2yy = constant. 2 = e 2 2 2 2 (y−a) 2 +z 2 2 (y y ) +2z =2R the , orequipotentials R ) 2yy k+1 0 =⇣0. 2y + z + (y y 2 + 2ay0 +Evidently a + z &= k(y' − 2ay + a2 0+are z 2 )circles, ⇒ y 2 (kwith − 1)y+ z⌘2a(k − 1) +and a2 (k − 1) − 2ay(k + 1) = 0, or 0 = k−1 k+1 Evidently the equipotentials are circles, with y = a and & ' 0 2 k+1 k 1 a circle, at (y20+2k−1) , 0) and=radius y2 + z 2 + a a22 − (k2 +2k+1−k k+1 4kR, is 2 k−1 ⇣a2 =⌘a22 for = 2ay y02 − R ⇒= R20.=The y02 −equation −(ka22with = a2center a2 (k−1) 2 , or 2 k−1 (k−1) +2k+1 k +2k 1) 2 2 4k k+1 2 2 2 2 2 2 2 2 2 2 2 )√R 2 = y2 2 =a2 2 a = y R a a = a = a , or 2 2 (y − y0 )0 + z = Rk , or 0 + z + (y0 −k R1 ) − 2yy0 =&0. ' (k 1) (k 1) 2a pR = |k−1| ; or, in terms of V0 : 2a kthe equipotentials and Evidently are circles, with y0 = a k+1 R = |k k−1 1| ; or, in terms of V0 : & '2 2 2 +2k−1) 2 2 k+1 ! = a2 (k−1) "4k 2 , or a2 = y02 − R2 ⇒ R2 4πϵ = y002V0− − a2 = a2 0(kV0+2k+1−k 2 2πϵ0 V0 /λ −2πϵ /λ (k−1) /λa = a k−1 ✓ ◆2πϵ0 V0 +0 /e +2⇡✏ 1 0 V0 / e 2⇡✏0 V √ e4⇡✏0 V0 / e+ 1 e =a + = 2⇡✏ y0 = a 4πϵ V /λ e a coth . 0 V0 /λ 0 a 0 /λ − e−2πϵ = ka ; 4⇡✏ = 0aV0coth . λ R =y02a or, in terms of V− e 0= e2πϵ0 V2⇡✏ 0: 1 |k−1| e 0 V0 / 0 V0 / 1 e2⇡✏0 V0 / e ! " 2πϵ0 V0 /λ ✓ ◆2πϵ0 V0 2 a 2⇡✏0 V0 / e e = 2a 2 V /λ ! a "( 2πϵ0 V0 ) = 2⇡✏ R a csch . = a −2πϵ = 0 V0 0 0 V0 /λ 0 /λ 0 V00 /λ R = 2a =0 V0 /λ )2πϵ a csch . λ −0 V10 /λ +(ee2πϵ − /e−2πϵ ea2πϵ V0 = e4πϵ4⇡✏ +e14πϵ0 V= 2⇡✏0sinh 0 V0 λ e 0 V0 / 1 a (e2⇡✏0 V0 / e 2⇡✏0 V0= ) a coth sinh = y0 = a 4πϵ . λ e 0 V0 /λ − 1 e2πϵ0 V0 /λ − e−2πϵ0 V0 /λ z ! " ✻ e2πϵ0 V0 /λ 2πϵ0 V0 2 a ( ) = a csch R = 2a 4πϵ V /λ . = a 2πϵ V /λ = λ e 0 0 −1 (e 0 0 − e−2πϵ0 V0 /λ ) sinh 2πϵλ0 V0 z

✒ R

✻

✒ R −λ

✲ y

λ y0

−λ

λ y0

✲ y

Problem 2.48 c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently No 1 portion of this material may be d2 exist. V ρ reproduced, in any any means, without permission inρ. writing from the publisher. (Eq. 2.24), so = − (a) ∇2 Vform = or − ϵby 0 dx2 ϵ0 * Problem 2.48 2qV 1 2 2 .1 (b) qV = 2 mv → v d= V ρ 2 =m− ρ. (a) ∇ V = − ϵ0 (Eq. 2.24), so dx2 ϵ0 dx (c) dq = Aρ dx*; dq dt = aρ dt = Aρv = I (constant). (Note: ρ, hence also I, is negative.) 2qV . (b) qV = 12 mv 2 → v = m Inc., Upper Saddle River, NJ. All rights reserved. This material is c ⃝2005 Pearson Education,

44


Problem 2.53 (a) r2 V = (b) qV =

⇢ ✏0

1 2 2 mv

(Eq. 2.24), so

! v=

r

d2 V = dx2

1 ⇢. ✏0

2qV 8 . m

! ! dq 2 d2 V (Note: −1/2 V A⇢ dx 1; I m 1 dx I = A⇢v I = I m(constant). (c)(d) dq d= =−a⇢ ⇢, hence also I, − is negative.) , where β = = βV = − ρ = = − ⇒ dt dt 2 dx ϵ0 ϵ0 Av ϵ0 A 2qV ϵ0 A 2q . dx2 q q (Note: I is negative, so β isI positive; q isd2positive.) 2 V 1/2 I m I m (d) ddxV2 = ✏10 ⇢ = ✏10 Av = ✏0 A 2qV ) = V , where = ✏ A 2q . 0 dx2 : (e) Multiply by V ′ = dV dx (Note: I is negative, so is positive; q is positive.) " " ′ dV 1 ′ 0 dV dV (e) Multiply by V V ′ = dx=:βV −1/2 ⇒ V ′ dV ′ = β V −1/2 dV ⇒ V 2 = 2βV 1/2 + constant. dx dx 2 Z Z 0 0 dV dV 1 V 1/2 is ) V 0 dV 0 = V 1/2 dVat)cathode V 2 =is2zero), V 1/2 so + constant. But V (0) = VV ′0(0) ==0 (cathode at potential zero, and field the constant is zero, and dx dx 2 # # ′ dV 2 V 0 (0) = 1/4 But V (0) (cathode zero,dV and V = = 4βV 1/20⇒ β Vpotential ⇒ V −1/4 = field 2 βat dx;cathode is zero), so the constant is zero, and = 2is at dx" " # # p 4 3/4 p 0 dV −1/4 1/2 x+ V )= = V 2V = 4 V dV )2 β = 2dx ⇒V 1/4 V 21/4βdV =constant. 2 dx; 3 dxZ Z p p 4 3/4 1/4 dVso=this 2 constant dx )is also V zero. =2 x + constant. But V V (0) = 0, 3 $ %4/3 $ %2/3 %1/3 $ 3 #constant is also zero. 3# 9 81I 2 m But V (0) = 0, 3/4so this 4/3 4/3 V = β x, so V (x) = β x , or V (x) = β x = x4/3 . 2 A2 q 2 2 4 32ϵ 0 ✓ ◆4/3 ✓ ◆2/3 ✓ ◆1/3 3p 3p 9 81I 2 m 4/3 4/3 V 3/4 = x, so V (x) = x , or V (x) = x = x4/3 . 2 2 & x '4/3 4 32✏20 A2 q (see graph). Interms of V0 (instead of I): V (x) = V0 d ( ) ⇣ x ⌘4/3 V✻ Without space-charge, V would increase linearly: V (x) = V0 xd . Interms of V0 (instead of I): V (x) = V0 (see graph). d V0 4ϵ0 V d2 V 4 1 −2/3 Without space-charge, V would1 increase linearly: V (x) =0 V0 xd . ρ = −ϵ0 2 = −ϵ0 V0 4/3 · x = − . 2 2/3 dx 3 3 d 9(d x) without 1 4 1 2/3 4✏0 V0 *d2 V ' & ⇢ = ✏0 2q 2√ = ✏0# V0 4/3 · xx 2/3 = . with 3 3 d 0 /m 9(d2 x)2/3 . v = dx V = 2qV d r m ✲ ⇣ x ⌘2/3 2q p & p ' x d v= V =81I 2 m 2qV1/3 /m . 2 2 0 32ϵ0 A q 3 81md4 2 2 (f) V (d) = m V0 = 32ϵ2 A2 q d4/3d⇒ V03 = 32ϵ I ; I = V ; 2 A2 q 81md4 0 0 0 * ⇣ ⌘ √ 1/3 √ 2 4 2q 32✏20 A2 q 3 4 2 ϵ A q 81I3/2 0A 3/2 81md4ϵ (f) V I(d) V03 = K32✏ . == 9V√0m0=d2 32✏ =20 A2 q I22 ; I 2 = V020 Am2 q= KVd04/3 ,) where 4 V0 ; 9d r m 81md p p 4✏0 A 2q 4 2 ✏ A q 3/2 3/2 Problem I = 9pm02.49 V0 = KV0 , where K = . d2 9d2 m " & ' 1 η −η/λ ρˆ ηˆ (a) E = 2.54 1+ e dτ. Problem 4πϵ0 η 2 λ ✓ ◆ Z r e r / d⌧. 1 ⇢ rˆ (a)(b)E Yes. = The field2 of 1a+point charge at the origin is radial and symmetric, so ∇×E = 0, and hence this is also r 4⇡✏0 true (by superposition) for any collection of charges. " r " r ' −r/λ 1 c 2012rPearson 1 & Education, (c) V = − 1+ e Inc., drUpper Saddle River, NJ. All rights reserved. This material is E·dl = − protected q 2 under all copyright laws as they currently exist. No portion of this material may be 4πϵ r λ 0 ∞ ∞ reproduced, in any form or+" by any means, without permission in writing ,from the publisher. " ∞ " ∞ 1 1 & r ' −r/λ q 1 ∞ 1 −r/λ 1 −r/λ = q 1+ e dr = e dr + e dr . 4πϵ0 r r2 λ 4πϵ0 r2 λ r r r

c ⃝2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


45

(b) Yes. The field of a point charge at the origin is radial and symmetric, so r⇥E = 0, and hence this is also true (by superposition) for any collection of charges. Z r Z r 1 1 ⇣ r ⌘ r/ 9 (c) V = E·dl = q 1 + e dr 4⇡✏0 1 r2 9 1 ⇢Z Z 1 Z 1 1 ⇣ ⌘ 1 1 r q 1 1 1 !ee−r/λ ! r/ −r/λ −r/λ q dr! = e r/ dr + e r/Therefore dr . − λ1−r/λ ←− exactly right term. Now r1= = −re2 r 1 + 2 e 4⇡✏ !0dr r 1dr 4⇡✏ r2 to kill the last r 0 e−r/λ 1 r−r/λ e r r dr ←− exactly right to kill the last term. Therefore Now r2 e dr = − r − λ R R e r/ "r r/ $ 1 # ∞ Now r12 e r/ dr = e r dr exactly right to kill the last term. Therefore −r/λ rq " ## e−r/λ #∞q$ e −r/λ V (r) = −(q = . −r/λ # e ) # r r/ #4πϵ0= r qr/ e V4πϵ (r)0 q= − r 1 q e4πϵ0 r . 4πϵ0e r #=r V (r) = . 4⇡✏0 r& r ' 4⇡✏0 r & ' % R −R/λ' q 1 1 R −R/λ' (d) IE·da%= !R !2 = 1✓+ q & e◆ . q 2 1 1✓+1 & e◆ R 4π (d) −R/λ 2 1 1 R q 4πϵ ! R ! λ ϵ λ R1 + RR/ e−R/λ . 0 = 0R R/ e 24π 1 + ! ! = q (d) S E·da E·da = q! 2 R 1 2+ eλ 4⇡ R = 1+ eλ . ( S ( R S 4⇡✏( 4πϵ R0 ! ✏0 ϵ0 ) −r/λ +,R * r q 0 R e−r/λ q q e ) ,R 2 −r/λ ( ( (  R R * ZV dτ = Z qR r/ Z re −r/λ −r/λ r e4π ! dr = − 1 dr = − Rq q 2r/ e ⇣λ r r⌘ R + 2 ϵ0 q −r/λ q e q e 4πϵ ! r ϵ (1/λ) 0 0 2 r/ V =0 ! dr0 = re redr = dr = −10 − 1 V d⌧ V= dτ r r4⇡'rdr4π =. 2 2 ! 0 0& ϵ0 )(1/λ) λ V 4⇡✏0 4πϵ r ✏0' 0 ϵ0. 0 ✏0 (1/ 0 V 0 R& 0 2 q −R/λ ⇢ q ◆1 R . =λ + −e 1✓+ ϵ q 2 λ R −e−R/λ = 02 = λ ϵ e R/ 1 + 1 ++λ1 + . 1 . 0 ✏0 ' & ' . -& % ( q 1 q ⇢✓ R ◆ ✓ R& ◆ I Z -& −R/λ' −R/λ' % ( 1+ = .. ∴ E·da + 2 1 V dτ = q 1+ Re RR/− −R/λ Re RR/+ 1−R/λ q qed 1 q λ+ V+ V d⌧ V ϵ=0 dτ = 1 + λ 1 +e λ 1 +e ϵ=01 .= qqed )S ∴ E·da E·da 1− + e e+ 1 + . qed 2 2 ✏0 ϵ0 ✏0 ϵ0 λ λ S Vλ S V (e)(e) Does thethe result in (d) hold forfor a nonspherical surface? Suppose we we R Does result in (d) hold a nonspherical surface? Suppose 2 2 S R (e) Does the result in (d) hold for a nonspherical surface? Suppose we make a “dent” in the sphere—pushing a patch (area R sin θ dθ dφ) make a “dent” in the sphere—pushing a patch (area R sin ✓2 d✓ d ) S q 2 2 make a “dent” in the sphere—pushing a patch (area R sin θ dθ dφ) from radius R out to radius S (area S sin θ dθ dφ). from radius R out to radius S (area S sin ✓2 d✓ d ). q from radius R out to radius S (area S sin θ dθ dφ). & ' & ' . % 1 1⇢ q . ✓+ S & e◆ ✓+ R & e◆ −S/λ ' 2 −R/λ' 2 % I 1 (S sin θ dθ dφ) − 1 (R sin θ dθ dφ) ∆ E·da = SS/ −S/λ 1λ R RR/ −R/λ 1λ S q S 2q1 2 1 2 2 2 2 4πϵ Rdφ) 1 + e (S sin θ dθ − 1 + e (R sin θ dθ dφ) ∆ E·da = E·da = 0 )& 1 + e (S sin ✓ d✓ d ) 1 + e (R sin ✓ d✓ d ) ' , λ& λ S02S 'S 2 R2 R 2 q 4⇡✏0 4πϵ R& ◆ ,dφ. )& ✓ ◆ ✓ −S/λ ' −R/λ' = e − 1 + e sin θ dθ 1 + q λ S S q 4πϵ λ R1 +eRR/ e−R/λ sindθ dθ = 0= 1 + 1 +e S/ e−S/λ1− + sin ✓ d✓ . dφ. λ λ 4⇡✏0 4πϵ0 ( SZ ( ( Z−r/λ S ( 1 1 Z 1 1q q ( 1 1q q e e r/ (r2 sin Sdr −r/λ θ dr dθ dφ = sin θ dθ dφ re−r/λ V dτ = 2 1 1 q e 1 q 2 2 2 V d⌧ = r sin ✓ dr d✓ d = sin ✓ d✓ d re r/ dr 2 λ 2∆ λ dτ 4πϵ r r λ dφ 4πϵ 2= 2= 0 0 r sin θ dr dθ sin θRdθRdφ re−r/λ dr V 4⇡✏ 4⇡✏ 0 0 2 2 2 * * ++# λ λ 4πϵ r λ 4πϵ 0 0 S R q q ⇣ ⇣ r ⌘⌘ −r/λ * + *r ## Sr ++#S =− θ qdθ✓ dφ # =4πϵ0 sinsin d✓ d e e r/ 1 −r/λ 1 λ+ R =)& − sin θ dθ dφ & e 1+ R # 4⇡✏ 0 ✓ 0S ' ◆ ✓ R ' ◆λ ,R q q 4πϵ )& ' ' θ dθ,dφ. −S/λ =− 1 q+1 + Se e SS/− 1 +1 +&Re−R/λ =4πϵ0 eRR/sin sin ✓ d✓ d . −S/λ λ λ =− 1+ e − 1+ e−R/λ sin θ dθ dφ. 4⇡✏ 0 4πϵ0 λ λ ! / H 1 1 R 1 1 exactly compensates forfor thethe change in inE·da, we we getget thethe total using So So thethe change in in change V 1d⌧ compensates change E·da, q for total using ! exactly /andand λ2 2V dτ ϵ0 q✏0for Vdid dτdid exactly compensates for Any theAny change in surface E·da, and we built getupϵ1up q for the total using So the change inasλ2as thethe dented sphere, justjust we we with thethe perfect sphere. closed surface cancan be be built by successive dented sphere, with perfect sphere. closed by successive 0 the of dented sphere, as we did with theall perfect sphere. Any closed surface be charges built upinside, byinside, successive distortions the sphere, sojust the result holds forfor all shapes. By superposition, if there arecan many distortions of the sphere, so the result holds shapes. By superposition, if there are many charges 1 distortions of .the sphere,outside so the do result for all(in shapes.argument By superposition, if therethat are many charges inside, total Qenc Charges not holds contribute ✏0 enc for this thethe total is isϵ10 Q do not contribute (in the the argument above above we wefound found that H. !1Charges1 outside R /the total is Q . Charges outside do not contribute (in the argument above we found that 1 E·da for this volume + V d⌧ = 0—and, again, the sum is not changed by distortions of the surface, as this enc 2 volume E·da +/ λ2 ϵ0V dτ =! 0—and, again, the sum is not changed by distortions of the surface, as long as q for 1 long as q remains outside). So the new “Gauss’s Law” holds for any charge configuration. volume E·da + V dτ = 0—and, again, the sum is not changed by distortions of the surface, as long as q remains outside. So the new λ2 “Gauss’s Law” holds for any charge configuration. remains outside. So the new “Gauss’s Law” holds for any charge configuration. 1 This material is 1 reserved. c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights V = (f) protected In differential “Gauss’s law”currently reads: exist. ∇·ENo+portion or,1putting under allform, copyright laws as they of this may be it all in terms of E: 1 ρ,material 2 λ ϵ 0 Vthe = publisher. ρ, or, putting it all in terms of E: (f) Inindifferential form, “Gauss’s law” reads: in∇·E + from reproduced, any means, without permission writing (any form or by 2 λ ϵ0 1 ( 1 ρ. Since E = −∇V , this also yields “Poisson’s equation”: −∇2 V + 1 V = 1 ρ. = ∇·E − 2 E·dl 1 1 λ λ22 V + ϵ10 V = 1 ρ. 0 = ρ. Since E = −∇V , this also yields “Poisson’s equation”: −∇ ∇·E − 2 ϵE·dl λ ϵ0 λ2 ϵ0 ∆

c ⃝2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected ⃝2005 under all copyright laws as Inc., they Upper currently exist.River, No portion this material c Pearson Education, Saddle NJ. Allofrights reserved.may Thisbematerial is reproduced, in any form bycopyright any means, without in writing from the protected underorall laws as theypermission currently exist. No portion of publisher. this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

=−

q 4πϵ0

1+

S λ

e−S/λ − 1 +

R λ

e−R/λ sin θ dθ dφ.

) * So the change in λ12 V dτ exactly compensates for the change in E·da, and we get ϵ10 q for the total using the dented sphere, just as we did with the perfect sphere. Any closed surface can be built up by successive distortions of the sphere, so the result holds for all shapes. By superposition, if there are many charges inside, the total is ϵ10 Qenc . Charges outside do not contribute (in the argument above we found that for this ) * volume E·da + λ12 V dτ = 0—and, again, the sum is not changed by distortions of the surface, as long as q 46new “Gauss’s Law” holds for any charge CHAPTER 2. ELECTROSTATICS remains outside. So the configuration. 1 1 (f) In differential form, “Gauss’s law” reads: ∇·E + 2 V = ρ, or, putting it all in terms of E: 1 1λ ϵ0 (f) In di↵erential form,!“Gauss’s law” reads: r·E + 2 V = ⇢, or, putting it all in terms of E: ✏0 1 1 1 1 Z − ∇·E E·dl = ρ. Since E = −∇V , this also yields “Poisson’s equation”: −∇2 V + 2 V = ρ. 1 1 1λ λ2 1 ϵ0 ϵ0 2 r·E E·dl = ⇢. Since E = rV , this also yields “Poisson’s equation”: r V + 2 V = ⇢. 2 ✏0 ✏0 ρ dτ

λ

r/

ϵ 0

)V

=−

ρ

R ρ r e−

ρ −

1 λ2

λ

−

(∇ 2

R

0 E=

×

V =−

✲

E·dl

dτ

∇

/ rλ

ρ ; ϵ0

E=−∇V

✛

r )e

= dl

✢ V

+ rˆ (1 r2

R E·

4π 1 ϵ 0

R

1 2 λ

=

1 0 ϵ 4π

−

·E

V

E=

∇

✣❪

❫ E

(g) ReferProblem to ”Gauss’s law” in di↵erential form (f). Since E is zero, inside a conductor (otherwise charge would 2.50 move, and in such a direction as to cancel the field), V is constant (inside), and hence ⇢ is uniform, throughout ∂ = ϵ0“extra” ∇·E = ϵcharge everywhere). = ϵreside 0 ∂x (ax) 0 a (constant the volume. ρAny must on the surface. (The fraction at the surface depends on , and The same charge density would be compatible (as far as Gauss’s law is concerned) with E = ayˆ y, for on the shape of the conductor.) a instance, or E = ( )r, etc. The point is that Gauss’s law (and ∇×E = 0) by themselves do not determine 3 Problem 2.55 the field —like any differential equations, they must be supplemented by appropriate boundary conditions. @ ⇢ = ✏Ordinarily, (ax)are = so ✏0 a“obvious” (constantthat everywhere). these we impose them almost subconsciously (“E must go to zero far from 0 r·E = ✏0 @x the source charges”)—or we appeal to symmetry ambiguity (“the with field E must beythe The same charge density would be compatible (as far to asresolve Gauss’sthe law is concerned) = ayˆ , forsame—in a magnitude—on both sides of an infinite plane of surface charge”). But in this case there are no natural instance, or E = ( 3 )r, etc. The point is that Gauss’s law (and r⇥E = 0) by themselves do not determine boundary conditions, and no persuasive symmetry conditions, to fix the answer. The question “What the field —like any di↵erential equations, they must be supplemented by appropriate boundary conditions. is the electric produced by athat uniform chargethem density filling all of space?”(“E is simply it does Ordinarily, thesefield are so “obvious” we impose almost subconsciously must goill-posed to zero: far from not give the source charges”)—or we appeal to symmetry to resolve the ambiguity (“the field must be the same—in magnitude—on both sides of an infinite of surface charge”). inRiver, this NJ. case are no This natural c plane ⃝2005 Pearson Education, Inc., Upper But Saddle All there rights reserved. material is underconditions, all copyright laws as they currently exist. No portion of this material boundary conditions, and no persuasiveprotected symmetry to fix the answer. The question “What is themay be reproduced, in any form or by any means, without permission in writing from the publisher. electric field produced by a uniform charge density filling all of space?” is simply ill-posed : it does not give us information to determine the answer. (Incidentally, it won’t help to appeal to Coulomb’s law ⇣ sufficient ⌘ R rˆ 1 E = 4⇡✏0 ⇢ r 2 d⌧ —the integral is hopelessly indefinite, in this case.) Problem 2.56

Compare Newton’s law of universal gravitation to Coulomb’s law: F= Evidently

1 4⇡✏0

G

m1 m2 ˆ r; r2

F=

1 q1 q2 ˆ r. 4⇡✏0 r2

! G and q ! m. The gravitational energy of a sphere (translating Prob. 2.34) is therefore Wgrav =

3 M2 G . 5 R


47


Now, G = 6.67 ⇥ 10 11 N m2 /kg2 , and for the sun M = 1.99 ⇥ 1030 kg, R = 6.96 ⇥ 108 m, so the sun’s gravitational energy is W = 2.28 ⇥ 1041 J. At the current rate this energy would be dissipated in a time t=

W 2.28 ⇥ 1041 = = 5.90 ⇥ 1014 s = 1.87 ⇥ 107 years. P 3.86 ⇥ 1026

Problem 2.57 First eliminate z, using the formula for the ellipsoid: (x, y) =

Q 1 p 2 2 4 2 2 4 4⇡ab c (x /a ) + c (y /b ) + 1

(x2 /a2 )

(y 2 /b2 )

.

Now (for parts (a) and (b)) set c ! 0, “squashing” the ellipsoid down to an ellipse in the x y plane: (x, y) =

1

Q p 2⇡ab 1

(x/a)2

(y/b)2

.

(I multiplied by 2 to count both surfaces.) (a) For the circular disk, set a = b = R and let r ⌘

p

x2 + y 2 .

(r) =

(b) For the ribbon, let Q/2b ⌘ ⇤, and then take the limit b ! 1: (c) Let b = c, r ⌘

Q 1 p . 2⇡R R2 r2

(x) =

⇤ 1 p . 2⇡ a2 x2

p y 2 + z 2 , making an ellipsoid of revolution: x2 r2 + = 1, a2 c2

with

=

Q 1 p . 4⇡ac2 x2 /a4 + r2 /c4

The charge on a ring of width dx is dq = 2⇡r ds, 2x dx 2r dr dr Now + 2 =0) = 2 a c dx

(x) =

where ds =

p p dx2 + dr2 = dx 1 + (dr/dx)2 .

r c2 x c4 x2 c2 p 2 4 , so ds = dx 1 + 4 2 = dx x /a + r2 /c4 . Thus 2 a r a r r

dq Q 1 c2 p 2 4 Q p = 2⇡r x /a + r2 /c4 = . (Constant!) dx 4⇡ac2 x2 /a4 + r2 /c4 r 2a


48


Problem 2.58

y ( _a2, ÷3 _a2 ) b (0,-a)

q

r

x

( _a2,- ÷3 _a2) (a) One such point is on the x axis (see diagram) at x = r. Here the field is  q 1 Ex = 4⇡✏0 (a + r)2

2

cos ✓ = 0, b2

Now, (a/2) cos ✓ = b Therefore

r

;

b = 2

⇣a

r

2

⌘2

+

or

2 cos ✓ 1 = . b2 (a + r)2

!2 3 a = (a2 2

p

2[(a/2) r] 1 . To simplify, let = (a + r)2 (a2 ar + r2 )3/2 (1 2u) 1 = , 2 3/2 (1 + u)2 (1 u + u )

or

(1

ar + r2 ).

r ⌘u: a

2u)2 (1 + u)4 = (1

u + u2 )3 .


49

CHAPTER 2. ELECTROSTATICS Multiplying out each side: 1

6u2

4u3 + 9u4 + 12u5 + 4u6 = 1

3u + 6u2

7u3 + 6u4

3u5 + u6 ,

or 3u

12u2 + 3u3 + 3u4 + 15u5 + 3u6 = 0.

u = 0 is a solution (of course—the center of the triangle); factoring out 3u we are left with a quintic equation: 1

4u + u2 + u3 + 5u4 + u5 = 0.

According to Mathematica, this has two complex roots, and one negative root. The two remaining solutions are u = 0.284718 and u = 0.626691. The latter is outside the triangle, and clearly spurious. So r = 0.284718 a. (The other two places where E = 0 are at the symmetrically located points, of course.)

y a a

_,_ ( ÷2 ÷2 )

b+

b_ q_

q+ r

x

(b) For the square: q Ex = 4⇡✏0 where cos ✓± = Thus

✓

cos ✓+ 2 2 b+

p (a/ 2) ± r ; b±

◆

cos ✓ 2 2 b

b2± =

✓

a p 2

◆2

=0

+

✓

) a p ±r 2

cos ✓+ cos ✓ = , b2+ b2 ◆2

= a2 ±

p

2 ar + r2 .

p p (a/ 2) + r (a/ 2) r p p = . (a2 + 2 ar + r2 )3/2 (a2 2 ar + r2 )3/2

To simplify, let w ⌘

p

2 r/a; then

1+w = (2 + 2w + w2 )3/2 (2

1 w , 2w + w2 )3/2

or

(1 + w)2 (2

2w + w2 )3 = (1

w)2 (2 + 2w + w2 )3 .

Multiplying out the left side: 8

8w

4w2 + 16w3

10w4

2w5 + 7w6

4w7 + w8 = (same thing with w !


w).

50


The even powers cancel, leaving 8w

16w3 + 2w5 + 4w7 = 0,

or

4

8v + v 2 + 2v 3 = 0,

where v ⌘ w2 . According to Mathematica, this cubic equation has one negative root, one root that is spurious (the point lies outside the square), and v = 0.598279, which yields r v r= a = 0.546936 a . 2

y (-a cos(2p/5), a sin(2p/5)) (a cos(p/5), a sin(p/5))

b a

For the pentagon: Ex = where cos ✓ =

q 4⇡✏0

c q r

✓

x

1 cos ✓ +2 2 (a + r)2 b

a cos(2⇡/5) + r , b

2

f

cos

2

=

cos c2

◆

= 0,

a cos(⇡/5) c

r

;

2

b2 = [a cos(2⇡/5) + r] + [a sin(2⇡/5)] = a2 + r2 + 2ar cos(2⇡/5), c2 = [a cos(⇡/5)

2

2

r] + [a sin(⇡/5)] = a2 + r2

2ar cos(⇡/5).

1 r + a cos(2⇡/5) r +2 +2 3/2 2 2 2 (a + r)2 [a + r + 2ar cos(2⇡/5)] [a + r2

a cos(⇡/5) 3/2

2ar cos(⇡/5)]

= 0.

Mathematica gives the solution r = 0.688917 a. For an n-sided regular polygon there are evidently n such points, lying on the radial spokes that bisect the sides; their distance from the center appears to grow monotonically with n: r(3) = 0.285, r(4) = 0.547, r(5) = 0.689, . . . . As n ! 1 they fill out a circle that (in the limit) coincides with the ring of charge itself. Problem 2.59 The theorem is false. For example, suppose the conductor is a neutral sphere and the external field is due to a nearby positive point charge q. A negative charge will be induced on the near side of the sphere (and a positive charge on the far side), so the force will be attractive (toward q). If we now reverse the sign of q, the induced charges will also reverse, but the force will still be attractive. If the external field is uniform, then the net force on the induced charges is zero, and the total force on the conductor is QEe , which does switch signs if Ee is reversed. So the “theorem” is valid in this very special case. c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

51


Problem 2.60 The initial configuration consists of a point charge q at the center, q induced on the inner surface, and +q on the outer surface. What is the energy of this configuration? Imagine assembling it piece-bypiece. First bring in q and place it at the origin—this takes no work. Now bring in q and spread it over the surface at a—using the method in Prob. 2.35, this takes work q 2 /(8⇡✏0 a). Finally, bring in +q and spread it over the surface at b—this costs q 2 /(8⇡✏0 b). Thus the energy of the initial configuration is Wi =

q2 8⇡✏0

✓

1 a

1 b

◆

.

The final configuration is a neutral shell and a distant point charge—the energy is zero. Thus the work necessary to go from the initial to the final state is W = Wf

Wi =

q2 8⇡✏0

✓

1 a

1 b

◆

.

Problem 2.61

y

R (2pj/n)

rj x

R

Suppose the n point charges are evenly spaced around the circle, with the jth particle at angle j(2⇡/n). According to Eq. 2.42, the energy of the configuration is 1 Wn = n qV, 2 where V is the potential due to the (n

1) other charges, at charge # n (on the x axis).

n X1 1 1 V = q , 4⇡✏0 j=1 r j

r

j

= 2R sin

✓

j⇡ n

◆

(see the figure). So Wn =

n 1 q2 n X 1 q2 = ⌦n . 4⇡✏0 R 4 j=1 sin(j⇡/n) 4⇡✏0 R


52


Mathematica says 9

⌦10 =

10 X 1 = 38.6245 4 j=1 sin(j⇡/10) 10

⌦11 =

11 X 1 = 48.5757 4 j=1 sin(j⇡/11) 11

⌦12 = If (n

12 X 1 = 59.8074 4 j=1 sin(j⇡/12)

1) charges are on the circle (energy ⌦n Wn = [⌦n

1q

1

2

/4⇡✏0 R), and the nth is at the center, the total energy is

+ (n

1)]

q2 . 4⇡✏0 R

For n = 11 :

⌦10 + 10 = 38.6245 + 10 = 48.6245 > ⌦11

n = 12 :

⌦11 + 11 = 48.5757 + 11 = 59.5757 < ⌦12

Thus a lower energy is achieved for 11 charges if they are all at the rim, but for 12 it is better to put one at the center.


53

CHAPTER 3. POTENTIAL

Chapter 3

Potential Problem 3.1 p The argument is exactly the same as in Sect. 3.1.4, except that since z < R, z 2 + R2 2zR = (R z), q 1 1 q instead of (z R). Hence Vave = [(z + R) (R z)] = . If there is more than one charge 4⇡✏0 2zR 4⇡✏0 R 1 Qenc inside the sphere, the average potential due to interior charges is , and the average due to exterior 4⇡✏0 R Qenc charges is Vcenter , so Vave = Vcenter + 4⇡✏ . X 0R Problem 3.2 A stable equilibrium is a point of local minimum in the potential energy. Here the potential energy is qV . But we know that Laplace’s equation allows no local minima for V . What looks like a minimum, in the figure, must in fact be a saddle point, and the box “leaks” through the center of each face. Problem 3.3 Laplace’s equation in spherical coordinates, for V dependent only on r, reads: ✓ ◆ 1 d dV dV dV c c r2 V = 2 r2 = 0 ) r2 = c (constant) ) = 2 ) V = + k. r dr dr dr dr r r Example: potential of a uniformly charged ✓ sphere.◆ 1 d dV dV dV c In cylindrical coordinates: r2 V = s =0)s =c) = ) V = c ln s + k. s ds ds ds ds s Example: potential of a long wire. Problem 3.4 Refer to Fig. 3.3, letting ↵ be the angle between r and the z axis. Obviously, Eave points in the direction, so I Z 1 1 q 1 Eave = E da = ˆ z 2 2 r 2 cos ↵ da. 4⇡R 4⇡R 4⇡✏0 By the law of cosines, R2 = z 2 +

r

2

r

2

= R2 + z 2

2 r z cos ↵ 2Rz cos ✓

) )

r 2 R2 , 2r z cos ↵ z 2 + r 2 R2 z = = 2 3 2 r 2z r (R + z 2

cos ↵ =

z2 +


R cos ✓ . 2Rz cos ✓)3/2

ˆ z

54 Eave = =

Z q z R cos ✓ R2 sin ✓ d✓ d 2 2 2 16⇡ R ✏0 (R + z 2 2Rz cos ✓)3/2 Z ⇡ Z qˆ z z R cos ✓ qˆ z sin ✓ d✓ = 8⇡✏0 0 (R2 + z 2 2Rz cos ✓)3/2 8⇡✏0

ˆ z

(where u ⌘ cos ✓). The integral is 1 I = p 2 R R + z2 ✓ 1 1 = R |z R|

So


1

2Rzu 1 ◆ 1 z+R

1 2Rz 2

1 1

z Ru du (R2 + z 2 2Rzu)3/2

✓p R2 + z 2

 1 |z 2Rz 2

◆ 1 R2 + z 2 2Rzu + p R2 + z 2 2Rzu 1 ✓ ◆ 1 1 (z + R) + (R2 + z 2 ) . |z R| z + R

R|

(a) If z > R, ✓ ◆  ✓ 1 1 1 1 1 2 2 I = (z R) (z + R) + (R + z ) R z R z+R 2Rz 2 z R ✓ ◆  1 2R 1 2R 2 = 2R + (R2 + z 2 ) 2 = 2. R z 2 R2 2Rz 2 z R2 z

1 z+R

◆

1 q ˆ z, 4⇡✏0 z 2 the same as the field at the center. By superposition the same holds for any collection of charges outside the sphere. (b) If z < R, ✓ ◆  ✓ ◆ 1 1 1 1 1 1 2 2 I = (R z) (z + R) + (R + z ) R R z z+R 2Rz 2 R z z+R ✓ ◆  1 2z 1 2z = 0. = 2z + (R2 + z 2 ) 2 R R2 z 2 2Rz 2 R z2 Eave =

So

Eave = 0.

By superposition the same holds for any collection of charges inside the sphere. Problem 3.5 H Same as proof of second uniqueness theorem, up to the equation S V3 E3 · da = each surface, either V3 = 0 (if V is specified on the surface), or else E3? = 0 (if @V @n = R (E3 )2 = 0, and hence E2 = E1 . qed V Problem 3.6 identity: Z Putting U = T = V3 into Green’s I ⇥ ⇤ 2 V3 r V3 + rV3 · rV3 d⌧ = V3 rV3 · da. But r2 V3 = r2 V1 V Z S I So E32 d⌧ = V3 E3 · da, and the rest is the same as before. V

r2 V 2 =

R

(E3 )2 d⌧ . But on E? is specified). So V

⇢ ⇢ + = 0, and rV3 = ✏0 ✏0

S

Problem 3.7 Place image charges +2q at z = d and q at z = 3d. Total force on +q is  ✓ ◆ ✓ ◆ q 2q 2q q q2 1 1 1 1 29q 2 F= + + ˆ z = + ˆ z = ˆ z. 4⇡✏0 (2d)2 (4d)2 (6d)2 4⇡✏0 d2 2 8 36 4⇡✏0 72d2 c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

E3 .

55


Problem 3.8 (a) From Fig. 3.13: q0

r

0

= =

Therefore:

r

=

p

r2 + a2

2ra cos ✓;

r

0

=

p

r 2 + b2

2rb cos ✓.

Therefore:

R q R2 p (Eq. 3.15), while b = (Eq. 3.16). a r2 + b2 2rb cos ✓ a q q q = q . 4 2 2 a R R ar 2+ 2 r 2r cos ✓ + R 2ra cos ✓ R a2 a R

1 V (r, ✓) = 4⇡✏0

✓

q

r

+

q0

r

0

◆

q = 4⇡✏0

(

p

1 r2 + a2

1

2ra cos ✓

p R2 + (ra/R)2

Clearly, when r = R, V ! 0. @V (b) = ✏0 @V (Eq. 2.49). In this case, @V @n @n = @r at the point r = R. Therefore, ✓ ◆⇢ q 1 2 (✓) = ✏0 (r + a2 2ra cos ✓) 3/2 (2r 2a cos ✓) 4⇡✏0 2 ✓ 2 ◆ 1 a 3/2 2 2 + R + (ra/R) 2ra cos ✓ 2r 2a cos ✓ 2 R2 r=R ⇢ q = (R2 + a2 2Ra cos ✓) 3/2 (R a cos ✓) + R2 + a2 2Ra cos ✓ 4⇡  q a2 = (R2 + a2 2Ra cos ✓) 3/2 R a cos ✓ + a cos ✓ 4⇡ R q = (R2 a2 )(R2 + a2 2Ra cos ✓) 3/2 . 4⇡R Z Z q qinduced = da = (R2 a2 ) (R2 + a2 2Ra cos ✓) 3/2 R2 sin ✓ d✓ d 4⇡R  ⇡ q 1 2 2 2 = (R a )2⇡R (R2 + a2 2Ra cos ✓) 1/2 4⇡R Ra 0  q 2 1 1 p = (a R2 ) p . 2a R2 + a2 + 2Ra R2 + a2 2Ra p But a > R (else q would be inside), so R2 + a2 2Ra = a R.  q 2 1 1 q q = (a R2 ) = [(a R) (a + R)] = ( 2R) 2a (a + R) (a R) 2a 2a =

2ra cos ✓

3/2

✓

a2 R

qR = q0 . a

(c) The force on q, due to the sphere, is the same as the force of the image charge q 0 , to wit: ✓ ◆ 1 qq 0 1 R 2 1 1 q 2 Ra F = = q = . 4⇡✏0 (a b)2 4⇡✏0 a (a R2 /a)2 4⇡✏0 (a2 R2 )2 c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

)

.

a cos ✓

◆

4 4

CHAPTER 3. SPECIAL TECHNIQUES CHAPTER 3. SPECIAL TECHNIQUES

(c) The force on q, due to the sphere, is the same as the force of the image charge q ′ , to wit: ! " ′ (c) The force on q, due charge CHAPTER POTENTIAL qq ′ sphere,1is the same R 2 as the 1force of the image q 2 Ra q , to3.wit: 1 to56the 1 − F = = q = − . ! " 2 2 2 2 2 2 ′ 2 4πϵ01 (a −qq b) 4πϵ01 a R (a − R 1/a) 4πϵ01 (a −q RRa) − q2 F = = =− . 2 2 2 − b)we 4πϵ a (a − R /a) 4πϵ0 (a2 − R2 )2 0 (a 0 To bring bring qq in in from from infinity infinity 4πϵ to a, a, then, we do do work To to then, work To bring q in from infinity %&a $ #aa to a, then, we do work 2 2 2 &a q R  11 R 2RZ q R #a aa qq2 R 11 qq2 R &%&a= − 11 W = da = − . $ 2 2 & 2 2 2 2R2 ). W = 4πϵ da = = 2 2 2 & 4πϵ 2 4πϵ 2(a − (a − R ) (a − R ) 2 2 q R a q R 1 1 1 q R2 ) 2 0 0 0 2 2 2 ∞& 4⇡✏ 4⇡✏ 2 4⇡✏ 2(a R (a R ) (a R ) 0 0 0 1 W = da = = − − . ∞ 4πϵ10 4πϵ0 2 (a2 − R2 ) &∞ 4πϵ0 2(a2 − R2 ) (a2 − R2 )2 ∞ Problem 3.8 Problem 3.9 2 CHAPTER 3. SPECIAL TECHNIQUES Place a 3.8 second image charge, q ′′ , at the center of the sphere; Problem 00 ′′at the this will alter thecharge, fact that is an Place a second image q , qthe center of equipotential, thethe sphere; a−b Place a not second image charge, , atsphere the center of sphere; *' ( ) 1 q ′′ this will not alter the fact that the sphere is an equipotential, this will not alter the fact that the sphere is an equipotential, a−b but merely increase that potential from zero to V0 = 00 ;′′ ′′ ′ *' ) 1 q q q q( 4πϵ01 Rq but′′but merely increase thatthat potential from zerozero to to V0 V= ; ; ' ′′ () * merely increase potential from 0 = ′ 4⇡✏4πϵ q qa q q = 4πϵ0 V0 R at center of sphere. 0 R0 R ' () * ′′ 4⇡✏0 V0 R at center of sphere. q 00 = q = 4πϵ V R at center of sphere. a 0 For a neutral0 sphere, q ′ + q ′′ = 0. ! ′′ ! " 0 ′ ′ 00 ′′" ′ ForFor a neutral q + 0. 0. qq a neutral sphere, q qq+ q= = 1 sphere, 1 q 1 F = q ✓ !2 ′′+ = − " ! 2+ " ◆ ✓ 2 ′ ′ 0 b) 0qq 4πϵ a00 q (a − 4πϵ a1 1 (a − b) q 1 2◆ 01 0 1 q q qq 1 q + − + + + 2 F F= =qq4πϵ = = 4πϵ ′ q 2 2b)2 2 a2−(a − −2b)2 −2 b)(a (a b) q(−Rq/a) /a)(2a R(a/a) 0b(2a2 a 4⇡✏(R b) 0 0 a = 4⇡✏0 ′ 2 a = 2 2 2 2 2 2 2a (a 4πϵ (a − b) 4πϵ0 − R /a) b(2a − b) q( q(−Rq/a) (R /a)(2a −2R 0 a qq 0qq b(2a b) R /a)/a) = Rq/a) (R /a)(2a = = 4πϵ 2 a2!(a − = 2 2 (a −2R2 /a) 2 " 2 2 2 b) 4πϵ a 3 4⇡✏0 qa02 (a Rb) (2a2 − 4⇡✏ R02 ) 0 a (a R /a) . =− " ! 3 2 2 2 22 2✓ ◆3 (a 2− R 4πϵ (2a −2)R q 2 q0 Ra R (2a R ) ). = − = 2 2. 2 a (a2(a R −2R 4⇡✏4πϵ )2 ) 0 0 abecause (Drop the minus sign, the problem asks for the force of attraction.) (Drop the minus sign, because the problem asks for the force of attraction.) Problem 3.9 (Drop the minus sign, because the problem asks for the force of attraction.) Problem3.9 3.9 Problem (a) Image 3.10 problem: λ above, −λ below. Potential was found in Prob. 2.47: Problem z (a)(a) Image problem: below. was found in in Prob. 2.52: Image λ above, −λ below. Potential was found Prob. 2.47: z 2λ above, λ Potential ✻problem: V (y, z) z= ln(s− /s+ ) = ln(s2− /s2+ ) ✻ zy 4πϵ 4πϵ z s+ (y, z) 2λ λ 0 0 ✻ 2 2 + 2 V (y,✻ z)✣ = y ln(s /s ) = ln(s /s ) ✻ − + λ 2 2 + − +/s ) = y4πϵ V (z (y,+ z)d) =2 , ln(s ln(s /s+ ) 4πϵ + ss+− (y, z) 0 d λ 0✲ y 2 + ✣ ✣ 4⇡✏ 4⇡✏ ✲y + 0 = lnλλx+2 2 ,0 ⇢ 2 2 d 2 2 d 4πϵ0λ ✲ y + (z − d) y + (z + d) s − ✲ y + (z + d) ✲xx − y = ln d 4πϵ0 y 2 + (z − d)2 = 4⇡✏0 ln y 2 + (z d)2 ∂V ∂V ∂V − (b) σ = −ϵ0 . Here = , evaluated at z = 0. ∂n∂V ∂n∂V ∂z∂V + ,& (b) σ = −ϵ@V . Here = , evaluated at z = 0. 0 & ∂n λ @V∂n @V1∂z &,& (b) = at − z = 0. 1 σ(y)✏0=@n−ϵ. 0Here @n+2= @z , evaluated 2(z + d) 2(z − d) & & 2 2 2 4πϵ0λ y + (z + y + (z − 1 d) 1 d) z=0 & σ(y) = −ϵ0+ ⇢ 2(z + d) − 2(z − d) & 4πϵ0d y 2 +1 (z−d + d)2, y 2 +1 (z − d)2 λd 2λ z=0 (y) = 2(z, + d) 2(z d) − = . = −✏0 4⇡✏ +2 y 2 2+−(z + 2 2 2 2 2 2 2 d) y + (z d) 4π2λ 0y + d y +−d d π(y +λdd ) z=0 = − − 2 . = − ⇢ 2 2 2 2 2 4π y + d y + d π(y + d ) dl parallel to the y axis: Check: = Total2 charge dinduced onda strip = of width . 2 + d2 2 + d2 2 + d2 ) 4⇡ y y ⇡(y ∞ $ of width %& # Check: Total charge induced on a strip l parallel to the y/ axis:- . &∞ 1 lλd 1 lλd π π .0 lλd −1 y & ∞ dy = − = − y axis: q = − tan − − $ %& # induced ind Total charge ∞to the . / & 2 2 Check: on a strip of width l parallel & πlλd y + 1d πlλd d 1 πlλd 2 π 2 π .0 −1 d y −∞ & dy = − = − − qind = − −∞ tan − & πZ1 y 2 + d2 π d 2 ⇣ ⌘d 1 −∞ l dπh ⇡ 2 ⇣ ⇡ ⌘i = −λl. λind = −λ, l d Therefore l d as1 it should −∞ 1 1 y be. qind = dy = tan = ⇡ y 2 + d2 λind = ⇡−λ, das it should d be. 1 ⇡ 2 2 = −λl. Therefore

Chapter 1

Special Techniques

=

l.

1

Therefore

c ⃝2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under laws as they currently exist. No portion of this material may be , as all it copyright should be. indc = ⃝2005 Pearson Upper Saddle NJ. All reserved. This material is reproduced, in anyEducation, form or by Inc., any means, withoutRiver, permission inrights writing from the publisher. protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


c ⃝2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be

CHAPTER 3. SPECIAL TECHNIQUES CHAPTER 3. POTENTIAL

5 57

Problem3.10 3.11 Problem Theimage imageconfiguration configurationisisasasshown. shown. The

y

✻

q !( q 1 1 q 1 1 p (x,y)y)== "p VV(x, ++" ✲x 4⇡✏0 0 +zz2 2 4πϵ (x(x− a)a)2 2++(y(y− b)b)2 2++zz2 2 (x(x++a)a)2 2++(y(y++b)b)2 2+ #) 11 11 q −q p − "p −" .. 2 + (y 2+z 2 2 + (y + b) 2 2 2 2 (x + a) b) (x a) (x + a) + (y − b) + z (x − a) + (y + b)2 2++zz2 2 CHAPTER 55 CHAPTER3.3.SPECIAL SPECIALTECHNIQUES TECHNIQUES For this to work, θ must be and integer divisor of 180◦. Thus 180◦, 90◦ , 60◦ , 45◦ , etc., are OK, but no ⇢ 2 others. It works for 45◦ , say, qwith the charges as1shown. 1 1 45◦ line F= x ˆ y ˆ+ p [cos ✓ x ˆ + sin ✓ y ˆ] (4) , (2) 2 2 2 2 )20), + − 4⇡✏0 the(2a) (Note the strategy: to make x axis an(2b) equipotential (2 a (V + b= Problem 3.10p Problem p reflection point. To make the you place the3.10 image charge (1) in the yy where cos ✓ = a/ a2 + b2 , is sin ✓asshown. =shown. b/ a2 + b2 . The image configuration as ◦ − The image configuration is + ✻✻ 45 line an equipotential, you place charge (2) at the image point. ✲x ⇢  −q qq + − (1) −q 2 But that screws up !the x axis, so you must now insert image (3) to q a 1 b 1 ! ◦ F = x ˆ + y ˆ . q 1 1 balance (2). Moreover, to make the 1452 line2 V = 0 ayou also need (4), 2 b2 (a + b )3/2 ++ (a2 +1 b2 )3/2 " VV (x,(x, y)y)== q "" 16⇡✏0 − + ✲✲x "0 you 2need 2+ 2 2V = 2+ 2 2 to balance (1).4πϵ But restore the x 2axis to (5) x (5) (3) 2 2 2 2 0 0now, (xto − a) (y − b) + z (x + a) + (y + b) z 4πϵ (x − a) + (y − b) + z (x + a) + (y + b) #+ z #  to balance (4), and so on.  2 ◦ 2 2 2 why1it works 1 q 1q1 1 1 q q 1 1q for θ = 45 −q−q q. − . . p W−" =" + 2 +2 −p = 1 −"" 2 2 2 2 2 2 2 2 ◦ 2 4⇡✏ (2a) (2b) 16⇡✏ a b ⇐No good 2+ 2+ 2 2+ 2+ 2+ b line (x + a) (y − b) + z (x − a) + (y + b) z (2 a + b ) a 0a)+ 0b)+ 135 (x + (y − b) z (x − a) (y + z ■ The reason this doesn’t work for arbitrary angles is that you are even+ (3) (0) ◦region ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ tually forced to place an image charge within the original of ◦ + are θ θmust divisor ofof180 . .Thus , etc., ForFor this to to work, ✓ must bebe an integer divisor of 180 . 180 Thus 180180 ,180 90, 90 ,, 60 ,, 60 45, 45 etc., are OK, butbut no Forthis this towork, work, must beand andinteger integer divisor Thus 90, 60 ,, 45 , etc., areOK, OK, butnono ✲x interest, and that’s not images must go outside the re◦ allowed—all ◦say, others. It works for 45 , with the charges as shown. others. It works for 45 , say, with the charges as shown. ◦ others. It works for 45 , say, with the charges as shown. ◦ 4545line − gion, or you’re no longer dealing with the same problem at all.) line (4)(4) (2)(2) (1) − (Note thethe strategy: to to make thethe xtheaxis an an equipotential (V (V= 0),=0),0), (Note x xaxis (Note thestrategy: strategy: tomake make axis anequipotential equipotential (V= (2) + +− − youyou place thethe image charge (1)(1) in the reflection point. ToTo make thethe charge point. youplace place theimage image charge (1)in inthe thereflection reflection point. Tomake make the why it doesn’t work for θ = 135◦ ◦ ◦ an equipotential, you place charge (2) at the image point. −− ++✲ 45 45line line an equipotential, you place charge (2) at the image point. 453.11 line an equipotential, you place charge (2) at the image point. ✲x x Problem ++ − (1) ButBut that screws up the x axis, so you must now insert image (3) to − (1) that screws up the x axis, so you must now insert image (3) to % $ But that screws up the x axis, so you must now 2insert 2 image (3) to √ λ ◦ line (x +you a) also + y need balance (2).(2). Moreover, make the = (4),(4), balance Moreover, to make V V0= From Prob. 2.47 (with y0to→ d): V the =45 where a2 = y0 2 − R2 ⇒ −a += d2 − R2 , , need balance (2). Moreover, to make the45line 45◦lnV line =0 0you youalso also need (4), 2 2 (x to a) y need − + 0axis to to balance (1).(1). ButBut now, to to restore the4πϵ xthe to V−to = (5)(5) now, the x xaxis V V0=+you (5)(5) (3)(3) tobalance balance (1). But now, torestore restore axis =0 0you youneed need (5) and to to balance (4), and so on. balance (4), and so on. ◦ to&balance (4), and so on. ' ◦ why it works forfor θ= 4545 ) ( why it works θ= d 2πϵ0 V0 2πϵ0 V0 a coth(2πϵ0 V0 /λ) = d No good ⇒ (dividing) λ= , or 135 = cosh ◦ ◦ −1 + ⇐ . No good ■■ areason csch(2πϵ Vdoesn’t =work R 135line line The reason angles is isthat evenRis that λeven0doesn’t 0 /λ) cosh (d/R) The this work forarbitrary arbitrary angles that youare are even+ ⇐ The reason thisthis doesn’t work forfor arbitrary angles youyou are (3) (3) (0)(0) tually forced charge within region ++ tually forced toplace place animage image charge within theoriginal original region tually forced to to place an an image charge within thethe original region of ofof ✲✲x interest, and that’s not allowed—all images must go outside the reinterest, and that’s not allowed—all images must go outside the reinterest, and that’s not allowed—all images must go outside the rex Problem 3.12 − gion, dealing with problem − gion, oryou’re you’re nolonger longer dealing with thesame same problem atall.) all.) gion, or or you’re no no longer dealing with thethe same problem at at all.) (1)(1) −− (2)(2) +a ∞ * 2 ◦ −nπx/a why it doesn’t for3.34). θ= 135 why it doesn’t work for θ= 135◦ V0 (y) sin(nπy/a) dy work (Eq. V (x, y) = Cn e sin(nπy/a) (Eq. 3.30), where Cn = a Problem 3.12 n=1  Problem 3.11 0 Problem 3.11 2 p (x$ + +2 y22 2 %2 % $ a) 2 2 √ R2 , (x + a) + y From Prob. 2.52&(with y0 ! d): V = ' λlnλ , where a2 =2 y20 2 2 R ) a = d√ (x + a) + y 2 2,2 2 2 2 R ⇒ 2 2− From (with V V= a a==y0y0− a a== d2d− RR ln(x 4⇡✏ +2 y 2 , ,where FromProb. Prob.2.47 2.47 d):a/2 =4πϵ where , ln (x a) − R ⇒ 0 +V , fory00y→ yd): < 0<→ 0(with 2 2 −− a)a)++ yy 0 0 4πϵ (x In this case V0 (y) = . Therefore, and −V , for a/2 < y < a 0 ✓ ◆ and and ⇢ a&coth(2⇡✏ V / ) = d ' 2⇡✏ 2⇡✏0 V0 ((0 V0 )) 0 0 &⎧ '(dividing) d ⎫= ) cosh , 0or = d 2πϵ 2πϵ V a coth(2πϵ V /λ) = d 0 V00V.03 0 0 0 0 ! # d 2πϵ V 1 2πϵ a/2 a coth(2πϵ V /λ) = d a 0 0 0 ⎪+ 3a/2λ λ= a csch(2⇡✏ + ⇒⇒(dividing) R⎪ 0 V0 / ) = R , ,oror cosh cosh (d/R) ⎬ == (dividing) = cosh −1 3 3a . . cos(nπy/a) 2V 2 a⎨csch(2πϵ cos(nπy/a) −1 V /λ) = R RR 0 λλ 0 00V0 /λ) = R cosh (d/R) a csch(2πϵ 3 3 cosh (d/R) sin(nπy/a) dy = − Cn = V0 sin(nπy/a) dy − + ⎪ ⎪ a ⎩ a (nπ/a) 30 (nπ/a) 3a/2 ⎭ Problem 3.13 0 a/2 Problem 3.12 Problem 3.12 4 5 nπ 6 5 nπ 67 2V 4 Za 5 nπ 67 1 2VX 0 0 2 n n⇡x/a + cos(0) + cos(nπ) − cos − cos = 1 + (−1) − 2 cos . V (x, = y) = C e sin(n⇡y/a) (Eq. 3.30), where C = V (y) sin(n⇡y/a) dy (Eq. 3.34). n n ∞∞ nπ * 2 2 nπ a 2 +a0+a 2 n=1 * 2 0 V0V(y) sin(nπy/a) dy (Eq. 3.34). VV (x,(x, y)y)== CnCen−nπx/a sin(nπy/a) e−nπx/a sin(nπy/a) (Eq. (Eq.3.30), 3.30), where where CnCn== 0 (y) sin(nπy/a) dy (Eq. 3.34). c ⃝2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This materialaisa n=1 n=1 −q

0 0 protected under all copyrightInc., laws Upper as theySaddle currently exist. of this material may beis c 2012 Pearson Education, River, NJ. No Allportion rights reserved. This material reproduced, in anyall form or by & any means, without permission in writing from the publisher. protected under copyright laws as they currently exist. No portion of this material may be '' & means, without permission reproduced, in any form or by any writing from the publisher. +V 0 , 0for +V , for0 0<
⎧⎧ ⎫⎫ !! a/2 a a ⎪ ⎪ 3a/2 3a3 ## a/2 + + ⎪ ⎨ ⎬⎪ + + ⎬ 2V2V 3 33a/2 cos(nπy/a) 3 3a cos(nπy/a) 22 ⎨ 0 0 cos(nπy/a) cos(nπy/a) 3 3 3 3 ++ dydy == −− CnCn== V0V0 sin(nπy/a) dydy −− sin(nπy/a) sin(nπy/a) sin(nπy/a) 3 3 3 3 ⎪⎪ ⎪ aa ⎩ aa (nπ/a) (nπ/a) ⎪ (nπ/a) (nπ/a) ⎭ 0 0 a/2 ⎩ ⎭ a/2 0 0 a/2 a/2 4 5 6 5 67 4 5 67 5 nπ 6 5 nπ 67 2V2V 4 5 nπ 67 2V0 4 nπ nπ nπ 0 nn == 2V0 −− cos ++ cos(0) ++ cos(nπ) −− cos == 0 1 + (−1) −− 2 cos . . cos cos(0) cos(nπ) cos 1 + (−1) 2 cos nπnπ 22 22 nπnπ 22

58 ⇢

In this case V0 (y) =


+V0 , for 0 < y < a/2 V0 , for a/2 < y < a

8 > Za/2 2 < Cn = V0 sin(n⇡y/a) dy a > :

Za

0

=

a/2

. Therefore,

2V0 sin(n⇡y/a) dy = > a ;

⇣ n⇡ ⌘ 2V0 n cos + cos(0) + cos(n⇡) n⇡ 2

The term in curly 8 > >n = 1 < n=2 n=3 > > : n=4

9 > =

cos

brackets is: : : : :

⇢

Cn = So

8V0 /n⇡, 0,

8V0 ⇡

V (x, y) =

2

=

a/2

cos(n⇡y/a) (n⇡/a)

0

2V0 n 1 + ( 1)n n⇡

cos(n⇡y/a) + (n⇡/a) 2 cos

⇣ n⇡ ⌘o 2

a a/2

)

.

9 2 cos(⇡/2) = 0, > > = 2 cos(⇡) = 4, etc. (Zero if n is odd or divisible by 4, otherwise 4.) 2 cos(3⇡/2) = 0, > > ; 2 cos(2⇡) = 0,

1 1 1+1 1 1 1+1

Therefore

⇣ n⇡ ⌘o

(

X

n = 2, 6, 10, 14, etc. (in general, 4j + 2, for j = 0, 1, 2, ...), otherwise. e

n⇡x/a

n=2,6,10,...

1 sin(n⇡y/a) 8V0 X e = n ⇡ j=0

sin[(4j + 2)⇡y/a] . (4j + 2)

(4j+2)⇡x/a

Problem 3.14 V (x, y) = So (y) =

✏0

@ @x

4✏0 V0 = a

4V0 ⇡

⇢

X

1 e n n=1,3,5,...

n⇡x/a

4V0 X 1 n⇡x/a e sin(n⇡y/a) ⇡ n X sin(n⇡y/a).

(Eq. 3.36);

=

✏0

x=0

=

✏0

@V @n

4V0 X 1 n⇡ ( )e ⇡ n a

(Eq. 2.49).

n⇡x/a

sin(n⇡y/a) x=0

n=1,3,5,...

Or, using the closed form 3.37: ✓ ◆ 2V0 sin(⇡y/a) 1 V (x, y) = tan ) ⇡ sinh(⇡x/a) =

sin(n⇡y/a)

=

2✏0 V0 sin(⇡y/a) cosh(⇡x/a) a sin2 (⇡y/a) + sinh2 (⇡x/a)

2V0 ✏0 ⇡ 1+ = x=0

1 sin2 (⇡y/a) sinh2 (⇡x/a)

✓

sin(⇡y/a) sinh2 (⇡x/a)

◆

⇡ cosh(⇡x/a) a

x=0

2✏0 V0 1 . a sin(⇡y/a)

[Comment: Technically, the series solution for is defective, since term-by-term di↵erentiation has produced a (naively) non-convergent sum. More sophisticated definitions of convergence permit one to work with series of this form, but it is better to sum the series first and then di↵erentiate (the second method.)]


59

CHAPTER 3. POTENTIAL Summation of series Eq. 3.36 V (x, y) = Now sin w = Im eiw , so where Z ⌘ e

⇡(x iy)/a

. Now

X 1 4V0 I, where I ⌘ e ⇡ n n=1,3,5,...

I = Im

X1 e n

n⇡x/a in⇡y/a

e

n⇡x/a

= Im

sin(n⇡y/a).

X1 Z n, n

8 9 Z Z
✓ ◆ 1 1 1+Z 1 1 = du = ln = ln Rei ✓ = (ln R + i ✓)7 , CHAPTER 3. SPECIAL 1 TECHNIQUES u2 2 1 Z 2 2 ZZ 0

1+Z i ✓i θ =1+Z whereReRe . Therefore where = 1 1−Z Z . Therefore

soso

# $# $ ⇢! " iy)/a −π(x−iy)/a −π(x+iy)/a iy)/a −π(x−iy)/a 1+ e e⇡(x 11− e e⇡(x+iy)/a 1+ 11 11 1+ ZZ 1 + e e⇡(x 1+ 1+ $# $ I I==Im RR ++ i ✓) == == # i θ) == ✓.θ. But But Im (ln(ln iy)/a −π(x−iy)/a iy)/a −π(x−iy)/a −π(x+iy)/a 22 22 11− ZZ 1 1 − e e⇡(x 11− e e⇡(x 11− e e⇡(x+iy)/a # $ −πx/a i⇡y/a −iπy/a −2πx/a −πx/a −2πx/a 1+ e e⇡x/a e eiπy/a − e ei⇡y/a e e2⇡x/a 1+ − 1+ 2ie sin(⇡y/a) e e2⇡x/a 1+ 2ie⇡x/a sin(πy/a) − == == , , % % %2 %2 2 2 iy)/a % iy)/a % −π(x−iy)/a −π(x−iy)/a 1%1 − e e⇡(x 1%1 − e e⇡(x

Therefore Therefore

−πx/a 2e2e⇡x/a sin(⇡y/a) 2 sin(⇡y/a) sin(⇡y/a) sin(πy/a) 2 sin(πy/a) sin(πy/a) tan ✓ θ== == ⇡x/a == . . tan −2πx/a −πx/a sinh(⇡x/a) 11− e e2⇡x/a e eπx/a − e e⇡x/a sinh(πx/a)

✓& ◆' ✓& ◆' 1 0 0 sin(πy/a) , and V (x, y) = 2V2V sin(πy/a) . 1−1 sin(⇡y/a) 1−1 sin(⇡y/a) I I== 1tan tan , and V (x, y) = ⇡ . tan sinh(⇡x/a) 2 2 tan sinh(⇡x/a) sinh(πx/a) π sinh(πx/a)

Problem Problem3.15 3.14 2 @2V @ 2V2 ∂ V with boundary conditions (a)(a) 2 + + ∂2 V= 0, = 0, with boundary conditions @x∂x2 @y∂y 2 8⎧ 9⎫ (i)(i) V V (x, 0)0)==0,0, > > (x, >⎪ > ⎪ ⎪ < =⎪ ⎨ ⎬ (ii) (x, a)a)==0,0, (ii)V V (x, (iii) (0,(0, y)y)==0,0, > > (iii)V V ⎪ >⎪ > ⎪ : ;⎪ ⎩ ⎭ (iv) (b,(b, y)y)==V0V(y). (iv)V V 0 (y).

y

✻

a

V0 (y)

V =0

b

z✙

✲x

AsAsininEx. Ex.3.4, 3.4,separation separationofofvariables variablesyields yields # kxkx kx $ VV (x, y)y)==AeAe ++ Be kyky++ DD cos ky) . . (x, Be−kx(C(Csinsin cos ky)

Here A, (ii)) Here(i)) (i)⇒DD==0,0,(iii)) (iii)⇒BB== −A, (ii)⇒kakais isananinteger integermultiple multipleofof⇡:π: ⇣/ ⌘ n⇡x/a n⇡x/a 0 nπx/a e −nπx/a sin(n⇡y/a) = (2AC) sinh(n⇡x/a) sin(n⇡y/a). VV (x, y) = AC e (x, y) = AC e −e sin(nπy/a) = (2AC) sinh(nπx/a) sin(nπy/a). c But 2012 Pearson Inc., Upper River,general NJ. All rights This material is (2AC) Education, is a constant, and Saddle the most linearreserved. combination of separable protected under all copyright laws as they currently exist. No portion of this material may be (ii), (iii) inisany form or by any means, without permission in writing from the publisher. reproduced,

V (x, y) =

∞ 1

solutions consistent with (i),

Cn sinh(nπx/a) sin(nπy/a).

n=1

It remains to determine the coefficients Cn so as to fit boundary condition (iv): 1

2 Cn sinh(nπb/a) sin(nπy/a) = V0 (y). Fourier’s trick ⇒ Cn sinh(nπb/a) = a

2a

V0 (y) sin(nπy/a) dy.

60


But (2AC) is a constant, and the most general linear combination of separable solutions consistent with (i), (ii), (iii) is 1 X V (x, y) = Cn sinh(n⇡x/a) sin(n⇡y/a). n=1

It remains to determine the coefficients Cn so as to fit boundary condition (iv): X

Cn sinh(n⇡b/a) sin(n⇡y/a) = V0 (y). Fourier’s trick ) Cn sinh(n⇡b/a) =

2 a

Za

V0 (y) sin(n⇡y/a) dy.

0

Therefore 2 Cn = a sinh(n⇡b/a)

Za

V0 (y) sin(n⇡y/a) dy.

0

2 (b) Cn = V0 a sinh(n⇡b/a)

Za 0

2V0 sin(n⇡y/a) dy = ⇥ a sinh(n⇡b/a)

V (x, y) =

4V0 ⇡

⇢

0, if n is even, if n is odd.

2a n⇡ ,

X

sinh(n⇡x/a) sin(n⇡y/a) . n sinh(n⇡b/a) n=1,3,5,...

Problem 3.16 Same format as Ex. 3.5, only the boundary conditions are: 8 9 (i) V = 0 when x = 0, > > > > > > > (ii) V = 0 when x = a, > > > > > < = (iii) V = 0 when y = 0, > (iv) V = 0 when y = a, > > > > > > (v) V = 0 when z = 0, > > > > > : ; (vi) V = V0 when z = a. This time we want sinusoidal fuctions in x and y, exponential in z: X(x) = A sin(kx) + B cos(kx),

Y (y) = C sin(ly) + D cos(ly),

p

Z(z) = Ee

k2 +l2 z

+ Ge

p

k2 +l2 z

.

(i)) B = 0; (ii)) k = n⇡/a; (iii)) D = 0; (iv)) l = m⇡/a; (v)) E + G = 0. Therefore p Z(z) = 2E sinh(⇡ n2 + m2 z/a). Putting this all together, and combining the constants, we have: V (x, y, z) =

1 X 1 X

n=1 m=1

Cn,m sin(n⇡x/a) sin(m⇡y/a) sinh(⇡

p n2 + m2 z/a).

It remains to evaluate the constants Cn,m , by imposing boundary condition (vi): i p XXh V0 = Cn,m sinh(⇡ n2 + m2 ) sin(n⇡x/a) sin(m⇡y/a). c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

61

CHAPTER 3. POTENTIAL According to Eqs. 3.50 and 3.51: ⇣ p ⌘ ✓ 2 ◆2 Za Za 2 2 Cn,m sinh ⇡ n + m = V0 sin(n⇡x/a) sin(m⇡y/a) dx dy = a 0

0

(

0, if n or m is even, 16V0 , if both are odd. ⇡ 2 nm

)

Therefore 16V0 V (x, y, z) = 2 ⇡

p sinh ⇡ n2 + m2 z/a 1 p sin(n⇡x/a) sin(m⇡y/a) . nm sinh ⇡ n2 + m2 n=1,3,5,... m=1,3,5,... X

X

Consider the superposition of six such cubes, one with V0 on each of the six faces. The result is a cube with V0 on its entire surface, so the potential at the center is V0 . Evidently the potential at the center of the original cube (with V0 on just one face) is one sixth of this: V0 /6. To check it, put in x = y = z = a/2: 16V0 V (a/2, a/2, a/2) = 2 ⇡

p sinh ⇡ n2 + m2 /2 1 p sin(n⇡/2) sin(m⇡/2) . nm sinh ⇡ n2 + m2 n=1,3,5,... m=1,3,5,... X

X

Let n ⌘ 2i + 1, m ⌘ 2j + 1, and note that sinh(2u) = 2 sinh(u) cosh(u). The double sum is then S=

1 1 i h p 1 XX ( 1)i+j sech ⇡ (2i + 1)2 + (2j + 1)2 /2 . 2 i=0 j=0 (2i + 1)(2j + 1)

Setting the upper limits at i = 3, j = 3 (or above) Mathematica returns S = 0.102808, which (to 6 digits) is equal to ⇡ 2 /96, confirming (at least, numerically) that V (a/2, a/2, a/2) = V0 /6. Problem 3.17 1 d3 1 d2 1 d2 3 2 2 2 2 x 1 = 3 x 1 2x = x x2 1 3 2 2 8 · 6 dx 48 dx 8 dx i 1 d ⇥ ⇤ 1 d h 2 2 2 2 = x 1 + 2x x 1 2x = x 1 x2 1 + 4x2 8 dx 8 dx ⇤ 1⇥ ⇤ 1 d ⇥ 2 2 = x 1 5x 1 = 2x 5x2 1 + x2 1 10x 8 dx 8 1 1 5 3 = 5x3 x + 5x3 5x = 10x3 6x = x3 x. 4 4 2 2

P3 (x) =

We need to show that P3 (cos ✓) satisfies 1 d sin ✓ d✓ where P3 (cos ✓) =

1 2

cos ✓ 5 cos2 ✓

✓

sin ✓

dP d✓

◆

=

l(l + 1)P, with l = 3,

3 .

⇤ dP3 1⇥ = sin ✓ 5 cos2 ✓ 3 + cos ✓(10 cos ✓( sin ✓) = d✓ 2 3 = sin ✓ 5 cos2 ✓ 1 . 2

1 sin ✓ 5 cos2 ✓ 2


3 + 10 cos2 ✓

62 @ @✓

✓

sin ✓

1 d sin ✓ d✓

dP3 d✓

✓

◆

3 d ⇥ 2 sin ✓ 5 cos2 ✓ 2 d✓ ⇥ 3 sin ✓ cos ✓ 5 cos2 ✓

= =

dP sin ✓ d✓

◆

1 1

⇥ 3 cos ✓ 5 cos2 1

= =

Z1


3·4·

P1 (x)P3 (x) dx =

1

Z1

⇤

3⇥ 2 sin ✓ cos ✓ 5 cos2 ✓ 2 ⇤ 5 sin2 ✓ . =

5 1

1 cos ✓ 5 cos2 ✓ 2

(x)

1 5x3 2

cos2 ✓ 3 =

3x dx =

⇤

=

⇤ 1 + sin2 ✓ ( 10 cos ✓ sin ✓)

3 cos ✓ 10 cos2 ✓

6

l(l + 1)P3 . qed

1 5 x 2

x3

1 1

=

1 (1 2

1+1

1) = 0. X

1

Problem 3.18 (a) Inside: V (r, ✓) =

1 X

Al rl Pl (cos ✓)

(Eq. 3.66) where

l=0

(2l + 1) Al = 2Rl

Z⇡

V0 (✓)Pl (cos ✓) sin ✓ d✓

(Eq. 3.69).

0

In this case V0 (✓) = V0 comes outside the integral, so (2l + 1)V0 Al = 2Rl

Z⇡

Pl (cos ✓) sin ✓ d✓.

0

But P0 (cos ✓) = 1, so the integral can be written Z⇡

P0 (cos ✓)Pl (cos ✓) sin ✓ d✓ =

0

Therefore Al = Plugging this into the general form:

⇢

⇢

0, if l 6= 0 2, if l = 0

0, if l 6= 0 V0 , if l = 0

(Eq. 3.68).

.

V (r, ✓) = A0 r0 P0 (cos ✓) = V0 . The potential is constant throughout the sphere. 1 X Bl Outside: V (r, ✓) = Pl (cos ✓) (Eq. 3.72), where rl+1 l=0

(2l + 1) l+1 Bl = R 2

Z⇡


(Eq. 3.73).

0

=

(2l + 1) l+1 R V0 2

Z⇡ 0

Pl (cos ✓) sin ✓ d✓ =

⇢

0, if l 6= 0 RV0 , if l = 0

.


63


Therefore V (r, ✓) = V0 (b)

R 1 (i.e. equals V0 at r = R, then falls o↵ like ). r r

V (r, ✓) =

8 1 9 X > > > l > > Al r Pl (cos ✓), for r  R (Eq. 3.78) > > > < = l=0

1 X > Bl > > Pl (cos ✓), for r > : rl+1

> > R (Eq. 3.79) > > ;

l=0

where

Bl = R2l+1 Al

,

(Eq. 3.81)

and 1 Al = 2✏0 Rl 1 = 2✏0 Rl

1

Z⇡

0 (✓)Pl (cos ✓) sin ✓ d✓

(Eq. 3.84)

0

1

0

Z⇡

Pl (cos ✓) sin ✓ d✓ =

0

⇢

0, R

Therefore V (r, ✓) =

Note: in terms of the total charge Q = 4⇡R2

if l 6= 0 if l = 0

8 R 0 > , > > < ✏0

.

9 for r  R > > > =

> 2 > > : R 0 1 , for r ✏0 r

0,

V (r, ✓) =

Problem 3.19

0 /✏0 ,

> > > R;

8 9 1 Q > > > > , for r  R > > < 4⇡✏0 R = > > > :

1 Q , for r 4⇡✏0 r

⇥ V0 (✓) = k cos(3✓) = k 4 cos3 ✓

> > ; R>

.

.

⇤ 3 cos ✓ = k [↵P3 (cos ✓) + P1 (cos ✓)] .

(I know that any 3rd order polynomial can be expressed as a linear combination of the first four Legendre polynomials; in this case, since the polynomial is odd, I only need P1 and P3 .)  ✓ ◆ 5↵ 3 1 3 3 3 4 cos ✓ 3 cos ✓ = ↵ 5 cos ✓ 3 cos ✓ + cos ✓ = cos ✓ + ↵ cos ✓, 2 2 2 so

4=

5↵ 8 )↵= ; 2 5

3=

Therefore V0 (✓) =

3 ↵= 2

3 8 · = 2 5

k [8P3 (cos ✓) 5

12 ) 5

3P1 (cos ✓)] .


=

12 5

3=

3 . 5

64


Now

V (r, ✓) =

8 1 9 X > > > l > > Al r Pl (cos ✓), for r  R (Eq. 3.66) > > > < = l=0

1 X > Bl > > Pl (cos ✓), for r > : rl+1 l=0

where (2l + 1) Al = 2Rl

Z⇡


> > R (Eq. 3.72) > > ;

,

(Eq. 3.69)

0

8 ⇡ 9 Z Z⇡ = (2l + 1) k < = 8 P (cos ✓)P (cos ✓) sin ✓ d✓ 3 P (cos ✓)P (cos ✓) sin ✓ d✓ 3 l 1 l ; 2Rl 5 : 0 0 ⇢ k (2l + 1) 2 2 k 1 = 8 3 = [8 l3 3 l1 ] l3 l1 5 2Rl (2l + 1) (2l + 1) 5 Rl ⇢ 8k/5R3 , if l = 3 = (zero otherwise). 3k/5R, if l = 1 Therefore V (r, ✓) =

 ⇣ ⌘ 3k 8k 3 k r 3 rP1 (cos ✓) + r P (cos ✓) = 8 P3 (cos ✓) 3 3 5R 5R 5 R

3

⇣r⌘ R

P1 (cos ✓) ,

or k 5

⇢ ⇣ ⌘ r 31⇥ 8 5 cos3 ✓ R 2

3 cos ✓

⇤

3

⇣r⌘ R

cos ✓

⇢ ⇣ ⌘ k r r 2⇥ ) V (r, ✓) = cos ✓ 4 5 cos2 ✓ 5R R

⇤ 3

3

(for r  R). Meanwhile, Bl = Al R2l+1 (Eq. 3.81—this follows from the continuity of V at R). Therefore Bl =

⇢

8kR4 /5, if l = 3 3kR2 /5, if l = 1

(zero otherwise).

So V (r, ✓) =

" ✓ ◆ 4 3kR2 1 8kR4 1 k R P (cos ✓) + P (cos ✓) = 8 P3 (cos ✓) 1 3 5 r2 5 r4 5 r

3

✓

R r

◆2

#

P1 (cos ✓) ,

or k V (r, ✓) = 5

✓

R r

◆2

( ✓ ◆ 2 R ⇥ cos ✓ 4 5 cos2 ✓ r

3

⇤

3

)


65

CHAPTER 3. POTENTIAL (for r

R). Finally, using Eq. 3.83: (✓) = ✏0

1 X

(2l + 1)Al Rl

1

l=0

⇥ ⇤ Pl (cos ✓) = ✏0 3A1 P1 + 7A3 R2 P3

 ✓ ◆ ✓ ◆ 3k 8k ✏0 k = ✏0 3 P1 + 7 R2 P3 = [ 9P1 (cos ✓) + 56P3 (cos ✓)] 5R 5R3 5R  ✏0 k 56 ✏0 k = 9 cos ✓ + 5 cos3 ✓ 3 cos ✓ = cos ✓[ 9 + 28 · 5 cos2 ✓ 28 · 3] 5R 2 5R =

⇥ ✏0 k cos ✓ 140 cos2 ✓ 5R

Problem 3.20

Use Eq. 3.83:

(✓) = ✏0

1 X

⇤ 93 .

(2l+1)Al R

l 1

l=0

Putting them together:

2l + 1 Pl (cos ✓). But Eq. 3.69 says: Al = 2Rl

Z⇡

V0 (✓)Pl (cos ✓) sin ✓ d✓.

0

1

✏0 X (✓) = (2l + 1)2 Cl Pl (cos ✓), 2R

with Cl =

l=0

Z⇡

V0 (✓)Pl (cos ✓) sin ✓ d✓. qed

0

Problem 3.21 Set V = 0 on the equatorial plane, far from the sphere. Then the potential is the same as Ex. 3.8 plus the potential of a uniformly charged spherical shell: ✓ E0 r

V (r, ✓) =

R3 r2

◆

cos ✓ +

1 Q . 4⇡✏0 r

Problem 3.22

1 1 1 hp X X X Bl Bl Bl (a) V (r, ✓) = P (cos ✓) (r > R), so V (r, 0) = P (1) = = r 2 + R2 l l rl+1 rl+1 rl+1 2✏0 l=0 l=0 l=0 p p 1 1 2 2 2 2 Since r > R in this region, r + R = r 1 + (R/r) = r 1 + (R/r) (R/r)4 + . . . , so 2 8

 1 X Bl 1 R2 = r 1 + rl+1 2✏0 2 r2

1 R4 + ... 8 r4

l=0

Comparing like powers of r, I see that B0 =  R2 1 V (r, ✓) = 4✏0 r " R2 = 1 4✏0 r

1 =

R2 , B1 = 0, B2 = 4✏0

2✏0

✓

R2 2r

◆ R4 + ... . 8r3

R4 , . . . . Therefore 16✏0

R2 P2 (cos ✓) + . . . , 4r3 1 8

✓

R r

◆2

3 cos ✓ 2

#

1 + ... ,


(for r > R).

i r .

66

(b) V (r, ✓) =

1 X l=0


Al rl Pl (cos ✓) (r < R). In the northern hemispere, 0  ✓  ⇡/2,

V (r, 0) =

1 X

Al rl =

l=0

Since r < R in this region,

p

r 2 + R2 = R 1 X l=0

Comparing like powers: A0 =

V (r, ✓) =

2✏0

 R

 R = 1 2✏0

2✏0

2✏0

 1 1 + (r/R)2 = R 1 + (r/R)2 2

p

 1 r2 Al r = R+ 2✏0 2R

1 r4 + ... 8 R3

l

R, A1 =

rP1 (cos ✓) +

i r .

hp r 2 + R2

2✏0

, A2 =

4✏0 R

1 ⇣ r ⌘2 cos ✓ + 3 cos2 ✓ R 4 R

r .

, . . . , so

1 2 r P2 (cos ✓) + . . . , 2R

⇣r⌘

1 (r/R)4 + . . . . Therefore 8

(for r < R, northern hemisphere).

1 + ... ,

In the southern hemisphere we’ll have to go for ✓ = ⇡, using Pl ( 1) = ( 1)l . V (r, ⇡) =

1 X

( 1)l Al rl =

l=0

2✏0

hp

r 2 + R2

i r .

(I put an overbar on Al to distinguish it from the northern Al ). The only di↵erence is the sign of A1 : A1 = +( /2✏0 ), A0 = A0 , A2 = A2 . So: V (r, ✓) =

 1 2 R + rP1 (cos ✓) + r P2 (cos ✓) + . . . , 2✏0 2R

 ⇣r⌘ R 1 ⇣ r ⌘2 = 1+ cos ✓ + 3 cos2 ✓ 2✏0 R 4 R Problem 3.23

V (r, ✓) =

(for r < R, southern hemisphere).

1 + ... ,

8 1 9 X > > > > > Al rl Pl (cos ✓), (r  R) (Eq. 3.78), > > > < = l=0

1 X > Bl > > Pl (cos ✓), (r > : rl+1 l=0

> > R) (Eq. 3.79), > > ;


67

CHAPTER 3. POTENTIAL where Bl = Al R2l+1 (Eq. 3.81) and 1 Al = 2✏0 Rl =

=

1 2✏0 Rl

1

0 (✓)Pl (cos ✓) sin ✓ d✓

(Eq. 3.84)

0

1

0

2✏0 Rl

Z⇡

1

8 9 > > Z⇡ < Z⇡/2 = Pl (cos ✓) sin ✓ d✓ Pl (cos ✓) sin ✓ d✓ 0 > > : ; 0 ⇡/2 8 1 9 Z0
(let x = cos ✓)

1

Now Pl ( x) = ( 1)l Pl (x), since Pl (x) is even, for even l, and odd, for odd l. Therefore Z0

Pl (x) dx =

1

Z0

Pl ( x) d( x) = ( 1)

l

1

Al =

2✏0

⇥ 1

0 Rl 1

( 1)l

⇤

Pl (x) dx =

0

8 0, > > <

> > : ✏0

0 Rl 1

Z1 0

So A0 = A2 = A4 = A6 = 0, and all we need are A1 , A3 , and A5 . Z1

P1 (x) dx =

Z1

1 P3 (x) dx = 2

Z1

1 P5 (x) dx = 8

0

Z1

x2 x dx = 2

0

0

Z1

1

= 0

5x

0

0

✓

x4 5 4

x2 3 2

◆

1 0

1 = 2

✓ 1 x6 63x 70x + 15x dx = 63 8 6 0 ✓ ◆ 1 21 35 15 1 1 = + = (36 35) = . 8 2 2 2 16 16 Z1

5

Therefore A1 =

3

Pl (x) dx, if l is odd > > ;

✓

5 4

3 2

◆

=

x4 x2 70 + 15 4 2

◆

1 . 8 1 0

✓ ◆ ✓ ◆ ✓ ◆ 1 1 1 0 0 ; A3 = ; A = ; etc. 5 2 4 ✏0 2 ✏0 R 8 ✏0 R 16 0

and B1 =

9 if l is even > > =

1 . 2

1 3x dx = 2

3

Pl (x) dx,

0

and hence Z1

Z1

0

✏0

R3

✓ ◆ ✓ ◆ ✓ ◆ 1 1 1 0 5 0 7 ; B3 = R ; B5 = R ; etc. 2 ✏0 8 ✏0 16


.

68 Thus V (r, ✓) =

Problem 3.24

8 > > > < > > > :


9  1 ⇣ r ⌘2 1 ⇣ r ⌘4 > P1 (cos ✓) P3 (cos ✓) + P5 (cos ✓) + ... , (r  R), > > = 2✏0 " 4 R 8 R # ✓ ◆ ✓ ◆ 2 4 3 1 R 1 R > 0R P1 (cos ✓) P3 (cos ✓) + P5 (cos ✓) + ... , (r R). > > ; 2✏0 r2 4 r 8 r 0r

1 @ s @s

✓

s

@V @s

◆

+

◆

+

1 @2V = 0. s2 @ 2

Look for solutions of the form V (s, ) = S(s) ( ): ✓ ◆ 1 d dS 1 d2 s + 2 S 2 = 0. s ds ds s d Multiply by s2 and divide by V = S : s d S ds

✓

s

dS ds

1 d2 = 0. d 2

Since the first term involves s only, and the second only, each is a constant: ✓ ◆ s d dS 1 d2 s = C1 , = C2 , with C1 + C2 = 0. S ds ds d 2 Now C2 must be negative (else we get exponentials for geometrically they must— when is increased by 2⇡). C2 =

k 2 . Then

d2 = d 2

k2

, which do not return to their original value—as

)

= A cos k + B sin k .

Moreover, since ( + 2⇡) = ( ), k must be an integer: k = 0, 1, 2, 3, . . . (negative integers are just repeats, but k =✓0 must ◆ be included, since = A (a constant) is OK). d dS s s = k 2 S can be solved by S = sn , provided n is chosen right: ds ds s

d snsn ds

1

= ns

d n (s ) = n2 ssn ds

1

= n2 sn = k 2 S ) n = ±k.

Evidently the general solution is S(s) = Csk + Ds k , unless k = 0, in which case we have only one solution to a second-order equation—namely, S = constant. So we must treat k = 0 separately. One solution is a constant—but what’s the other? Go back to the diferential equation for S, and put in k = 0: ✓ ◆ d dS dS dS C ds s =0)s s = constant = C ) = ) dS = C ) S = C ln s + D (another constant). ds ds ds ds s s So the second solution in this case is ln s. [How about ? That too reduces to a single solution, case k = 0. What’s the second solution here? Well, putting k = 0 into the equation: d2 d =0) d 2 d

= constant = B )

= A, in the

= B + A.


1

69


Contents

But a term of the form B is unacceptable, since it does not return to its initial value when by 2⇡.] Conclusion: The general solution with cylindrical symmetry is 1 X ⇥ k V (s, ) = a0 + b0 ln s + s (ak cos k + bk sin k ) + s k=1 1 Special Techniques

k

is augmented

⇤ (ck cos k + dk sin k ) .

Yes: the potential of a line charge goes like ln s, which is included. Problem Problem3.25 3.24 Picking V V= = 0 on thethe yz yz plane, with E0 E in0the direction, we have (Eq. 3.74): Picking 0 on plane, with in x the x direction, ⇢! " (i)(i)V V== 0,0, when whens = s =R,R, (ii) == −E E0 s0cos R.R. 0 x0 x (ii)V V!→ E −E s cos,φ,forfors s ≫

y

− − − − −

2

✻ + + φ+ + +

s

✲x Evidently a0 = b0 = bk = dk = 0, and ak = ck = 0 except for k = 1: $ # c1 cos φ. V (s, φ) = a1 s + s =⇒E0 Evidently a0 = b0 = bk = dk = 0, and ak = ck = 0 except for k = 1: ✢ ⇣ (i)⇒ c1 = −a1 R2 ; (ii)→ a1 = −E0 . Therefore c1 ⌘ z V (s, ) = a1 s + cos . s ( '% & & % 2 2 E R R 0 − 1 cos φ. (s, φ)a= (i)) c1 = a1 R2 ; V(ii)! E0 s. + Therefore cos φ, or V (s, φ) = −E0 s 1 = −E s s "✓ ◆ # ✓ ◆ 2 R ) E0 R 2 ) % & ) s V (s, ) = E0∂V s +) cos , or 2 V (s, ) = E 1 cos . ) s = −ϵ0 E0 − R − 1 cos φ) 0 = s2ϵ0 E0 cos φ. σ = −ϵ0 ) ∂s )s=R s2 s=R ✓ ◆ @V R2 Problem 3.25 = ✏* = ✏0 E0 1 cos = 2✏0 E0 cos . 0∞ @sk s=R s2 s=R Inside: V (s, φ) = a0 + s (ak cos kφ + bk sin kφ) . (In this region ln s and s−k are no good—they blow k=1 Problem 3.26 1 up at s = 0.) X ∞k k Inside: V (s, ) = a0 + * s (a 1 k cos k + bk sin k ) . (In this region ln s kand s are no good—they blow + Outside: V (s, φ) = a0 k=1 (c cos kφ + d sin kφ). (Here ln s and s are no good at s → ∞). k k sk k=1 up at s = 0.) 1 X &) % 1 ) ln s and sk are no good at s ! 1). Outside: V (s, ) = a0 + (ck cos k +∂V dkout sin k ∂V ).in(Here ) − (Eq. 2.36). sk σ = −ϵ0 k=1 ∂s ∂s )s=R ◆ ✓ @Vout @Vin Thus = ✏ (Eq. 2.36). " 0 ∞ ! * @s @s k s=R a sin 5φ = −ϵ0 − k+1 (ck cos kφ + dk sin kφ) − kRk−1 (ak cos kφ + bk sin kφ) . R k=1 Thus ⇢ 1 X k 1 a sin 5 = Inc., ✏0 Upper Saddle (ck cos dk sin k ) This kRkmaterial (ak is cos k + bk sin k ) . c ⃝2005 Pearson Education, River, NJ.kAll + rights reserved. k+1 Rcurrently exist. No portion of this material may be protected under all copyright laws as they k=1 reproduced, in any form or by any means, without permission in writing from the publisher. ✓ ◆ 1 4 Evidently ak = ck = 0; bk = dk = 0 except k = 5; a = 5✏0 d5 + R b5 . Also, V is continuous at s = R: R6 1 a0 +R5 b5 sin 5 = a0 + 5 d5 sin 5 . So a0 = a0 (might as well choose both zero); R5 b5 = R 5 d5 , or d5 = R10 b5 . R


70

CHAPTER 3. POTENTIAL a aR6 ; d5 = . Therefore 4 10✏0 R 10✏0

Combining these results: a = 5✏0 R4 b5 + R4 b5 = 10✏0 R4 b5 ; b5 = ⇢

a sin 5 V (s, ) = 10✏0

s5 /R4 , for s < R, R6 /s5 , for s > R.

Problem 3.27 Since r is on the z axis, the angle ↵ is just the polar angle ✓ (I’ll drop the primes, for simplicity). Monopole term: Z Z  1 ⇢ d⌧ = kR (R 2r) sin ✓ r2 sin ✓ dr d✓ d . r2 But the r integral is

ZR (R

2r) dr = Rr

R 0

r2

= R2

R2 = 0.

0

So the monopole term is zero. Dipole term:  Z Z 1 r cos ✓⇢ d⌧ = kR (r cos ✓) 2 (R r But the ✓ integral is

Z⇡

sin3 ✓ 3

sin2 ✓ cos ✓ d✓ =

0

2r) sin ✓ r2 sin ✓ dr d✓ d .

⇡

=

1 (0 3

1



1 (R r2

=

R4 3

0

So the dipole contribution is likewise zero. Quadrupole term: ✓ ◆ Z Z 3 1 1 r2 cos2 ✓ ⇢ d⌧ = kR r2 3 cos2 ✓ 2 2 2 r integral:

Z

R

r2 (R

2r) dr =

0

✓ integral:

Z⇡

3 cos ✓ {z 2

|

1

✓

r3 R 3

0 3(1 sin2 ✓) 1=2 3 sin2 ✓

integral:

= 2

⇣⇡⌘ 2

◆

sin ✓ d✓ = 2 2

}

r4 2

3

R 0

Z⇡

0) = 0.

2r) sin ✓ r2 sin ✓ dr d✓ d .

R4 = 2

sin ✓ d✓ 2

0

✓

3⇡ 8

Z2⇡

◆

✓

=⇡ 1

3

Z⇡

R4 . 6

sin4 ✓ d✓

0

9 8

◆

=

⇡ . 8

d = 2⇡.

0

The whole integral is:

1 kR 2

✓

R4 6

◆⇣

⇡⌘ k⇡ 2 R5 (2⇡) = . 8 48


71

CHAPTER 3. POTENTIAL For point P on the z axis (r ! z in Eq. 3.95) the approximate potential is 1 k⇡ 2 R5 . 4⇡✏0 48z 3

V (z) ⇠ =

(Quadrupole.)

Problem 3.28

z r

R f'

r'

y

x For a line charge, ⇢(r0 ) d⌧ 0 ! (r0 ) dl0 , which in this case becomes R d 0 . r r0 r · r0 cos ↵ n=0:

= = = =

r sin ✓ cos x ˆ + r sin ✓ sin y ˆ + r cos ✓ ˆ z, 0 0 R cos + R sin , so rR sin ✓ cos cos 0 + rR sin ✓ sin sin 0 = rR cos ↵, sin ✓(cos cos 0 + sin sin 0 ). Z

⇢(r ) d⌧ ! R 0

0

Z

2⇡

d

0

= 2⇡R ;

V0 =

0

n=1: Z Z r0 cos ↵ ⇢(r0 ) d⌧ 0 ! R cos ↵ R d

0

= R2 sin ✓

Z

2⇡

1 2⇡R R = . 4⇡✏0 r 2✏0 r

(cos cos

0

+ sin sin

0

)d

0

= 0; V1 = 0.

0

n=2: ✓ ◆ Z Z Z h i 3 1 R3 2 0 2 0 0 2 2 Rd 0 = (r ) P2 (cos ↵) ⇢(r )d⌧ ! R cos ↵ 3 sin2 ✓ (cos cos 0 + sin sin 0 ) 1 d 2 2 2  ✓ ◆ Z 2⇡ Z 2⇡ Z 2⇡ Z 2⇡ R3 = 3 sin2 ✓ cos2 cos2 0 d 0 + sin2 sin2 0 d 0 + 2 sin cos sin 0 cos 0 d 0 d 0 2 0 0 0 0 ✓ ◆ ⇤ ⇡ R3 R3 ⇥ 3 1 = 3 sin2 ✓ ⇡ cos2 + ⇡ sin2 + 0 2⇡ = 3 sin2 ✓ 2 = ⇡ R3 cos2 ✓ . 2 2 2 2

So

V2 =

R3 3 cos2 ✓ 8✏0 r3

1 =

R3 P2 (cos ✓). 4✏0 r3


0

72 Problem 3.29 p = (3qa qa) ˆ z + ( 2qa


2q( a)) y ˆ = 2qa ˆ z. Therefore V ⇠ =

1 p·ˆ r , 2 4⇡✏0 r

and p · ˆ r = 2qa ˆ z·ˆ r = 2qa cos ✓, so 1 2qa cos ✓ . 4⇡✏0 r2

V ⇠ =

(Dipole.)

Problem 3.30 R R (a) By symmetry, p is clearly in the z direction: p = p ˆ z; p = z⇢ d⌧ ) z da. p =

=

Z

(R cos ✓)(k cos ✓)R sin ✓ d✓ d = 2⇡R k 2

3

Z⇡

cos ✓ sin ✓ d✓ = 2⇡R k 2

3

0

2 3 ⇡R k[1 3

4⇡R3 k ; 3

( 1)] =

p=

✓

cos3 ✓ 3

◆

⇡ 0

4⇡R3 k ˆ z. 3

(b) V ⇠ =

1 4⇡R3 k cos ✓ kR3 cos ✓ = . 2 4⇡✏0 3 r 3✏0 r2

(Dipole.)

This is also the exact potential. Conclusion: all multiple moments of this distribution (except the dipole) are exactly zero. Problem 3.31 Using Eq. 3.94 with r0 = d/2 and ↵ = ✓ (Fig. 3.26): 1

r for

r

, we let ✓ ! 180 + ✓, so cos ✓ !

cos ✓:

1

r

+

◆n 1 ✓ 1X d = Pn (cos ✓); r n=0 2r

=

◆n 1 ✓ 1X d Pn ( cos ✓). r n=0 2r

But Pn ( x) = ( 1)n Pn (x), so V =

1 q 4⇡✏0

✓

1

r

+

1

r

◆

=

◆n 1 ✓ 1 1X d q [Pn (cos ✓) 4⇡✏0 r n=0 2r

Therefore Vdip = Voct =

2q 4⇡✏0 r

✓

d 2r

◆3

Pn ( cos ✓)] =

2q 1 d qd cos ✓ P1 (cos ✓) = , 4⇡✏0 r 2r 4⇡✏0 r2

P3 (cos ✓) =

2q d3 1 5 cos3 ✓ 4⇡✏0 8r4 2

while

3 cos ✓ =

2q 4⇡✏0 r

X

n=1,3,5,...

✓

d 2r

◆n

Pn (cos ✓).

Vquad = 0. qd3 1 5 cos3 ✓ 4⇡✏0 8r4

3 cos ✓ .


73


Problem 3.32 (a) (i) Q = 2q,

(ii) p = 3qa ˆ z,

(b) (i) Q = 2q,

(ii) p =

(c) (i) Q = 2q,

(iii) V ⇠ =

1 4⇡✏0

Q r

+

p·ˆ r r2

i

 1 2q 3qa cos ✓ = + . 4⇡✏0 r r2

 1 2q qa cos ✓ + . 4⇡✏0 r r2  1 2q 3qa sin ✓ sin (iii) V ⇠ + = 4⇡✏0 r r2 (iii) V ⇠ =

qa ˆ z,

(ii) p = 3qa y ˆ,

Problem 3.33 (a) This point is at r = a, ✓ =

h

= 0, so E =

⇡ 2,

(from Eq. 1.64, y ˆ ·ˆ r = sin ✓ sin ).

p p ✓ˆ = ( ˆ z); F = qE = 3 4⇡✏0 a 4⇡✏0 a3

pq ˆ z. 4⇡✏0 a3

p 2p 2pq (2ˆ r) = ˆ z. F = ˆ z. 4⇡✏0 a3 4⇡✏0 a3 4⇡✏0 a3 ⇣ ⇡ ⌘i qp h pq V (a, 0, 0)] = cos(0) cos = . 4⇡✏0 a2 2 4⇡✏0 a2

(b) Here r = a, ✓ = 0, so E = (c) W = q [V (0, 0, a) Problem 3.34 Q=

1 q ; 4⇡✏0 r

q, so Vmono =

q 4⇡✏0

V (r, ✓) ⇠ =

✓

p = qa ˆ z,

1 a cos ✓ + r r2

◆

.

so Vdip =

E(r, ✓) ⇠ =

1 qa cos ✓ . Therefore 4⇡✏0 r2 q 4⇡✏0



⌘ 1 a ⇣ ˆ . ˆ r + 2 cos ✓ ˆ r + sin ✓ ✓ r2 r3

Problem 3.35 The total charge is zero, so the dominant term is the dipole. We need the dipole moment of this configuration. It obviously points in the z direction, and for the southern hemisphere (✓ : ⇡2 ! ⇡) ⇢ switches sign but so does z, so p =

Z

z⇢ d⌧ = 2⇢0 R4 sin2 ✓ 4 2

= 4⇡⇢0

Z

⇡/2

✓=0 ⇡/2

= 0

r cos ✓ r sin ✓ dr d✓ d = 2⇢0 (2⇡) 2

Z

R 3

r dr

0

⇡⇢0 R4 . 2

Z

⇡/2

cos ✓ sin ✓ d✓

0

Therefore (Eq. 3.103) E⇡

⌘ ⇡⇢0 R4 ⇣ ˆ . 2 cos ✓ ˆ r + sin ✓ ✓ 8⇡✏0 r3

Problem 3.36 ˆ ✓ˆ = p cos ✓ ˆ p = (p · ˆ r) ˆ r + (p · ✓) r p sin ✓ ✓ˆ (Fig. 3.36). So 3(p · ˆ r) ˆ r ˆ 2p cos ✓ ˆ r + p sin ✓ ✓. So Eq. 3.104 ⌘ Eq. 3.103. X Problem 3.37 Vave (R) =

1 4⇡R2

Z

p = 3p cos ✓ ˆ r

p cos ✓ ˆ r + p sin ✓ ✓ˆ =

V (r) da, where the integral is over the surface of a sphere of radius R. Now da =


74 R2 sin ✓ d✓ d , so Vave (R) =

1 4⇡

Z

CHAPTER 3. POTENTIAL V (R, ✓, ) sin ✓ d✓ d .

Z Z Z dVave 1 @V 1 1 = sin ✓ d✓ d = (rV · ˆ r) sin ✓ d✓ d = (rV ) · (R2 sin ✓ d✓ d ˆ r) dR 4⇡ @R 4⇡ 4⇡R2 Z Z 1 1 = (rV ) · da = (r2 V ) d⌧ = 0. 4⇡R2 4⇡R2 5 (The final integral, from the divergence theorem, is over the volume of the sphere, where by assumption the Laplacian of V is zero.) So Vave is independent of R—the same for all spheres, regardless of their radius—and 2 hence (takingcan thealso limit R ! 0), Vdirectly. V (0). ave (R) =Let This integral beas integrated x = uqed ; dx = 2u du. Problem 3.38 At a point (x, y) on the plane the field of q is " $ %&'0 !0 √ !0 ' u2 u# u x d −1 2 1 q ' = −d sin−1 (1) = −d π . √ √ E =du = 2 − rˆ , d and √ dx = 2 − u +r = sinx x '√ ˆ + y y ˆ d ˆ z, 2 q 2 2 2 d−x d − u 4⇡✏0 r 3 d d √ d

d

q d . Meanwhile, the field of (just below the surface) is , 2 2 3/2 4⇡✏0 (x(+ y + d2 )) 2✏0 ) (Eq. 2.17). (Of course, this is for a duniform surface but as long is πd d π 2 d2 charge, 2π 3as d3 ϵwe 0 mare infinitesimally far away 16πϵ . t =field inside=the conductor 0 m =so e↵ectively uniform.) The total is zero, 2 2 2A 2 4 2q q

so its z component is Therefore

Problem 3.35

Problem 3.39 x✛

q d 4⇡✏0 (x2 + y 2 + d2 )3/2

+

−

+

−

2✏0

+

=0

+

−

qd . X 2⇡(x2 + y 2 + d2 )3/2

(x, y) =

)

+

−

q

+

−

−

The image configuration is shown in the figure; the positive image charge forces cancel in pairs. The net force of the negative image charges is: *( 1 2 11 1 11 11 + + 2+ 2+ 2 + ... FF == 4⇡✏0qq2 2 2 2 + ... [2(a x)] [2a + 2(a x)] [4a + 2(a x)] 4πϵ0 [2(a − x)] [2a + 2(a − x)] [4a + 2(a − x)] & 1 1 1 1 1 1 − (2x)22− (2a + 2x)22− (4a + 2x)22− . .. .. . (2x) (2a + 2x) (4a + 2x) ⇢ 2"+ , + 1 ,& 1 11 11 11 11 11 qq2 = + + + . . . − 1 2++ 1 2++ +. .. .. . . . 2 2 2 2 + + + . . . + = 4⇡✏0 4 (a− x) x)2 (2a− x) x)2 (3a− x) x)2 (a++x) x)2 (2a++x) x)2 4πϵ0 4 (a (2a (3a xx2 (a (2a

1 1 q2 2 When a ! 1 (i.e. a x) only the 1 2 term survives: F = X (same as for only one plane— 1 q When a → ∞ (i.e. a ≫ x) only the x2 term survives: F = − 4⇡✏0 (2x)22 ! (same as for only one plane— x 4πϵ0 (2x) Eq. 3.12). When x = a/2, Eq. 3.12). When x = a/2, ⇢  1 q 2"+ 1 1 1 1 1 1 , + F = 1 q2 + 1 2 + 1 2 + ... + 1 2 + 1 2 + . . ,& . = 0. X 1 1 2 2 (a/2) + (3a/2) + (5a/2) + . . . − (a/2) + (3a/2) + (5a/2) + . . . F = 4⇡✏0 4 = 0. ! 2 2 2 2 2 2 4πϵ0 4 (a/2) (3a/2) (5a/2) (a/2) (3a/2) (5a/2)

Problem 3.40 Problem 3.36Prob. 2.52, we place image line charges Following at y = b and + at y = b (here y is the horizontal axis, z vertical). Following Prob. 2.47, we place image line charges −λ at y = b and +λ at y = −b (here y is the horizontal axis, z vertical). z Upper Saddle River, NJ. All rights reserved. This material is c 2012 Pearson Education, Inc., protected under all copyright✻ laws as they currently exist. No portion of this material may be P reproduced, in any form or by any means, s1 without permission in writing from the publisher. s2 −λ

−b

s3

b

+λ R

☛ -

- ./ 0 - ./ 0 λ a−b a−b 2 2 ./ 0 y0 ./ 0 a

−λ

c ⃝2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is

s4 ✲y

When a → ∞ (i.e. a ≫ x) only the

1 1 q2 term survives: F = − ! (same as for only o 2 x 4πϵ0 (2x)2

Eq. 3.12). When x = a/2, F =

1 q2 4πϵ0 4

"+

, + ,& 1 1 1 1 1 1 + + + . . . − + + + . . . = (a/2)2 (3a/2)2 (5a/2)2 (a/2)2 (3a/2)2 (5a/2)2

Problem 3.36 CHAPTER 3. POTENTIAL 75 Following Prob. 2.47, we place image line charges −λ at y = b and +λ at y = −b (here y is the axis, z vertical). z ✻ P s1 s2 −b

−λ

b

+λ R

- ./ 0 - ./ 0 λ a−b a−b 2 2 ./ 0 y0 ./ 0 a

−λ

☛ -

In the solution to Prob. 2.52 substitute:

s3

s4 ✲y

c ⃝2005 Pearson Education, Inc., Upper Saddle material is ✓ ◆2 River, ✓ NJ. All ◆2 rights reserved. This protected all copyright may be a under b a + b laws asa theyb currentlyaexist. + b No portion R2 2 of this material areproduced, ! ,inyany = permission in writing R ) from b = the publisher. . form or byso any means, without 0 !

2

2

2

2

a

 ✓ 2◆ ✓ 2◆ ✓ 2 2◆ s s s s ln 32 + ln 12 = ln 12 32 4⇡✏0 s4 s2 4⇡✏0 s4 s2 ⇢ 2 2 2 2 [(y + a) + z ][(y b) + z ] = ln , or, using y = s cos , z = s sin , 4⇡✏0 [(y a)2 + z 2 ][(y + b)2 + z 2 ] ⇢ 2 (s + a2 + 2as cos )[(as/R)2 + R2 2as cos ] ln . = 4⇡✏0 (a2 + a2 2as cos )[(as/R)2 + R2 + 2as cos ]

V =

Problem 3.41 Same as Problem 3.9, only this time we want q 0 + q 00 = q, so q 00 = q q 0 : ✓ 00 ◆ ✓ ◆ q q q0 q2 qq 0 1 1 F = + = + + . 4⇡✏0 a2 (a b)2 4⇡✏0 a2 4⇡✏0 a2 (a b)2 The second term is identical to Problem 3.9, and I’ll just quote the answer from there:  2 q2 R2 ) 3 (2a F = a R . 4⇡✏0 a3 (a2 R2 )2 (a) F = 0 ) a(a2

R2 )2 = R3 (2a2

R2 ), or (letting x ⌘ a/R), x(x2 1)2 2x2 + 1 = 0. We want a real p root greater than 1; Mathematica delivers x = (1 + 5)/2 = 1.61803, so a = 1.61803R = 5.66311 ˚ A. (b) Let a0 = x0 R be the minimum value of a. The work necessary is  Z a0 Z 1 Z 1 2 q2 1 R2 ) q2 1 (2x2 1) 3 (2a W = F da = a R da = dx 3 2 2 2 2 4⇡✏0 a0 a (a R ) 4⇡✏0 R x0 x x3 (x2 1)2 1  q2 1 + 2x0 2x30 = . 4⇡✏0 R 2x20 (1 x20 ) p Putting in x0 = (1 + 5)/2, Mathematica says the term in square brackets is 1/2 (this is not an accident; see q2 footnote 6 on page 127), so W = . Numerically, 8⇡✏0 R W =

(1.60 ⇥ 10 19 )2 8⇡(8.85 ⇥ 10 12 )(5.66 ⇥ 10

10 )

J = 2.03 ⇥ 10

19


J = 1.27 eV.

76


Problem 3.42 y

y V1

V1

a

a V0 -b

y 0

0

b

V0

=

V0

a

x

+

V0 0

-b

x

b

0 -b

0 0

b

x

The first configuration on the right is precisely Example 3.4, but unfortunately the second configuration is not the same as Problem 3.15: y 0

a 0

V0 0

x

b

We could reconstruct Problem 3.15 with the modified boundaries, but let’s see if we can’t twist it around by an astute change of variables. Suppose we let x ! y, y ! u, a ! c, b ! a, and V0 ! V1 : y V1

a 0

0 0

u

c

This is closer; making the changes in the solution to Problem 3.15 we have (for this configuration) V (u, y) =

4V1 ⇡

X

sinh(n⇡y/c) sin(n⇡u/c) . n sinh(n⇡a/c) n=1,3,5...

Now let c ! 2b and u ! x + b, and the configuration is just what we want: y V1 a 0

0 0

-b

b

x

The potential for this configuration is V (x, y) =

4V1 ⇡

X

sinh(n⇡y/2b) sin(n⇡(x + b)/2b) . n sinh(n⇡a/2b) n=1,3,5...


77

CHAPTER 3. POTENTIAL (If you like, write sin(n⇡(x + b)/2b) as ( 1)(n V (x, y) =

4 ⇡

1)/2

cos(n⇡x/2b).) Combining this with Eq. 3.42,

 1 cosh(n⇡x/a) sin(n⇡y/a) sinh(n⇡y/2b) sin(n⇡(x + b)/2b) V0 + V1 . n cosh(n⇡b/a) sinh(n⇡a/2b) n=1,3,5... X

Here’s a plot of this function, for the case a = b = 1, V0 = 1/2, V1 = 1:

1.0 1.0 0.5

0.0 -1.0

0.5 -0.5 0.0 0.5 1.0

0.0

Problem 3.43

X✓

◆ Bl Since the configuration is azimuthally symmetric, V (r, ✓) = Al r + l+1 Pl (cos ✓). r X Bl (a) r > b: Al = 0 for all l, since V ! 0 at 1. Therefore V (r, ✓) = Pl (cos ✓). rl+1 ◆ X✓ Dl a < r < b : V (r, ✓) = Cl rl + l+1 Pl (cos ✓). r < a : V (r, ✓) = V0 . r We need to determine Bl , Cl , Dl , and V0 . To ✓ do ◆this, invoke boundary conditions as follows: (i) V is @V 1 continuous at a, (ii) V is continuous at b, (iii) 4 = (✓) at b. @r ✏ 0 ◆ X Bl X✓ Dl Bl Dl (ii) ) Pl (cos ✓) = Cl bl + l+1 Pl (cos ✓); = Cl bl + l+1 ) Bl = b2l+1 Cl + Dl . (1) bl+1 b8 bl+1 b 9 Dl > > ✓ ◆ l < X Cl a + l+1 = 0, if l 6= 0, = D = a2l+1 C , l 6= 0, Dl l l a (i) ) Cl al + l+1 Pl (cos ✓) = V0 ; (2) D0 D = aV aC > > a 0 0 0 0. : C0 a + ; = V , if l = 0; 0 a1 Putting (2) into (1) gives Bl = b2l+1 Cl a2l+1 Cl , l 6= 0, B0 = bC0 + aV0 aC0 . Therefore l

Bl = b2l+1 a2l+1 Cl , l 6= 0, (10 ) B0 = (b a)C0 + aV0 . (iii) )

X

Bl [ (l + 1)]

1

b

P (cos ✓) l+2 l

X✓

(l + 1) Bl bl+2

or (l + 1)Bl

Cl lbl

✓

Cl lb

1

l 1

+ Dl

(l + 1) bl+2

+ Dl

(l + 1) bl+2

lCl b2l+1 + (l + 1)Dl = 0;

◆

◆

Pl (cos ✓) =

k ✏0

P1 (cos ✓). So

= 0, if l 6= 1;

(l + 1)(Bl

Dl ) =


lb2l+1 Cl .

78 B1 (+2) Therefore


✓ ◆ 1 2 k + C + D = , for l = 1; 1 1 2 b2 b ✏0

C1 +

2 (B1 b3

D1 ) = k.

(l + 1)(Bl Dl ) + lb2l+1 Cl = 0, for l 6= 1, 2 k (3) C1 + 3 (B1 D1 ) = . b ✏0 Plug (2) and (10 ) into (3): For l 6= ⇥ 0 or 1: ⇤ (l+1) b2l+1 a2l+1 Cl + a2l+1 Cl +lb2l+1 Cl = 0; (l+1)b2l+1 Cl +lb2l+1 Cl = 0; (2l+1)Cl = 0 ) Cl = 0. Therefore (10 ) and (2) ) Bl = Cl = Dl = 0 for l > 1. ⇤ 2 ⇥ For l = 1: C1 + 3 b3 a3 C1 + a3 C1 = k; C1 + 2C1 = k ) C1 = k/3✏0 ; D1 = a3 C1 ) b D1 = a3 k/3✏0 ; B1 = b3 a3 C1 ) B1 = b3 a3 k/3✏0 . For l = 0: B0 D0 = 0 ) B0 = D0 ) (b a)C0 +aV0 = aV0 aC0 , so bC0 = 0 ) C0 = 0; D0 = aV0 = B0 . ✓ ◆ b3 a3 k aV0 aV0 k a3 Conclusion: V (r, ✓) = + cos ✓, r b. V (r, ✓) = + r cos ✓, a  r  b. r 3r2 ✏0 r 3✏0 r2  ✓ ◆ ✓ ◆ @V aV0 k a3 V0 k ✏0 (b) i (✓) = ✏0 = ✏0 + 1 + 2 cos ✓ = ✏ + cos ✓ = k cos ✓ + V0 . 0 2 3 @r a 3✏ a a ✏ a 0 0 a Z V0 ✏0 aV 1 Q 1 4⇡a✏ V aV ? 0 0 0 0 (c)qi = 4⇡a2 = 4⇡a✏0 V0 = Qtot . At large r: V ⇡ = = = .X i da = a r 4⇡✏0 r 4⇡✏0 r r Problem 3.44

dr' r' q'=q

Use multipole expansion (Eq. 3.95): ⇢ d⌧ 0 ! dr0 ; Q = ; the r0 integral breaks into two pieces: 2a

r

q'=p-q

2 a 3 Z Za 1 1 X 1 4 V (r) = (r0 )n Pn (cos ✓0 ) dr0 + (r0 )n Pn (cos ✓0 ) dr0 5 . 4⇡✏0 n=0 rn+1 0

0

In the first integral ✓ = ✓ (see diagram); in the second integral ✓ = ⇡ ✓, so cos ✓0 = ( 1)n Pn (z), so the integrals cancel when n is odd, and add when n is even. 0

0

cos ✓. But Pn ( z) =

Z 1 X 1 Q 1 V (r) = 2 Pn (cos ✓) xn dx. 4⇡✏0 2a n=0,2,4,... rn+1 a

0

The integral is

an+1 , so n+1

 Q 1 X 1 ⇣ a ⌘n V = Pn (cos ✓) . 4⇡✏0 r n=0,2,4,... n + 1 r c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

79

CHAPTER 3. POTENTIAL Problem 3.45 Use separation of variables in cylindrical coordinates (Prob. 3.24): V (s, ) = a0 + b0 ln s +

1 X ⇥ k s (ak cos k + bk sin k ) + s

k

k=1

⇤ (ck cos k + dk sin k ) .

P1 s < R : V (s, ) = k=1 sk (ak cos k + bk sin k ) (ln s and s k blow up at s = 0); P1 s > R : V (s, ) = k=1 s k (ck cos k + dk sin k ) (ln s and sk blow up as s ! 1).

(We X may as well pick constantsXso V ! 0 as s ! 1, and hence a0 = 0.) Continuity at s = R ) Rk (ak cos k + bk sin k ) = R k (ck cos k + dk sin k ), so ck = R2k ak , dk = R2k bk . Eq. 2.36 says: @V @V 1 = . Therefore @s R+ @s R ✏0 X

or:

X

k

R

(c cos k + dk sin k ) k+1 k X

2kRk

1

kRk

1

(ak cos k + bk sin k ) =

⇢

1 , ✏0

(ak cos k + bk sin k ) = 0 /✏0 0 /✏0

(0 < (⇡ <

< ⇡) < 2⇡)

.

Fourier’s trick: multiply by (cos l ) d and integrate from 0 to 2⇡, using Z2⇡

Z2⇡

sin k cos l d = 0;

0

Then 2lRl

1

⇡al =

0

✏0

cos k cos l d =

0

2 ⇡ Z 4 cos l d 0

Z2⇡ ⇡

Multiply by (sin l ) d and integrate, using

3

cos l d 5 =

2⇡ R

0

✏0

(

sin k sin l d =

0

2lR

l 1

0

2 ⇡ Z 4 sin l d

Z2⇡

3

0

(

⇢

cos l l

sin l d 5 = ✏0 ✏0 ⇡ 0 ⇢ ⇢ 0, 0, if l is even = ) bl = 4 0 /l✏0 , if l is odd 2 0 /⇡✏0 l2 Rl

⇡bl =

Conclusion: V (s, ) =

2 0R ⇡✏0

X

k=1,3,5,...

1 sin k k2

⇢

0

0, k 6= l ⇡, k = l

sin l l

.

2⇡ ⇡

⇡ 0

cos l + l

2⇡ ⇡

)

= .

(s/R)k (s < R) (R/s)k (s > R)

.

In =

)

= 0;

al = 0.

0, k 6= l : ⇡, k = l

if l is even , if l is odd

Problem 3.46 1 1 X Pn (cos ✓) Use Eq. 3.95, in the form V (r) = In ; 4⇡✏0 n=0 rn+1

⇡

sin l l

⇢

1

Za

z n (z) dz.

a


0

l✏0

(2

2 cos l⇡)

Contents 1

Contents Problem 3.40

80

Contents Z ⇣ ⌘ a

⇡z Problem (a) I0 = k 3.40 cos 2a

dz = k

a

CHAPTER 3. POTENTIAL 

λ(z) ✻

⇣ ⇡z ⌘ 2a sin ⇡ 2a

V (r, ✓) ⇠ =

1 4⇡✏0

a

= a

✓

2ak h ⇣ ⇡ ⌘ sin ⇡ 2

4ak ⇡

◆

1 . r

sin

⇣ ⇡ ⌘i 4ak = . Therefore: 2 ⇡

(Monopole.)

✲z Problem 3.40 −a a (b) I0 = 0. a ⇢⇣ ⌘ ⇣ ⇡z ⌘ az ⇣ ⇡z ⌘ a λ(z) Z a 2 ✻ I1 = kλ(z)z sin(⇡z/a) dz = k sin cos ⇡ a ⇡ a ✻ a a ✲ z ⇢ ⌘2 z −a a ⇣ a✲ a2 a2 2a2 −a = k a [sin(⇡) sin( ⇡)] cos(⇡) cos( ⇡) = k ; ⇡ ⇡ ⇡ ⇡ λ(z) λ(z) ✻✻ ✓ 2 ◆ 1 2a k 1 ⇠ λ(z) V (r, ✓) cos ✓. (Dipole.) = ✲✲ ✻ z 4⇡✏0 ⇡ r2 −a−a aa z −a a✲ z (c) I0 = I1 = 0. λ(z) a ⇢ Z✻ ⇣ ⇡z ⌘ ⇣ ⇡z ⌘ 2z cos(⇡z/a) (⇡z/a)2 2 2 I2 =λ(z) k z cos + sin ✲ z a dz = k (⇡/a)2 (⇡/a)3 a −a ✻ a a ⇣✲ −a a a ⌘z2 4a3 k = 2k [a cos(⇡) + a cos( ⇡)] = . ⇡ ⇡2 λ(z) ✻

a✲

−a

z

V (r, ✓) ⇠ =

1 4⇡✏0

✓

4a3 k ⇡2

◆

1 3 cos2 ✓ 2r3

1 .

a a

(Quadrupole.)

2

Problem 3.47

c ⃝2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright currently exist. qNo (a) The average field laws dueastothey a point charge atportion r is of this material may be Problem 3.41 reproduced, in any form or by any means, without permission in writing from the publisher.

r

✮

r q

Eave =

1 4 3 3 ⇡R

Z

E d⌧, Z

where E =

rˆ r

1 q rˆ , 4⇡✏0 r 2

1 1 dτ c ⃝2005 Pearson Education, Inc., Upper Saddle NJ. All rights reserved. This material is so E q 2 d⌧. ave =River, 4 3 No 4⇡✏ protected under all copyright laws as they currently⇡R exist. portion of this material may be 0 3 reproduced, in any form or by any means, without permission in writing from the publisher.

(Here r is the source point, d⌧ is the field point, so r goes from r to d⌧ .) The field at r due to uniform Z rˆ 1 charge ⇢ over the sphere is E⇢ = ⇢ 2 d⌧. This time d⌧ is the source point and r is the field 4⇡✏ 0 c ⃝2005 Pearson Education, Inc., Upper Saddle River, NJ.rAll rights reserved. This material is protected under copyright as they currently exist. No portion of this material r allgoes point, so from laws d⌧ to r, and hence carries the opposite sign.may So be with ⇢ = q/ 43 ⇡R3 , the two reproduced, in any form or by any means, without permission in writing from the publisher. expressions agree: Eave = E⇢ .

(b) From Prob. 2.12: E⇢ =

1 ⇢r = 3✏0

q r = 4⇡✏0 R3

p . 4⇡✏0 R3


81


(c) If there are many charges inside the sphere, Eave is the sum of the individual averages, and ptot is the p sum of the individual dipole moments. So Eave = . qed 4⇡✏0 R3 (d) The same argument, only with q placed at r outside the sphere, gives Eave = E⇢ =

1 4⇡✏0

4 3 3 ⇡R ⇢ r2

ˆ r (field at r due to uniformly charged sphere) =

1 q ˆ r. 2 4⇡✏0 r

But this is precisely the field produced by q (at r) at the center of the sphere. So the average field (over the sphere) due to a point charge outside the sphere is the same as the field that same charge produces at the center. And by superposition, this holds for any collection of exterior charges. Problem 3.48 (a) p ˆ (2 cos ✓ ˆ r + sin ✓ ✓) 4⇡✏0 r3 p = [2 cos ✓(sin ✓ cos x ˆ + sin ✓ sin y ˆ + cos ✓ ˆ z) 4⇡✏0 r3 + sin ✓(cos ✓ cos x ˆ + cos ✓ sin y ˆ sin ✓ ˆ z)] 2

Edip =

= Eave = =

But

Z2⇡ 0

3

p 6 7 ˆ + 3 sin ✓ cos ✓ sin y ˆ + 2 cos2 ✓ sin2 ✓ ˆ z5 . 43 sin ✓ cos ✓ cos x 4⇡✏0 r3 | {z } 1

4 3 3 ⇡R

1 4 3 ⇡R 3

cos d =

Z2⇡

Z

✓

=3 cos2 ✓ 1

Edip d⌧

p 4⇡✏0

◆Z

1 ⇥ 3 sin ✓ cos ✓(cos x ˆ + sin y ˆ) + 3 cos2 ✓ r3

sin d = 0, so the x ˆ and y ˆ terms drop out, and

2⇡ R

⇤ 1 ˆ z r2 sin ✓ dr d✓ d .

d = 2⇡, so

0

0

Eave =

1 4 3 ⇡R 3

✓

p 4⇡✏0

◆

2⇡

ZR

Z⇡

1 dr r

0

0

(

|

3 cos2 ✓

1 sin ✓ d✓ .

{z

}

cos3 ✓+cos ✓)|⇡ 0 =1 1+1 1=0

Z

1 dr, r blows up, since ln r ! 1 as r ! 0. If, as suggested, we truncate the r integral at r = ✏, then it is finite, and the ✓ integral gives Eave = 0.] (b) We want E within the ✏-sphere to be a delta function: E = A 3 (r), with A selected so that the average field is consistent with the general theorem in Prob. 3.47: Z 1 A p p p 3 Eave = 4 3 A 3 (r) d⌧ = 4 3 = )A= , and hence E = (r). 3 4⇡✏0 R 3✏0 3✏0 3 ⇡R 3 ⇡R Evidently Eave = 0, which contradicts the result of Prob. 3.47. [Note, however, that the r integral,

0


R

82


Problem 3.49 We need to show that the field inside the sphere approaches a delta-function with the right coefficient (Eq. 3.106) in the limit as R ! 0. From Eq. 3.86, the potential inside is V =

k k r cos ✓ = z, 3✏0 3✏0

so

E=

rV =

k ˆ z. 3✏0

From Prob. 3.30, the dipole moment of this configuration is p = (4⇡R3 k/3) ˆ z, so k ˆ z = 3p/(4⇡R3 ), and hence the field inside is 1 E= p. 4⇡✏0 R3 Clearly E ! 1 as R ! 0 (if p is held constant); its volume integral is Z 1 4 1 E d⌧ = p ⇡R3 = p, 4⇡✏0 R3 3 3✏0 which matches the delta-function term in Eq. 3.106. X Problem 3.50 Z (a) I = (rV1 ) · (rV2 ) d⌧ . But r·(V1 rV2 ) = (rV1 ) · (rV2 ) + V1 (r2 V2 ), so I= Problem 3.41

Z

2 r·(V1 rV2 ) d⌧

Z

I

1 V1 (r V2 ) = V1 (rV2 ) · da + ✏ 0 S 2

Z

V1 ⇢2 d⌧.

Z 1 But the surface integral is over a huge sphere “at infinity”, where V1 and V2 ! 0. So I = V1 ⇢2 d⌧ . By ✏0 Z Z Z r 1 the same with 1 and 2 reversed, I = V2 ⇢1 d⌧ . So V1 ⇢2 d⌧ = V2 ⇢1 d⌧ . qed q ✮ argument, ✏0 R 9 R dτ 8 < Situation (1 ) : Qa = a ⇢1 d⌧ = Q; Qb = b ⇢1 d⌧ = 0; V1b ⌘ Vab . = (b) R R : ; Situation (2 ) : Qa = a ⇢2 d⌧ = 0; Qb = b ⇢2 d⌧ = Q; V2a ⌘ Vba . 8R 9 R R < V1 ⇢2 d⌧ = V1a a ⇢2 d⌧ + V1b b ⇢2 d⌧ = Vab Q. = :R

V2 ⇢1 d⌧ = V2a

R

a

⇢1 d⌧ + V2b

R

b

⇢1 d⌧ = Vba Q.

Green’s reciprocity theorem says QVab = QVba , so Vab = Vba . qed Problem 3.51

;

(a) Situation (1): actual. Situation (2): right plate at V0 , left plate at V = 0, no charge at x. Z V =0 V =0 x V1 ⇢2 d⌧ = Vl1 Ql2 + Vx1 Qx2 + Vr1 Qr2 . ✲x 0 q d R But Vl1 = Vr1 = 0 and Qx2 = 0, so V1 ⇢2 d⌧ = 0. Z V2 ⇢1 d⌧ = Vl2 Ql1 + Vx2 Qx1 + Vr2 Qr1 . But Vl2 = 0 Qx1 = q, Vr2 = V0 , Qr1 = Q2 , and Vx2 = V0 (x/d). So 0 = V0 (x/d)q + V0 Q2 , and hence Q2 =

qx/d.


83


Situation (1): actual. Situation (2): left plate at V0 , right plate at V = 0, no charge at x. Z Z V1 ⇢2 d⌧ = 0 = V2 ⇢1 d⌧ = Vl2 Ql1 + Vx2 Qx1 + Vr2 Qr1 = V0 Q1 + qVx2 + 0.

⇣ But Vx2 = V0 1

x⌘ , so d

Q1 =

q(1

x/d).

(b) Situation (1): actual. Situation (2): inner sphere at V0 , outer sphere at zero, no charge at r. Z V1 ⇢2 d⌧ = Va1 Qa2 + Vr1 Qr2 + Vb1 Qb2 .

R But Va1 = Vb1 = 0, Qr2 = 0. So V1 ⇢2 d⌧ = 0. Z V2 ⇢1 d⌧ = Va2 Qa1 + Vr2 Qr1 + Vb2 Qb1 = Qa V0 + qVr2 + 0.

But Vr2 is the potential at r in configuration 2: V (r) = A + B/r, with V (a) = V0 ) A + B/a = V0 , or aA + B = aV0 , and V (b) = 0 ) A + B/b = 0, or bA + B = 0. Subtract: (b a)A = aV0 ) A = aV0 /(b a); B a1 1b = V0 = B (baba) ) B = abV0 /(b a). So V (r) = (baV0a) rb 1 . Therefore aV0 Qa V0 + q (b a)

✓

b r

◆ 1 = 0;

qa

Qa =

(b

a)

✓

b r

◆ 1 .

Now let Situation (2) be: inner sphere at zero, outer at V0 , no charge at r. Z Z V1 ⇢2 d⌧ = 0 = V2 ⇢1 d⌧ = Va2 Qa1 + Vr2 Qr1 + Vb2 Qb1 = 0 + qVr2 + Qb V0 . B This time V (r) = A + with V (a) = 0 ) A + B/a = 0; V (b) = V0 ) A + B/b = V0 , so r ⇣ ⌘ bV0 a bV0 ⇣ a⌘ qb ⇣ a⌘ V (r) = 1 . Therefore q 1 + Qb V0 = 0; Qb = 1 . (b a) r (b a) r (b a) r Problem 3.52 8 Z 3 3 3 X 1 < X 0X (a) ˆ riˆ rj Qij = 3 ˆ ri ri ˆ rj rj0 : 2 i,j=1 i=1 j=1 But

3 X i=1

ˆ ri ri0 = ˆ r · r0 = r0 cos ↵ =

3 X j=1

ˆ rj rj0 ;

X i,j

(r0 )2

X

ˆ riˆ rj

i,j

ˆ riˆ rj

ij

=

X

ij

9 = ;

⇢ d⌧ 0

ˆ rj ˆ rj = ˆ r·ˆ r = 1.

Z ⌘ 1 1 1 ⇣ 02 2 02 Vquad = 3r cos ↵ r ⇢ d⌧ 0 = the third term in Eq. 3.96. 4⇡✏0 r3 2 ⇥ (b) Because x2 = y 2 = (a/2)2 for all four charges, Qxx = Qyy = 12 3(a/2)2 p Because z = 0 for all four charges, Qzz = 12 [ ( 2a/2)2 ](q q q + q) = 0 and This leaves only Qxy = Qyx =

So X p ⇤ ( 2a/2)2 (q q q + q) = 0. Qxz = Qyz = Qzx = Qzy = 0.

⇣a⌘ ⇣ a⌘ ⇣ a⌘ ⇣a⌘ ⇣ a⌘ ⇣ a⌘ i 3 h⇣ a ⌘ ⇣ a ⌘ 3 q+ ( q) + ( q) + q = a2 q. 2 2 2 2 2 2 2 2 2 2


84


(c) 2Qij = =

Z

⇥ 3(ri

Z

di )(rj

dj )

(r

⇥ ⇤ 3ri rj r2 ij ⇢ d⌧ Z 2 d ij ⇢ d⌧ = Qij

d)2

3di

Z

ij

⇤

(I0 ll drop the primes, for simplicity.) Z Z Z 3dj ri ⇢ d⌧ + 3di dj ⇢ d⌧ + 2d · r⇢ d⌧

⇢ d⌧

rj ⇢ d⌧

3(di pj + dj pi ) + 3di dj Q + 2

So if p = 0 and Q = 0 then Qij = Qij . qed (d) Eq. 3.95 with n = 3: Z 1 1 Voct = (r0 )3 P3 (cos ↵)⇢ d⌧ 0 ; 4⇡✏0 r4

ij d

P3 (cos ✓) =

·p

d2

ij

ij Q.

1 5 cos3 ✓ 2

3 cos ✓ .

1 1 X ˆ riˆ rj ˆ rk Qijk . 4⇡✏0 r4

Voct =

i,j,k

Define the “octopole moment” as 1 2

Qijk ⌘ Problem 3.53 ⇢ ✓ 1 1 V = q r1 4⇡✏0

r r r r

1

=

2

=

3

=

1

=

1

r

2

1

r

1

3

But q 0 =

+q

0

⇥ 0 0 0 5ri rj rk ✓

1

r

r

4

◆

✓

1

r

jk

+ rj0

ik

+ rk0

⇤

ij )

Problem 3.46

r

3

4

⇢(r0 ) d⌧ 0 .

◆

r ✯ ❃ ✒ ✼▼

r4

r2

r −q′

!

r

2

◆

r3

b

"# a

2r ⇠ = 2 cos ✓ (we want a a

r1

θ

+q

!"#$ !"#$+q′

−q

1

1

(r0 )2 (ri0

1

p r2 + a2 2ra cos ✓, p r2 + a2 + 2ra cos ✓, p r2 + b2 2rb cos ✓, p r2 + b2 + 2rb cos ✓.

Expanding as in Ex. 3.10: ✓

◆

Z

b

$!

r, not r

"# a

$

a, this time).

2b ⇠ = 2 cos ✓ (here we want b ⌧ r, because b = R2 /a, Eq. 3.16) r 2 R2 = cos ✓. a r2

R q (Eq. 3.15), so a V (r, ✓) ⇠ =

 1 2r q 2 cos ✓ 4⇡✏0 a

R 2 R2 1 q cos ✓ = 2 a ar 4⇡✏0

✓

2q a2

◆✓ r

R3 r2

◆

cos ✓.


85

CHAPTER 3. POTENTIAL Set E0 =

1 2q (field in the vicinity of the sphere produced by ±q): 4⇡✏0 a2 ✓ E0 r

V (r, ✓) =

R3 r2

◆

cos ✓

(agrees with Eq. 3.76).

Problem 3.54 The boundary conditions are 9 (i) V = 0 when y = 0, > > = (ii) V = V0 when y = a, (iii) V = 0 when x = b, > > ; (iv) V = 0 when x = b.

Go back to Eq. 3.26 and examine the case k = 0: d2 X/dx2 = d2 Y /dy 2 = 0, so X(x) = Ax + B, Y (y) = Cy + D. But this configuration is symmetric in x, so A = 0, and hence the k = 0 solution is V (x, y) = Cy + D. Pick D = 0, C = V0 /a, and subtract o↵ this part: V (x, y) = V0

y + V¯ (x, y). a

The remainder (V¯ (x, y)) satisfies boundary conditions similar to Ex. 3.4: 9 (i) V¯ = 0 when y = 0, > > = (ii) V¯ = 0 when y = a, ¯ (iii) V = V0 (y/a) when x = b, > > ; (iv) V¯ = V0 (y/a) when x = b.

(The point of peeling o↵ V0 (y/a) was to recover (ii), on which the constraint k = n⇡/a depends.) The solution (following Ex. 3.4) is V¯ (x, y) =

1 X

Cn cosh(n⇡x/a) sin(n⇡y/a),

n=1

and it remains to fit condition (iii): V¯ (b, y) = Invoke Fourier’s trick: X

Cn cosh(n⇡b/a)

Z

X a

Cn cosh(n⇡b/a) sin(n⇡y/a) =

sin(n⇡y/a) sin(n0 ⇡y/a) dy =

0

a Cn cosh(n⇡b/a) = 2

V0 a

Z

a

V0 (y/a).

V0 a

Z

a

y sin(n0 ⇡y/a) dy,

0

y sin(n⇡y/a) dy.

0

⇣ ⌘ ⇣ ay ⌘ 2V0 a 2 Cn = sin(n⇡y/a) cos(n⇡y/a) a2 cosh(n⇡b/a) n⇡ n⇡ ✓ 2◆ 2V0 a 2V0 ( 1)n = 2 cos(n⇡) = . a cosh(n⇡b/a) n⇡ n⇡ cosh(n⇡b/a)

a 0


86


V (x, y) = V0

"

# 1 y 2 X ( 1)n cosh(n⇡x/a) + sin(n⇡y/a) . a ⇡ n=1 n cosh(n⇡b/a)

Alternatively, start with the separable solution V (x, y) = (C sin kx + D cos kx) Aeky + Be

ky

.

Note that the configuration is symmetric in x, so C = 0, and V (x, 0) = 0 ) B = constants)

A, so (combining the

V (x, y) = A cos kx sinh ky. But V (b, y) = 0, so cos kb = 0, which means that kb = ±⇡/2, ±3⇡/2, . . . , or k = (2n 1)⇡/2b ⌘ ↵n , with n = 1, 2, 3, . . . (negative k does not yield a di↵erent solution—the sign can be absorbed into A). The general linear combination is 1 X V (x, y) = An cos ↵n x sinh ↵n y, n=1

and it remains to fit the final boundary condition: V (x, a) = V0 =

1 X

An cos ↵n x sinh ↵n a.

n=1

Use Fourier’s trick, multiplying by cos ↵n0 x and integrating: V0

Z

V0

b b

cos ↵n0 x dx =

1 X

An sinh ↵n a

n=1

1 2 sin ↵n0 b X = An sinh ↵n a(b ↵n0 n=1

2V0 sin ↵n b So An = . But sin ↵n b = sin b ↵n sinh ↵n a V (x, y) =

✓

2n

1 2

⇡

◆

Z

b b

n0 n )

=

cos ↵n0 x cos ↵n x dx,

= bAn0 sinh ↵n0 a.

( 1)n , so

1 2V0 X sinh ↵n y ( 1)n cos ↵n x. b n=1 ↵n sinh ↵n a

Problem 3.55 (a) Using Prob. 3.15b (with b = a): V (x, y) =

4V0 X sinh(n⇡x/a) sin(n⇡y/a) . ⇡ n sinh(n⇡) n odd


87

CHAPTER 3. POTENTIAL (y) =

✏0

@V @x

x=0

=

✏0

4V0 X ⇣ n⇡ ⌘ cosh(n⇡x/a) sin(n⇡y/a) ⇡ a n sinh(n⇡) n odd

x=0

4✏0 V0 X sin(n⇡y/a) = . a sinh(n⇡) n odd Z a Z a 4✏0 V0 X 1 = (y) dy = sin(n⇡y/a) dy. a sinh(n⇡) 0 0 n odd Z a a a 2a a But sin(n⇡y/a) dy = cos(n⇡y/a) 0 = [1 cos(n⇡)] = (since n is odd). n⇡ n⇡ n⇡ 0 8✏0 V0 X 1 ✏0 V0 = = ln 2. ⇡ n sinh(n⇡) ⇡ n odd

[Summing the series numerically (using Mathematica) gives 0.0866434, which agrees precisely with ln 2/8. The series can be summed analytically, by manipulation of elliptic integrals—see “Integrals and Series, Vol. I: Elementary Functions,” by A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (Gordon and Breach, New York, 1986), p. 721. I thank Ram Valluri for calling this to my attention.] Using Prob. 3.54 (with b = a/2): " # y 2 X ( 1)n cosh(n⇡x/a) sin(n⇡y/a) V (x, y) = V0 + . a ⇡ n n cosh(n⇡/2) (x) = = =

= =

" # @V 1 2 X ⇣ n⇡ ⌘ ( 1)n cosh(n⇡x/a) cos(n⇡y/a) ✏0 = ✏0 V0 + @y y=0 a ⇡ n a n cosh(n⇡/2) y=0 " # " # X ( 1)n cosh(n⇡x/a) 1 2 X ( 1)n cosh(n⇡x/a) ✏0 V0 ✏0 V0 + = 1+2 . a a n cosh(n⇡/2) a cosh(n⇡/2) n " # Z a/2 X ( 1)n Z a/2 ✏0 V0 (x) dx = a+2 cosh(n⇡x/a) dx . a cosh(n⇡/2) a/2 a/2 n Z a/2 a/2 a 2a cosh(n⇡x/a) dx = sinh(n⇡x/a) = sinh(n⇡/2). But n⇡ n⇡ a/2 a/2 " # " # ✏0 V0 4a X ( 1)n tanh(n⇡/2) 4 X ( 1)n tanh(n⇡/2) a+ = ✏0 V0 1 + a ⇡ n n ⇡ n n ✏0 V0 ln 2. ⇡

[The numerical value is -0.612111, which agrees with the expected value (ln 2 ⇡)/4.] (b) From Prob. 3.24: ◆ 1 ✓ X 1 k V (s, ) = a0 + b0 ln s + ak s + bk k [ck cos(k ) + dk sin(k )]. s k=1

In the interior (s < R) b0 and bk must be zero (ln s and 1/s blow up at the origin). Symmetry ) dk = 0. So V (s, ) = a0 +

1 X

ak sk cos(k ).

k=1 c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

4 Problem 3.48

88 y


✻ V0 ✲x

At the surface: V (R, ) =

X

ak R cos(k ) = k

k=0

⇢

V0 , if ⇡/4 < 0, otherwise.

< ⇡/4,

Fourier’s trick: multiply by cos(k 0 ) and integrate from ⇡ to ⇡: Problem 3.49 8 ⇡/4 Z ⇡ Z ⇡/4 1 < X V0 sin(k 0 )/k 0 = (V0 /k 0 ) sin(k 0 ⇡/4), if k 0 6= 0, k 0 0 ak R cos(k ) cos(k ) d = V0 cos(k ) d = ⇡/4 : V ⇡/2, if k 0 = 0. ⇡ ⇡/4 k=0 0 z

But

✻ θ φ

Z

l

T❦

8 < 0, if k 6= k 0 0 cos(k ) cos(k ) d = 2⇡, if k = k 0 = 0, : ⇡ ⇡, if k = k 0 6= 0.

⇡

mg ❄ So 2⇡a0 = V0 ⇡/2 ) a0 = V0 /4; ⇡ak Rk = (2V0 /k) sin(k⇡/4) ) ak = (2V0 /⇡kRk ) sin(k⇡/4) (k 6= 0); hence

V (s, ) = V0

"

# 1 1 2 X sin(k⇡/4) ⇣ s ⌘k + cos(k ) . 4 ⇡ k R k=1

Using Eq. 2.49, and noting that in this case n ˆ= ˆ s: ( ) = ✏0

@V @s

s=R

= ✏0 V0

1 2 X sin(k⇡/4) k ks ⇡ kRk

1

cos(k )

k=1

s=R

We want the net (line) charge on the segment opposite to V0 ( ⇡ < = =

Z

( )R d = 2R

4✏0 V0 ⇡

Z

⇡

( )d =

3⇡/4

1 X

k=1

sin(k⇡/4)



sin(k ) k

=

1 2✏0 V0 X sin(k⇡/4) cos(k ). ⇡R k=1

<

3⇡/4 and 3⇡/4 <

< ⇡):

Z ⇡ 1 4✏0 V0 X sin(k⇡/4) cos(k ) d ⇡ 3⇡/4 k=1

⇡ 3⇡/4

=

1 4✏0 V0 X sin(k⇡/4) sin(3k⇡/4) . ⇡ k k=1

k sin(k⇡/4) sin(3k⇡/4) product p p 1 1/ 2 1/ 2 1/2 2 1p -1 -1 p 3 1/ 2 1/ 2 1/2 4 0p 0p 0 5 -1/ 2 -1/ 2 1/2 c 6Pearson Education, -1p 1p Saddle River, -1 NJ. All rights reserved. This material is ⃝2005 Inc., Upper protected under all copyright laws as they currently exist. No portion of this material may be 7 -1/ 2 -1/ 2 1/2 reproduced, in any form or by any means, without permission in writing from the publisher. 8 0 0 0 c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

89


" # 4✏0 V0 1 X 1 1 X 1 = = = 0. ⇡ 2 1,3,5... k 2 1,3,5,... k P Ouch! What went wrong? The problem is that the series (1/k) is divergent, so the “subtraction” 1 1 is suspect. One way to avoid this is to go back to V (s, ), calculate ✏0 (@V /@s) at s 6= R, and save the limit s ! R until the end: 1 @V 2✏0 V0 X sin(k⇡/4) ksk 1 ( , s) ⌘ ✏0 = cos(k ) 4 @s ⇡ k Rk 4✏0 V0 ⇡

"

1 X 1 2 1,3,5... k

X

1 k 2,6,10,...

#

k=1

1 2✏0 V0 X k x Problem =3.48 ⇡R

1

sin(k⇡/4) cos(k )

k=1

(x) ⌘

( , s)R d =

1 4✏0 V0 X 1 k x ⇡ k

1

(where x ⌘ s/R ! 1 at the end). sin(k⇡/4) sin(3k⇡/4)

k=1 3 5

 ✓ ◆ ✓ ◆ ✻ 4✏0 V0 1 x x 1 x2 x6 x10 = x+ + + ··· + + + ··· ⇡ V 2x 3 5 x 2 6 10 0 ✓ ◆ ✓ ◆ 2✏0 V✲ x3 x5 x6 x10 0 x x+ = + + ··· x2 + + + ··· . ⇡x 3 5 3 5 ✓ ◆ ✓ ◆ 1+x x3 x5 But (see math tables) : ln =2 x+ + + ··· . 1 x 3 5  ✓ ◆ ✓ ◆ ✓ ◆✓ ◆ 2 2✏0 V0 1 1+x 1 1+x ✏0 V0 1+x 1 + x2 = ln ln = ln ⇡x 2 1 x 2 1 x2 ⇡x 1 x 1 x2  Problem 3.49 ✏0 V0 (1 + x)2 ✏0 V0 = ln ; = lim (x) = ln 2. 2 x!1 ⇡x 1+x ⇡ y

Problem 3.56 z

✻ θ φ

F = qE = l

T❦

qp ˆ (2 cos ✓ ˆ r + sin ✓ ✓). 4⇡✏0 r3

mg ❄

Now consider the pendulum: F = mg ˆ z Tˆ r, where T mg cos = mv 2 /l and (by conservation of 2 2 energy) mgl cos = (1/2)mv ) v = 2gl cos (assuming it started from rest at = 90 , as stipulated). But cos = cos ✓, so T = mg( cos ✓) + (m/l)( 2gl cos ✓) = 3mg cos ✓, and hence F=

mg(cos ✓ ˆ r

ˆ + 3mg cos ✓ ˆ ˆ sin ✓ ✓) r = mg(2 cos ✓ ˆ r + sin ✓ ✓).

This total force is such as to keep the pendulum on a circular arc, and it is identical to the force on q in the field of a dipole, with mg $ qp/4⇡✏0 l3 . Evidently q also executes semicircular motion, as though it were on a tether of fixed length l. Problem 3.57 Symmetry suggests that the plane of the orbit is perpendicular to the z axis, and since we need a centripetal force, pointing in toward the axis, the orbit must lie at the bottom of the field loops (Fig. 3.37a), where the z component of the field is zero. Referring to Eq. 3.104, c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

90


p p E·ˆ z = 0 ) 3(p·ˆ r)(ˆ r ·ˆ z) p·ˆ z = 0, or 3 cos2 ✓ 1 = 0. So cos2 ✓ = 1/3, cos ✓ = 1/ 3, sin ✓ = 2/3, z/s = p tan ✓ ) z = 2 s. The field at the orbit is (Eq. 3.103) ! r 1 p 2ˆ E= 2p ˆ r+ ✓ 4⇡✏0 r3 3 3 r h i p p 2 = 2(sin ✓ cos x ˆ + sin ✓ sin y ˆ + cos ✓ ˆ z ) + (cos ✓ cos x ˆ + cos ✓ sin y ˆ sin ✓ ˆ z ) 4⇡✏0 r3 3 r h⇣ ⌘ ⇣ p ⌘ ⇣ p ⌘ i p p 2 = 2 sin ✓ + cos ✓ cos x ˆ+ 2 sin ✓ + cos ✓ sin y ˆ+ 2 cos ✓ sin ✓ ˆ z 3 4⇡✏0 r 3 " ! ! # r r r p p 1 p 2 2 1 2 p = 2 (cos x ˆ + sin y ˆ) + 2p ˆ z 4⇡✏0 r3 3 3 3 3 3 r h i p p 2 p p p p = 3 (cos x ˆ + sin y ˆ ) = 2ˆ s= ˆ s. 3 3 4⇡✏0 r 3 4⇡✏0 r 3 3⇡✏0 s3 p (I used s = r sin ✓ = r 2/3, in the last step.) The centripetal force is F = qE =

qp = 3 3⇡✏0 s3 p

mv 2 s

qp v2 = p 3 3⇡✏0 ms2

)

The angular momentum is L = smv =

r

v=

)

1 s

r

qp p . 3 3⇡✏0 m

qpm p , 3 3⇡✏0

the same for all orbits, regardless of their radius (!), and the energy is W = Problem 3.58 Potential of q:

1 1 qp q p cos ✓ qp p = p mv 2 + qV = + 2 2 2 2 3 3⇡✏0 s 4⇡✏0 r 6 3⇡✏0 s2

Vq (r) =

1 q , 4⇡✏0 r

where

Equation 3.94, with r0 ! a and ↵ ! ✓:

= a2 + r2 2ar cos ✓. 1 1 1 X ⇣ a ⌘n = r r n=0 r Pn (cos ✓). So

Vq (r, ✓) =

Meanwhile, the potential of

is (Eq. 3.79)

r

qp p = 0. 4⇡✏0 3(3/2)s2

2

1 q 1 X ⇣ a ⌘n Pn (cos ✓). 4⇡✏0 r n=0 r

V (r, ✓) =

1 X Bl Pl (cos ✓). rl+1 l=0

Comparing the two (Vq = V ) we see that Bl = (q/4⇡✏0 )al , and hence (Eq. 3.81) Al = (q/4⇡✏0 )al /R2l+1 . Then (Eq. 3.83) " 1 # 1 1 ⇣ ⌘l ⇣ a ⌘l X ⇣ a ⌘l X q X q a (✓) = (2l + 1) Pl (cos ✓) = 2 l Pl (cos ✓) + Pl (cos ✓) . 4⇡R2 R 4⇡R2 R R l=0

l=0

l=0


91

CHAPTER 3. POTENTIAL Now (second line above, with r ! R) p

1 a2 + R2

2aR cos ✓

=

1 X ⇣ a ⌘l Pl (cos ✓). R R 1

l=0

Di↵erentiating with respect to a: d da Thus

✓

p

1 a2 + R2

2aR cos ✓

◆

=

(a R cos ✓) 1 X ⇣ a ⌘l = l Pl (cos ✓). aR R (a2 + R2 2aR cos ✓)3/2 l=0 1

 q (a R cos ✓) R 2aR 2 + 2 4⇡R2 (a + R2 2aR cos ✓)3/2 (a + R2 2aR cos ✓)1/2 ⇥ ⇤ 2a(a R cos ✓) + (a2 + R2 2aR cos ✓) q q (R2 a2 ) = = . 2 2 3/2 2 4⇡R 4⇡R (a + R2 2aR cos ✓)3/2 (a + R 2aR cos ✓)

(✓) =


92

CHAPTER 4. ELECTRIC FIELDS IN MATTER

Chapter 4

Electric Fields in Matter Problem 4.1 E = V /x = 500/10 3 = 5⇥105 . Table 4.1: ↵/4⇡✏0 = 0.66⇥10 30 , so ↵ = 4 ⇡(8.85⇥10 12 )(0.66⇥10 7.34 ⇥ 10 41 . p = ↵E = ed ) d = ↵E/e = (7.34 ⇥ 10 41 )(5 ⇥ 105 )/(1.6 ⇥ 10 19 ) = 2.29 ⇥ 10 16 m. d/R = (2.29 ⇥ 10

Rex/↵ = (0.5 ⇥ 10 Problem 4.2

16

)/(0.5 ⇥ 10

10

10

)(1.6 ⇥ 10

) = 4.6 ⇥ 10

19

)(10

3

6

30

)=

. To ionize, say d = R. Then R = ↵E/e = ↵V /ex ) V =

)/(7.34 ⇥ 10

First find the field, at radius r, using Gauss’ law:

R

41

) = 108 V.

E·da =

 Z 4⇡q r 2r/a 2 4q a = ⇢ d⌧ = e r dr = 3 e 3 ⇡a 0 a 2 0  ✓ ◆  2q a2 a2 2r/a 2 = e r + ar + = q 1 a2 2 2

1 ✏0

1 1 4⇡✏0 r 2 Qenc .

◆ r a2 Qenc r + ar + 2 0 ✓ ◆ 2 r r e 2r/a 1 + 2 + 2 2 . a a h ⇣ ⌘i 1 q r r2 2r/a [Note: Qenc (r ! 1) = q.] So the field of the electron cloud is Ee = 4⇡✏ 1 e 1 + 2 + 2 . The 2 2 a a 0 r proton will be shifted from r = 0 to the point d where Ee = E (the external field):  ✓ ◆ 1 q d d2 2d/a E= 1 e 1+2 +2 2 . 4⇡✏0 d2 a a Z

r

2r/a

✓

Qenc , or E =

2

Expanding in powers of (d/a): e 1

e

2d/a

✓

2d/a

=1

◆

=1

d d2 1+2 +2 2 a a

✓

2d a 1

◆

◆2 2d a ✓ ◆2 d d 2 +2 a a 1 + 2

✓

◆3 ✓ ◆2 2d d d + ··· = 1 2 + 2 a a a !✓ ✓ ◆3 ◆ 4 d d d2 + ··· 1+2 +2 2 3 a a a

1 3!

✓

d d2 d d2 d3 1/ 2/ 2 /2 + 2/ + 4 /2 + 4 /3 a a a a a ✓ ◆3 4 d = + higher order terms. 3 a = 1/

d2 2 /2 a

4 3

✓ ◆3 d + ··· a

d3 4 d3 4 /3 + + ··· a 3 a3


1

93

CHAPTER 4. ELECTRIC FIELDS IN MATTER E=

1 q 4⇡✏0 d2

✓

4 d3 3 a3

◆

=

1 4 1 (qd) = p. 4⇡✏0 3a3 3⇡✏0 a3

↵ = 3⇡✏0 a3 .

[Not so di↵erent from the uniform sphere model of Ex. 4.1 (see Eq. 4.2). Note that this result predicts 1 3 3 3 10 3 = 0.09 ⇥ 10 30 m3 , compared with an experimental value (Table 4.1) of 4⇡✏0 ↵ = 4 a = 4 0.5 ⇥ 10 0.66 ⇥ 10 30 m3 . Ironically the “classical” formula (Eq. 4.2) is slightly closer to the empirical value.]

Contents

Problem 4.3 H Rr ⇢(r) = Ar. Electric field (by Gauss’s Law): E·da = E 4⇡r2 = ✏10 1Qenc = ✏10 0 Ar 4⇡r2 dr, or E = 1 4⇡A r4 Ar2 = . This “internal” field balances the external field E when nucleus is “o↵-center” an amount 2 4⇡r ✏0 4 4✏0 p p p Problem d: ad2 /4✏4.4 4✏0 E/A. So the induced dipole moment is p = ed = 2e ✏0 /A E. Evidently 0 = E ) d = p is proportional to E 1/2 . For Eq. 4.1 to hold in the weak-field limit, E must be proportional to r, for small r, which means that ⇢ must go to a constant (not zero) at the origin: ⇢(0) 6= 0 (nor infinite).

Problem 4.4 Contents r

1 q Field of q: 4⇡✏ r. Induced dipole moment of atom: p = ↵ E = 2 ˆ 0 r ↵q ˆ r . 2 4⇡✏0 r ✓ ◆ q 1 1 2↵ q Hello Field of this dipole, at location of q (✓ = ⇡, in Eq. 3.103): E = (to the right). 4⇡✏0 r3 4⇡✏0 r2 ✓ ◆2 q 1 Problem Force on 4.6 q due to this field: F = 2↵ (attractive). 5 4⇡✏ r θ+ 0 Problem 4.4✸ p − 4.5 Problem z p1 ˆ Field of p1 at p2 (✓ = ⇡/2 in Eq. 3.103): E1 = ✓ (points down). 4⇡✏0 r3 z p1 p2 Torque (points into the page). + ❦on p2 : N2 = p2 ⇥ E1 = p2 E1 sin 90 = p2 E1 = 4⇡✏0 r3 pi − θ p2 Field of p2 at p1 (✓ = ⇡ in Eq. 3.103): E2 = ( 2ˆ r) (points to the right). r 4⇡✏0 r3 q 2p1 p2 Torque on p1 : N1 = p1 ⇥ E2 = (points into the page). 4⇡✏0 r3

Problem 4.6 Problem 4.6 (a) θ+ ✸ p − z +❦ pi

z θ

(b)

θ pi ✻

✲θ p

Use image dipole as shown in Fig. (a). Redraw, placing pi at the origin, Fig. (b).

2z

−

Ei =

p ˆ (2 cos ✓ ˆ r + sin ✓ ✓); 4⇡✏0 (2z)3

ˆ p = p cos ✓ ˆ r + p sin ✓ ✓.

h i p2 ˆ ⇥ (2 cos ✓ ˆ ˆ (cos ✓ ˆ r + sin ✓ ✓) r + sin ✓ ✓) 4⇡✏0 (2z)3 h i 2 p ˆ + 2 sin ✓ cos ✓( ˆ) = cos ✓ sin ✓ 4⇡✏0 (2z)3 p2 sin ✓ cos ✓ ˆNJ. All rights reserved. This material is c ⃝2005 Pearson Education, Inc., = Upper Saddle3 River, ( ) (out of the page). protected under all copyright laws as they currently exist. No portion of this material may be 4⇡✏ 0 (2z) N = p ⇥ Ei =

reproduced, in any form or by any means, without permission in writing from the publisher.


c ⃝2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is

1

94


p2 sin 2✓ (out of the page). 4⇡✏0 (16z 3 ) For 0 < ✓ < ⇡/2, N tends to rotate p counterclockwise; for ⇡/2 < ✓ < ⇡, N rotates p clockwise. Thus the stable orientation is perpendicular to the surface—either " or #. But sin ✓ cos ✓ = (1/2) sin 2✓, so N =

Problem 4.7 If the potential is zero at infinity, the energy of a point charge Q is (Eq. 2.39) W = QV (r). For a physical dipole, with q at r and +q at r + d, " Z # U = qV (r + d)

qV (r) = q[V (r + d)

r+d

V (r)] = q

r

E · dl .

For an ideal dipole the integral reduces to E · d, and U=

qE · d =

p · E,

since p = qd. If you do not (or cannot) use infinity as the reference point, the result still holds, as long as you bring the two charges in from the same point, r0 (or two points at the same potential). In that case W = Q[V (r) V (r0 )], and U = q[V (r + d)

V (r0 )]

q[V (r)

V (r0 )] = q[V (r + d)

V (r)],

as before. Problem 4.8 U=

p1 ·E2 , but E2 =

Problem 4.9

1 1 4⇡✏0 r 3

[3 (p2 ·ˆ r) ˆ r

p2 ]. So U =

1 1 4⇡✏0 r 3

[p1 ·p2

3 (p1 ·ˆ r) (p2 ·ˆ r)]. qed

1 q q xx ˆ+yy ˆ +zˆ z ˆ r= . 4⇡✏0 r2 4⇡✏0 (x2 + y 2 + z 2 )3/2 ✓ ◆ @ @ @ q x Fx = px + py + pz 2 2 @x @y @z 4⇡✏0 (x + y + z 2 )3/2 ⇢   q 1 3 2x 3 2y = px x 2 + py x 2 2 2 2 3/2 2 2 5/2 2 4⇡✏0 2 (x + y + z ) 2 (x + y + z 2 )5/2 (x + y + z )    3 2z q px 3x q p + pz x 2 = (px x + py y + pz z) = 3 5 2 2 5/2 2 (x + y + z ) 4⇡✏0 r r 4⇡✏0 r3

(a) F = (p · r)E (Eq. 4.5); E =

F=

1 q [p 4⇡✏0 r3

3r(p · r) r5

. x

3(p · ˆ r) ˆ r] .

1 1 1 1 {3 [p · ( ˆ r)] ( ˆ r) p} = [3(p · ˆ r) ˆ r 3 4⇡✏0 r 4⇡✏0 r3 are because r points toward p, in this problem.) (b) E =

F = qE =

p] . (This is from Eq. 3.104; the minus signs

1 q [3(p · ˆ r) ˆ r 4⇡✏0 r3

p] .

[Note that the forces are equal and opposite, as you would expect from Newton’s third law.]


✲θ p θ pi ✻

z Problem 4.7 CHAPTER 4. ELECTRIC FIELDS IN MATTER

95

Problem 4.10 (a)

b

✻ y = P·ˆ n = kR; ⇢b = E ✻

r·P =

1 @ 2 (r kr) = r2 @r

1 3kr2 = r2

3k.

(b) For r < R, E = 3✏10θ⇢r ˆ r (Prob. 2.12), so E = (k/✏0 ) r. ✲ ✲ ✒p p x For r > R, same as if all charge at center; but Qtot = (kR)(4⇡R2 ) + ( 3k)( 43 ⇡R3 ) = 0, so E = 0. Problem 4.11 ⇢b = 0; b = P·ˆ n = ±P (plus sign at one end—the one P points toward ; minus sign at the other—the one P points away 4.11 from). Problem (i) L a. Then the ends look like point charges, and the whole thing is like a physical dipole, of length L and charge P ⇡a2 . See Fig. (a). (ii) L ⌧ a. Then it’s like a circular parallel-plate capacitor. Field is nearly uniform inside; nonuniform “fringing field” at the edges. See Fig. (b). (iii) L ⇡ a. See Fig. (c). ✛ ✛ ✛ −

✛ ✛ ✛

(a) Like a dipole

+

✲ P

✛ ✻✻ ✛ ✛ ✛ ✛ ✛ ✛ ✛ ✛ ✛ ✛ ✛ − +❄ ❄

✛ ✛ ✛ ✛ ✛ ✛ ✛ ✛ ✛

✲ P

(b) Like a parallel-plate capacitor

✲ P

(c)

Problem 4.12 n o R P· rˆ R rˆ 1 1 V = 4⇡✏ . But the term in curly brackets is precisely the field of a uniformly 2 d⌧ = P· 2 d⌧ 4⇡✏ 0 0 r r charged sphere, divided by ⇢. The integral was done explicitly in Probs. 2.7 and 2.8:

1 4⇡✏0

Z

8 9 1 (4/3)⇡R3 ⇢ > > > ˆ r, (r > R), > > > = r2 1 < 4⇡✏0

rˆ r 2 d⌧ = ⇢ > > >

Problem 4.13

> 3 > : 1 (4/3)⇡R ⇢ r, (r < R). > ; 3 4⇡✏0 R

So V (r, ✓) =

8 9 > > R3 R3 P cos ✓ > > > P·ˆ r= , (r > R), > > > 2 2 > > 3✏0 r < 3✏0 r = > > > 1 P r cos ✓ > > P·r = , : 3✏0 3✏0

> > > > (r < R). > ;


Think of it as two cylinders of opposite uniform charge density ±⇢. Inside, the field at a distance s from the axis of a uniformly charge cylinder is given by Gauss’s law: E2⇡s` = ✏10 ⇢⇡s2 ` ) E = (⇢/2✏0 )s. For two such cylinders, one plus and one minus, the net field (inside) is E = E+ + E = (⇢/2✏0 ) (s+ s ). But s+ s = d, so E = ⇢d/(2✏0 ), where d is the vector from the negative axis to positive axis. In this case the total dipole moment of a chunk of length ` is P ⇡a2 ` = ⇢⇡a2 ` d. So ⇢d = P, and E = s < a. c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

P/(2✏0 ), for

96


Outside, Gauss’s law gives E2⇡s` = ⇣ ⌘ 2 ˆ s+ ˆ s E+ + E = ⇢a , where 2✏0 s+ s

✓

ˆ s+ s+

1 2 ✏0 ⇢⇡a `

)E=

⇢a2 ˆ s 2✏0 s ,

for one cylinder. For the combination, E =

d s± = s ⌥ ; 2 ✓ ◆✓ ◆ 1 ✓ ◆✓ ◆ 1 ✓ ◆✓ ◆ s± d d2 1 d s·d 1 d s·d 2 ⇠ ⇠ = s⌥ s + ⌥s·d 1⌥ 2 1± 2 = 2 s⌥ = 2 s⌥ s2± 2 4 s 2 s s 2 s ✓ ◆ 1 (s · d) d = 2 s±s 2 ⌥ (keeping only 1st order terms in d). s s 2 ◆ ✓ ◆ ✓ ◆ ✓ ◆ ˆ s 1 (s · d) d (s · d) s 1 s(s · d) = 2 s+s 2 s s 2 + = 2 2 d . s s s 2 s 2 s s2 E(s) =

a2 1 [2(P · ˆ s) ˆ s 2✏0 s2

P] ,

for s > a.

Problem 4.14

H H R Total charge is Qtot = S b da + V ⇢b d⌧ = S P · da H on the dielectric R theorem says S P · da = V r·P d⌧ , so Qenc = 0. qed Problem 4.15

V

r·P d⌧ . But the divergence

◆ ⇢ k k +P · ˆ r = k/b (at r = b), = ; = P·ˆ n = b P·ˆ r = k/a (at r = a). r r2 1 Qenc Gauss’s law ) E = 4⇡✏ ˆ r . For r < a, Q = 0, so E = 0. For r > b, Qenc = 0 (Prob. 4.14), so E = 0. 2 enc 0 r R r k k 2 For a < r < b, Qenc = a 4⇡a2 + a r2 4⇡r dr = 4⇡ka 4⇡k(r a) = 4⇡kr; so E = (k/✏0 r) ˆ r. H (b) D·da = Qfenc = 0 ) D = 0 everywhere. D = ✏0 E + P = 0 ) E = ( 1/✏0 )P, so E = 0 (for r < a and r > b); E = (k/✏0 r) ˆ r (for a < r < b). (a) ⇢b =

r·P =

1 @ r2 @r

✓

R

r2

Problem 4.16 (a) Same as E0 minus the field at the center of a sphere with uniform polarization P. The latter (Eq. 4.14) 1 is P/3✏0 . So E = E0 + P. D = ✏0 E = ✏0 E0 + 13 P = D0 P + 13 P, so D = D0 23 P. 3✏0 (b) Same as E0 minus the field of ± charges at the two ends of the “needle”—but these are small, and far away, so E = E0 . D = ✏0 E = ✏0 E0 = D0 P, so D = D0 P. (c) Same as E0 minus the field of a parallel-plate capacitor with upper plate at (1/✏0 )P , so E = E0 +

1 ✏0 P.

= P . The latter is

D = ✏0 E = ✏0 E0 + P, so D = D0 .


97

CHAPTER 4. ELECTRIC FIELDS IN MATTER Problem 4.17

✻✻✻✻✻✻✻

❄ ❄ ❄❄ ❄ ❄❄ ❄ ❄

❄ ❄❄ ❄❄ ❄ ✻✻✻ ✻✻✻

6

P (uniform) Problem 4.18

E (field of two circular plates)

D (same as E outside, but lines continuous, since ∇·D = 0)

For more detailed figures see the solution to Problem 6.14, reading P for M, E for H, and D for B. Problem 4.18 R (a) Apply D · da = Qfenc to the gaussian surface shown. DA = A ) D = . (Note: D = 0 inside the metal plate.) This is true in both slabs; D points down. 6

+σ

❖ (b) D = ✏E ) E = /✏ in slab 1, E = /✏ in slab 2. But ✏ = ✏0 ✏r , so ✏1 = 2✏0 ; ✏2 = 32 ✏0 . E1 = /2✏0 , 1 2 Problem 4.18 E2 = 2 /3✏0 .

(c) P = ✏0

e E,

so P = ✏0

e d/(✏0 ✏r )

=(

e /✏r )

;

e

(d) V = E1 a + E2 a = ( a/6✏0 )(3 + 4) = 7 a/6✏0 .

= ✏r

1 ) P = (1

✏r 1 ) .

P1 = /2, P2 = /3.

+σ −σ/2

⃝ 1 = +P1 at bottom of slab (1) = /2, bottom of slab (2) = /3, b = +P 2 at +σ/2 +σ P1 at top of slab (1) = /2; top of slab (2) = /3. b = b = ❖ P2 at −σ/3 ⃝ 2 ⇢ +σ/3 total surface charge above: ( /2) = /2, −σ (f) In slab 1: =) E1 = .X total surface charge below : ( /2) ( /3) + ( /3) = /2, 2✏0 ⇢ 2 total surface charge above: ( /2) + ( /2) ( /3) = 2 /3, In slab 2: =) E2 = .X total surface charge below : ( /3) = 2 /3, 3✏0

(e) ⇢b = 0;

b

+σ −σ/2

⃝ 1 c ⃝2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is +σ/2 Problem 4.22 protected under all copyright laws as they currently exist. No portion of this material may be −σ/3 reproduced, in any form or by any means, without permission in writing from the publisher. ⃝ 2 +σ/3 −σ

Problem 4.19

E0 ✻ ✛

φ

x ✻

With no dielectric, C0 = A✏0 /d (Eq. 2.54). y In configuration (a), with + on upper plate, on lower, D = between the plates. ✙ z E = /✏0 (in air) and E = /✏ (in dielectric). So V = ✏0 d2 + ✏ d2 = 2✏Qd 1 + ✏✏0 . 0A Problem 4.22 c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

y

E0 ✻ ✛ ✙ z

φ

x ✻

98

Ca =

Q V

=

✏0 A d

⇣

2 1+1/✏r

⌘

=)


Ca 2✏r = . C0 1 + ✏r

In configuration (b), with potential di↵erence V : E = V /d, so = ✏0 E = ✏0 V /d (in air). P = ✏0 e E = ✏0 e V /d (in dielectric), so b = ✏0 e V /d (at top surface of dielectric). ✏0 e V /d, so f = ✏0 V (1 + e )/d = ✏0 ✏r V /d (on top plate above dielectric). tot = ✏0 V /d = f + b = f ✓ ◆ ✓ ◆ ✓ ◆ Q 1 A A A V V A✏0 1 + ✏r Cb 1 + ✏r =) Cb = = + f = ✏0 + ✏0 ✏r = . = . V V 2 2 2V d d d 2 C0 2 Cb [Which is greater? C 0 If the x axis points down:

Ca C0

=

1+✏r 2

2✏r 1+✏r

=

(1+✏r )2 4✏r 2(1+✏r )

E (a) air (a) dielectric

2 V (✏r +1) d V ˆ d x V ˆ d x

(b) dielectric b

(a) (b)

V =

Z

0

1

E · dl =

x ˆ

2✏r ✏0 V (✏r +1) d

x ˆ

x ˆ

2✏r ✏0 V (✏r +1) d ✏0 V ˆ d x ✏0 V ✏r d x ˆ

x ˆ

(top surface) 2(✏r 1) ✏0 V (✏r +1) d (✏r 1) ✏0dV

Problem 4.20 R D·da = Qfenc ) D4⇡r2 = ⇢ 43 ⇡r3 ) D = ⇢R3 /3r2 ) E = (⇢R3 /3✏0 r2 ) ˆ r, for r > R. ⇢R3 1 3✏0 r

R 1

1+2✏r +4✏2r 4✏r 2(1+✏r )

D

2✏r V (✏r +1) d

(b) air

=

f

=

(1 ✏r )2 2(1+✏r )

> 0. So Cb > Ca .]

P 0 2(✏r 1) ✏0 V (✏r +1) d

x ˆ

0 (✏r

1) ✏0dV x ˆ

(top plate)

2✏r ✏0 V (✏r +1) d ✏r ✏0dV (left); ✏0dV

(right)

1 3 ⇢r

) E = (⇢r/3✏) ˆ r, for r < R; D4⇡r2 = ⇢ 43 ⇡R3 ) D =

Z

rdr =

⇢ 3✏

0

R

⇢R2 ⇢ R2 ⇢R2 + = 3✏0 3✏0 3✏ 2

✓

1+

1 2✏r

◆

.

Problem 4.21 Let Q be the charge on a length ` of the inner conductor. I Q Q Q D · da = D2⇡s` = Q ) D = ; E= (a < s < b), E = (b < r < c). 2⇡s` 2⇡✏0 s` 2⇡✏s` ◆ ◆  ✓ ◆ Z a Z b✓ Z c✓ Q ds Q ds Q b ✏0 ⇣ c ⌘ V = E · dl = + = ln + ln . 2⇡✏0 ` s 2⇡✏` s 2⇡✏0 ` a ✏ b c a b C Q 2⇡✏0 = = . ` V` ln(b/a) + (1/✏r ) ln(c/b)


+σ/2 −σ/3

⃝ 2

+σ/3 −σ

99

CHAPTER 4. ELECTRIC FIELDS IN MATTER Problem 4.22 Problem 4.22

Same method as Ex. 4.7: solve Laplace’s equation for Vin (s, ) (s < a) and Vout (s, ) (s > a), subject to x the boundary conditions ✻ 8 φ at s = a, < (i) Vin = Vout ✻ @Vin @Vout E 0 (ii) ✏ @s = ✏0 @s at s = a, : ✛ (iii) Vout ! E0 s cos for s a. y ✙ z From Prob. 3.24 (invoking boundary condition (iii)): Vin (s, ) =

1 X

sk (ak cos k + bk sin k ),

Vout (s, ) =

1 X

E0 s cos +

k=1

s

k

(ck cos k + dk sin k ).

k=1

(I eliminated the constant terms by setting V = 0 on the y z plane.) Condition (i) says X X ak (ak cos k + bk sin k ) = E0 a cos + a k (ck cos k + dk sin k ), while (ii) says

✏r

X

kak

1

X 1 (ak cos k⃝2005 sin k Education, ) = E0 cos ka kRiver, (ckNJ. cosAll k rights + dkreserved. sin k ).This material is c + bkPearson Inc., Upper Saddle

protected under all copyright laws as they currently exist. No portion of this material may be

in any form or by any means, without permission in writing from the publisher. Evidently bk = dk = 0 for all k, areproduced, k = ck = 0 unless k = 1, whereas for k = 1,

aa1 =

E0 a + a

1

c1 ,

✏r a1 =

E0

a

2

c1 .

Solving for a1 , a1 = and hence Ein (s, ) =

E0 , (1 + e /2)

so Vin (s, ) =

E0 s cos (1 + e /2)

=

E0 x, (1 + e /2)

@Vin E0 x ˆ= . As in the spherical case (Ex. 4.7), the field inside is uniform. @x (1 + e /2)

Problem 4.23 P0 = ✏0 e E0 ; E1 = ⇣ ⌘n e En = E0 , so 3

1 P0 = 3✏0

e

3

E0 ; P1 = ✏0

e E1

E = E0 + E1 + E2 + · · · =

The geometric series can be summed explicitly: 1 X

n=0

xn =

1 1

x

,

so

✏0 2e E0 ; E2 = 3

= "

1 ⇣ X

e

3

n=0

E=

⌘n

1 (1 +

e /3)

#

2 1 P1 = e E0 ; . . .. Evidently 3✏0 9

E0 .

E0 ,

which agrees with Eq. 4.49. [Curiously, this method formally requires that e < 3 (else the infinite series diverges), yet the result is subject to no such restriction, since we can also get it by the method of Ex. 4.7.] c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

100


Problem 4.24 Potentials: 8 > Vout (r, ✓) = <

P l E0 r cos ✓ + ⌘rB (r > b); l+1 Pl (cos ✓), P⇣ ¯l Vmed (r, ✓) = Al rl + rB P (cos ✓), (a < r < b); l l+1 > : Vin (r, ✓) = 0, (r < a). Boundary Conditions: 8 < (i) Vout = Vmed , (r = b); med (ii) ✏ @V@r = ✏0 @V@rout , (r = b); : (iii) Vmed = 0, (r = a).

X Bl X✓ ¯l ◆ B l (i) ) E0 b cos ✓ + Pl (cos ✓) = Al b + l+1 Pl (cos ✓); bl+1 b X X ¯l B Bl (ii) ) ✏r lAl bl 1 (l + 1) l+2 Pl (cos ✓) = E0 cos ✓ (l + 1) l+2 Pl (cos ✓); b b ¯l B ¯l = a2l+1 Al . (iii) ) Al al + l+1 = 0 ) B a

For l 6= 1 : ✓ ◆ Bl a2l+1 Al l (i) l+1 = Al b ) Bl = Al b2l+1 a2l+1 ; b bl+1  ✓ ◆ a2l+1 Al Bl l (ii) ✏r lAl bl 1 + (l + 1) l+2 = (l + 1) l+2 ) Bl = ✏r Al b2l+1 + a2l+1 b b l+1

) Al = Bl = 0.

For l = 1 : B1 a3 A1 E0 b + 2 = A1 b ) B1 E0 b3 = A1 b3 a3 ; 2 b b ✓ ◆ a3 A1 B1 (ii) ✏r A1 + 2 3 = E0 2 3 ) 2B1 E0 b3 = ✏r A1 b3 + 2a3 . b b (i)

So

⇥ 3E0 b3 = A1 2 b3 Vmed (r, ✓) = E(r, ✓) =

2[1

a3 + ✏r b3 + 2a3

⇤

;

3E0 3 (a/b) ] + ✏r [1 + 2(a/b)3 ]

rVmed =

2[1

A1 = ✓

r

3E0 . (a/b)3 ] + ✏r [1 + 2(a/b)3 ]

2[1 a3 r2

◆

cos ✓,

3E0 (a/b)3 ] + ✏r [1 + 2(a/b)3 ]

⇢✓ ◆ 2a3 1 + 3 cos ✓ ˆ r r

✓

1

a3 r3

◆

sin ✓ ✓ˆ .

Problem 4.25 There are four charges involved: (i) q, (ii) polarization charge surrounding q, (iii) surface charge ( b ) on the top surface of the lower dielectric, (iv) surface charge ( b0 ) on the lower surface of the upper dielectric. c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

101

CHAPTER 4. ELECTRIC FIELDS IN MATTER In view of Eq. 4.39, the bound charge (ii) is qp = qt = q + qp = q/(1 + 0e ) = q/✏0r . As in Ex. 4.8, (a)

b

= ✏0

(b)

0 b

=

Solve for

b,

e

✏0 0e

0 b:

" "

1 qd/✏0r 4⇡✏0 (r2 + d2 ) 32 1 qd/✏0r 4⇡✏0 (r2 + d2 ) 32

first divide by 0 b 0 e

b e

Plug this into (a) and solve for b

0 b

b,

=

2✏0

b

0 b

2✏0

2✏0

#

+

0 e ),

so the total (point) charge at (0, 0, d) is

(here

b

= P·ˆ n = +Pz = ✏0

(here

b

=

Pz =

✏0

e Ez );

0 e Ez ).

0 e

using ✏0r = 1 +

0 e:

1 qd/✏0r 1 qd b e 0 ( e + 0e ), so b = ; e (1 + e ) 3 4⇡ (r2 + d2 ) 2 2 4⇡ (r2 + d2 ) 32 [1 + ( e + 0e )/2] ( ) 1 qd 1 1 qd/✏0r 1 qd ✏r 0e /✏0r 0 = e + , so b0 = . 3 3 3 0 4⇡ (r2 + d2 ) 2 [1 + ( e + e )/2] 2⇡ (r2 + d2 ) 2 4⇡ (r2 + d2 ) 2 [1 + ( e + 0e )/2] =

The total bound surface charge is 0 e

2✏0

#

0 e /(1

(respectively) and subtract: " # 1 qd/✏0r 1 qd/✏0r b 0 0 = ) b= e + . 2⇡ (r2 + d2 ) 32 2⇡ (r2 + d2 ) 32 e e

and

0 b

b

q(

e ).

t

=

b+

0 b

=

( 0e qd e) 1 4⇡ (r 2 +d2 ) 32 ✏0r [1+( e + 0e )/2]

(which vanishes, as it should, when

The total bound charge is (compare Eq. 4.51): ✓ 0 ◆ ( 0e ✏r ✏r q e )q qt = 0 = , and hence 2✏r [1 + ( e + 0e )/2] ✏0r + ✏r ✏0r 1 V (r) = 4⇡✏0

Meanwhile, since

(

q/✏0r

p x2 + y 2 + (z

qt

d)2

+p x2 + y 2 + (z + d)2

 q q ✏0 ✏r 2q + q = 1 + r0 = 0 , t 0 0 ✏r ✏r ✏r + ✏r ✏r + ✏r

V (r) =

Problem 4.26 From Ex. 4.5:

)

(for z > 0).

1 [2q/(✏0r + ✏r )] p (for z < 0). 4⇡✏0 x2 + y 2 + (z d)2

9 (r < a) > > > = 0, (r < a) ˆ r , (a < r < b) D= , E = 4⇡✏r2 . Q > > ˆ r, (r > a) > > Q 2 > > 4⇡r : ˆ r, (r > b) ; 4⇡✏0 r2 ( Z ) ( ✓ ◆ Z Z 1 1 Q2 1 b 1 1 2 1 1 1 Q2 1 1 W = D·E d⌧ = 4⇡ r dr + dr = 2 2 (4⇡)2 ✏ a r2 r2 ✏0 b r2 8⇡ ✏ r ⇢ ✓ ◆ ✓ ◆ Q2 1 1 1 1 Q2 1 e = + = + . 8⇡✏0 (1 + e ) a b b 8⇡✏0 (1 + e ) a b (

)

8 0, > > > Q <


b

1 + ✏ 0 a

✓

1 r

◆

1 b

)

102


Problem 4.27

R Using Eq. 4.55: W = ✏20 E 2 d⌧ . From Ex. 4.2 and Eq. 3.103, 8 9 1 > Pˆ z, (r < R) > < = 3✏0 E= , so 3 > ˆ (r > R) > : R P (2 cos ✓ ˆ ; r + sin ✓ ✓), 3 3✏0 r ✓ ◆2 ✏0 P 4 3 2⇡ P 2 R3 WrR = 4 cos2 ✓ + sin2 ✓ r2 sin ✓ dr d✓ d 2 3✏0 r6 Z ⇡ Z 1 (R3 P )2 1 ⇡(R3 P )2 = 2⇡ (1 + 3 cos2 ✓) sin ✓ d✓ dr = ( cos ✓ 4 18✏0 9✏0 0 R r ✓ ◆ ⇡(R3 P )2 4 4⇡R3 P 2 = = . 9✏0 3R3 27✏0

cos3 ✓)

⇡ 0

✓

1 3r3

◆

1 R

2⇡R3 P 2 . 9✏0

Wtot =

This is the correct electrostatic energy of the configuration, but it is not the “total work necessary to assemble the system,” because it leaves out the mechanical energy involved in polarizing the molecules. R Using Eq. 4.58: W = 12 D·E d⌧ . For r > R, D = ✏0 E, so this contribution is the same as before. ✏0 2 1 2 1 For r < R, ⇣ D = ⌘✏0 E + P = 3 P + P = 3 P = 2✏0 E, so 2 D·E = 2 2 E , and this contribution is 2

3

3

2

P R R P now ( 2) 2⇡ = 4⇡ 27 ✏0 27 ✏0 , exactly cancelling the exterior term. Conclusion: Wtot = 0. This is not surprising, since the derivation in Sect. 4.4.3 calculates the work done on the free charge, and in this problem there is no free charge in sight. Since this is a nonlinear dielectric, however, the result cannot be interpreted as the “work necessary to assemble the configuration”—the latter would depend entirely on how you assemble it. Problem 4.28

First find the capacitance, as a function of h: Air part: E = Oil part: D = Q=

0

h + (`

2 4⇡✏0 s 2 0 4⇡s

=) V =

=) E =

h) = ✏r h

2 4⇡✏0

2 0 4⇡✏s

ln(b/a),

=) V =

2 0 4⇡✏

h + ` = [(✏r C=

ln(b/a),

9 = ;

=)

1)h + `] = (

✏0

=

eh

0

✏

;

0

=

✏ ✏0

= ✏r .

+ `), where ` is the total height.

Q ( e h + `) ( e h + `) = 4⇡✏0 = 2⇡✏0 . V 2 ln(b/a) ln(b/a)

1 2 2⇡✏0 e The net upward force is given by Eq. 4.64: F = 12 V 2 dC dh = 2 V ln(b/a) . 2 2 The gravitational force down is F = mg = ⇢⇡(b a )gh.

h=

⇢(b2

✏0 e V 2 . a2 )g ln(b/a)


7 Problem 4.29

z

✻

p1

✻

✰ x


p ✲ 2✲ ❄ y

103 E1

7

Problem 4.29

z @ ✻ (a) Eq. 4.5 )Problem F2 = (p2 4.29 · r) E1 = p2 (E1 ); @y p1 p p1 ˆ p1 ✲ 2✲ ✻ Eq. 3.103 ) E1 = ✓ = ˆ z . Therefore ❄ y 4⇡✏0 r3 4⇡✏0 y 3 ✰ E 1 x  ✓ ◆ z✻ p1 p2 d 1 3p1 p2 3p1 p2 F2 = ˆ z= ˆ z, or F2 = ˆ z (upward). 4⇡✏0 dy y 3 4⇡✏0 y 4 4⇡✏0 r4 ✲ ✻ 7 p2 y To calculate F1 , put p2 at thez origin, pointing in the z direction; then p1 ✰ ✛ ✻ x is at r ˆ z, and it points in the y ˆ direction. So F1 = (p1 · r) E2 = p1 Problem 4.29 p p 1 @E 2 ✲ 2✲of x, y, and z. ✻ a function z✻ p1 ; we need E2 as ❄ y @y x=y=0, z= r ✰ E1 x  1 ✻1 3(p ✲2 · r)r From Eq. 3.104: E = p2 , where r = x x ˆ +y y ˆ +z ˆ z, p2 = p2 ˆ z, and hence p2 ·r = p2 z. 2 p y r2 Problem 4.30 2 3 4⇡✏0 r ✰ ✛  p2x 3z(x x ˆ ˆ +zˆ z) (x2 + y 2 + z 2 ) ˆ z p2 3xz x ˆ + 3yz y ˆ (x2 + y 2 2z 2 ) ˆ z p1+ y y E2 = = 2 2 2 5/2 2 4⇡✏0 4⇡✏0 (x + y + z ) (x + y 2 + z 2 )5/2 ⇢ @E2 p2 5 2y 1 = [3xz x ˆ + 3yz y ˆ (x2 + y 2 2z 2 ) ˆ z] + 5 (3z y ˆ 2y ˆ z) ; @y 4⇡✏0 2 r7 r z✻ ✓ ◆ @E2 p2 3z 4.30 p2 3r 3p1 p2 Problem = y ˆ; F1 = p1 y ˆ = y ˆ. @y (0,0) 4⇡✏0 r5 4⇡✏0 r5 4⇡✏0 r4 ✲ ✻ p y + But 2y ˆ in these coordinates corresponds to ˆ z in the original system, so these results are consistent with F ◆ ✰ E Newton’s third law: F = F . 1 2 ✛ x p(b) 1 From the remark following Eq. 4.5, ✌ N2 = (p2−⇥✍FE1 ) + (r ⇥ F2 ). The first term was calculated in ✌ ry Prob. 4.5; the second we get from (a), using r = ˆ: ✌ ✌ ✓ ◆ ✌ p1 p2 3p1 p2 3p1 p2 2p1 p2 ( x ˆ); r ⇥ F2 = (r y ˆ) ⇥ ˆ z = x ˆ; so N2 = x ˆ. p2 ⇥ E1 = 4⇡✏0 r3 4⇡✏0 r4 4⇡✏0 r3 4⇡✏0 r3 Problem 4.30 + F ◆ center of E This is equal and opposite to the torque on p1 due to p2 , with respect to the p1 (see Prob. 4.5). ✍F ✌ − Problem 4.30 ✌ Net force is to the right (see diagram). Note that the field lines must bulge to the right, as shown, because ✌ ✌ E is perpendicular to the surface of each conductor. ✌

+

✌

F

◆ ✍F

E

− c ⃝2005 Pearson Education, Inc., Upper Saddle River, ✌ NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be ✌ reproduced, in any form or by any means, without permission in writing from the publisher.

✌

✌

c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. c ⃝2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

104


1 @ 1 Q Problem 4.31 In cylindrical coordinates (in the z = 0 plane), p = p ˆ, p · r = p , and E = ˆ s, so s@ 4⇡✏0 s2 ✓ ◆ p @ 1 Q pQ @ˆ s pQ ˆ Q F = (p · r)E = ˆ s= = = p. X s@ 4⇡✏0 s2 4⇡✏0 s3 @ 4⇡✏0 s3 4⇡✏0 R3 Qualitatively, the forces on the negative and positive ends, though equal in magnitude, point in slightly di↵erent directions, and they combine to make a net force in the “forward” direction:

y F+ p F_ f

R x

To keep the dipole going in a circle, there must be a centripetal force exerted by the track (we may as well take it to act at the center of the dipole, and it is irrelevant to the problem), and to keep it aiming in the tangential direction there must be a torque (which we could model by radial forces of equal magnitude acting at the two ends). Indeed, if the dipole has the orientation indicated in the figure, and is moving in the ˆ direction, the torque exerted by Q is clockwise, whereas the rotation is counterclockwise, so these constraint forces must actually be larger than the forces exerted by Q, and the net force will be in the “backward” direction—tending to slow the dipole down. [If the motion is in the ˆ direction, then the electrical forces will dominate, and the net force will be in the direction of p, but this again will tend to slow it down.] Problem 4.32 (a) According to Eqs. 4.1 and 4.5, F = ↵(E · r)E. From the product rule, rE 2 = r(E · E) = 2E ⇥ (r ⇥ E) + 2(E · r)E. But in electrostatics r ⇥ E = 0, so (E · r)E = 12 r(E 2 ), and hence F=

1 ↵r(E 2 ). X 2

[It is tempting to start with Eq. 4.6, and write F = rU = r(p · E) = ↵r(E · E) = ↵r(E 2 ). The error occurs in the third step: p should not have been di↵erentiated, but after it is replaced by ↵E we are di↵erentiating both E’s.] (b) Suppose E 2 has a local maximum at point P . Then there is a sphere (of radius R) about P such that E (P 0 ) < E 2 (P ), and hence |E(P 0 )| < |E(P )|, for all points on the surface. But if there is no charge inside the 2


105


sphere, then Problem 3.4a says the average field over the spherical surface is equal to the value at the center: Z 1 E da = E(P ), 4⇡R2 or, choosing the z axis to lie along E(P ),

1 4⇡R2

Z

Ez da = E(P ).

But if E 2 has a maximum at P , then Z Z Z Ez da  |E| da < |E(P )| da = 4⇡R2 E(P ), and it follows that E(P ) < E(P ), a contradiction. Therefore, E 2 cannot have a maximum in a charge-free region. [It can have a minimum, however; at the midpoint between two equal charges the field is zero, and this is obviously a minimum.] Problem 4.33 P = kr = k(x x ˆ+yy ˆ +zˆ z) =) ⇢b = Total volume bound charge: Qvol = b

r·P =

k(1 + 1 + 1) =

3k.

= ka/2. Clearly,

b

3ka3 .

= P·ˆ n. At top surface, n ˆ=ˆ z, z = a/2; so

b

= ka/2 on all six surfaces.

Total surface bound charge: Qsurf = 6(ka/2)a2 = 3ka3 . Total bound charge is zero. X Problem 4.34 Say the high voltage is connected to the bottom plate, so the electric field points in the x direction, while the free charge density ( f ) is positive on the lower plate and negative on the upper plate. (If you connect x the battery the other way, all the signs will switch.) The susceptibility is e = , and the permittivity is d ⇣ x⌘ ✏ = ✏0 1 + . Between the plates d Z 0 Z d ⇣ 1 1 x⌘ d f f f fd D= fx ˆ, E = D = x ˆ; V = E · dl = dx = d ln 1 + = ln 2. ✏ ✏0 (1 + x/d) ✏ (1 + x/d) ✏ d ✏ 0 0 0 0 0 d So

f

=

✏0 V , d ln 2

E=

V 1 x ˆ, d ln 2 (1 + x/d)

P = ✏0

eE

=

✏0 V x x ˆ. d2 ln 2 (1 + x/d)

The bound charges are therefore ⇢b =

r·P=



✏0 V 1 2 d ln 2 (1 + x/d)

The total bound charge is Z Z Qb = ⇢b d⌧ + b da = ✏0 V A = d ln 2

✓

1 2

1 1+ 2

◆

x/d = (1 + x/d)2

✏0 V d2 ln 2 = 0,

Z

0

d

✏0 V 1 ; 2 d ln 2 (1 + x/d)2

b

=P·n ˆ=

1 ✏0 V ✏0 V A A dx + A= (1 + x/d)2 2d ln 2 d ln 2

X


"

8 > <0

(x = 0),

✏0 V > : 2d ln 2

1 d d (1 + x/d)

d

(x = d).

1 + 2 0

#

106


where A is the area of the plates. Problem 4.35 I q 1 q ˆ r q e ˆ r D·da = Qfenc ) D = ˆ r; E = D = ; P = ✏0 e E = . 2 2 4⇡r ✏ 4⇡✏0 (1 + e ) r 4⇡(1 + e ) r2 ✓ ◆ q e ˆ r q e e 3 ⇢b = r·P = r· 2 = q (r) (Eq. 1.99); b = P·ˆ r= ; 4⇡(1 + e ) r 1+ e 4⇡(1 + e )R2 Qsurf =

b (4⇡R

2

)= q

e

1+

The compensating negative charge is at the center:

. e

Z

⇢b d⌧ =

q e 1+ e

Z

3

(r) d⌧ =

q

e

1+

. e

Problem 4.36 Ek is continuous (Eq. 4.29); D? is continuous (Eq. 4.26, with ✏1 Ey1 = ✏2 Ey2 , and hence

f

= 0). So Ex1 = Ex2 , Dy1 = Dy2 )

tan ✓2 Ex2 /Ey2 Ey 1 ✏2 = = = . qed tan ✓1 Ex1 /Ey1 Ey 2 ✏1 If 1 is air and 2 is dielectric, tan ✓2 / tan ✓1 = ✏2 /✏0 > 1, and the field lines bend away from the normal. This is the opposite of light rays, so a convex “lens” would defocus the field lines. Problem 4.37 In view of Eq. 4.39, the net dipole moment at the center is p0 = p 1+e e p = 1+1 e p = potential produced by p0 (at the center) and b (at R). Use separation of variables: 8 9 1 X > > B l > > > Pl (cos ✓) (Eq. 3.72) > > > < Outside: V (r, ✓) = = rl+1 l=0

1 > 1 p cos ✓ X > + Al rl Pl (cos ✓) (Eqs. 3.66, 3.102) > > 2 ; 4⇡✏0 ✏r r l=0 9 Bl l 2l+1 > = Al R , or Bl = R Al (l 6= 1) > > = Rl+1 . > B1 1 p p > 3> = + A1 R, or B1 = 4⇡✏0 ✏r + A1 R ; R2 4⇡✏0 ✏r R2

> > > > : Inside:

V continuous at R ) @V @r

R+

@V @r

= R

= (l + 1) For l = 1:

Bl Rl+2

2

8 > > > < > > > :

X

We want the

.

V (r, ✓) =

(l + 1)

1 P·ˆ r= ✏0

lAl Rl

1 ✏r p.

1

B1 1 2p + 3 R 4⇡✏0 ✏r R3

=

Bl 1 2p cos ✓ Pl (cos ✓) + l+2 R 4⇡✏0 ✏r R3 1 (✏0 ✏0

e lAl R

A1 =

l 1

e

✓

r) e E·ˆ

=

e

(l 6= 1); or

@V @r

= R

X

e

⇢

(2l + 1)Al Rl

1 2p + A1 4⇡✏0 ✏r R3

◆

B1 +

lAl Rl

1

Pl (cos ✓) =

1 ✏0

b

1 2p cos ✓ X + lAl Rl 4⇡✏0 ✏r R3 1

=

p 4⇡✏0 ✏r

e lAl R

l 1

A1 R3 = 2

1

Pl (cos ✓) .

) Al = 0 (` 6= 1). 1 ep + 4⇡✏0 ✏r

e

A1 R3 ; 2


107


A1 R3 A1 R3 1 ep ) (3 + e ) = . 2 2 4⇡✏0 ✏r  1 2 ep 1 2(✏r 1)p p 2(✏r 1) p 3✏r ) A1 = = ; B1 = 1+ = . 3 3 4⇡✏0 R ✏r (3 + e ) 4⇡✏0 R ✏r (✏r + 2) 4⇡✏0 ✏r (✏r + 2) 4⇡✏0 ✏r ✏r + 2 ✓ ◆✓ ◆ p cos ✓ 3 V (r, ✓) = (r R). 4⇡✏0 r2 ✏r + 2 p 4⇡✏0 ✏r

A1 R3 +

A1 R3 = 2

p 4⇡✏0 ✏r

Meanwhile, for r  R, V (r, ✓) =

1 ep + 4⇡✏0 ✏r

e

1 p cos ✓ 1 pr cos ✓ 2(✏r 1) + 2 4⇡✏0 ✏r r 4⇡✏0 R3 ✏r (✏r + 2) =

 ✓ ◆ p cos ✓ ✏r 1 r3 1+2 4⇡✏0 r2 ✏r ✏r + 2 R3

(r  R).

Problem 4.38 Given two solutions, V1 (and E1 = rV1 , D1 = ✏E1 ) and V2 (E2 = rV2 , D2 = ✏E2 ), define V3 ⌘ V2 V1 (E3 = E2 E1 , D3 = D2 D1 ). R R R R r·(V3 D3 ) d⌧ = S V3 D3 · da = 0, (V3 = 0 on S), so (rV3 ) · D3 d⌧ + V3 (r·D3 ) d⌧ = 0. V R But r·D3 = r·D2 r·D1 = ⇢f ⇢f = 0,Rand rV3 = rV2 rV1 = E2 + E1 = E3 , so E3 · D3 d⌧ = 0. But D3 = D2 D1 = ✏E2 ✏E1 = ✏E3 , so ✏(E3 )2 d⌧ = 0. But ✏ > 0, so E3 = 0, so V2 V1 = constant. But at surface, V2 = V1 , so V2 = V1 everywhere. qed Problem 4.39 R R ˆ r, in which case P = ✏0 e V0 2 ˆ r, 2 r r R ✏0 e V0 in the region z < 0. (P = 0 for z > 0, of course.) Then b = ✏0 e V0 2 (ˆ r·ˆ n) = . (Note: n ˆ points out R R of dielectric ) n ˆ= ˆ r.) This b is on the surface at r = R. The flat surface z = 0 carries no bound charge, since n ˆ =ˆ z ?ˆ r. Nor is there any volume bound charge (Eq. 4.39). If V is to have the required spherical symmetry, the net charge must be uniform: 2 tot 4⇡R = Qtot = 4⇡✏0 RV0 (since V0 = Qtot /4⇡✏0 R), so tot = ✏0 V0 /R. Therefore (a) Proposed potential: V (r) = V0

f

=

⇢

R . If so, then E = r

rV = V0

(✏0 V0 /R), on northern hemisphere (✏0 V0 /R)(1 + e ), on southern hemisphere

.

(b) By construction, tot = b + f = ✏0 V0 /R is uniform (on the northern hemisphere b = 0, f = ✏0 V0 /R; on the southern hemisphere b = ✏0 e V0 /R, so f = ✏V0 /R). The potential of a uniformly charged sphere is V0 =

Qtot = 4⇡✏0 r

tot (4⇡R

4⇡✏0 r

2

)

=

✏0 V0 R2 R = V0 . X R ✏0 r r

(c) Since everything is consistent, and the boundary conditions (V = V0 at r = R, V ! 0 at 1) are met, Prob. 4.38 guarantees that this is the solution. (d) Figure (b) works the same way, but Fig. (a) does not: on the flat surface, P is not perpendicular to n ˆ, so we’d get bound charge on this surface, spoiling the symmetry.


108


Problem 4.40 ✏0 e Eext (Ex. 4.7). 1 + e /3 ◆✓ ◆ ✓ ◆ ✓ ◆✓ ◆2 ✓ ◆ ✓ ◆ Z ✏0 e d ✏0 e 1 1 F= ˆ s d⌧ = ˆ s d⌧ 1 + e /3 2⇡✏0 s ds 2⇡✏0 s 1 + e /3 2⇡✏0 s s2 ✓ 2 ◆ ✓ ◆ 2 3 1 4 3 R e e = ⇡R ˆ s = ˆ s. 2 3 1 + e /3 4⇡ ✏0 s 3 3 + e ⇡✏0 s3

Eext =

2⇡✏0 s Z ✓

ˆ s. Since the sphere is tiny, this is essentially constant, and hence P =

Problem 4.41 1 The density of atoms is N = (4/3)⇡R 3 . The macroscopic field E is Eself + Eelse , where Eself is the average field over the sphere due to the atom itself.

p = ↵Eelse ) P = N ↵Eelse . [Actually, it is the field at the center, not the average over the sphere, that belongs here, but the two are in fact equal, as we found in Prob. 3.47d.] Now Eself = (Eq. 3.105), so E=

✓

1 ↵ Eelse + Eelse = 4⇡✏0 R3

So P=

e

e

or

✏0 ↵= N (1 +

N↵ 3✏0

e

3✏0 e = /3) N (3 + e

=

e

=

◆

Eelse =

✓

1

N↵ 3✏0

◆

Eelse .

e E,

N ↵/✏0 . (1 N ↵/3✏0 )

⌘ N↵ N↵ ⇣ e ) 1+ = ✏0 ✏0 3

. But e

↵ 4⇡✏0 R3

N↵ E = ✏0 N ↵/3✏0 )

(1

and hence

Solving for ↵:

1

1 p 4⇡✏0 R3

e

= ✏r

e,

3✏0 1, so ↵ = N

✓

✏r 1 ✏r + 2

◆

. qed

Problem 4.42 For an ideal gas, N = Avagadro’s number/22.4 liters = (6.02 ⇥ 1023 )/(22.4 ⇥ 10 3 ) = 2.7 ⇥ 1025 . N ↵/✏0 = (2.7 ⇥ 1025 )(4⇡✏0 ⇥ 10 30 ) /✏0 = 3.4 ⇥ 10 4 , where is the number listed in Table 4.1. 9 H: = 0.667, N ↵/✏0 = (3.4 ⇥ 10 4 )(0.67) = 2.3 ⇥ 10 4 , e = 2.5 ⇥ 10 4 > > = He: = 0.205, N ↵/✏0 = (3.4 ⇥ 10 4 )(0.21) = 7.1 ⇥ 10 5 , e = 6.5 ⇥ 10 5 agreement is quite good. Ne: = 0.396, N ↵/✏0 = (3.4 ⇥ 10 4 )(0.40) = 1.4 ⇥ 10 4 , e = 1.3 ⇥ 10 4 > > ; Ar: = 1.64, N ↵/✏0 = (3.4 ⇥ 10 4 )(1.64) = 5.6 ⇥ 10 4 , e = 5.2 ⇥ 10 4 c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

109

CHAPTER 4. ELECTRIC FIELDS IN MATTER Problem 4.43 (a) Doing the (trivial)

hui =

R pE

ue

pE R pE pE

= kT = kT

e

(⇥ e

integral, and changing the remaining integration variable from ✓ to u (du = pE sin ✓ d✓), u/kT

u/kT

du du

=

(kT )2 e

u/kT

[ (u/kT )

kT e

pE pE

1]|

u/kT pE pE

⇤ ⇥ ⇤) epE/kT + (pE/kT )e pE/kT + (pE/kT )epE/kT e pE/kT epE/kT  pE/kT ✓ ◆ e + e pE/kT pE pE pE/kT = kT pE coth . kT e e pE/kT pE/kT

⇢ ✓ ◆ hui pE kT = N p coth . pE kT pE ⇣ ⌘ 8 3 1 x x3 Let y ⌘ P/N p, x ⌘ pE/kT . Then y = coth x 1/x. As x ! 0, y = x1 + x3 x45 + · · · x = 3 45 +· · · ! 0, so the graph starts at the origin, with an initial slope of 1/3. As x ! 1, y ! coth(1) = 1, so the graph Problem 4.40 goes asymptotically to y = 1 (see Figure). ˆ = hp · Ei(E/E) ˆ P = N hpi; p = hp cos ✓iE =

P Np

1

ˆ hui(E/E); P = Np

✻

✲ pE/kT

(b) For small x, y ⇡ 13 x, so

P Np

⇡

pE 3kT

, or P ⇡

For water at 20 = 293 K, p = 6.1 ⇥ 10

30

N p2 3kT E

= ✏0

6.0 ⇥ 1023 = 2.11 ⇥ 1025 ; 2.85 ⇥ 10 2

Table 4.2 gives 5.9 ⇥ 10

3

e

) P is proportional to E, and

C m; N =

molecules molecules ⇥ moles volume = mole gram (0.33⇥1029 )(6.1⇥10 30 )2 (3)(8.85⇥10 12 )(1.38⇥10 23 )(293)

1 N = 6.0 ⇥ 1023 ⇥ 18 ⇥ 106 = 0.33 ⇥ 1029 ; e = experimental value of 79, so it’s pretty far o↵. For water vapor at 100 = 373 K, treated as an ideal gas,

N=

eE

=

volume mole

= (22.4 ⇥ 10

(2.11 ⇥ 1025 )(6.1 ⇥ 10 (3)(8.85 ⇥ 10 12 )(1.38 ⇥ 10

, so this time the agreement is quite good.


3

⇥

e

=

N p2 . 3✏0 kT

grams volume .

= 12. Table 4.2 gives an

)⇥

)

30 2 23 )(373)

373 293

= 2.85 ⇥ 10

= 5.7 ⇥ 10

3

.

2

m3 .

2

CHAPTER 5. MAGNETOSTATICS

110


Chapter 1

Chapter 5

Magnetostatics Magnetostatics Problem 5.1

Problem 5.1 Since v ⇥ B points upward, and that is also the direction of the force, q must be positive. To find R, in terms of a and d, use the pythagorean theorem: (R

2Rd + d2 + a2 = R2 ) R =

a2 + d2 . 2d

!

d)2 + a2 = R2 ) R2

R R

$

(a2 + d2 ) p = qBR = qB . 2d

"#

The cyclotron formula then gives

a

}d

Problem 5.2 The general solution is (Eq. 5.6): Problem 5.2 E y(t) = C1 cos(!t) + C2 sin(!t) + t + C3 ; B

z(t) = C2 cos(!t)

C1 sin(!t) + C4 .

(a) y(0) = z(0) = 0; y(0) ˙ = E/B; z(0) ˙ = 0. Use these to determine C1 , C2 , C3 , and C4 . y(0) = 0 ) C1 + C3 = 0; y(0) ˙ = !C2 + E/B = E/B ) C2 = 0; z(0) = 0 ) C2 + C4 = 0 ) C4 = 0; z(0) ˙ = 0 ) C1 = 0, and hence also C3 = 0. So y(t) = Et/B; z(t) = 0. Does this make sense? The magnetic force is q(v ⇥ B) = q(E/B)B ˆ z = qE, which exactly cancels the z electric force; since there is no net force, ✻ the particle moves in a straight line at constant speed. X z β ✻ (b) Assuming it starts from the origin, so C3 = C1 , C4 = C2 , we have z(0) ˙ = 0 ) C1✸ = 0 ) C3 = 0; β ✲y E E E E E E y(0) ˙ = ) C2 ! + = ) C2 = ◆ = C4 ; y(t) = sin(!t) + t; 2B B 2B 2!B B ✲ y 2!B E E E E −β z(t) = cos(!t) + (b) , or y(t) = [2!t sin(!t)] ; z(t) = [1 (c)cos(!t)] . Let ⌘ E/2!B. 2!B 2!B 2!B 2!B Then y(t) = [2!t sin(!t)] ; z(t) = [1 cos(!t)] ; (y 2 !t) = sin(!t), (z )= cos(!t) ) (y 2 !t)2 + (z )2 = 2 . This is a circle of radius whose center moves to the right at constant speed: c ⃝2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is y0 = 2 !t; z0 = . protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any in writing from the publisher. E E E means, without E permission E (c) z(0) ˙ = y(0) ˙ = ) C1 ! = ) C1 = C3 = ; C2 ! + = ) C2 = C4 = 0. B B !B B B c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

R "#

R

$

Problem 5.2 CHAPTER 5. MAGNETOSTATICS

a

}d

111

E E E E E E cos(!t) + t + ; z(t) = sin(!t). y(t) = [1 + !t cos(!t)] ; z(t) = sin(!t). !B B !B !B !B !B Let ⌘ E/!B; then [y (1 + !t)] = cos(!t), z = sin(!t); [y (1 + !t)]2 + z 2 = 2 . This is a circle of radius whose center is at y0 = (1 + !t), z0 = 0. z ✻ z β ✻ ✸ β ✲y ◆ ✲y y(t) =

(b)

−β

(c)

Problem 5.3 c E River, NJ. All rights reserved. This material is ⃝2005 Pearson Education, Inc., Upper Saddle (a) From Eq. 5.2, F = q[E + (v ⇥ B)] = all 0) E = vB v = currently . protected under copyright laws) as they exist. No portion of this material may be B reproduced, in any form or by any means, without permission in writing from the publisher. q v E (b) From Eq. 5.3, mv = qBR ) = = . m BR B2R

Problem 5.4 Suppose I flows counterclockwise (if not, change the sign of the answer). The force on the left side (toward the left) cancels the force on the right side (toward the right); the force on the top is IaB = Iak(a/2) = Ika2 /2, (pointing upward), and the force on the bottom is IaB = Ika2 /2 (also upward). So the net force is F = Ika2 ˆ z. Problem 5.5 I , because the length-perpendicular-to-flow is the circumference. 2⇡a Z Z Z ↵ 1 I I (b) J = ) I = J da = ↵ s ds d = 2⇡↵ ds = 2⇡↵a ) ↵ = ;J = . s s 2⇡a 2⇡as (a) K =

Problem 5.6 (a) v = !r, so K = !r.

(b) v = !r sin ✓ ˆ ) J = ⇢!r sin ✓ ˆ, where ⇢ ⌘ Q/(4/3)⇡R3 .

Problem 5.7Z Z ✓ ◆ Z dp d @⇢ = ⇢r d⌧ = r d⌧ = (r · J)r d⌧ (by the continuity equation). Now product rule #5 dt dt V @t R says r · (xJ) = x(r · J) + J · (rx). But rx = x ˆ, so r · (xJ) = x(r · J) + Jx . Thus V (r · J)x d⌧ = Z Z R r · (xJ) d⌧ Jx d⌧ . The first term is S xJ · da (by the divergence theorem), and since J is entirely V V R R inside V, it is zero on the surface S. Therefore V (r · J)x d⌧ = J d⌧ , or, combining this with the y and V x Z R R dp z components, V (r · J)r d⌧ = J d⌧. Or, referring back to the first line, = J d⌧ . qed V dt Here’s a quicker method, if the distribution consists of a collection of point charges. Use Eqs. 5.30 and 3.100: Z X d X dp J d⌧ = qi vi = qi ri = . dt dt


112


Problem 5.8 (a) Use Eq. 5.37, with z = R, ✓2 =

p

2µ0 I . ⇡R

✓1 = 45 , and four sides: B =

⇡ nµ0 I , and n sides: B = sin(⇡/n). n 2⇡R nµ0 I ⇣ ⇡ ⌘ µ0 I (c) For small ✓, sin ✓ ⇡ ✓. So as n ! 1, B ! = (same as Eq. 5.41, with z = 0). 2⇡R n 2R (b) z = R, ✓2 =

✓1 =

Problem 5.9

✓ ◆ µ0 I 1 1 (a) The straight segments produce no field at P . The two quarter-circles give B = (out). 8 a b µ0 I µ0 I (b) The two half-lines are the same as one infinite line: ; the half-circle contributes . 2⇡R 4R ✓ ◆ µ0 I 2 So B = 1+ (into the page). 4R ⇡ Problem 5.10 (a) The forces on the two sides cancel. At the bottom, B = top, B =

µ0 I )F = 2⇡s

✓

µ0 I 2⇡s

µ0 I µ0 I 2 a µ0 I 2 a2 )F = (down). The net force is (up). 2⇡(s + a) 2⇡(s + a) 2⇡s(s + a)

◆

Ia =

µ0 I 2 a (up). At the 2⇡s

"#

$

µ0 I (b) The force on the bottom is the same as before, µ0 I 2 a/2⇡s (up). On the left side, B = ˆ z; 2⇡y ✓ ◆ µ0 I µ0 I 2 CHAPTER 3 dF = I(dl ⇥ B) = I(dx x ˆ +5.dyMAGNETOSTATICS y ˆ + dz ˆ z) ⇥ ˆ z = ( dx y ˆ + dy x ˆ). But the x component cancels the 2⇡y 2⇡y p Z p µ0 I 2 (s/ 3+a/2) 1 corresponding Problem term from 5.10 the right side, and Fy = dx. Here y = 3x, so p 2⇡ s/ 3 y ! p p ! 2 2 µ0 I s/ 3 + a/2 µ0 I 3a p ln p p ln 1 + Fy = = . The force on the right side is the same, so the net 2s 2 3⇡ s/ 3 2 3⇡ y " p !# ✻ µ0 I 2 a 2 3a p ln 1 + force on the triangle is . a 2⇡ s 2s 3 a s

◦

✲x

!

60

√s 3

✰ z

Problem 5.11 c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

η

✲z

θ a }

❄

Problem 5.11

113


}

Problem 5.11 Use Eq. 5.41 of width dz, with I ! nI dz: r Z for a ring µ0 nI a2 θ B= dz. But z = a cot ✓, 3/2 2 a (a2 + z 2 ) ❄ a 1 sin3 ✓ ! "# $ so dz = d✓, and = . dz 3/2 z a3 sin2 ✓ (a2 + z 2 ) So Z 2 3 Z µ0 nI a sin ✓ µ0 nI µ0 nI µ0 nI ✓2 B= sin ✓ d✓ = cos ✓ ✓1 = (cos ✓2 cos ✓1 ). 2 ( a d✓) = 3 2 2 2 2 a sin ✓ For an infinite solenoid, ✓2 = 0, ✓1 = ⇡, so (cos ✓2 Problem 5.12

cos ✓1 ) = 1

( 1) = 2, and B = µ0 nI. X

z Rsinq

Rcosq

q

R


Field (at center of sphere) due to the ring at ✓ (see figure) is (Eq. 5.41): dB =

µ0 dI (R sin ✓)2 µ0 = sin2 ✓ dI. 2 [(R sin ✓)2 + (R cos ✓)2 ]3/2 2R

dI = K R d✓, so

K = v,

=

Q , 4⇡R2

v = !R sin ✓,

Q Q! !R sin ✓R d✓ = sin ✓ d✓. 4⇡R2 4⇡ ✓ ◆ Z µ0 Q! ⇡ 3 µ0 Q! 4 µ0 Q! B= sin ✓ d✓ = . B= ˆ z. 2R 4⇡ 0 8⇡R 3 6⇡R dI =

Problem 5.13

µ0 2 v 2 Magnetic attraction per unit length (Eqs. 5.40 and 5.13): fm = . 2⇡ d 1 Electric field of one wire (Eq. 2.9): E = . Electric repulsion per unit length on the other wire: 2⇡✏0 s 2 1 1 1 fe = . They balance when µ0 v 2 = , or v = p . Putting in the numbers, 2⇡✏0 d ✏0 ✏0 µ0 c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

✲z

114


1

v=p

= 3.00 ⇥ 108 m/s. This is precisely the speed of light(!), so in fact you could (8.85 ⇥ 10 12 )(4⇡ ⇥ 10 7 ) never get the wires going fast enough; the electric force always dominates. Problem 5.14 ( ) I 0, for s < a; (a) B · dl = B 2⇡s = µ0 Ienc ) B = µ0 I ˆ , for s > a. 2⇡s Z a Z a 2⇡ka3 (b) J = ks; I = J da = ks(2⇡s) ds = )k = 3 0 0 8 µ0 Is2 ˆ > > > < 2⇡a3 , 2⇡ks3 s3 = I 3 , for s < a; Ienc = I, for s > a. So B = > 3 a > 4 > : µ0 I ˆ, 2⇡s Problem 5.15Problem 5.14 By the right-hand-rule, the field points in the At z = 0, B = 0. Use the amperian loop shown: I B · dl = Bl = µ0 Ienc = µ0 lzJ ) B = so B =

⇢

3I . Ienc = 2⇡a3 9 > for s < a; > > =

Z

s

J da =

0

Z

s

k¯ s(2⇡¯ s) d¯ s=

0

> > > for s > a. ;

CHAPTER 5. MAGNETOSTAT

y ˆ direction for z > 0, and in the +ˆ y direction for z < 0.

µ0 Jz y ˆ ( a < z < a). If z > a, Ienc = µ0 laJ,

µ0 Ja y ˆ, for z > +a; +µ0 Ja y ˆ, for z > a.

z ✛

z{ !

✻ ✲

"# l

amperian loop

✠

✲y

$

Problem 5.16 z to the right (ˆ The field inside a solenoid is µ0 nI, and outside it is zero. The outer solenoid’s field points z), ✻ whereas the inner one points to the left ( ˆ z). So: (i) B = µ0 I(n2 n1 ) ˆ z, (ii) B = µ0 In2 ˆ z, (iii) B = 0. loop 1 Problem 5.17 I ✿ it, and into From Ex. 5.8, the top plate produces a field µ0 K/2 (aiming out of the page, for points above the page, for points below). The bottom plate produces a field µ0 K/2 (aiming into the page, for points ηabove 1 it, and out of the page, for points below). Above and below both plates the two fields cancel; between the plates ❥r ✣ they add up to Problem µ0 K, pointing 5.17in.

(a) B = µ0 v (in) betweem the plates, B = 0 elsewhere. R ✰ η (b) The Lorentz force law says F = (K ⇥ B) da, so the force per unit area xis f = K ⇥ B. Here K =2 v,

to the right, and B (the field of the lower plate) is µ0 v/2, into the page. So fm = µ0

v /2 (up).

2 2

(c) The electric field of the lower plate is fe =

2

/2✏0 ; the electric force per unit area on the upper plate loop is 2 I p ✿in Prob. 5.13. /2✏0 (down). They balance if µ0 v = 1/✏0 , or v = 1/ ✏0 µ0 = c (the speed of light), as 2

Problem 5.18 We might as well orient the axes so the field point r lies on the y axis: r = (0, y, 0). Consider a source point at (x0 , y 0 , z 0 ) on loop #1:

r

=

x0 x ˆ + (y

y0 ) y ˆ

z0 ˆ z; dl0 = dx0 x ˆ + dy 0 y ˆ;


✲y

4


Problem 5.14

115


dl0 ⇥

r

z ✻ amperian loop x ˆ y ˆ ˆ z ✠ 0 0 0 0 0 ✛ 0 0 dy 0 = ( z dy ) x = dx ˆ + (z zdx ˆ + [(y y ) dx0 + x0 dy 0 ] ˆ z. { )y ✲ ✲y 0 0 0 x (y y ) z

dB1 =

"# $y 0 ) dx0 + x0 dy 0 ] ˆ µ0 I dl0 ⇥ r µ0 I ( z 0 dy 0 ) x ˆ + (z 0 dx0 ) y ˆ! + [(y z = . l 3 3/2 0 2 0 2 0 2 r 4⇡ 4⇡ [(x ) + (y y ) + (z ) ]

Now consider the symmetrically placed source element on loop #2, at (x0 , y 0 , z 0 ). Since z 0 changes sign, while everything else is the same, the x ˆ and y ˆ components from dB1 and dB2 cancel, leaving only a ˆ z component. qed With this, Ampére’s law yields immediately:

z

✻ loop 1 I ✿

( µ0 nI ˆ z, inside the solenoid; B= 0, outside Problem 5.17

r

1

❥r ✣

(the same as for a circular solenoid—Ex. 5.9). For the toroid, N/2⇡s = n (the number of turns per unit length), so Eq. 5.60 yields B = µ0 nI inside, and zero outside, consistent with the solenoid. [Note: N/2⇡s = n applies only if the toroid is large in circumference, so that s is essentially constant over the cross-section.]

✰ x

r I ✿

✲y

2

loop 2

Problem 5.19 R It doesn’t matter. According to Theorem 2, in Sect. 1.6.2, J · da is independent of surface, for any given boundary line, provided that J is divergenceless, which it is, for steady currents (Eq. 5.33). Problem 5.20 ✓ ◆ charge charge atoms moles grams 1 (a) ⇢ = = · · · = (e)(N ) (d), where volume atom mole gram volume M e N M d

= = = =

charge of electron Avogadro0 s number atomic mass of copper density of copper

= = = =

1.6 ⇥ 10 19 C, 6.0 ⇥ 1023 mole, 64 gm/mole, 3 9.0 gm/cm .

◆ 9.0 = 1.4 ⇥ 104 C/cm3 . 64 I I 1 (b) J = 2 = ⇢v ) v = 2 = = 9.1 ⇥ 10 3 cm/s, or about 33 cm/hr. This ⇡s ⇡s ⇢ ⇡(2.5 ⇥ 10 3 )(1.4 ⇥ 104 ) is astonishingly small—literally ✓ slower ◆than a snail’s pace. µ0 I1 I2 (4⇡ ⇥ 10 7 ) (c) From Eq. 5.40, fm = = = 2 ⇥ 10 7 N/cm. 2⇡ ✓ d ◆ 2⇡ ✓ ◆ ✓ 2◆ ✓ ◆ 1 1 1 1 I1 I2 c µ0 I1 I2 c2 1 2 (d) E = ; fe = = 2 = = 2 fm , where 2 2⇡✏0 d 2⇡✏0 d v 2⇡✏0 d v 2⇡ d v ✓ ◆ c ⃝2005 Pearson2 Education, Inc., Upper 10 2 Saddle River, NJ. All rights reserved. This material is fe under c all copyright 3.0 ⇥ laws 10 as they currently exist. p protected No portion of this material may be c ⌘ 1/ ✏0 µ0 = 3.00 ⇥ 108 m/s. Here = = =without 1.1 ⇥ permission 1025 . reproduced, or by any 3means, in writing from the publisher. fm inv 2any form9.1 ⇥ 10 ⇢ = (1.6 ⇥ 10

19

)(6.0 ⇥ 1023 )

fe = (1.1 ⇥ 1025 )(2 ⇥ 10

7

✓

) = 2 ⇥ 1018 N/cm.


116


Problem 5.21 Ampére’s law says r ⇥ B = µ0 J. Together with the continuity equation (5.29) this gives r · (r ⇥ B) = µ0 r · J = µ0 @⇢/@t, which is inconsistent with div(curl)=0 unless ⇢ is constant (magnetostatics). The other Maxwell equations are OK: r ⇥ E = 0 ) r · (r ⇥ E) = 0 (X), and as for the two divergence equations, there is no relevant vanishing second derivative (the other one is curl(grad), which doesn’t involve the divergence). Problem 5.22 At this stage I’d expect no changes in Gauss’s law or Ampére’s law. The divergence of B would take the form r · B = ↵0 ⇢m , where ⇢m is the density of magnetic charge, and ↵0 is some constant (analogous to ✏0

and µ0 ). The curl of E becomes r ⇥ E = 0 Jm , where Jm is the magnetic current density (representing the flow of magnetic charge), and 0 is another constant. Presumably magnetic charge is conserved, so ⇢m and Jm satisfy a continuity equation: r · Jm = @⇢m /@t. As for the Lorentz force law, one might guess something of the form qm [B + (v ⇥ E)] (where qm is the magnetic charge). But this is dimensionally impossible, since E has the sameCHAPTER units as vB. Evidently we 5. MAGNETOSTATICS need to divide (v ⇥ E) by something with the dimensions of velocity-squared. The natural candidate is  1 c2 = 1/✏0 µ0 : F = qe [E + (v ⇥ B)] + qm B (v ⇥ E) . In this form the magnetic analog Problem 5.22 to Coulomb’s c2 ↵0 qm1 qm2 law reads F = ˆ r, so to determine ↵0 we would first introduce (arbitrarily) a unit of magnetic charge, 4⇡ r2 then measure the force between unit charges at a given separation. [For further details, and an explanation of the minus sign in the force law, see Prob. 7.38.] Problem 5.23

r

µ0 I dz = ˆ z 4⇡

z1

z2

z1

dz p z 2 + s2

dl=dz

s

"

# p z2 + (z2 )2 + s2 µ0 I p = ln ˆ z 4⇡ z1 + (z1 )2 + s2

}

Iˆ z

Z

}

µ0 A= 4⇡

Z

✲z

r

! "# $ ⌘i z2 p µ0 I h ⇣ z2 ˆ z ln z + z 2 + s2 4⇡ z1 " # @A ˆ µ0 I 1 s 1 s ˆ p p p p B = r⇥A= = @s 4⇡ z2 + (z2 )2 + s2 (z2 )2 + s2 z1 + (z1 )2 + s2 (z1 )2 + s2 " # p p (z2 )2 + s2 z1 (z1 )2 + s2 µ0 Is z2 1 1 Problem ˆ 5.26 p p = 4⇡ (z2 )2 [(z2 )2 + s2 ] (z2 )2 + s2 z12 [(z1 )2 + s2 ] (z1 )2 + s2 # " # ✓ ◆" µ0 Is 1 z2 z1 µ0 I z2 z1 ˆ ˆ, p p p = 1 p +1 = 4⇡ s2 4⇡s (z2 )2 + s2 (z1 )2 + s2 (z2 )2 + s2 (z1 )2 + s2 z1 z2 or, since sin ✓1 = p and sin ✓2 = p , 2 2 (z1 ) + s (z2 )2 + s2

=

=

µ0 I (sin ✓2 4⇡s

Problem 5.24

sin ✓1 ) ˆ (as in Eq. 5.37).

1 @ k 1 1 A =k )B=r⇥A= (sk) ˆ z= ˆ z; J = (r ⇥ B) = s @s s µ0 µ0



@ @s

✓ ◆ k s

ˆ=

k ˆ . µ0 s2

Problem 5.36 c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Problem 5.37

117


Problem 5.25 1 1 r·A = r · (r ⇥ B) = [B · (r ⇥ r) r · (r ⇥ B)] = 0, since r ⇥ B = 0 (B is uniform) and 2 2 1 1 r ⇥ r = 0 (Prob. 1.63). r ⇥ A = r ⇥ (r ⇥ B) = [(B · r)r (r · r)B + r(r · B) B(r · r)]. But 2 2 @x @y @z (r · r)B = 0 and r · B = 0 (since B is uniform), and r · r = + + = 1 + 1 + 1 = 3. Finally, @x @y @z ✓ ◆ @ @ @ 1 (B·r)r = Bx + By + Bz (x x ˆ +y y ˆ +z ˆ z) = Bx x ˆ +By y ˆ +Bz ˆ z = B. So r⇥A = (B 3B) = B. @x @y @z 2 qed

}

}

Problem 5.26 MAGNETOSTATICS 5 wire). In cylindrical (a) A isCHAPTER parallel (or5.antiparallel) to I, and is a function only of s (the distance from the @A ˆ µ0 I ˆ coordinates, then, A = A(s) ˆ z, so B = r ⇥ A = = (the field of an infinite wire). Therefore @s 2⇡s 5.22 @A µProblem µ0 I 0I = , and A(r) = ln(s/a) ˆ z (the constant a is arbitrary; you could use 1, but then the units @s 2⇡s 2⇡ @Az @Az ˆ µ0 I ˆ look fishy). r · A = = 0. X r ⇥ A = = = B. X @z @s 2⇡s I I µ0 Is2 (b) Here Ampére’s law gives B · dl = B 2⇡s = µ0 Ienc = µ0 J ⇡s2 = µ0 ⇡s2 = . 2 ⇡R R2 z1 dl=dz µ0 Is ˆ @A µ0 I s ✲ µ0 I 2 B= . = ) zA = (s b2 ) ˆ z. Here b is again arbitrary, except that since A 2 2⇡ R @s 2⇡ R2 4⇡R2 s η µ0 I µ0 I must be continuous at R, ln(R/a) = (R2 b2 ), which means that we must pick a and b such that 2 2⇡ 4⇡R ! "# $ 8 µ I 9 z 0 2 2 > > (s R ) ˆ z , for s  R; > > < 4⇡R2 = 2 2 ln(R/b) = 1 (b/R) . I’ll use a = b = R. Then A = > > > > : µ0 I ln(s/R) ˆ z, for s R. ; 2⇡ Problem Problem 5.27 5.26 µ0 K K=Kx ˆ)B=± y ˆ (plus for z < 0, minus for z > 0). 2 A is parallel (or antiparallel) to K, and depends only on z, so A = A(z) x ˆ. x ˆ y ˆ ˆ z @A µ0 K B = r ⇥ A = @/@x @/@y @/@z = y ˆ=± y ˆ. @z 2 A(z) 0 0 A=

µ0 K |z| x ˆ will do the job—or this plus any constant. 2

Problem 5.28Z ✓ ◆ µ0 J 0 (a) r·A = r· r d⌧ ; 4⇡

✓

J

◆

1

r

r

(r·J)+J·r

✲y

✠ ✠ ✠ ✠ ✠ ✠K ✠ x

✓

1

◆

. But the first term is zero, because J(r0 ) ✓ ◆ ✓ ◆ 1 1 0 0 is a function of the source coordinates, not the field coordinates. And since r = r r , r = r . r r ✓ ◆ ✓ ◆ ✓ ◆ ✓ ◆ J 1 J 1 1 0 Problem 5.36 So r · = J·r . But r0 · = (r0 · J) + J · r0 , and r0 · J = 0 in magnetostatics r r r r r ✓ ◆ ✓ ◆ ✓ ◆ Z J J µ0 J 0 (Eq. 5.33). So r· = r0 · , and hence, by the divergence theorem, r·A = r0 · r r r d⌧ = 4⇡ r·

=

z ✻

r


Problem 5.37

118


I µ0 J 0 r · da , where the integral is now over the surface surrounding all the currents. But J = 0 on this 4⇡ surface, so r · A = 0. X ✓ ◆ ✓ ◆ Z Z  µ0 J µ0 1 1 0 (b) r ⇥ A = r⇥ d⌧ = (r ⇥ J) J ⇥ r d⌧ 0 . But r ⇥ J = 0 (since J is r r r 4⇡ 4⇡ ✓ ◆ Z rˆ 1 µ0 J ⇥ rˆ 0 not a function of r), and r = (Eq. 1.101), so r ⇥ A = 2 r r r 2 d⌧ = B. X 4⇡ ✓ ◆ ✓ ◆ ✓ ◆ Z µ0 J J 1 (c) r2 A = r2 d⌧ 0 . But r2 = Jr2 r r ✓ ◆ r (once again, J is a constant, as far as 4⇡ 1 di↵erentiation with respect to r is concerned), and r2 = 4⇡ 3 ( r ) (Eq. 1.102). So r Z ⇥ ⇤ µ0 r2 A = J(r0 ) 4⇡ 3 ( r ) d⌧ 0 = µ0 J(r). X 4⇡ Problem I5.29 µ0 I =

B · dl =

Z

a

b

rU · dl =

For an infinite straight wire, B =

[U (b) µ0 I ˆ . 2⇡s

U (a)] (by the gradient theorem), so U (b) 6= U (a). qed U=

µ0 I 2⇡

would do the job, in the sense that

µ0 I µ0 I 1 @ ˆ r( ) = = B. But when advances by 2⇡, this function does not return to its initial 2⇡ 2⇡ s @ value; it works (say) for 0  < 2⇡, but at 2⇡ it “jumps” back to zero. rU =

Problem 5.30 Use Eq. 5.69, with R ! r¯ and

! ⇢ d¯ r:

Z Z R µ0 !⇢ sin ✓ ˆ r 4 µ0 !⇢ ˆ r ¯ d¯ r + r sin ✓ r¯ d¯ r 3 r2 3 0 r  ✓ 5◆ ✓ 2 ◆ ⇣ µ !⇢ ⌘ 1 r r r2 ˆ 0 ˆ = µ0 !⇢ r sin ✓ R = sin ✓ 2 + R2 r 2 . 3 r 5 2 2 3 5 ⇢  ✓ 2 ◆  ✓ 2 µ0 !⇢ 1 @ R r2 1 @ R 2 B = r⇥A= sin ✓ r sin ✓ ˆ r r sin ✓ 2 r sin ✓ @✓ 3 5 r @r 3 ✓ 2 ◆ ✓ ◆ R r2 R2 2r2 = µ0 !⇢ cos ✓ ˆ r sin ✓ ✓ˆ . 3 5 3 5

A =

r2 5

◆

✓ˆ

Problem 5.31 8 9 Rx @Wz > > 0 0 > > = F ) W (x, y, z) = F (x , y, z) dx + C (y, z). > > y z y 1 0 < = @x (a) > > Rx > > @Wy > : ; = Fz ) Wy (x, y, z) = + 0 Fz (x0 , y, z) dx0 + C2 (y, z). > @x These satisfy (ii) and (iii), for any C1 and C2 ; it remains to choose these functions so as to satisfy (i): Z x Z x @Fy (x0 , y, z) 0 @C1 @Fz (x0 , y, z) 0 @C2 @Fx @Fy @Fz dx + dx = Fx (x, y, z). But + + = 0, so @y @y @z @zZ @x @y @z 0 Z x0 x @Fx (x0 , y, z) 0 @C1 @C2 @Fx (x0 , y, z) 0 dx + = Fx (x, y, z). Now dx = Fx (x, y, z) Fx (0, y, z), so 0 @x @y @z @x0 0 0 c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

119

CHAPTER 5. MAGNETOSTATICS @C1 @y

Z y @C2 = Fx (0, y, z). We may as well pick C2 = 0, C1 (y, z) = Fx (0, y 0 , z) dy 0 , and we’re done, with @z 0 Z x Z y Z x Wx = 0; Wy = Fz (x0 , y, z) dx0 ; Wz = Fx (0, y 0 , z) dy 0 Fy (x0 , y, z) dx0 . 0

✓

◆

✓

◆

0

✓

◆

0

@Wz @Wy @Wx @Wz @Wy @Wx x ˆ+ y ˆ+ ˆ z @y @z @z @x @x @y Z x x @Fy (x0 , y, z) 0 @Fz (x0 , y, z) 0 dx dx x ˆ + [0 + Fy (x, y, z)] y ˆ + [Fz (x, y, z) 0] ˆ z. @y 0 0  Z x@z 0 @Fx (x , y, z) 0 But r · F = 0, so the x ˆ term is Fx (0, y, z) + dx = Fx (0, y, z) + Fx (x, y, z) Fx (0, y, z), @x0 0 so r ⇥ W = F. X Z x Z y Z x @Wx @Wy @Wz @Fz (x0 , y, z) 0 @Fx (0, y 0 , z) 0 @Fy (x0 , y, z) 0 r·W = + + = 0+ dx + dy dx 6= 0, @x @y @z @y @z @z 0 0 0 in general. Z Z y Z x x x2 y2 (c) Wy = x0 dx0 = ; Wz = y 0 dy 0 z dx0 = zx. 2 2 0 0 0 (b) r ⇥ W =  Z = Fx (0, y, z)

W=

x2 y ˆ+ 2

✓

y2 2

◆ zx ˆ z.

x ˆ y ˆ ˆ z @/@z r ⇥ W = @/@x @/@y = yx ˆ+zy ˆ + xˆ z = F. X 0 x2 /2 (y 2 /2 zx)

Problem 5.32 (a) At the surface of the solenoid, Babove = 0, Bbelow = µ0 nI ˆ z = µ0 K ˆ z; n ˆ =ˆ s; so K ⇥ n ˆ = Kˆ z. Evidently Eq. 5.76 holds. X (b) In Eq. 5.69, ✓ both expressiions reduce to (µ0 R2 ! /3) sin ✓ ˆ at the surface, so Eq. 5.77 is satisfied. ◆ 4 @A µ0 R ! 2 sin ✓ ˆ 2µ0 R! @A µ0 R! = = sin ✓ ˆ; = sin ✓ ˆ. So the left side of @r R+ 3 r3 3 @r 3 R R ! ⇥ r) = !R sin ✓ ˆ, so the right side of Eq. 5.78 is Eq. 5.78 is µ0 R! sin ✓ ˆ. Meanwhile K = v = (! ˆ µ0 !R sin ✓ , and the equation is satisfied. Problem 5.33 @A @A Because Aabove = Abelow at every point on the surface, it follows that and are the same above @x @y and below; any discontinuity is confined to the ✓ ◆ normal ✓ derivative. ◆ @Ayabove @Aybelow @Axabove @Axbelow Babove Bbelow = x ˆ+ y ˆ. But Eq. 5.76 says this equals + @z @z @z @z @Ayabove @Aybelow @Axabove @Axbelow µ0 K( y ˆ). So = , and = µ0 K. Thus the normal derivative of the com@z @z @z @z @Aabove @Abelow ponent of A parallel to K su↵ers a discontinuity µ0 K, or, more compactly: = µ0 K. @n @n Problem 5.34 ˆ ✓ˆ = m cos ✓ ˆ (Same idea as Prob. 3.36.) Write m = (m · ˆ r) ˆ r + (m · ✓) r m sin ✓ ✓ˆ (Fig. 5.54). Then ˆ and Eq. 5.89 , Eq. 5.88. qed 3(m · ˆ r) ˆ r m = 3m cos ✓ ˆ r m cos ✓ ˆ r + m sin ✓ ✓ˆ = 2m cos ✓ ˆ r + m sin ✓ ✓, Problem 5.35 (a) m = Ia = I⇡R2 ˆ z. ⌘ 2 ⇣ µ0 I⇡R ˆ . (b) B ⇡ 2 cos ✓ ˆ r + sin ✓ ✓ 4⇡ r3


120


µ0 IR2 ˆ z (for z < 0, ✓ = ⇡, ˆ r= ˆ z, so the field 2z 3 is the same, with |z|3 in place of z 3 ). The exact answer (Eq. 5.41) reduces (for z R) to B ⇡ µ0 IR2 /2|z|3 , so they agree. (c) On the z axis, ✓ = 0, r = z, ˆ r=ˆ z (for z > 0), so B ⇡

Problem 5.36 The field of one side is given by Eq. 5.37, with s ! p (w/2) z 2 + (w/2)2 and sin ✓2 = sin ✓1 = p ; z 2 + w2 /2 µ0 I w p p B= . To pick o↵ the vertical 2 2 4⇡ z + (w /4) z 2 + (w2 /2) (w/2) component, multiply by sin = p ; for all four z 2 + (w/2)2 sides, multiply by 4: B =

µ0 I w2 p ˆ z. 2⇡ (z 2 + w2 /4) z 2 + w2 /2

µ0 Iw2 ˆ z. The field of a dipole m = Iw2 , 2⇡z 3 µ0 m for points on the z axis (Eq. 5.88, with r ! z, ˆ r!ˆ z, ✓ = 0) is B = ˆ z. X 2⇡ z 3 Problem 5.37 RR (a) For a ring, m = I⇡r2 . Here I ! v dr = !r dr, so m = 0 ⇡r2 !r dr = ⇡ !R4 /4. For z

w, B ⇡

(b) The total charge on the shaded ring is dq = (2⇡R sin ✓)R d✓. The time for one revolution is dt = 2⇡/!. dq So the current in the ring is I = = !R2 sin ✓ d✓. The dt area of the ring is ⇡(R sin ✓)2 , so the magnetic moment of the ring is dm = ( !R2 sin ✓ d✓)⇡R2 sin2 ✓, and the total dipole moment of the shell is R⇡ 4⇡ m = !⇡R4 0 sin3 ✓ d✓ = (4/3) !⇡R4 , or m = !R4 ˆ z. 3

µ0 4⇡ sin ✓ µ0 !R4 sin ✓ ˆ !R4 2 ˆ = , 4⇡ 3 r 3 r2 which is also the exact potential (Eq. 5.69); evidently a spinning sphere produces a perfect dipole field, with no higher multipole contributions. The dipole term in the multipole expansion for A is therefore Adip =

Problem 5.38 (a) I dl ! J d⌧, so

A=

Z 1 µ0 X 1 (r0 )n Pn (cos ✓)J d⌧. 4⇡ n=0 rn+1

Z µ0 µ0 dp J d⌧ = (Prob. 5.7), where p is the total electric dipole moment. In mag4⇡r 4⇡r dt netostatics, p is constant, so dp/dt = 0, and hence Amon = 0. qed H R (c) m = Ia = 12 I (r ⇥ dl) ! m = 12 (r ⇥ J) d⌧. qed (b) Amon =

Problem 5.39 (a) Yes. (Magnetic forces do no work.)


121


µ0 ˆ (b) B = ; 2⇡s

ˆ s ˆ ˆ z ⇣µ ⌘ 0 ˙ ˆ v = s˙ ˆ s+s + z˙ ˆ z; (v ⇥ B) = s˙ s ˙ z˙ = ( z˙ ˆ s + s˙ ˆ z). 2⇡s µ0 0 2⇡s 0 F = q(v ⇥ B) =

(c) a = (¨ s

0

2⇡s

( z˙ ˆ r + s˙ ˆ z).

s ˙ 2) ˆ s + (s ¨ + 2s˙ ˙ ) ˆ + z¨ ˆ z (see, for example J. R. Taylor, Classical Mechanics, Eq. 1.47). F = ma

where ↵ ⌘

⇣ qµ ⌘

⇣ qµ ⌘ 0 . Thus 2⇡m

)

s¨

(¨ s

s ˙2 =

↵ s ˙ 2) ˆ s + (s ¨ + 2s˙ ˙ ) ˆ + z¨ ˆ z = ( z˙ ˆ s + s˙ ˆ z), s

z˙ ↵ , s

s ¨ + 2s˙ ˙ = 0,

s˙ z¨ = ↵ . s

(d) If z˙ is constant, then z¨ = 0, and it follows from the third equation that s˙ = 0, and hence s is constant. 1p Then the first equation says ˙ 2 = ↵(z/s ˙ 2 ), so ˙ = ± ↵z˙ is also constant (the second equation holds s automatically). The charge moves in a helix around the wire. Problem 5.40 The mobile charges do pull in toward the axis, but the resulting concentration of (negative) charge sets up an electric field that repels away further accumulation. Equilibrium is reached when the electric repulsion on a mobile charge q balances the magnetic attraction: F = q[E + (v ⇥ B)] = 0 ) E = (v ⇥ B). Say the current is in the z direction: J = ⇢ v ˆ z (where ⇢ and v are both negative). I µ0 ⇢ vs ˆ B · dl = B 2⇡s = µ0 J⇡s2 ) B = ; 2 Z 1 1 E · da = E 2⇡sl = (⇢+ + ⇢ )⇡s2 l ) E = (⇢+ + ⇢ )s ˆ s. ✏0 2✏0 ✓ 2◆ h ⇣ µ ⇢ vs ⌘i µ 1 v 0 0 2 2 ˆ (⇢+ + ⇢ )s ˆ s= (v ˆ z) ⇥ = ⇢ v sˆ s ) ⇢+ + ⇢ = ⇢ (✏0 µ0 v ) = ⇢ . 2✏0 2 2 c2 ✓ ◆ v2 ⇢ 2 Evidently ⇢+ = ⇢ 1 = 2 , or ⇢ = ⇢+ . In this naive model the mobile negative charges fill a c2 smaller inner cylinder, leaving a shell of positive (stationary) charge at the outside. But since v ⌧ c, the e↵ect is extremely small. Problem 5.41 (a) If positive charges flow to the right, they are deflected down, and the bottom plate acquires a positive charge. (b) qvB = qE ) E = vB ) V = Et = vBt, with the bottom at higher potential. (c) If negative charges flow to the left, they are also deflected down, and the bottom plate acquires a negative charge. The potential di↵erence is still the same, but this time the top plate is at the higher potential. Problem 5.42 R RFrom Eq. 5.17,R F = I (dl ⇥ B). But B is constant, in this case, so it comes outside the integral: F = I dl ⇥ B, and dl = w, the vector displacement from the point at which the wire first enters the field to the point where it leaves. Since w and B are perpendicular, F = IBw, and F is perpendicular to w. c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

122


Problem 5.43 The angular momentum acquired by the particle as it moves out from the center to the edge is L=

Z

dL dt = dt

Z

N dt =

Z

(r ⇥ F) dt =

Z

r ⇥ q(v ⇥ B) dt = q

Z

r ⇥ (dl ⇥ B) = q

Z

(r · B) dl

Z

B(r · dl) .

But r is perpendicular to B, so r · B = 0, and r · dl = r · dr = 12 d(r · r) = 12 d(r2 ) = r dr = (1/2⇡)(2⇡r dr). Z R Z R q q q So L = B 2⇡r dr = B da. It follows that L = , where = B da is the total flux. 2⇡ 0 2⇡ 2⇡ In particular, if = 0, then L = 0, and the charge emerges with zero angular momentum, which means it is going along a radial line. qed Problem 5.44 R From Eq. 5.24, F = (K ⇥ Bave ) da. Here K = v, v = !R sin ✓ ˆ, da = R2 sin ✓ d✓ d , and Bave = 12 (Bin + Bout ). From Eq. 5.70, Bin = Bout = = Bave = K ⇥ Bave =

2 2 ˆ From Eq. 5.69, µ0 R! ˆ z = µ0 R!(cos ✓ ˆ r sin ✓ ✓). 3 3 ✓ ◆  ✓ ◆ ✓ ◆ µ0 R4 ! sin ✓ ˆ µ0 R4 ! 1 @ sin2 ✓ 1 @ sin ✓ ˆ r⇥A=r⇥ = ˆ r ✓ 3 r2 3 r sin ✓ @✓ r2 r @r r µ0 R4 ! ˆ = µ0 R! (2 cos ✓ ˆ ˆ (since r = R). (2 cos ✓ ˆ r + sin ✓ ✓) r + sin ✓ ✓) 3r3 3 µ0 R! ˆ (4 cos ✓ ˆ r sin ✓ ✓). 6 ✓ ◆h i µ0 R! ˆ ⇥ (4 cos ✓ ˆ ˆ = µ0 ( !R)2 (4 cos ✓ ✓ˆ + sin ✓ ˆ ( !R sin ✓) r sin ✓ ✓) r) sin ✓. 6 6

Picking out the z component of ✓ˆ (namely, µ0 (K ⇥ Bave )z = ( !R)2 sin2 ✓ cos ✓, so 2 Fz =

µ0 ( !R)2 R2 2

Z

sin ✓) and of ˆ r (namely, cos ✓), we have

sin ✓ cos ✓ d✓ d = 3

µ0 ( !R2 )2 2⇡ 2

✓

sin4 ✓ 4

◆

⇡/2

, or F = 0

µ0 ⇡ ( !R2 )2 ˆ z. 4

Problem 5.45

µ0 qe qm µ0 qe qm (v ⇥ ˆ r); a = (v ⇥ r). 2 4⇡ r 4⇡ mr3 1 d 1 d 2 dv dv (b) Because a ? v, a · v = 0. But a · v = (v · v) = (v ) = v . So = 0. qed 2 dt 2 dt dt dt ✓ ◆ ⇣ ⌘ dQ µ0 qe qm d r µ0 qe qm µ0 qe qm v r dr (c) = m(v ⇥ v) + m(r ⇥ a) =0+ [r ⇥ (v ⇥ r] dt 4⇡ dt r 4⇡r3 4⇡ r r2 dt ⇢  µ0 qe qm 1 2 v r d p µ0 qe qm v (ˆ r · v) v ˆ r 2(r · v) [r v (r · v)r] = 0. X = + 2 ( r · r) = ˆ r + 4⇡ r3 r r dt 4⇡ r r r 2r r µ0 qe qm (d) (i) Q · ˆ = Q(ˆ z · ˆ) = m(r ⇥ v) · ˆ (ˆ r · ˆ). But ˆ z· ˆ =ˆ r · ˆ = 0, so (r ⇥ v) · ˆ = 0. But 4⇡ (a) F = ma = qe (v ⇥ B) =


123

CHAPTER 5. MAGNETOSTATICS r = rˆ r, and v =

dl = r˙ ˆ r + r✓˙ ✓ˆ + r sin ✓ ˙ ˆ (where dots denote di↵erentiation with respect to time), so dt ˆ ˆ r ✓ˆ ˙ ˆ. 0 r⇥v = r 0 = ( r2 sin ✓ ˙ ) ✓ˆ + (r2 ✓) ˙ ˙ r˙ r✓ r sin ✓

Therefore (r ⇥ v) · ˆ = r2 ✓˙ = 0, so ✓ is constant. qed µ0 qe qm (ii) Q · ˆ r = Q(ˆ z·ˆ r) = m(r ⇥ v) · ˆ r (ˆ r·ˆ r). But ˆ z·ˆ r = cos ✓, and (r ⇥ v) ? r ) (r ⇥ v) · ˆ r = 0, so 4⇡ µ0 qe qm µ0 qe qm Q cos ✓ = , or Q = . And since ✓ is constant, so too is Q. qed 4⇡ 4⇡ cos ✓ ˆ = m(r ⇥ v) · ✓ˆ µ0 qe qm (ˆ ˆ But ˆ (iii) Q · ✓ˆ = Q(ˆ z · ✓) r · ✓). z · ✓ˆ = sin ✓, ˆ r · ✓ˆ = 0, and (r ⇥ v) · ✓ˆ = r2 sin ✓ ˙ 4⇡ Q k Q µ0 qe qm (from (i)), so Q sin ✓ = mr2 sin ✓ ˙ ) ˙ = = 2 , with k ⌘ = . mr2 r m 4⇡m cos ✓ k k2 k 2 sin2 ✓ (e) v 2 = r˙ 2 + r2 ✓˙2 + r2 sin2 ✓ ˙ 2 , but ✓˙ = 0 and ˙ = 2 , so r˙ 2 = v 2 r2 sin2 ✓ 4 = v 2 . r r r2 r⇣ ⌘ ✓ ◆2 ⇣ ⌘ dr r˙ 2 v 2 (k sin ✓/r)2 vr 2 dr vr 2 2 2 = = = r sin ✓ ; = r sin2 ✓. ˙2 d (k 2 /r4 ) k d k Z Z ⇣ vr ⌘ dr 1 vr q (f) = d ) sec 1 ; sec[( , or 0 = 0 ) sin ✓] = sin ✓ k sin ✓ k sin ✓ 2 r (vr/k)2 sin ✓ r( ) =

A

cos[(

0 ) sin ✓]

, where A ⌘

µ0 qe qm tan ✓ . 4⇡mv

Problem 5.46 Put the field point on the x axis, so r = (s, 0, 0). Then B = Z µ0 (K ⇥ rˆ ) ˆ r 2 da; da = R d dz; K = K = K( sin xˆ + 4⇡ cos y ˆ); r = (s R cos ) x ˆ R sin y ˆ zˆ z. x ˆ y ˆ ˆ z sin cos 0 K ⇥ r = K = (s R cos ) ( R sin ) ( z) K [( z cos ) x ˆ + ( z sin ) y ˆ + (R s cos ) ˆ z] ; r 2 = z 2 + R2 + s2 2Rs cos . The x and y components integrate to zero (z integrand is odd, as in Prob. 5.18). ⇢Z 1 Z (R s cos ) µ0 KR 2⇡ dz d dz = (R s cos ) d , 2 + d2 )3/2 4⇡ (z 2 + R2 + s2 2Rs cos )3/2 (z 0 1 Z 1 1 dz 2z 2 where d2 ⌘ R2 + s2 2Rs cos . Now = p = 2. 2 2 3/2 2 z 2 + d2 0 d (z + d ) d 1 Z ⇤ µ0 KR 2⇡ (R s cos ) 1 ⇥ 2 = d ; (R s cos ) = (R s2 ) + (R2 + s2 2Rs cos ) . 2 2 2⇡ (R + s 2Rs cos ) 2R 0  Z 2⇡ Z 2⇡ µ0 K d = (R2 s2 ) + d . 4⇡ (R2 + s2 2Rs cos ) 0 0

Bz =

µ0 KR 4⇡

Z


124 Z

2⇡

0

d a + b cos

=2

Z

2Rs, so a2

d a + b cos

0

= p b=

⇡


4 a2

b2

tan

=p

1

"p

4 a2

a2

b2

tan

1

"p

a2

#

b2 tan( /2) a+b

tan(⇡/2) 4 =p 2 a+b a b2 b2

#

⇣⇡⌘

⇡ 0

=p

2

2⇡ a2

b2 = R4 + 2R2 s2 + s4 4R2 s2 = R4 2R2 s2 + s4 = (R2 s2 )2 ;  ✓ ◆ µ0 K (R2 s2 ) µ0 K R2 s2 Bz = 2⇡ + 2⇡ = + 1 . 4⇡ |R2 s2 | 2 |R2 s2 |

b2

p a2

. Here a = R2 + s2 , b2 = |R2

d2 |.

µ0 K µ0 K (1+1) = µ0 K. Outside the solenoid, s > R, so Bz = ( 1+1) = 0. 2 2 Here K = nI, so B = µ0 nI ˆ z(inside), and 0(outside) (as we found more easily using Ampére’s law, in Ex. 5.9). Inside the solenoid, s < R, so Bz = Problem 5.47 µ0 IR2 (a) From Eq. 5.41, B = 2 @B µ0 IR2 = @z 2

@B @z

z=0

(

(

1 3/2

[R2 + (d/2 + z)2 ]

( 3/2)2(d/2 + z)

+

1 3/2

[R2 + (d/2

( 3/2)2(d/2

z)( 1)

z)2 ]

)

.

)

+ 5/2 5/2 [R2 + (d/2 + z)2 ] [R2 + (d/2 z)2 ] ) ( 3µ0 IR2 (d/2 + z) (d/2 z) = + . 5/2 5/2 2 [R2 + (d/2 + z)2 ] [R2 + (d/2 z)2 ] ( ) 3µ0 IR2 d/2 d/2 = + = 0. X 5/2 5/2 2 [R2 + (d/2)2 ] [R2 + (d/2)2 ]

(b) Di↵erentiating again: @2B 3µ0 IR2 n 1 (d/2 + z)( 5/2)2(d/2 + z) = + 2 5/2 7/2 2 2 @z 2 [R + (d/2 + z) ] [R2 + (d/2 + z)2 ] 1 (d/2 z)( 5/2)2(d/2 z)( 1) o + + . 5/2 7/2 [R2 + (d/2 z)2 ] [R2 + (d/2 z)2 ] ( ) ✓ @2B 3µ0 IR2 2 2(5/2)2(d/2)2 2 3µ0 IR2 = + = R2 5/2 7/2 7/2 @z 2 z=0 2 [R2 + (d/2)2 ] [R2 + (d/2)2 ] [R2 + (d/2)2 ] =

3µ0 IR2

d2 5d2 + 4 4

◆

d2 R2 . Zero if d = R, in which case 7/2 [R2 + (d/2)2 ] ( ) µ0 IR2 1 1 1 8µ0 I B(0) = + = µ0 IR2 = 3/2 . 3/2 3/2 2 3/2 2 (5R /4) 5 R [R2 + (R/2)2 ] [R2 + (R/2)2 ] Problem 5.48 The total charge on the shaded ring is dq = (2⇡r) dr. The time for one revolution is dt = 2⇡/!. So the current in the dq ring is I = = !r dr. From Eq. 5.41, the magnetic field of dt c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

125


µ0 r2 !r 2 dr ˆ z, and the total field of the disk is 2 (r + z 2 )3/2

this ring (for points on the axis) is dB = Z

R

r3 dr ˆ z. Let u ⌘ r2 , so du = 2r dr. Then 2 2 3/2 0 (r + z )  ✓ ◆ R2  Z 2 µ0 ! R u du µ0 ! u + 2z 2 µ0 ! (R2 + 2z 2 ) p = = 2 p = 4 4 2 (u + z 2 )3/2 u + z2 R2 + z 2 0 0

B =

µ0 ! 2

When z

so

R, the term in square brackets is ✓ ◆✓ ◆ 2z 2 (1 + R2 /2z 2 ) R2 1 R2 3 R4 []= 2z ⇡ 2z 1 + 2 1 + 2z 2 z2 8 z4 z[1 + (R/z)2 ] 1/2 ✓ ◆ ✓ ◆ 2 4 2 4 4 R 3R R R 1R R4 ⇡ 2z 1 + + 1 = 2z = , 2z 2 8 z4 2z 2 4z 4 8 z4 4z 4

2z ˆ z.

B⇡

1

µ0 ! R4 µ0 !R4 ˆ z= ˆ z. 4 2 4z 8z 3

Meanwhile, from Eq. 5.88, the dipole field is Bdip =

⌘ µ0 m ⇣ ˆ , 2 cos ✓ ˆ r + sin ✓ ✓ 4⇡r3

and for points on the +z axis ✓ = 0, r = z, ˆ r=ˆ z, so Bdip =

µ0 m ˆ z. In this case (Problem 5.37a) m = ⇡ !R4 /4, 2⇡z 3

µ0 !R4 ˆ z, in agreement with the approximation. X 8z 3 Problem 5.49 Z µ0 I dl0 ⇥ r B = r 3 ; r = R cos xˆ + (y R sin ) yˆ + z ˆz. (For simplicity I’ll drop the prime on 4⇡ .) r 2 = R2 cos2 + y 2 2Ry sin + R2 sin2 + z 2 = R2 + y 2 + z 2 2Ry sin . The source coordinates (x0 , y 0 , z 0 ) satisfy x0 = R cos ) dx0 = R sin d ; y 0 = R sin ) dy 0 = R cos d ; z 0 = 0 ) dz 0 = 0. So dl0 = R sin d x ˆ + R cos d y ˆ. x ˆ y ˆ ˆ z 0 r R sin d R cos d 0 = (Rz cos d ) x dl ⇥ = ˆ + (Rz sin d ) y ˆ + ( Ry sin d + R2 d ) ˆ z. R cos (y R sin ) z so Bdip =

Bx =

µ0 IRz 4⇡

Z

cos

2⇡

0

d 3/2

(R2 + y 2 + z 2

2Ry sin )

=

µ0 IRz 1 1 p 4⇡ Ry R2 + y 2 + z 2

2⇡

2Ry sin

= 0, 0

since sin = 0 at both limits. The y and z components are elliptic integrals, and cannot be expressed in terms of elementary functions. Bx = 0;

µ0 IRz By = 4⇡

Z

0

sin

2⇡

(R2

+

y2

+

z2

d

µ0 IR ; Bz = 3/2 4⇡ 2Ry sin )

Z

(R

2⇡

0


(R2

+

y2

y sin ) d

+ z2

3/2

2Ry sin )

.

126


Problem 5.50 From the Biot-Savart law, the field of loop #1 is B = F = I2

I

2

dl2 ⇥ B =

µ0 I1 I2 4⇡

I I 1

dl2 ⇥ (dl1 ⇥ rˆ )

r

2

µ0 I1 I2 4⇡

F=

2

⇢I I

µ0 I1 4⇡

I

dl1 ⇥ rˆ

r

1

2

; the force on loop #2 is

. Now dl2 ⇥ (dl1 ⇥ rˆ ) = dl1 (dl2 · rˆ )

rˆ r 2 (dl1 · dl2 )

I

dl1

I

rˆ (dl1 · dl2 ),

so

(dl2 · rˆ )

r

2

The first term is what we want. It remains to show that the second term is zero: ⇥ ⇤ r = (x2 x1 ) xˆ + (y2 y1 ) yˆ + (z2 z1 ) ˆz, so r2 (1/ r ) = @ (x2 x1 )2 + (y2 y1 )2 + (z2 z1 )2 1/2 xˆ @x2 ⇤ 1/2 ⇤ 1/2 @ ⇥ @ ⇥ + (x2 x1 )2 + (y2 y1 )2 + (z2 z1 )2 y ˆ+ (x2 x1 )2 + (y2 y1 )2 + (z2 z1 )2 ˆ z @y2 @z2 ✓ ◆ I I r rˆ rˆ (x2 x1 ) (y2 y1 ) (z2 z1 ) 1 = x ˆ y ˆ ˆ z= = . So · dl2 = r2 · dl2 = 0 (by 3 3 3 3 2 2

r

r

r

Corollary 2 in Sect. 1.3.3). qed Problem 5.51 I µ0 I dl ⇥ rˆ (a) B = r2 = I4⇡ µ0 I d✓ so B = . X 4⇡ r

µ0 I 4⇡

r

I

r

dl ⇥ ˆ r , dl ⇥ ˆ r= 2 r

µ0 I 1 (b) For a circular loop, r = R is constant; B = 4⇡ R (c)

r

r

dl sin ˆ z, dl sin I

d✓ =

= r d✓ (see diagram in the book),

µ0 I , which agrees with Eq. 5.41 (z = 0). 2R

0.8 0.6 0.4 0.2 -0.5

0.5

1.0

1.5

-0.2 -0.4

µ0 I B= 4⇡a (d) B =

µ0 I 4⇡p

Problem 5.52 (a)

Z

2⇡

Z

0

µ0 4⇡

p

0

(1 + e cos ✓) d✓ =

B=

2⇡

Z

 µ0 I 2 3/2 ✓ d✓ = ✓ 4⇡a 3

µ0 I µ0 I (2⇡) = . 4⇡p 2p

J ⇥ rˆ

r

2

d⌧

)

2⇡

0

p µ0 I 2⇡ = . 3a

X

A=

1 4⇡

Z

B ⇥ rˆ

r

2

d⌧.

1 ⇢. For dielectrics (with no free charge), ⇢b = r · P ✏0Z 1 P(r0 ) · rˆ 0 (Eq. 4.12), and the resulting potential is V (r) = r 2 d⌧ . In general, ⇢ = ✏0 r · E (Gauss’s law), 4⇡✏0 (b) Poisson’s equation (Eq. 2.24) says r2 V =


127


Z 1 E(r0 ) · rˆ 0 r 2 d⌧ . qed 4⇡ [There are many other ways to obtain this result. For example, using Eq. 1.100: ✓ ◆ ✓ ◆ rˆ rˆ 0 r· = r · = 4⇡ 3 ( r ) = 4⇡ 3 (r r0 ), 2 2

so the analogy is P !

✏0 E, and hence V (r) =

r

V (r) =

Z

V (r ) (r r ) d⌧ = 0

0

3

0

(Eq. 1.59). But r0 V (r0 ) =

1 4⇡

r

Z

0

V (r )r · 0

✓

rˆ r2

◆

d⌧ 0 =

1 4⇡

Z

rˆ ⇥ 0 0 ⇤ 0 r 2 · r V (r ) d⌧

E(r0 ), and the surface integral ! 0 at 1, so V (r) =

before. You can also check the result, by computing its gradient—but it’s not easy.]

1 4⇡

1 4⇡ Z

I

V (r0 )

rˆ 0 r 2 ·da

E(r0 ) · rˆ

d⌧ 0 , as

r

2

Problem 5.53 Rr Rr (a) For uniform B, 0 (B ⇥ dl) = B ⇥ 0 dl = B ⇥ r 6= A = 12 (B ⇥ r). ✓ ◆ ✓ ◆ I µ0 I ˆ µ0 I µ0 I µ0 Iw 1 1 (b) B = , so B ⇥ dl = ˆ s ˆ s w= ˆ s 6= 0. 2⇡s 2⇡a 2⇡b 2⇡ a b R1 1 (c) A = r ⇥ B 0 d = 2 (r ⇥ B). Z 1 µ0 I ˆ µ0 I ˆ µ0 I 1 µ0 I (d) B = ; B( r) = ; A = (r ⇥ ˆ) d = (r ⇥ ˆ). But r here is the 2⇡s 2⇡ s 2⇡s 2⇡s 0 i µ0 I h vector from the origin—in cylindrical coordinates r = s ˆ s + zˆ z. So A = s(ˆ s ⇥ ˆ) + z(ˆ z ⇥ ˆ) , and 2⇡s µ0 I ˆ ˆ (ˆ s⇥ )=ˆ z, (ˆ z⇥ )= ˆ s. So A = (z ˆ s sˆ z. 2⇡s The examples in (c) and (d) happen to be divergenceless, but this is not in general the case. For (letting R1 L ⌘ 0 B( r) d , for short) r · A = r · (r ⇥ L) = [L · (r ⇥ r) r · (r ⇥ L)] = r · (r ⇥ L), and R1 R1 R1 R1 r ⇥ L = 0 [r ⇥ B( r)] d = 0 2 [r ⇥ B( r)] d = µ0 0 2 J( r) d , so r · A = µ0 r · 0 2 J( r) d , and it vanishes in regions where J = 0 (which is why the examples in (c) and (d) were divergenceless). To construct an explicit counterexample, we need the field at a point where J 6= 0—say, inside a wire with uniform current. µ0 J ˆ Here Ampére’s law gives B 2⇡s = µ0 Ienc = µ0 J⇡s2 ) B = s , so 2 Z

1

✓

µ0 J 2

◆

s ˆd = r⇥ 0  µ0 J 1 @ 2 @ r·A = (s z) + ( s2 ) = 6 s @s @z A =

µ0 J µ0 Js s(r ⇥ ˆ) = (z ˆ s sˆ z). 6 6 ✓ ◆ µ0 J 1 µ0 Jz 2sz = 6= 0. 6 s 3

Conclusion: (ii) does not automatically yield r · A = 0. Problem 5.54 (a) Exploit the analogy with the electrical case:

1 1 1 p·ˆ r [3(p · ˆ r) ˆ r p] (Eq. 3.104) = rV, with V = 4⇡✏0 r3 4⇡✏0 r2 µ0 1 B = [3(m · ˆ r) ˆ r m] (Eq. 5.89) = rU, (Eq. 5.67). 4⇡ r3 E=


(Eq. 3.102).

128


µ0 m · ˆ r . 2 4⇡ r (b) Comparing Eqs. 5.69 and 5.87, the dipole moment of the shell is m = (4⇡/3)! R4 ˆ z (which we also got

Evidently the prescription is p/✏0 ! µ0 m : U (r) =

µ0 ! R4 cos ✓ for r > R. 3 r2 Inside the shell, the field is uniform (Eq. 5.70): B = 23 µ0 !R ˆ z, so U (r) = 23 µ0 !Rz + constant. We may

in Prob. 5.37). Using the result of (a), then, U (r) = as well pick the constant to be zero, so U (r) =

2 3 µ0

!Rr cos ✓ for r < R.

[Notice that U (r) is not continuous at the surface (r = R): Uin (R) = 23 µ0 !R2 cos ✓ 6= Uout (R) = !R2 cos ✓. As I warned you on p. 245: if you insist on using magnetic scalar potentials, keep away from places where there is current!] (c) ✓ ◆ ✓ ◆ µ0 !Q 3r2 6r2 @U 1 @U ˆ 1 @U ˆ B = 1 cos ✓ ˆ r 1 sin ✓ ✓ˆ = rU = ˆ r ✓ . 4⇡R 5R2 5R2 @r r @✓ r sin ✓ @ @U = 0 ) U (r, ✓, ) = U (r, ✓). @ ✓ ◆✓ ◆ ✓ ◆✓ ◆ 1 @U µ0 !Q 6r2 µ0 !Q 6r2 = 1 sin ✓ ) U (r, ✓) = 1 r cos ✓ + f (r). r @✓ 4⇡R 5R2 4⇡R 5R2 ✓ ◆✓ ◆ ✓ ◆✓ ◆ @U µ0 !Q 3r2 µ0 !Q r3 = 1 cos ✓ ) U (r, ✓) = r cos ✓ + g(✓). @r 4⇡R 5R2 4⇡R 5R2 1 3 µ0

Equating the two expressions: ✓ ◆✓ µ0 !Q 1 4⇡R or

6r2 5R2

◆

r cos ✓ + f (r) =

✓

µ0 !Q 4⇡R3

◆

✓

µ0 !Q 4⇡R

◆✓ 1

r2 5R2

◆

r cos ✓ + g(✓),

r3 cos ✓ + f (r) = g(✓).

But there is no way to write r3 cos ✓ as the sum of a function of ✓ and a function of r, so we’re stuck. The reason is that you can’t have a scalar magnetic potential in a region where the current is nonzero. Problem 5.55 Z µ0 J 0 (a) r · B = 0, r ⇥ B = µ0 J, and r · A = 0, r ⇥ A = B ) A = r d⌧ , so Z 4⇡ 1 B 0 r · A = 0, r ⇥ A = B, and r · W = 0 (we’ll choose it so), r ⇥ W = A ) W = r d⌧ . 4⇡ (b) W will be proportional to B and to two factors of r (since di↵erentiating twice must recover B), so I’ll try something of the form W = ↵r(r · B) + r2 B, and see if I can pick the constants ↵ and in such a way that r · W = 0 and r ⇥ W = A. ⇥ ⇤ @x @y @z r · W = ↵ [(r · B)(r · r) + r · r(r · B)] + r2 (r · B) + B · r(r2 ) . rr = + + = 1 + 1 + 1 = 3; @x @y @z r(r · B) = r ⇥ (r ⇥ B) + B ⇥ (r ⇥ r) + (r · r)B + (B · r)r; but B is constant, so all derivatives of B vanish, and r ⇥ r = 0 (Prob. 1.63), so ✓ ◆ @ @ @ r(r · B) = (B · r)r = Bx + By + Bz (x x ˆ+yy ˆ +zˆ z) = Bx x ˆ + By y ˆ + Bz ˆ z = B; @x◆ @y @z ✓ @ @ @ r(r2 ) = x ˆ +y ˆ +ˆ z (x2 + y 2 + z 2 ) = 2x x ˆ + 2y y ˆ + 2z ˆ z = 2r. So @x @y @z c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

129


r · W = ↵ [3(r · B) + (r · B)] + [0 + 2(r · B)] = 2(r · B)(2↵ + ), which is zero if 2↵ + = 0. ⇥ ⇤ r ⇥ W = ↵ [(r · B)(r ⇥ r) r ⇥ r(r · B)] + r2 (r ⇥ B) B ⇥ r(r2 ) = ↵ [0 (r ⇥ B)] + [0 2(B ⇥ r)] 1 = (r ⇥ B)(↵ 2 ) = (r ⇥ B) (Prob. 5.25). So we want ↵ 2 = 1/2. Evidently ↵ 2( 2↵) = 5↵ = 1/2, 2 ⇤ 1 ⇥ or ↵ = 1/10; = 2↵ = 1/5. Conclusion: W = r(r · B) 2r2 B . (But this is certainly not unique.) 10 R R H (c) R r ⇥ W = A ) (r ⇥ W) · da = A · da. Or W · dl = A · da. Integrate around the amperian loop shown, taking W to point parallel to the axis, and choosing W = 0 on the axis: ◆ Z s✓ µ0 nI µ0 nI s2 l Wl = l¯ s d¯ s= (using Eq. 5.72 for A). 2 2 2 0 µ0 nIs2 ˆ z (s < R). 4 ◆ Z s✓ µ0 nIR2 l µ0 nI R2 µ0 nIR2 l µ0 nIR2 l For s > R, W l = + l d¯ s= + ln(s/R); 4 2 s¯ 4 2 R W=

W=

µ0 nIR2 [1 + 2 ln(s/R)] ˆ z (s > R). 4

Problem 5.56 Apply the divergence theorem to the function [U ⇥ (r ⇥ V)], noting (from the product rule) that r · [U ⇥ (r ⇥ V)] = (r ⇥ V) · (r ⇥ U) U · [r ⇥ (r ⇥ V)]: Z Z I r · [U ⇥ (r ⇥ V)] d⌧ = {(r ⇥ V) · (r ⇥ U) U · [r ⇥ (r ⇥ V)]} d⌧ = [U ⇥ (r ⇥ V)] · da.

As always, suppose we have two solutions, B1 (and A1 ) and B2 (and A2 ). Define B3 ⌘ B2 B1 (and A3 ⌘ A2 A1 ), so that r ⇥ A3 = B3 and r ⇥ B3 = r ⇥ B1 r ⇥ B2 = µ0 J µ0 J = 0. Set U = V = A3 in Z the above identity: Z Z {(r ⇥ A3 ) · (r ⇥ A3 ) A3 · [r ⇥ (r ⇥ A3 )]} d⌧ = {(B3 ) · (B3 ) A3 · [r ⇥ B3 ]} d⌧ = (B3 )2 d⌧ I I = [A3 ⇥ (r ⇥ A3 )] · da = (A3 ⇥ B3 ) · da. But either A is specified (in which case A3 = 0), or else B is Z H specified (in which case B3 = 0), at the surface. In either case (A3 ⇥ B3 ) · da = 0. So (B3 )2 d⌧ = 0, and hence B1 = B2 . Problem 5.57

qed

µ0 m0 ˆ ThereFrom Eq. 5.88, Btot = B0 ˆ z (2 cos ✓ ˆ r + sin ✓ ✓). 4⇡r3 ⇣ µ0 m0 µ0 m0 ⌘ fore B · ˆ r = B0 (ˆ z·ˆ r) 2 cos ✓ = B0 cos ✓. 3 4⇡r 2⇡r3 µ0 m0 This is zero, for all ✓, when r = R, given by B0 = , or 2⇡R3 ✓ ◆1/3 µ0 m0 R= . Evidently no field lines cross this sphere. 2⇡B0 c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

130


Problem 5.58 Q Q! Q! Q (a) I = = ; a = ⇡R2 ; m = ⇡R2 ˆ z = !R2 ˆ z. L = RM v = M !R2 ; L = M !R2 ˆ z. (2⇡/!) 2⇡ 2⇡ 2 ✓ ◆ m Q !R2 Q Q Q = = . m = L, and the gyromagnetic ratio is g = . 2 L 2 M !R 2M 2M 2M (b) Because g is independent of R, the same ratio applies to all “donuts”, and hence to the entire sphere Q (or any other figure of revolution): g = . 2M (c) m =

e ~ e~ (1.60 ⇥ 10 19 )(1.05 ⇥ 10 = = 2m 2 4m 4(9.11 ⇥ 10 31 )

34

)

= 4.61 ⇥ 10

24

A m2 .

⇡

cos ✓ sin ✓ p d✓. R2 + (z 0 )2 2Rz 0 cos ✓

Problem 5.59

Z Z 1 3 (a) Bave = B d⌧ = (r ⇥ A) d⌧ = 3 (3/4)⇡R3 ⇢Z I I4⇡R 3 3 µ0 J A ⇥ da = d⌧ 0 ⇥ da = 3 4⇡ r 4⇡R3 4⇡R ⇢I Z 3µ0 1 0 J⇥ r da d⌧ . Note that J depends on (4⇡)2 R3 the source point r0 , not on the field point r. To do the surface 0 integral, choose the (x, y, z) coordinates p so that r lies on the 2 0 2 z axis (see diagram). Then r = R + (z ) 2Rz 0 cos ✓, 2 while da = R sin ✓ d✓ d ˆ r. By symmetry, the x and y components must integrate to zero; since the z component of ˆ r is cos ✓, we have I

1

r

da = ˆ z

= = = =

=

Z

p

cos ✓

R2 sin ✓ d✓ d = 2⇡R2 ˆ z

Z

R2 + (z 0 )2 2Rz 0 cos ✓ 0 Let u ⌘ cos ✓, so du = sin ✓ d✓. Z 1 u p 2⇡R2 ˆ z du 2 0 R + (z )2 2Rz 0 u 1 ( ) ⇥ ⇤ 1 2 2(R2 + (z 0 )2 ) + 2Rz 0 u p 2 2 0 )2 0u 2⇡R ˆ z R + (z 2Rz 3(2Rz 0 )2 1 o p ⇤ ⇥ ⇤p 2⇡R2 ˆ z n⇥ 2 0 2 0 2 0 2 0 2 + (z 0 )2 0 2 + (z 0 )2 + 2Rz 0 R + (z ) + Rz R 2Rz R + (z ) Rz R 3(Rz 0 )2  ⇥ 2 ⇤ ⇥ 2 ⇤ 2⇡ ˆ z R + (z 0 )2 + Rz 0 |R z 0 | R + (z 0 )2 Rz 0 (R + z 0 ) 0 2 3(z ) 8 9 4⇡ 4⇡ 0 > > > z0 ˆ z= r, (r0 < R); > > > < 3 = 3 > > > > 4⇡ R3 0 4⇡R3 > : ; ˆ z= r , (r0 > R). > 0 2 0 3 3(z ) 3 (r )


131

CHAPTER 5. MAGNETOSTATICS For now we want r0 < R, so Bave = (Eq. 5.90), so Bave =

µ0 2m . 4⇡ R3

3µ0 4⇡ (4⇡)2 R3 3

qed

Z

(J⇥r0 ) d⌧ 0 =

µ0 4⇡R3

Z

(J⇥r0 ) d⌧ 0 . Now m =

1 2

R

(r⇥J) d⌧

◆ Z ✓ Z 3µ0 4⇡ 3 r0 µ0 J ⇥ rˆ 0 0 (b) This time r > R, so Bave = R J ⇥ 0 3 d⌧ = r 2 d⌧ , where (4⇡)2 R3 3 (r ) 4⇡ goes from the source point to the center ( r = r0 ). Thus Bave = Bcen . qed 0

r

now

Problem 5.60 4⇡ (a) Problem 5.53 gives the dipole moment of a shell: m = !R4 ˆ z. Let R ! r, ! ⇢ dr, and integrate: 3 Z R 4⇡ 4⇡ R5 Q 1 m= !⇢ ˆ z r4 dr = !⇢ ˆ z. But ⇢ = , so m = Q!R2 ˆ z. 3 3 3 5 (4/3)⇡R 5 0 (b) Bave =

µ0 2m µ0 2Q! = ˆ z. 4⇡ R3 4⇡ 5R

µ0 m sin ✓ ˆ µ0 Q!R2 sin ✓ ˆ = . 4⇡ r2 4⇡ 5 r2 (d) Use Eq. 5.69, with R ! r¯, ! ⇢ d¯ r, and integrate: (c) A ⇠ =

µ0 !⇢ sin ✓ ˆ A= 3 r2

Z

R

0

r¯4 d¯ r=

µ0 ! 3Q sin ✓ R5 ˆ µ0 Q!R2 sin ✓ ˆ = . 3 4⇡R3 r2 5 4⇡ 5 r2

This is identical to (c); evidently the field is pure dipole, ✓for points ◆ outside the ✓ sphere. ◆ µ0 !Q 3r2 6r2 (e) According to Prob. 5.30, the field is B = 1 cos ✓ ˆ r 1 sin ✓ ✓ˆ . The average 4⇡R 5R2 5R2 obviously points in the z direction, so take the z component of ˆ r (cos ✓) and ✓ˆ ( sin ✓): ✓ ◆ ✓ ◆ Z µ0 !Q 1 3r2 6r2 2 Bave = 1 cos ✓ + 1 sin2 ✓ r2 sin ✓ dr d✓ d 4⇡R (4/3)⇡R3 5R2 5R2 ◆ ✓ 3 ◆ Z ⇡ ✓ 3 3µ0 !Q r 3 R5 R 6 R5 2 = 2⇡ cos ✓ + sin2 ✓ sin ✓ d✓ (4⇡R2 )2 3 5 5R2 3 5 5R2 0 ◆ Z ✓ Z ⇡ 3µ0 !Q 3 ⇡ 16 7 3µ0 !Q 1 2 2 = R cos ✓ + sin ✓ sin ✓ d✓ = 7 + 9 cos2 ✓ sin ✓ d✓ 8⇡R4 75 75 8⇡R 75 0 0 ⇡ µ0 !Q µ0 !Q µ0 !Q = 7 cos ✓ 3 cos3 ✓ = (20) = (same as (b)). X 200⇡R 200⇡R 10⇡R 0 Problem 5.61 The issue (and the integral) is identical to the one in Prob. 3.48. The resolution (as before) is to regard Eq. 5.89 as correct outside an infinitesimal sphere centered at the dipole. Inside this sphere the field is a delta-function, A 3 (r), with A selected so as to make the average field consistent with Prob. 5.59: Bave =

1 (4/3)⇡R3

Z

A 3 (r) d⌧ =

3 µ0 2m 2µ0 m 2µ0 A= )A= . The added term is m 3 (r). 3 3 4⇡R 4⇡ R 3 3


132


Problem 5.62 For a dipole at the origin and a field point in the x z plane ( = 0), we have µ0 m ˆ = µ0 m [2 cos ✓(sin ✓ x (2 cos ✓ ˆ r + sin ✓ ✓) ˆ + cos ✓ ˆ z) + sin ✓(cos ✓ x ˆ 4⇡ r3 4⇡ r3 µ0 m = [3 sin ✓ cos ✓ x ˆ + (2 cos2 ✓ sin2 ✓) ˆ z]. 4⇡ r3

B =

Here we have a stack of such dipoles, running from z = L/2 to z = +L/2. Put the field point at s on the x axis. The x ˆ components cancel (because of symmetrically placed dipoles above and below z = 0), leaving B = Z L/2 (3 cos2 ✓ 1) µ0 2M ˆ z dz, where M is the dipole mo4⇡ r3 0 ment per unit length: m = I⇡R2 = ( vh)⇡R2 = !R⇡R2 h ) m s 1 sin3 ✓ M = = ⇡ !R3 . Now sin ✓ = , so 3 = ; z = h r r s3 s s cot ✓ ) dz = d✓. Therefore sin2 ✓ Z ✓m Z ✓m µ0 sin3 ✓ s µ0 !R3 3 2 B = (⇡ !R ) ˆ z (3 cos ✓ 1) 3 d✓ = ˆ z (3 cos2 ✓ 2⇡ s sin2 ✓ 2s2 ⇡/2 ⇡/2 =

µ0 !R3 ˆ z 2s2

cos3 ✓ + cos ✓

✓m

⇡/2

=

µ0 !R3 cos ✓m 1 2s2

cos2 ✓m ˆ z=

s (L/2) But sin ✓m = p , and cos ✓m = p , so B = 2 2 2 s + (L/2) s + (L/2)2

sin ✓ ˆ z)]

1) sin ✓ d✓

µ0 !R3 cos ✓m sin2 ✓m ˆ z. 2s2

µ0 !R3 L ˆ z. 4[s2 + (L/2)2 ]3/2


133

CHAPTER 6. MAGNETIC FIELDS IN MATTER

Chapter 6

Magnetic Fields in Matter Problem 6.1 N = m2 ⇥B1 ; B1 = Problem 6.3µ0 m1 m2 N= (ˆ y⇥ˆ z) = 4⇡ r3 downward ( ˆ z).

µ0 1 [3(m1 ·ˆ r)ˆ r 4⇡ r3

m1 ] ; ˆ r=y ˆ; m1 = m1ˆ z; 1m2 = m2 y ˆ . B1 =

µ0 m1 m2 x ˆ. Here m1 = ⇡a2 I, m2 = b2 I. So N = 4⇡ r3

µ0 (abI)2 x ˆ. 4 r3

µ0 m1 ˆ z. 4⇡ r3 Final orientation :

Problem 6.2 dF = I dl⇥B; dN = r⇥dF = I r⇥(dl⇥B). Now (Prob. 1.6): r⇥(dl⇥B) + dl⇥(B⇥r) + B⇥(r⇥dl) = 0. But d [r⇥(r⇥B)] = dr⇥(r⇥B) + r⇥(dr⇥B) (since B is constant), and dr = dl, so dl⇥(B⇥r) = r⇥(dl⇥B) 1 d [r⇥(r⇥B)]. dN H Hence 2r⇥(dl⇥B) H = d [r⇥(r⇥B)] B⇥(r⇥dl). H = 2 I {d [r⇥(r⇥B)] B⇥(r⇥dl)}. 1 ) N = 2 I d [r⇥(r⇥B)] B⇥ (r⇥dl) . But the first term is zero ( d(· · · ) = 0), and the second integral is 2a (Eq. 1.107). So N = I(B⇥a) = m⇥B. qed Problem 6.3 ˆ z✻ (a) I

ˆ r B ✕✸ θ✲ y ˆ

R

❥

According to Eq. 6.2, F = 2⇡IRB cos ✓. But B = r)ˆ r m1 ] µ0 [3(m1 ·ˆ , and B cos ✓ = B · y ˆ, so B cos ✓ = 4⇡ r3 µ0 1 [3(m · ˆ r )(ˆ r · y ˆ ) (m · y ˆ )]. But m1 · y ˆ = 0 and 3 1 1 4⇡ r ˆ r·y ˆ = sin , while m1 · ˆ r = m1 cos ✓. ) B cos ✓ = µ0 1 4⇡ r 3 3m1 sin cos .

r φ m1 ✻ µ0 1 F = 2⇡IR 4⇡ r 3 3m1 sin cos . Now sin

p

R2 /r, so F = 3 µ20 m1 IR2 p 3µ0 m1 m2 r 2 R2 0 But IR2 ⇡ = m2 , so F = 3µ , while for a dipole, R ⌧ r, so F = . 2⇡ m1 m2 r5 2⇡ r4 ⇥ ⇤ µ0 1 µ0 d d 1 (b) F = r(m2 ·B) = (m2 ·r)B = m2 dz z)ˆ z m1 ) = 2⇡ m1 m2ˆ z dz 3 , 1 ·ˆ 4⇡ z 3 (3(m | {z } | {zz } 2m1 3 z14

Problemor, 6.5 since z = r: F =

=

R r

, cos

3µ0 m1 m2 ˆ z. 2⇡ r4

=

r2

z

✻

! J = J0ˆ z


m ❂ ❂ x

✲ y

p

r 2 R2 . r5

1 Problem 6.3

134


Problem 6.4 dF = I {(dy y ˆ)⇥B(0, y, 0) + (dz ˆ z)⇥B(0, ✏, z) (dy y ˆ)⇥B(0, y, ✏) (dz ˆ z)⇥B(0, 0, z)} n o =I (dy y ˆ)⇥ [B(0, y, ✏) B(0, y, 0)] +(dz ˆ z)⇥ [B(0, ✏, z) B(0, 0, z)] | {z } | {z } @B @B ⇡ˆ ✏ ⇡ ✏ z✻ ˆ r @z @y B ⇢  ✕ ✸ R @B @B R θ✲ R @B y ˆ dy @B I y ) I✏2 ˆ z⇥ ˆ⇥ . Note that dz @B ⇡✏ @z 0,y,0 ⇡ ✏ @z 0,0,0 and @y @y ❥ @z 0,0,z ( x ˆ 0 F=m

@B @y

0,0,0

.

r x ˆφ y ˆ m 0✻ 1

⇢ ˆ z ) @Bx @By @Bz @Bx 0 =m y ˆ x ˆ x ˆ +ˆ z 1 @y @y @z @z @B @Bx @By @Bz @B @B y z x @y @y @y @z @z @z  ✓ ◆ @Bx @Bx @Bx @By @Bz @Bx =m x ˆ +y ˆ +ˆ z using r·B = 0 to write + = . @x @y @z @y @z @x ⇣ ⌘ @Bx @Bx x But m·B = mBx (since m = mˆ x, here), so r(m·B) = mr(Bx ) = m @B x ˆ + y ˆ + ˆ z . @x @y @z Therefore F = r(m·B). qed y ˆ 0

ˆ z 1

Problem 6.5 Problem 6.5 (a) B = µ0 J0 xˆ y (Prob. 5.15). m·B = 0, so Eq. 6.3 says F = 0.

z

✻

! J = J0ˆ z

(b) m·B = m0 µ0 J0 x, so F = m0 µ0 J0 x ˆ. (c) Use product rule #4: r(p · E) = p ⇥ (r ⇥ E) + E ⇥ (r ⇥ p) + (p · r)E + (E · r)p. But p does not depend on (x, y, z), so the second and fourth terms vanish, and r ⇥ E = 0, so the first term is zero. Hence r(p · E) = (p · r)E. qed ❂ x This argument does not apply to the magnetic analog, since r ⇥ B 6= 0. In fact, r(m · B) = (m · r)B + µ0 (m ⇥ J). @ @ (m·r)Ba = m0 @x (B) = m0 µ0 J0 y ˆ, (m·r)Bb = m0 @y (µ0 J0 xˆ y) = 0.

m ❂

✲ y

Problem 6.6 1 Aluminum, copper, copper chloride, and sodium all have an odd number of electrons, so we expect them to be paramagnetic. The rest (having an even number) should be diamagnetic. Problem Problem 6.7 6.7 ✻ Jb = r⇥M = 0; Kb = M⇥ˆ n = M ˆ. ˆ M ✲n ˆ The field is that of a surface current Kb = M , but that’s just a solenoid, so the field c ⃝2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is outside is zero, and inside B = µ0 Klaws µ0they M . currently Moreover, it points upward (in the drawing), so B = µ0 M. b =as protected under all copyright exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


135

CHAPTER Problem 6.9 6. MAGNETIC FIELDS IN MATTER Problem 6.8

ˆ Kb = M×ˆ n = M φ. ▼ ✿ 1✲@ 1 B2 (Essentially a long solenoid) 2 r⇥M = Jb =M (s ksK )ˆ z = (3ks z = 3ksˆ z, Kb = M⇥ˆ n = ks2 ( ˆ⇥ˆ s) = kR2ˆ z. ③ )ˆ ✎✎✎ s @s ✎ ✎ ✎ s ✌ ✎ 1 So the bound current flows up the cylinder, and returns down the surface. [Incidentally, the total current should R RR R 3 be zero. . . is it? Yes, for Jb da = 0 (3ks)(2⇡s ds) = 2⇡kR , while Kb dl = ( kR2 )(2⇡R) = 2⇡kR3 .] Since these currents Problem 6.9have cylindrical symmetry, we can get the field by Ampère’s law: Z s B · 2⇡s = µ0 Ienc = µ0 Jb da = 2⇡kµ0 s3 ) B = µ0 ks2 ˆ = µ0 M. 0

Outside the cylinder Ienc = 0, so B = 0. Problem 6.9 ▼ ✲ M

✎✎✎

✎✎✎

✿ ③B

K ✌

ˆ Kb = M×ˆ n = M φ. (Essentially a long solenoid)

✎

✻✻ ✶ B '

K

! ❄❄

✛

(Essentially a physical dipole) 3

c Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is ❖ ⃝2005 protected under all copyright laws as they currently exist. No portion of this material may be

reproduced, in any form or by any means, without permission in writing from the publisher. M ✲ ✶ B (Intermediate case) " K [The external fields are the same as in the electrical ✎ ✎ case; the internal fields (inside the bar) are completely ✛ ✎ di↵erent—in fact, opposite in direction.]

Problem 6.10 Kb = M , so the field inside a complete ring would be µ0 M . The field of a square loop, at the center, is p given by Prob. 5.8: Bsq = 2 µ0 I/⇡R. Here I = M w, and R = a/2, so Bsq =

p

p 2 µ0 M w 2 2 µ0 M w = ; ⇡(a/2) ⇡a

net field in gap : B = µ0 M 1

c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be Problem 6.12 reproduced, in any form or by any means, without permission in writing from the publisher.

! p 2 2w . ⇡a

z ✻

.

R✲

Jb ■ c ⃝2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Kb

❃

}

l

136

1 CHAPTER 6. MAGNETIC FIELDS IN MATTER

Problem Problem6.12 6.11 As in Sec. 4.2.3, we want the average of B = Bout + Bin , where Bout is due to molecules outside a small sphere around point P , and Bin is due to molecules inside the sphere. The average of Bout is same as field at center (Prob. 5.59b), and for this it is OK to use Eq. 6.10, since the center is “far” from all the molecules in question: Z µ0 M⇥ rˆ Aout = r 2 d⌧ 4⇡

.

outside

µ0 4⇡

2m 4 3 The average of Bin is R3 —Eq. 5.93—where m = 3 ⇡R M. Thus the average Bin is 2µ0 M/3. But what is left out of the integral Aout is the contribution of a uniformly magnetized sphere, to wit: 2µ0 M/3 (Eq. 6.16), and this is precisely what Bin puts back in. So we’ll get the correct macroscopic field using Eq. 6.10. qed Problem 6.12 z ✻ R✲ (a) M = ksˆ z; Jb = r⇥M = k ˆ; Kb = M⇥ˆ n = kR ˆ. Jb B is in the z direction (this is essentially a superposition of solenoids). So 1 ■ l loop shown (shaded)—inner side at radius s: ⇥R ⇤ HB = 0 outside. Use the amperian Kb B · dl = Bl = µ0 Ienc = µ0 Jb da + Kb l = µ0 [ kl(R s) + kRl] = µ0 kls. ❃ Problem 6.13 ) B = µ0 ksˆ z inside. H (b) By symmetry, H points in the z direction. That same amperian loop gives H·dl = Hl = µ0 Ifenc = 0, since there is no free current here. So H = 0 , and hence B = µ0 M. Outside M = 0, so B = 0; inside M = ksˆ z, so B = µ0 ksˆ z. 1 Problem 6.13 2 2 (a) The field µ0 M, with the sphere removed. Problem 6.13 of a magnetized sphere is 3 µ0 M (Eq. 6.16), so B = B0 3

In the cavity, H = (b)

1 µ0 B,

so H =

1 µ0

B0

2 3 µ0 M

= H0 + M

2 3M

}

1 ) H = H0 + M. 3

The field inside a long solenoid is µ0 K. Here K = M , so the field of the bound current on the inside surface of the cavity is µ0 M , pointing down. Therefore

Kb ▼

B = B0 H=

µ0 M;

1 (B0 µ0

µ0 M) =

1 B0 µ0

M ) H = H0 .

(c) ✸

✰Kb

This time the bound currents are small, and far away from the center, so B = B0 , while H =

1 µ0 B0

= H0 + M ) H = H0 + M.

[Comment: In the wafer, B is the field in the medium; in the needle, H is the H in the medium; in the sphere (intermediate case) both B and H are modified.]

c ⃝2005 Pearson Education, Inc., Upper Saddle NJ.Education, All rights reserved. This material is NJ. All rights reserved. This material is c 2012River, Pearson Inc., Upper Saddle River, protected under all copyright laws as they currently exist.allNo portion of this may beexist. No portion of this material may be protected under copyright laws asmaterial they currently reproduced, in any form or by any means, without permission in writing from the publisher. reproduced, in any form or by any means, without permission in writing from the publisher.

1

137 1

CHAPTER MAGNETIC FIELDS IN MATTER Problem6.6.14 ✲ Problem ✲ ✲ M:

❯

◆

6.14 ✲ ✲ ✲

❄

+ ✲ ✶ ✯

✻

M. ✛

B ✛

H ✛

✛

✛ ✍

1 µ0 B

✛

✻ ✍

✲ ✲ ✲ ✲ ✲ ✲ ✲ ✲ ✲ ✲ ✲ ✲ ✲ ✲ ✲ ✲ ✲

❥

✕

; B is the same as the field of a short solenoid; H =

✕ ✯ ✶ ✲ + ❥

◆ ❄

❯

✛

❯

◆

❄

❥

✶

✛

✛ ✍

✻

(The continuity of W follows from the gradient theorem: W (b) W (a) = if the two points are infinitesimally separated, this last integral ! 0.) ⇢ l (i) ) Al Rl = RBl+1 ) Bl = R2l+1 Al , P P Bl (ii) ) (l + 1) Rl+2 Pl (cos ✓) + lAl Rl 2 Pl (cos ✓) = M cos ✓. Combining these:

◆ ❄

✛ ✛

Rb a

rW · dl =

c ⃝2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is X all copyright laws as they currently exist. No portion of this material may be protected under 1)Rorl by1 Aany (cos ✓)without = M cos ✓, so A = 0 (lfrom 6= 1), 3A1 = reproduced, in (2l any+ form permission in lwriting the and publisher. l Plmeans,

Thus Win (r, ✓) =

❥

✯ ✕


c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is c ⃝2005 Pearson Inc., Upper Saddle River,exist. NJ. All reserved. This material is protected under Education, all copyright laws as they currently No rights portion of this material may be protected under all form copyright they without currentlypermission exist. No portion of from this material may be reproduced, in any or bylaws any as means, in writing the publisher. reproduced, in any form or by any means, without permission in writing from the publisher.

Rb a

M ) A1 =

M M M 1 r cos ✓ = z, and hence Hin = rWin = ˆ z= M, so 3 3 3 3 ✓ ◆ 1 2 B = µ0 (H + M) = µ0 M + M = µ0 M. X 3 3

✕ ✯ ✶ ✲ *

✛

* ✲

✛

Problem 6.15 “Potentials”: ⇢ P Win (r, ✓) = A rl P (cos ✓), (r < R); P Bl l l Wout (r, ✓) = P (cos ✓), (r > R). r l+1 l Boundary Conditions: ⇢ (i) Win (R, ✓) = Wout (R, ✓), ? out in (ii) @W + @W =Mˆ z·ˆ r = M cos ✓. @r @r R = M R

✻ ✍

✛

H · dl;

M . 3

❯

138 Problem 6.16 H H·dl = Ifenc = I, so H =

I 2⇡s


ˆ. B = µ0 (1 +

1 @ s @s

Jb = r⇥M =

✓

s

mI

2⇡s

◆

m )H

ˆ z = 0.

I ˆ mI ˆ . M = mH = . 2⇡s 2⇡s ( mI z, at s = a; 2⇡a ˆ Kb = M⇥ˆ n= mI z, at s = b. 2⇡b ˆ = µ0 (1 +

m)

Total enclosed current, for an amperian loop between the cylinders: I µ0 (1 + m )I ˆ mI I+ 2⇡a = (1 + m )I, so B · dl = µ0 Ienc = µ0 (1 + m )I ) B = .X 2⇡a 2⇡s Problem 6.17 From Eq. 6.20:

H

H·dl = H(2⇡s) = Ifenc H=

Jb =

m Jf

⇢

Is 2⇡a2 , I 2⇡s ,

(Eq. 6.33), and Jf =

Kb = M⇥ˆ n=

n m H⇥ˆ

I ⇡a2 ,

) Kb =

Ib = Jb (⇡a2 ) + Kb (2⇡a) =

mI

(s < a) (s > a)

( I(s2 /a2 ), (s < a); = I (s > a). ( ,

so Jb =

so B = µH =

mI ⇡a2

µ0 (1+ m )Is , 2⇡a2 µ0 I 2⇡s ,

(s < a); (s > a).

(same direction as I).

mI

(opposite direction to I). 2⇡a m I = 0 (as it should be, of course).

Problem 6.18 By the method of Prob. 6.15: For large r, we want B(r, ✓) ! B0 = B0 ˆ z, so H = 1 µ0 B0 r cos ✓.

1 µ0 B

!

1 z, µ0 B0 ˆ

and hence W !

1 µ0 B0 z

=

“Potentials”: P ⇢ Win (r, ✓) = Al rl Pl (cos ✓), (r < R); P Bl 1 Wout (r, ✓) = µ0 B0 r cos ✓ + P (cos ✓), (r > R). l l+1 r

Boundary Conditions: ⇢ (i) Win (R, ✓) = Wout (R, ✓), out in (ii) µ0 @W + µ @W @r @r R = 0. R (The latter follows from Eq. 6.26.)  X X 1 Bl (ii) ) µ0 B0 cos ✓ + (l + 1) l+2 Pl (cos ✓) + µ lAl Rl µ0 R

1

Pl (cos ✓) = 0.

For l 6= 1, (i) ) Bl = R2l+1 Al , so [µ0 (l + 1) + µl]Al Rl 1 = 0, and hence Al = 0. For l = 1, (i) ) A1 R = µ10 B0 R + B1 /R2 , and (ii) ) B0 + 2µ0 B1 /R3 + µA1 = 0, so A1 = Win (r, ✓) =

3B0 /(2µ0 + µ).

3B0 r cos ✓ = (2µ0 + µ)

3B0 z 3B0 3B0 . Hin = rWin = ˆ z= . (2µ0 + µ) (2µ0 + µ) (2µ0 + µ) ✓ ◆ 3µB0 1+ m B = µH = = B0 . (2µ0 + µ) 1 + m /3


139

CHAPTER 6. MAGNETIC FIELDS IN MATTER By the method of Prob. 4.23: Step 1 : B0 magnetizes the sphere: M0 = the sphere given by Eq. 6.16:

m H0

=

m

µ0 (1+

m)

2 2 2 m µ0 M0 = B0 = B0 3 31+ m 3

B1 =

B0 . This magnetization sets up a field within (where  ⌘

m

1+

m

).

Step 2 : B1 magnetizes the sphere an additional amount M1 = µ0 B1 . This sets up an additional field in the sphere: ✓ ◆2 2 2 2 B2 = µ0 M1 = B1 = B0 , etc. 3 3 3 The total field is:

⇥ ⇤ B = B0 + B1 + B2 + · · · = B0 + (2/3)B0 + (2/3)2 B0 + · · · = 1 + (2/3) + (2/3)2 + · · · B0 = 1

1 = 2/3 3

3 2 m /(1 +

Problem 6.19 2 2 m = e4mre B; M =

m V

3+3 = 3+3 m m)

e2 r 2 4me V B, e2 r 2 m = 4me V

=

m

2

m

3(1 + = 3+

m)

, so B =

m

✓

1+ 1+

m m /3

where V is the volume per electron. M =

◆

B0 . (1 2/3)

B0 .

mH

(Eq. 6.29)

= µ0 (1+ m ) B (Eq. 6.30). So µ0 . [Note: m ⌧ 1, so I won’t worry about the (1 + m ) term; for the same reason we need not distinguish B from Belse , as we ⇣ did ⌘in deriving the Clausius-Mossotti µ0 3e2 10 ˚ equation in Prob. 4.41.] Let’s say V = 43 ⇡r3 . Then m = 4⇡ m for r. 4me r . I’ll use 1 A= 10 ⇣ ⌘ 19 2 3(1.6⇥10 ) Then m = (10 7 ) 4(9.1⇥10 = 2 ⇥ 10 5 , which is not bad—Table 6.1 says m = 1 ⇥ 10 5 . 31 )(10 10 ) However, I used only one electron per atom (copper has 29) and a very crude value for r. Since the orbital radius is smaller for the inner electrons, they count for less ( m ⇠ r2 ). I have also neglected competing paramagnetic e↵ects. But never mind. . . this is in the right ball park. Problem 6.20 Place the object in a region of zero magnetic field, and heat it above the Curie point—or simply drop it on a hard surface. If it’s delicate (a watch, say), place it between the poles of an electromagnet, and magnetize it back and forth many times; each time you reverse the direction, reduce the field slightly. Problem 6.21 m

(a) The magnetic force on the dipole is given by Eq. 6.3; to move the dipole in from infinity we must exert an opposite force, so the work done is Z r Z r U= F · dl = r(m · B) · dl = m · B(r) + m · B(1) 1

1

(I used the gradient theorem, Eq. 1.55). As long as the magnetic field goes to zero at infinity, then, U = m·B. If the magnetic field does not go to zero at infinity, one must stipulate that the dipole starts out oriented perpendicular to the field. (b) Identical to Prob. 4.8, but starting with Eq. 5.89 instead of 3.104. (c) U = U=

µ0 1 4⇡ r 3 [3 cos ✓1 cos ✓2 µ0 m1 m2 (sin ✓1 sin ✓2 4⇡ r3

cos(✓2

✓1 )]m1 m2 . Or, using cos(✓2

✓1 ) = cos ✓1 cos ✓2 + sin ✓1 sin ✓2 ,

2 cos ✓1 cos ✓2 ) .


140


Stable position occurs at minimum energy: (

= =

@U @✓1 @U @✓2

µ0 m1 m2 4⇡r 3 (cos ✓1 sin ✓2 µ0 m1 m2 4⇡r 3 (sin ✓1 cos ✓2

@U @✓1

=

@U @✓2

=0

+ 2 sin ✓1 cos ✓2 ) = 0 ) 2 sin ✓1 cos ✓2 = + 2 cos ✓1 sin ✓2 ) = 0 ) 2 sin ✓1 cos ✓2 =

8
1

! ! or " " or 3

cos ✓1 sin ✓2 ; 4 cos ✓1 sin ✓2 . 2

! " # 4

Which of these is the stable minimum? Certainly not 2 or 3 —for these m2 is not parallel to B1 , whereas we know m2 will line up along B1 . It remains to compare 1 (with ✓1 = ✓2 = 0) and 4 (with ✓1 = ⇡/2, ✓2 = ⇡/2): m1 m2 m1 m2 U1 = µ04⇡r ( 2); U2 = µ04⇡r ( 1). U1 is the lower energy, hence the more stable configuration. 3 3 Conclusion: They line up parallel, along the line joining them: (d) They’d line up the same way: Problem 6.22

F=I

I

(because

dl ⇥ B = I H

✓I

! !

! ! ! ! ! !

◆

dl ⇥ B0 + I

I

dl ⇥ [(r · r0 )B0 ]

I

✓I

◆

dl ⇥ [(r0 · r0 )B0 ] = I

I

dl ⇥ [(r · r0 )B0 ]

dl = 0). Now

Fi = I

(dl ⇥ B0 )i =

X

✏ijk

j,k,l

I

X

✏ijk dlj (B0 )k ,

j,k

rl dlj [(r0 )l (B0 )k ]

and (r · r0 ) =

(

Lemma 1 :

I

X

rl dlj =

rl (r0 )l , so

l

X m

)

✏ljm am (proof below).

8 9 < = X =I ✏ijk ✏ljm am (r0 )l (B0 )k Lemma 2 : ✏ijk ✏ljm = il km (proof below). im kl : ; j j,k,l,m X X =I ( il km [ak (r0 )i (B0 )k ai (r0 )k (B0 )k ] im kl ) am (r0 )l (B0 )k = I X

k,l,m

= I [(r0 )i (a · B0 )

k

ai (r0 · B0 )] .

But r0 · B0 = 0 (Eq. 5.50), and m = Ia (Eq. 5.86), so F = r0 (m · B0 ) (the subscript just reminds us to take the derivatives at the point where m is located). qed Proof of Lemma 1: H P H P Eq. 1.108 says (c · r) dl = a ⇥ c = c ⇥ a. The jth component is p cp rp dlj = ✏jpm cp am . Pick p,mP H P cp = pl (i.e. 1 for the lth component, zero for the others). Then rl dlj = m ✏jlm am = m ✏ljm am . qed Proof of Lemma 2: ✏ijk ✏ljm = 0 unless ijk and ljm are both permutations of 123. In particular, i must either be l or m, and k must be the other, so X ✏ijk ✏ljm = A il km + B im kl . j


141


To determine the constant A, pick i = l = 1, k = m = 3; the only contribution comes from j = 2: ✏123 ✏123 = 1 = A

11 33

+B

13 31

= A ) A = 1.

13 31

+B

11 33

=B)B=

To determine B, pick i = m = 1, k = l = 3: ✏123 ✏321 = So

X

1=A

✏ijk ✏ljm =

qed

im kl .

il km

1.

j

Problem 6.23 (a) B1 =

µ0 2m z 4⇡ z 3 ˆ

2

3µ0 m z. This 2⇡z 4 ˆ ( md gˆ z):

(Eq. 5.88, with ✓ = 0). So m2 ·B1 =

µ0 m2 2⇡ z 3 .

F = r(m·B) (Eq. 6.3) ) F =

@ @z

h

µ0 m2 2⇡ z 3

i

ˆ z=

is the magnetic force upward (on the upper magnet); it balances the gravitational force downward 3µ0 m2 2⇡z 4

md g = 0 ) z =



1/4

3µ0 m2 2⇡md g

.

(b) The middle magnet is repelled upward by lower magnet and downward by upper magnet: 3µ0 m2 2⇡x4

3µ0 m2 2⇡y 4

md g = 0.

The top magnet is repelled upward by middle magnet, and attracted downward by lower magnet:

Subtracting:

3µ0 m2 2⇡

h

1 x4

Let ↵ ⌘ x/y; then 2 =

1 ↵4

+

1 y4

1 y4

+

1 (↵+1)4 .

3µ0 m2 2⇡y 4 i

1 (x+y)4

3µ0 m2 2⇡(x + y)4

md g = 0.

md g +md g = 0, or

1 x4

2 y4

1 + (x+y) 4 = 0, so: 2 =

1 (x/y)4

1 + (x/y+1) 4.

Mathematica gives the numerical solution ↵ = x/y = 0.850115 . . .

Problem 6.24 (a) Forces on the upper charge: 1 q2 Fq = ˆ z, 4⇡✏0 z 2 At equilibrium,

Fm

✓

2µ0 m = r(m · B) = r m 4⇡z 3

1 q2 3µ0 m2 = 2 4⇡✏0 z 2⇡z 4

) z2 =

6µ0 ✏0 m2 q2

)

◆

µ0 m2 = 2⇡

z=

✓

3 z4

◆

ˆ z.

p m 6 , qc

p where 1/ ✏0 µ0 = c, the speed of light. (b) For electrons, q = 1.6 ⇥ 10 19 C (actually, it’s the magnitude of the charge we want in the expression above), and m = 9.22 ⇥ 10 24 A m2 (the Bohr magneton—see Problem 5.58), so z=

p

6

9.22 ⇥ 10 24 = 4.72 ⇥ 10 (1.6 ⇥ 10 19 )(3 ⇥ 108 )

13


m.

142


(For comparison, the Bohr radius is 0.5 ⇥ 10 10 m, so the equilibrium separation is about 1% of the size of a hydrogen atom.) (c) Good question! Certainly the answer is no. Presumably this is an unstable equilibrium, so unless you could find a way to maintain the orientation of the dipoles, and keep them on the z axis, the structure would fall apart. Problem 6.25 (a) The electric field inside a uniformly polarized sphere, E = 3✏10 P (Eq. 4.14) translates to H = 1 3 M. But B = µ0 (H+M). So the magnetic field inside a uniformly magnetized sphere is B = µ0 ( 2 µ0 M (same as Eq. 6.16). 3

1 3µ0 (µ0 M) 1 3 M+M)

= =

(b) The electric field inside a sphere of linear dielectric in an otherwise uniform electric field is E = 1+ 1e /3 E0 (Eq. 4.49). Now e translates to m , for then Eq. 4.30 (P = ✏0 e E) goes to µ0 M = µ0 m H, or M = m H (Eq. 6.29). So Eq. 4.49 ) H = 1+ 1m /3 H0 . But B = µ0 (1 + m )H, and B0 = µ0 H0 (Eqs. 6.31 and 6.32), so the magnetic field inside a sphere of linear magnetic material in an otherwise uniform magnetic field is ✓ ◆ B 1 B0 1+ m = , or B = B0 (as in Prob. 6.18). µ0 (1 + m ) (1 + m /3) µ0 1 + m /3 p 1 (c) The average electric field over a sphere, due to charges within, is Eave = 4⇡✏ 3 . Let’s pretend the charges 0 R are all due to the frozen-in polarization of some medium (whatever ⇢ might be, we can solve r·PR= ⇢ to find R 1 1 the appropriate P). In this case there are no free charges, and p = P d⌧ , so Eave = 4⇡✏ P d⌧ , which 3 0 R translates to Z 1 1 1 Have = µ0 M d⌧ = m. 3 4⇡µ0 R 4⇡R3

µ0 2m , in agreement 4⇡ R3 with Eq. 5.93. (We must assume for this argument that all the currents are bound, but again it doesn’t really matter, since we can model any current configuration by an appropriate frozen-in magnetization. See G. H. Goedecke, Am. J. Phys. 66, 1010 (1998).) But B = µ0 (H + M), so Bave =

µ0 m 4⇡ R3

+ µ0 Mave , and Mave =

m

4 3 3 ⇡R

, so Bave =

Problem 6.26 n o R rˆ 1 0 Eq. 2.15 : E = ⇢ 4⇡✏ d⌧ (for uniform charge density); 2 n 0 VR r o ˆ r 2 d⌧ 0 1 Eq. 4.9 : V = P · 4⇡✏ (for uniform polarization); 0 nV r R o ˆ r 1 0 Eq. 6.11 : A = µ0 ✏0 M ⇥ 4⇡✏0 V r 2 d⌧ (for uniform magnetization). 8 ⇣ ⌘ < Ein = ⇢ 1 r (Prob. 2.12), ⇣ 3✏0 3 ⌘ For a uniformly charged sphere (radius R): : Eout = ⇢ 1 R2 ˆ 3✏0 r r (Ex. 2.3). ( Vin = 3✏10 (P · r), 3 So the scalar potential of a uniformly polarized sphere is: Vout = 3✏10 Rr2 (P · ˆ r), ⇢ µ0 Ain = 3 (M ⇥ r), 3 and the vector potential of a uniformly magnetized sphere is: Aout = µ30 Rr2 (M ⇥ ˆ r), (confirming the results of Ex. 4.2 and of Exs. 6.1 and 5.11).


143


Problem 6.27 At the interface, the perpendicular component of B is continuous (Eq. 6.26), and the parallel component of k k k k H is continuous (Eq. 6.25 with Kf = 0). So B1? = B2? , H1 = H2 . But B = µH (Eq. 6.31), so µ11 B1 = µ12 B2 . k

k

Now tan ✓1 = B1 /B1? , and tan ✓2 = B2 /B2? , so k

k

tan ✓2 B2 B1? B µ2 = ? = 2k = k tan ✓1 µ1 B2 B1 B1 (the same form, though for di↵erent reasons, as Eq. 4.68). Problem 6.28 In view of Eq. 6.33, there is a bound dipole at the center: mb = m m. So the net dipole moment at the center is mcenter = m + mb = (1 + m )m = µµ0 m. This produces a field given by Eq. 5.89: Bcenter = dipole

µ 1 [3(m·ˆ r)ˆ r 4⇡ r3

m] .

This accounts for the first term in the field. The remainder must be due to the bound surface current (Kb ) at r = R (since there can be no volume bound current, according to Eq. 6.33). Let us make an educated guess (based either on the answer provided or on the analogous electrical Prob. 4.37) that the field due to the surface bound current is (for interior points) of the form B surface = Am (i.e. a constant, proportional to m). In that current

case the magnetization will be: M=

mH

=

m

µ

B=

1 [3(m·ˆ r)ˆ r 4⇡ r3 m

m] +

m

µ

Am.

This will produce bound currents Jb = r⇥M = 0, as it should, for 0 < r < R (no need to calculate this curl—the second term is constant, and the first is essentially the field of a dipole, which we know is curl-less, except at r = 0), and ✓ ◆ 1 A m mA Kb = M(R)⇥ˆ r= ( m⇥ˆ r) + (m⇥ˆ r) = m m + sin ✓ ˆ. 4⇡R3 µ 4⇡R3 µ ˆ But this ⇣ is exactly ⌘ the surface current produced by a spinning sphere: K = v = !R sin ✓ , with ( !R) $ A 1 mm µ 4⇡R3 . So the field it produces (for points inside) is (Eq. 5.70): ◆ 1 . 4⇡R3 current ⇣ ⌘ ⇣ 1 Everything is consistent, therefore, provided A = 23 µ0 m A , or A 1 3 µ ⇣ ⌘ ⇣ ⌘ ⇣ ⌘4⇡R (µ µ ) (µ µ) µ µ 2µ 0 1, so A 1 23 + 23 µ0 = 23 4⇡R30 , or A 1 + µ0 = 2 4⇡R 3 ; A = m = µ0 B surface =

B=

µ 4⇡

⇢

2 2 µ0 ( ! R) = µ0 3 3

1 [3(m·ˆ r)ˆ r r3

m] +

mm

✓

A µ

2(µ0 µ)m R3 (2µ0 + µ)

2µ0 3µ

m

⌘

=

µ 2(µ0 µ) 4⇡ R3 (2µ0 +µ) ,

2 µ0 m 3 4⇡R3 .

But

and hence

. qed

The exterior field is that of the central dipole plus that of the surface current, which, according to Prob. 5.37, is also a perfect dipole field, of dipole moment ✓ ◆ 4 4 3 2⇡R3 µ 2(µ0 µ)m µ(µ0 µ)m m surface = ⇡R3 ( ! R) = ⇡R3 B surface = = . 3 3 3 2µ0 current µ0 4⇡ R (2µ0 + µ) µ0 (2µ0 + µ) current c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

144


So the total dipole moment is: mtot =

µ µ (µ0 µ) 3µm m+ m = , µ0 µ0 (2µ0 + µ) (2µ0 + µ)

and hence the field (for r > R) is µ0 B= 4⇡

✓

3µ 2µ0 + µ

◆

1 [3(m·ˆ r)ˆ r r3

m] .

Problem 6.29 The problem is that the field inside a cavity is not the same as the field in the material itself. (a) 8 Ampére type. The 2field deep1inside the magnet is that of a long solenoid, B0 ⇡ µ0 M. From Prob. 6.13: < Sphere : B = B0 3 µ0 M = 3 µ0 M; Needle : B = B0 µ0 M = 0; : Wafer : B = µ0 M.

(b) Gilbert type. This is analogous to the electric case. The field at the center is approximately that midway between B0 ⇡ 0. From Prob. 4.16 (with E ! B, 1/✏0 ! µ0 , P ! M): 8 two distant pointµcharges, < Sphere : B = B0 + 30 M = 13 µ0 M; Needle : B = B0 = 0; : Wafer : B = B0 + µ0 M = µ0 M. In the cavities, then, the fields are the same for the two models, and this will be no test at all. Yes. Fund it with $1 M from the Office of Alternative Medicine.


145

CHAPTER 7. ELETRODYNAMICS

Chapter 7

Electrodynamics Problem 7.1 (a) Let Q be the charge on the inner shell. Then E = Ra Ra Q 1 1 1 E·dr = 4⇡✏ Q b r12 dr = 4⇡✏ a b . b 0 0 I=

(b) R =

Va

Vb I

1 = 4⇡

Z

Z

J·da = ✓

1 a

1 b

◆

E·da =

1 Q r 4⇡✏0 r 2 ˆ

in the space between them, and (Va

Q 4⇡✏0 (Va Vb ) (Va = = 4⇡ ✏0 ✏0 (1/a 1/b) (1/a

Vb ) =

Vb ) . 1/b)

.

(c) For large b (b a), the second term is negligible, and R = 1/4⇡ a. Essentially all of the resistance is in the region right around the inner sphere. Successive shells, as you go out, contribute less and less, because the cross-sectional area (4⇡r2 ) gets larger and larger. For the two submerged spheres, R = 4⇡2 a = 2⇡1 a (one R as the current leaves the first, one R as it converges on the second). Therefore I = V /R = 2⇡ aV. Problem 7.2 (a) V = Q/C = IR. Because positive I means the charge on the capacitor is decreasing, dQ 1 = I= Q, so Q(t) = Q0 e t/RC . But Q0 = Q(0) = CV0 , so Q(t) = CV0 e t/RC . dt RC dQ 1 V0 t/RC Hence I(t) = = CV0 e t/RC = e . dt RC R Z 1 Z 1 Z V 2 1 2t/RC (b) W = 12 CV02 . The energy delivered to the resistor is P dt = I 2 R dt = 0 e dt = R 0 0 0 ✓ ◆ 1 V02 RC 2t/RC 1 e = CV02 . X R 2 2 0 dQ 1 dQ (c) V0 = Q/C + IR. This time positive I means Q is increasing: =I= (CV0 Q) ) = dt RC Q CV0 1 1 dt ) ln(Q CV0 ) = t + constant ) Q(t) = CV0 + ke t/RC . But Q(0) = 0 ) k = CV0 , so RC RC ✓ ◆ ⇣ ⌘ dQ 1 V0 t/RC t/RC t/RC Q(t) = CV0 1 e . I(t) = = CV0 e = e . dt RC R c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

146 (d) Energy from battery:

Z

1


V0 I dt =

0

V02 R

Z

1

e

t/RC

dt =

0

V02 ⇣ RCe R

t/RC

Since I(t) is the same as in (a), the energy delivered to the resistor is again

⌘

1 0

=

V02 RC = CV02 . R

1 2 2 CV0 .

The final energy in

the capacitor is also 12 CV02 , so half the energy from the battery goes to the capacitor, and the other half to the resistor. Problem 7.3 R (a) I = J · da, where the integral R is taken over a surfaceR enclosing the positively charged conductor. But J = E, and Gauss’s law says E · da = ✏10 Q, so I = E · da = ✏0 Q. But Q = CV , and V = IR, so ✏0 I = ✏0 CIR, or R = . qed C (b) Q = CV = CIR )

dQ dt

=

I=

time constant is ⌧ = RC = ✏0 / .

1 RC Q

) Q(t) = Q0 e

t/RC

, or, since V = Q/C, V (t) = V0 e

t/RC

. The

Problem 7.4 I = J(s) 2⇡sL ) J(s) = I/2⇡sL. E = J/ = I/2⇡s L = I/2⇡kL. Z a I b a V = E · dl = (a b). So R = . 2⇡kL 2⇡kL b Problem 7.5

 E E 2R dP 1 2 2 I= ; P =I R= ; =E r+R (r + R)2 dR (r + R)2

2R = 0 ) r + R = 2R ) R = r. (r + R)3

Problem H 7.6 H E = E·dl = zero for all electrostatic fields. It looks as though E = E · dl = ( /✏0 )h, as would indeed be the case if the field were really just /✏0 inside and zero outside. But in fact there is always a “fringing field” at the edges (Fig. 4.31), and this is evidently just right to kill o↵ the contribution from the left end of the loop. The current is zero. Problem 7.7 Blv . (Never mind the minus sign—it just tells you the R direction of flow: (v⇥B) is upward, in the bar, so downward through the resistor.) (a) E =

d dt

=

(b) F = IlB =

Bl dx dt =

Blv; E = IR ) I =

B 2 l2 v , to the left. R

B 2 l2 dv B 2 l2 dv B 2 l2 = v) = v ) v = v0 e mR t . dt R dt Rm (d) The energy goes into heat in the resistor. The power delivered to resistor is I 2 R, so

(c) F = ma = m

dW B 2 l2 v 2 B 2 l2 2 = I 2R = R = v e dt R2 R 0 The total energy delivered to the resistor is W =

2↵t

↵mv02

, where ↵ ⌘

Z

0

1

e

2↵t

B 2 l2 ; mR

dt = ↵mv02

dW = ↵mv02 e dt e

2↵t 1

2↵

0

2↵t

= ↵mv02

.

1 1 = mv02 . X 2↵ 2


147


1

Problem 7.8 µ0 I ˆ (a) The field of long wire is B = , so 2⇡s

=

Z

µ0 I B·da = 2⇡

s+a Z

1 µ0 Ia (a ds) = ln s 2⇡

s

✓

s+a s

◆

.

✓ ◆ ✓ ◆ d µ0 Ia d s+a ds µ0 Ia 1 ds 1 ds µ0 Ia2 v (b) E = = ln , and = v, so = . dt 2⇡ dt s dt 2⇡ s + a dt s dt 2⇡s(s + a) The field points out of the page, so the force on a charge in the nearby side of the square is to the right. In the far side it’s also to the right, but here the field is weaker, so the current flows counterclockwise.

Contents

(c) This time the flux is constant, so E = 0.

Problem 7.9 R Since r·B = 0, Theorem 2(c) (Sect. 1.6.2) guarantees that B·da is the same for all surfaces with a given boundary line. Problem 7.10Problem 7.10 ✲ Φ = B · a = Ba2 cos θ ✲ = B · a = Ba2 cos ✓ B (view from above) θ ✲ Here ✓ = !t, so E = ddt = Ba2 ( sin !t)!; ❘a E = B!a2 sin !t. Problem 7.11 E = Blv = IR ) I = downward: mg

Bl Rv

) upward magnetic force = IlB =

B 2 l2 dv dv v=m ; =g R dt dt

dv = dt ) g ↵v

1 ln(g ↵

↵v, where ↵ ⌘

↵v) = t + const. ) g

↵v = g(1

e

↵t

);

B 2 l2 R v.

B 2 l2 . g mR ↵v = Ae

This opposes the gravitational force

↵vt = 0 ) vt = ↵t

g mgR = . ↵ B 2 l2

; at t = 0, v = 0, so A = g.

g (1 e ↵t ) = vt (1 e ↵t ). ↵ ) e ↵t = 1 0.9 = 0.1; ln(0.1) =

v=

At 90% of terminal velocity, v/vt = 0.9 = 1 e ↵t ↵t; ln 10 = ↵t; v t t = ↵1 ln 10, or t90% = ln 10. g Now the numbers: m = 4⌘Al, where ⌘ is the mass density of aluminum, A is the cross-sectional area, and l is the length of a side. R = 4l/A , where is the conductivity of aluminum. So 8 9 ⇢ = 2.8 ⇥ 10 8 ⌦ m > > > > < = 4⌘Alg4l 16⌘g 16g⌘⇢ g = 9.8 m/s2 vt = = = , and . 3 3 2 2 2 2 ⌘ = 2.7 ⇥ 10 kg/m > > A B l B B > > : ; B = 1T So vt =

(16)(9.8)(2.7⇥103 )(2.8⇥10 1

8

)

= 1.2 cm/s; t90% =

1.2⇥10 9.8

2

ln(10) = 2.8 ms.

If the loop were cut, it would fall freely, with acceleration g.

c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any⃝2005 by any Education, means, without in writing fromAll the publisher. cform orPearson Inc., permission Upper Saddle River, NJ. rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

148


Problem 7.12 =⇡

⇣ a ⌘2

2 Problem 7.14

B=

2 ⇡a2 B0 cos(!t); E = 4

d ⇡a2 E ⇡a2 ! = B0 ! sin(!t). I(t) = = B0 sin(!t). dt 4 R 4R

Problem 7.13 =

Z

B dx dy = kt

2

Z

a

dx

0

Z

a

y 3 dy =

0

1 2 5 kt a . E = 4

d = dt

1 5 2 kta .

Problem 7.14

✿

Iind

pipe ✲ falling magnet

❑

B✕ ✻ ✿ ✿ I✿

✲ ✗

ring ✲

✻ B❖

②

Iind

Suppose the current (I) in the magnet flows counterclockwise (viewed from above), as shown, so its field, near the ends, points upward. A ring of pipe below the magnet experiences an increasing upward flux, as the magnet approaches, and hence (by Lenz’s law) a current (Iind ) will be induced in it such as to produce a downward flux. Thus Iind must flow clockwise, which is opposite to the current in the magnet. Since opposite currents repel, the force on the magnet is upward. Meanwhile, a ring above the magnet experiences a decreasing (upward) flux, so its induced current is parallel to I, and it attracts the magnet upward. And the flux through rings next to the magnet is constant, so no current is induced in them. Conclusion: the delay is due to forces exerted on the magnet by induced eddy currents in the pipe.

Problem 7.15 In the quasistatic approximation, B =

⇢

µ0 nI ˆ z, (s < a); 0, (s > a).

Inside: for an “amperian loop” of radius s < a, = B⇡s = µ0 nI⇡s ; 2

2

I

E · dl = E 2⇡s =

d = dt

µ0 n⇡s2

dI ; dt

E=

µ0 ns dI ˆ . 2 dt

Outside: for an “amperian loop” of radius s > a: = B⇡a2 = µ0 nI⇡a2 ; E 2⇡s =

µ0 n⇡a2

dI ; dt

E=

µ0 na2 dI ˆ . 2s dt 3

Problem 7.16 (a) The magnetic field (in the quasistatic is “circumferential”. This is analogous to the current Problemapproximation) 7.16 in a solenoid, and hence the field is longitudinal. ✛ ✲ l ❄ (b) Use the “amperian loop” shown. Outside, B = 0, so here E = 0R (like B outside H R a aµ0solenoid). I d d So E·dl = El = ddt = dt B·da = dt l ds0 s 2⇡s0 µ0 dI a dI ) E = 2⇡ dt ln s . But dt = I0 ! sin !t, ⇣a⌘ µ0 I0 ! so E = sin(!t) ln ˆ z. 2⇡ s

I

✲

✻ a

s ✻

✻

✲

z



149


Problem 7.17 (a) The field inside the solenoid is B = µ0 nI. So

= ⇡a2 µ0 nI ) E =

⇡a2 µ0 n(dI/dt).

⇡a2 µ0 nk . R B is to the right and increasing, so the field of the loop is to the left, so the current is counterclockwise, or to the right, through the resistor. In magnitude, then, E = ⇡a2 µ0 nk. Now E = Ir R, so Iresistor =

(b)

= 2⇡a2 µ0 nI; I =

Problem 7.18 =

Z

dQ E = = dt R

1d ) R dt

µ0 I ˆ B·da; B = ; 2⇡s E = Iloop R =

dQ =

Q=

µ0 Ia = 2⇡

dQ R= dt

1 R

Z

s+a

s

d = dt

, in magnitude. So

Q=

2⇡a2 µ0 nI . R

ds0 µ0 Ia s + a = ln ; s0 2⇡ s

µ0 a dI ln(1 + a/s) . 2⇡ dt

µ0 a Iµ0 a ln(1 + a/s) dI ) Q = ln(1 + a/s). 2⇡R 2⇡R

The field of the wire, at the square loop, is out of the page, and decreasing, so the field of the induced current must point out of page, within the loop, and hence the induced current flows counterclockwise. Problem 7.19 In the quasistatic approximation, B =

⇢

µ0 N I 2⇡s

0, (Eq. 5.60). The flux around the toroid is therefore =

µ0 N I 2⇡

Z

a+w

a

ˆ, (inside toroid); (outside toroid)

1 µ0 N Ih ⇣ w ⌘ µ0 N hw h ds = ln 1 + ⇡ I. s 2⇡ a 2⇡a

d µ0 N hw dI µ0 N hwk = = . dt 2⇡a dt 2⇡a

The electric field is the same as the magnetic field of a circular current (Eq. 5.41): B=

µ0 I a2 ˆ z, 2 (a2 + z 2 )3/2

with (Eq. 7.19) I!

1 d = µ0 dt

N hwk µ0 . So E = 2⇡a 2

✓

N hwk 2⇡a

◆

a2 ˆ z= (a2 + z 2 )3/2

Problem 7.20 @B/@t is nonzero along the left and right edges of the shaded rectangle: c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

µ0 N hwka ˆ z. 4⇡ (a2 + z 2 )3/2

150


E

The (inward) flux through the strip on the left is increasing; the (inward) flux through the strip on the right is decreasing. This is analogous to two current sheets under Ampère’s law, with B ! E and µ0 Ienc ! d /dt (Eq. 7.19). The one on the left is like a current flowing out (taking account of the minus sign), so its field is counterclockwise and the one on the right is like a current flowing in, so its field is clockwise. Problem 7.21 The answer is indeterminate, until some boundary conditions are supplied. We know that r ⇥ E = (dB0 /dt) ˆ z (and r · E = 0), but this is insufficient information to determine E. Ordinarily we would invoke some symmetry of the configuration, or require that the field go to zero at infinity, to resolve the ambiguity, 0 but neither is available in this case. E = 12 dB ˆ xy ˆ) would do the job, in which case the force would be dt (y x zero, but we could add any constant vector to this, and make the force anything we like. Problem 7.22 (a) From Eq. 5.41, the field (on the axis) is B = is

=

µ0 I b2 z, 2 (b2 +z 2 )3/2 ˆ

so the flux through the little loop (area ⇡a2 )

µ0 ⇡Ia2 b2 . 2(b2 + z 2 )3/2

µ0 m ˆ where m = I⇡a2 . Integrating over the spherical “cap” (b) The field (Eq. 5.88) is B = 4⇡ r + sin ✓ ✓), r 3 (2 cos ✓ ˆ (bounded by the big loop and centered at the little loop):

=

where r =

p

Z

B·da =

µ0 I⇡a2 4⇡ r3

Z

(2 cos ✓)(r2 sin ✓ d✓ d ) =

b2 + z 2 and sin ✓¯ = b/r. Evidently

(c) Dividing o↵ I (

1

= M12 I2 ,

2

=

µ0 I⇡a2 sin2 ✓ r 2

= M21 I1 ): M12 = M21 =

✓¯ 0

µ0 Ia2 2⇡ 2r =

Z

✓¯

cos ✓ sin ✓ d✓

0

µ0 ⇡Ia2 b2 , the same as in (a)!! 2(b2 + z 2 )3/2

µ0 ⇡a2 b2 . 2(b2 + z 2 )3/2

Problem 7.23 E=

d = dt

M

dI = dt

M k. c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

CHAPTER 7. ELETRODYNAMICS 4

151

Problem 7.21 E =−

dΦ dI = −M = −M k. dt dt

I ❖

✻ a ❄ ✻ a ❄

✻ a ❄

It’s hard to calculate M using a current in the little loop, so, exploiting the equality of the mutual inductances, I’ll find the flux through the Rlittle loop when a current I flows in the big loop: = M I. The field of one long 2a 1 0I 0I 0 Ia wire is B = µ2⇡s ) 1 = µ2⇡ a ds = µ2⇡ ln 2, so the total flux is a s =2

1

=

µ0 Ia ln 2 µ0 a ln 2 µ0 ka ln 2 ) M= ) E= , ⇡ ⇡ ⇡

in magnitude.

Direction: The net flux (through the big loop), due to I in the little loop, is into the page. (Why? Field lines point in, for the inside of the little loop, and out everywhere outside the little loop. The big loop encloses all of the former, and only part of the latter, so net flux is inward.) This flux is increasing, so the induced current in the big loop is such that its field points out of the page: it flows counterclockwise. Problem 7.24 B = µ0 nI ) 1 = µ0 nI⇡R2 (flux through a single turn). In a length l there are nl such turns, so the total flux is = µ0 n2 ⇡R2 Il. The self-inductance is given by = LI, so the self-inductance per unit length is L = µ0 n2 ⇡R2 . Problem 7.25

The field of one wire is B1 =

µ0 I 2⇡ s ,

so

= 2·

µ0 I 2⇡

·l

dR ✏ ✏

ds s

=

µ0 Il ⇡

ln

d ✏ ✏

. The ✏ in the numerator is

negligible (compared to d), but in the denominator we cannot let ✏ ! 0, else the flux is infinite. µ0 l L= ln(d/✏). Evidently the size of the wire itself is critical in determining L. ⇡ Problem 7.26 (a) In the quasistatic approximation B =

µ0 ˆ . So 2⇡s

1

This is the flux through one turn; the total flux is N times E=

d µ0 N h (4⇡ ⇥ 10 )(10 )(10 = ln(b/a)I0 ! sin(!t) = dt 2⇡ 2⇡ 7

= 2.61 ⇥ 10

4

= 5.22 ⇥ 10

7

3

=

µ0 I 2⇡

1: 2

= )

Z

b

1

h ds =

a s µ0 N h

2⇡

µ0 Ih ln(b/a). 2⇡

ln(b/a)I0 cos(!t). So

ln(2)(0.5)(2⇡ 60) sin(!t)

sin(!t) (in volts), where ! = 2⇡ 60 = 377/s. Ir =

E 2.61 ⇥ 10 = R 500

4

sin(!t)

sin(!t) (amperes). 7 6 dIr )(10 2 ) N 2h (b) Eb = L ; where (Eq. 7.28) L = µ02⇡ ln(b/a) = (4⇡⇥10 )(10 ln(2) = 1.39 ⇥ 10 2⇡ dt Therefore Eb = (1.39 ⇥ 10 3 )(5.22 ⇥ 10 7 !) cos(!t) = 2.74 ⇥ 10 7 cos(!t) (volts). Ratio of amplitudes:

2.74 ⇥ 10 2.61 ⇥ 10

7 4

= 1.05 ⇥ 10

Problem 7.27 With I positive clockwise, E = 2

d Q dt2

=

1 LC Q

=

! Q, where ! = 2

3

=

3

(henries).

µ0 N 2 h! ln(b/a). 2⇡R

L dI dt = Q/C, where Q is the charge on the capacitor; I = p1 . LC

dQ dt ,

so

The general solution is Q(t) = A cos !t + B sin !t. At t = 0,

c ⃝2005 Pearson Education, Inc., Upper Saddle c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is River, NJ. All rights reserved. This material is protected all copyright laws as they exist. No portion of this material may be protected under all copyright laws as they currently exist.under No portion of this material maycurrently be reproduced, in in any form or by the any publisher. means, without permission in writing from the publisher. reproduced, in any form or by any means, without permission writing from

1

152


Q = CV , so A = CV ; I(t) = dQ A! sin !t + B! sin !t. At t = 0, I = 0, so B = 0, and dt = r ✓ ◆ C t I(t) = CV ! sin !t = V sin p . L LC

Contents

2

Q d Q dQ 1 If you put in a resistor, the oscillation is “damped”. This time L dI dt = C + IR, so L dt2 + R dt + C Q = 0. For an analysis of this case, see Purcell’s Electricity and Magnetism (Ch. 8) or any book on oscillations and waves. Problem 7.28 1 (a) W = 12 LI 2 . L = µ0 n2 ⇡R2 l (Prob. 7.24) W = µ0 n2 ⇡R2 lI 2 . Problem 7.26 2 H 1 ˆ (b) W = 2 (A·I)dl. A = (µ0 nI/2)R , at the surface (Eq. 5.72 or 5.73). So W1 = 12 µ02nI RI · 2⇡R, for one turn. There areR nl such turns in length l, so W = 12 µ0 n2 ⇡R2 lI 2 . X R (c) W = 2µ1 0 B 2 d⌧. B = µ0 nI, inside, and zero outside; d⌧ = ⇡R2 l, so W = 2µ1 0 µ20 n2 I 2 ⇡R2 l = 1 2 2 2 5 .X 2 µ0 n ⇡R lI ⇥R ⇤ H R (d) W = 2µ1 0 B 2 d⌧ (A⇥B)·da . This time B 2 d⌧ = µ20 n2 I 2 ⇡(R2 a2 )l. Meanwhile, A⇥B = 07.26 outside (at s = b). Inside, A = µ02nI a ˆ (at s = a), while B = µ0 nI ˆ z. Problem 1 2 2 2 ˆ A⇥B = 2 µ0 n I a( ⇥ˆ z) ˆ s points inward (“out” of the volume) | {z } ✕ ☛ ˆ s ✲ ˆ z H R (A⇥B) · da = ( 12 µ20 n2 I 2 aˆ s) · [a d dz( ˆ s)] = 12 µ20 n2 I 2 a2 2⇡l. ✲ˆ z ⇥ ⇤ ˆ W = 2µ1 0 µ20 n2 I 2 ⇡(R2 a2 )l + µ20 n2 I 2 ⇡a2 l = 12 µ0 n2 I 2 R2 ⇡l. X φ ❖

Problem 7.29 ✓ ◆ Z Z µ0 nI 1 1 µ20 n2 I 2 1 µ0 n2 I 2 b 1 B= ; W = B 2 d⌧ = hs d ds = h2⇡ ln = µ0 n2 I 2 h ln(b/a). 2⇡s 2µ0 2µ0 4⇡ 2 s2 8⇡ 2 a 4⇡ µ0 2 L= n h ln (b/a) (same as Eq. 7.28). 2⇡

→

Problem 7.30 I ✛ ✲ l µ0 Is B·dl = B(2⇡s) = µ0 Ienc = µ0✕I(s2 /R2 ) ) B = . 2 →I2⇡R w Z 2 2 Z R 2 1 1 ❄ µ0 I µ0 I l s4 R Bµ0 l 2 1 ☛ 2 W = B 2 d⌧ = s (2⇡s)l ds = = I = LI 2 . h 2 4 4 2µ0 2µ0 4⇡ R 0 4⇡R 4 0 16⇡ 2 ✛ µ0 ✻ So L = l, and L = L/l = µ0 /8⇡, independent ofI R! 8⇡ Problem 7.31 dI dI R E0 (a) Initial current: I0 = E0 /R. So L = IR ) = I ) I = I0 e Rt/L , or I(t) = e dt dt L R E02 2Rt/L dW 2 2 2Rt/L (b) P = I R = (E0 /R) e R= e = . R dt ✓ ◆ Z 1 E 2 1 2Rt/L E2 L E2 1 2 W = 0 e dt = 0 e 2Rt/L = 0 (0 + L/2R) = L (E0 /R) . R 0 R 2R R 2 0 2 (c) W0 = 12 LI02 = 12 L (E0 /R) . X Problem 7.32 µ0 1 ˆ ) rˆ (a) Pearson B1 = 4⇡ a1 ], River, since NJ. m1All = rights I1 a1 .reserved. The flux through loop 2 is then 1· r r 3 I1 [3(a c ⃝2005 Education, Inc., Upper Saddle This material is protected under all copyright µ0 1 laws as they currently exist. No portion of this material may µ0 be ˆwithout )(a2 · rˆpermission ) a1 ·ain = M Ifrom M publisher. = [3(a1 · rˆ )(a2 · rˆ ) reproduced, any form or by3 I any means, the 2 = Bin 1 ·a 2 = 1 [3(a 1· r 2 ] writing 1. 4⇡ r 4⇡ r 3

Rt/L

.

a1 ·a2 ].


153


dW 2 2 (b) E1 = M dI E1 I1 = M I1 dI dt , dt 1 = dt . (This is the work done per unit time against the mutual emf in loop 1—hence the minus sign.) So (since I1 is constant) W1 = M I1 I2 , where I2 is the final current in loop 2: µ0 W = [3(m1 · rˆ )(m2 · rˆ ) m1 ·m2 ]. 4⇡ r 3 Notice that this is opposite in sign to Eq. 6.35. In Prob. 6.21 we assumed that the magnitudes of the dipole moments were fixed, and we did not worry about the energy necessary to sustain the currents themselves—only the energy required to move them into position and rotate them into their final orientations. But in this problem we are including it all, and it is a curious fact that this merely changes the sign of the answer. For commentary on this subtle issue see R. H. Young, Am. J. Phys. 66, 1043 (1998), and the references cited there. Problem 7.33 (a) The (solenoid) magnetic field is ( µ0 K ˆ z = µ0 !R ˆ z (s < R), B= 0 (s > R).

From Example 7.7, the electric field is 8 > s dB sR > > µ0 !˙ ˆ < 2 dt ˆ = 2 E= > R2 dB ˆ R3 > > = µ0 !˙ ˆ : 2s dt 2s At the surface (s = R) E =

1 2 2 µ0 R

N=

(s < R), (s > R).

!˙ ˆ, so the torque on a length ` of the cylinder is ✓ ◆ 1 2 R( 2⇡R`) µ0 R !˙ ˆ z = ⇡µ0 2 R4 !` ˙ ˆ z, 2

and the work done per unit length is

W = `

But d = ! dt, and the integral becomes Z !f 1 ! d! = !f2 ; 2 0

⇡µ0

)

Z

d! d . dt

W = `

µ0 ⇡ 2

2

R4

!f R 2

2

.

(This is the work done by the field; the work you must do does not include the minus sign.) (b) Because B = µ0 K ˆ z = µ0 !f R ˆ z is uniform inside the solenoid (and zero outside), W = W 1 µ0 ⇡ = (µ0 !f R)2 ⇡R2 = ` 2µ0 2 Problem 7.34 The displacement current density (Sect. 7.3.2) is Jd = ✏0 @E @t = radius s, I

B·dl = B · 2⇡s = µ0 Idenc = µ0

I A

1 2 2 B ⇡R `. 2µ0

!f R 2

2

=

ˆ z. Drawing an “amperian loop” at

I ⇡a2

.

I s2 µ0 Is2 2 · ⇡s = µ I ) B = ; 0 ⇡a2 a2 2⇡sa2


B=

µ0 Is ˆ . 2⇡a2

154

Problem 7.35 (t) (a) E = ˆ z; ✏0


Q(t) It It = ; ˆ z. ⇡a2 ⇡a2 ⇡✏0 a2 I dE 2 s2 s2 µ0 I ˆ (b) Idenc = Jd ⇡s2 = ✏0 ⇡s = I 2 . B · dl = µ0 Idenc ) B 2⇡s = µ0 I 2 ) B = s . dt a a 2⇡a2 (c) A surface current flows radially outward over the left plate; let I(s) be the total current crossing a circle of radius s. The charge density (at time t) is (t) =

(t) =

[I

I(s)]t . ⇡s2

Since we are told this is independent of s, it must be that I so a2 = I, or = I/a2 . Therefore I(s) = I(1 s2 /a2 ). B 2⇡s = µ0 Ienc = µ0 [I

I(s)] = µ0 I

I(s) = s2 , for some constant . But I(a) = 0, s2 µ0 I ˆ ) B= s . X a2 2⇡a2

Problem 7.36 µ0 I0 ! 2 µ0 ✏0 2 (a) Jd = ✏0 cos(!t) ln (a/s) ˆ z. But I0 cos(!t) = I. So Jd = ! I ln(a/s) ˆ z. 2⇡ 2⇡ Z Z Z a µ0 ✏0 ! 2 I a (b) Id = Jd · da = ln(a/s)(2⇡s ds) = µ0 ✏0 ! 2 I (s ln a s ln s)ds 2⇡ 0 0 h ia h 2 i 2 µ0 ✏0 ! 2 Ia2 s2 s2 a a2 a2 2 = µ0 ✏0 ! 2 I (ln a) s2 ln s + = µ ✏ ! I ln a ln a + = . 0 0 2 4 2 2 4 4 0 (c)

Id µ0 ✏0 ! 2 a2 = . I 4

Since µ0 ✏0 = 1/c2 , Id /I = (!a/2c)2 . If a = 10

8

m/s 10 2c 11 ! = 10a = 3⇥10 5⇥10 3 m , or ! = 0.6 ⇥ 10 /s = 6 ⇥ 10 /s; ⌫ = microwave region, way above radio frequencies.)

! 2⇡

3

m, and

Id I

=

1 100 ,

so that

!a 2c

=

1 10 ,

⇡ 1010 Hz, or 104 megahertz. (This is the

Problem 7.37 Physically, this is the field of a point charge q at the origin, out to an expanding spherical shell of radius vt; outside this shell the field is zero. Evidently the shell carries the opposite charge, q. Mathematically, using product rule #5 and Eq. 1.99: ✓ ◆ 1 q 1 q q 3 1 q @ r · E = ✓(vt r)r · ˆ r + ˆ r · r[✓(vt r)] = (r)✓(vt r) + (ˆ r·ˆ r) ✓(vt r). 2 2 2 4⇡✏0 r 4⇡✏0 r ✏0 4⇡✏0 r @r But

3

(r)✓(vt

r) =

3

(r)✓(t), and

@ @r ✓(vt

r) =

(vt

⇢ = ✏0 r · E = q 3 (r)✓(t)

r) (Prob. 1.46), so q (vt 4⇡r2

r).

(For t < 0 the field and the charge density are zero everywhere.) Clearly r · B = 0, and r ⇥ E = 0 (since E has only an r component, and it is independent of ✓ and ). There remains only the Ampére/Maxwell law, r ⇥ B = 0 = µ0 J + µ0 ✏0 @E/@t. Evidently ⇢ @E q @ q J = ✏0 = ✏0 [✓(vt r)] ˆ r= v (vt r) ˆ r. 2 @t 4⇡✏0 r @t 4⇡r2 c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

155


(The stationary charge at the origin does not contribute to J, of course; for the expanding shell we have J = ⇢v, as expected—Eq. 5.26.) Problem 7.38 µ0 qm ˆ From r·B = µ0 ⇢m it follows that the field of a point monopole is B = 4⇡ r 2 r . The force law has the 1 2 form F / qm B c2 v⇥E (see Prob. 5.22—the c is needed on dimensional grounds). The proportionality ✓ ◆ 1 constant must be 1 to reproduce “Coulomb’s law” for point charges at rest. So F = qm B v⇥E . c2 Problem 7.39 Integrate the “generalized Faraday law” (Eq. 7.44iii), r ⇥ E = µ0 Jm @B @t , over the surface of the loop: Z I Z Z d d (r ⇥ E) · da = E · dl = E = µ0 Jm · da B · da = µ0 Imenc . dt dt

dI dI µ0 1d µ0 1 But E = L , so = Im + , or I = Qm + , where Qm is the total magnetic charge dt dt L enc L dt L L passing through the surface, and is the change in flux through the surface. If we use the flat surface, then Qm = qm and = 0 (when the monopole is far away, = 0; the flux builds up to µ0 qm /2 just before it passes through the loop; then it abruptly drops to µ0 qm /2, and rises back up to zero as the monopole disappears into the distance). If we use a huge balloon-shaped surface, so that qm remains inside it on the far side, then Qm = 0, but rises monotonically from 0 to µ0 qm . In either case, I=

µ0 qm . L

[The analysis is slightly di↵erent for a superconducting loop, but the conclusion is the same.] Problem 7.40  V 1 V @D @ @ V0 cos(2⇡⌫t) ✏V0 E= ) Jc = E = E = . Jd = = (✏E) = ✏ = [ 2⇡⌫ sin(2⇡⌫t)]. d ⇢ ⇢d @t @t @t d d The ratio of the amplitudes is therefore: ⇥ Jc V0 d 1 = = = 2⇡(4 ⇥ 108 )(81)(8.85 ⇥ 10 Jd ⇢d 2⇡⌫✏V0 2⇡⌫✏⇢

12

Problem 7.41 Begin with7.39 a di↵erent problem: two parallel Problem wires carrying charges + and as shown. Field of one wire: E =

s; 2⇡✏0 s ˆ

potential: V =

⇤ )(0.23) 9

1

= 2.41. z✻

2⇡✏0

b

ln(s/a).

Potential of combination: V = 2⇡✏0 ln(s /s+ ), n o (y+b)2 +z 2 or V (y, z) = 4⇡✏0 ln (y b)2 +z 2 .

−λ ✰ x

b

✲y

+λ

Find the locus of points of fixed V (i.e. equipotential surfaces): e4⇡✏0 V / ⌘ µ = y (µ 2

(y

(y + b)2 + z 2 =) µ(y 2 (y b)2 + z 2

1) + b (µ 2

b )2 + z 2 + b2

1) + z (µ b2

2

1)

2

= 0 =) (y

2yb + b2 + z 2 ) = y 2 + 2yb + b2 + z 2 ;

2yb(µ + 1) = 0 =) y + z + b b

2

) + z 2 = b2 (

2

2

2

1).

2


2yb = 0

✓

µ+1 ⌘ µ 1

◆

;

156


q 2 p p 2b µ (µ2 2µ+1) 2 This is a circle, with center at y0 = b = b µµ+11 and radius = b 1 = b (µ +2µ+1) = 2 (µ 1) µ 1 . This suggests an image solution to the problem at hand. We want y0 = d, radius = a, and V = V0 . These determine the parameters b, µ, and of the image solution: b µµ+11 d y0 µ+1 d = = 2bpµ = p . Call ⌘ ↵. a radius 2 µ a µ 1

4↵ µ = (µ + 1) = µ2 + 2µ + 1 =) µ2 + (2 4↵2 )µ + 1 = 0; p p p 4↵2 2 ± 4(1 2↵2 )2 4 µ= = 2↵2 1 ± 1 4↵2 + 4↵4 1 = 2↵2 1 ± 2↵ ↵2 1; 2 4⇡✏0 V0 4⇡✏0 V0 p = ln µ =) = . That’s the line charge in the image problem. 2 ln 2↵ 1 ± 2↵ ↵2 1 Z Z 1 I = J·da = E·da = Qenc = l. ✏0 ✏0 2

2

I 4⇡ V0 p = = . Which sign do we want? Suppose l ✏0 ln 2↵2 1 ± 2↵ ↵2 1 the cylinders are far apart, d a, so that ↵ 1.  p 1 1 ( ) = 2↵2 1 ± 2↵2 1 1/↵2 = 2↵2 1 ± 2↵2 1 + ··· 2 2↵ 8↵4 ( 4↵2 2 1/2↵2 + · · · ⇡ 4↵2 (+ sign), 1 2 = 2↵ (1 ± 1) (1 ± 1) ⌥ 2 ± · · · = 4↵ 1/4↵2 ( sign). The current per unit length is i =

The current must surely decrease with increasing ↵, so evidently the + sign is correct: i= Problem 7.42

From Prob. 3.24,

4⇡ V0 p 1 + 2↵ ↵2

ln 2↵2

8 1 X > > > V (s, ) = sk bk sin(k ), > in > > < k=1 > 1 > X > > > V (s, ) = s > out :

k

1

, where ↵ =

d . a

(s < a);

dk sin(k ), (s > a).

k=1

(We don’t need the cosine terms, because V is clearly an odd function of .) At s = a, Vin = Vout = V0 /2⇡. Let’s start with Vin , and use Fourier’s trick to determine bk : Z ⇡ Z 1 1 X X V0 ⇡ V0 ak bk sin(k ) = ) ak bk sin(k ) sin(k 0 ) d = sin(k 0 ) d . But 2⇡ 2⇡ ⇡ ⇡ k=1 k=1 Z ⇡ 0 sin(k ) sin(k ) d = ⇡ kk0 , and ⇡  Z ⇡ ⇡ 0 1 2⇡ 2⇡ 0 0 0 sin(k ) d = sin(k ) cos(k ) = cos(k 0 ) = ( 1)k . So 0 2 0 0 0 (k ) k k k ⇡  ✓ ◆k ⇡ 1 V 2⇡ V 1 V 0 X 1 ⇣ s ⌘k 0 0 ⇡ak bk = ( 1)k , or bk = , and hence Vin (s, ) = sin(k ). 2⇡ k ⇡k a ⇡ k a k=1


157


1 1 X V0 X 1 ⇣ a ⌘k 1 sin(k ). Both sums are of the form S ⌘ ( x)k sin(k ) (with ⇡ k s k k=1 k=1 x = s/a for r < a and x = a/s for r > a). This series can be summed explicitly, using Euler’s formula 1 1 X X 1 1 k (ei✓ = cos ✓ + i sin ✓): S = Im ( x)k eik = Im xei . k k k=1 k=1 1 X ⇥ ⇤ 1 2 1 3 1 4 1 But ln(1 + w) = w w + w w ··· = ( w)k , so S = Im ln 1 + xei . 2 3 4 k

Similarly, Vout (s, ) =

k=1

Now ln Rei✓ = ln R + i✓, so S = Im 1 + xei tan ✓ = = Re (1 + xei )

✓, where

1 2i 1 2

⇥

8 V0 > > Vin (s, ) = tan > > < ⇡

Conclusion:

> > V0 > > tan : Vout (s, ) = ⇡

= @Vin = @s = @Vin @s

1

1

✓ ✓

s sin a + s cos a sin s + a cos

◆ ◆

i i

)]

⇤

=

x ei e i i [2 + x (ei + e

i

)]

=

x sin . 1 + x cos

, (s < a); , (s > a).

@Vin . @s s=a s=a 8 9 > > > >  V0 < 1 ( a sin ) = V0 a sin  ⇣ ⌘2 (s + a cos )2 > = ⇡ (s + a cos )2 + (a sin )2 ⇡ > a sin > > : 1 + s+a ; cos ✓ ◆ V0 a sin ; 2 ⇡ s + 2as cos + a2 8 9 > > > >  V0 < 1 [(a + s cos ) sin s sin cos ] = V0 a sin  = ⇣ ⌘ 2 2 > > ⇡ > (a + s cos ) ⇡ (a + s cos )2 + (s sin )2 s sin > : 1 + a+s ; cos ✓ ◆ V0 a sin . ⇡ s2 + 2as cos + a2

(b) From Eq. 2.36, ( ) =

@Vout = @s

⇢

1 + xei 1 + xe i [(1 + xe ) + (1 + xe

= s=a

@Vout @s

✏0

s=a

@Vout @s

V0 = 2⇡a

✓

sin 1 + cos

◆

, so ( ) =

✏0 V0 sin = ⇡a (1 + cos )

Problem 7.43

✏0 V0 tan( /2). ⇡a

✓ ◆ ✓ ◆ ✓ ◆ 1 @ @(zf ) @ 2 (zf ) z d df d df df (a) r V = s + = s =0) s =0)s = A (a constant) ) s @s @s @z 2 s ds ds ds ds ds ds A = df ) f = A ln(s/s0 ) (s0 another constant). But (ii) ) f (b) = 0, so ln(b/s0 ) = 0, so s0 = b, and s I⇢ 1 I⇢z ln(s/b) V (s, z) = Az ln(s/b). But (i) ) Az ln(a/b) = (I⇢z)/(⇡a2 ), so A = ; V (s, z) = . ⇡a2 ln(a/b) ⇡a2 ln(a/b) 2


158

(b) E =

rV =


⇣z ⇣s⌘ ⌘ @V I⇢z 1 I⇢ ln(s/b) I⇢ ˆ z= ˆ s + ˆ z = ˆ s + ln ˆ z . @z ⇡a2 s ln(a/b) ⇡a2 ln(a/b) ⇡a2 ln(a/b) s b  ⇣z ⌘ ⇤ I⇢ ✏0 I⇢z Es (a ) = ✏0 0 = . 2 ⇡a ln(a/b) a ⇡a3 ln(a/b)

@V ˆ s @s

⇥ (c) (z) = ✏0 Es (a+ )

Problem 7.44 @B (a) Faraday’s law says r⇥E = @B @t , so E = 0 )H @t = 0 ) B(r) is independent of t. (b) Faraday’s law in integral form (Eq. 7.19) says E · dl = d /dt. In the wire itself E = 0, so through the loop is constant. (c) Ampère-Maxwell) r⇥B = µ0 J + µ0 ✏0 @E @t , so E = 0, B = 0 ) J = 0, and hence any current must be at the surface. (d) From Eq. 5.70, a rotating shell produces a uniform magnetic field (inside): B = 23 µ0 !aˆ z. So to cancel 3 B0 3B 0 such a field, we need !a = . Now K = v = !a sin ✓ ˆ, so K = sin ✓ ˆ. 2 µ0 2µ0 Problem 7.45 (a) To make the field parallel to the plane, we need image monopoles of the same sign (compare Figs. 2.13 and 2.14), so the image dipole points down (-z). (b) From Prob. 6.3 (with r ! 2z): F =

3µ0 m2 . 2⇡ (2z)4

3µ0 m2 1 = Mg ) h = 4 2⇡ (2h) 2

Problem 7.43 (c) Using Eq. 5.89, and referring to the figure:

B= = = = =

✓

3µ0 m2 2⇡M g

◆1/4

.

11

m✻ ✻ θ r1 h

µ0 1 {[3(m ˆ z·ˆ r1 ) ˆ r1 mˆ z] + [3( m ˆ z·ˆ r2 ) ˆ r2 + mˆ z]} 4⇡ (r1 )3 ˆ r2 3µ0 m [(ˆ z·ˆ r1 ) ˆ r1 (ˆ z·ˆ r2 ) ˆ r2 ] . But ˆ z·ˆ r1 = ˆ z·ˆ r2 = cos ✓. r ✲ ❯ ✕✲ ˆ r 4⇡(r1 )3 ✕❯ 3µ0 m ˆ r1 cos ✓(ˆ r1 + ˆ r2 ). But ˆ r1 + ˆ r2 = 2 sin ✓ ˆ r. 4⇡(r1 )3 r2 h p 3µ0 m r h 2 + h2 . sin ✓ cos ✓ ˆ r . But sin ✓ = , cos ✓ = , and r = r 1 θ 2⇡(r1 )3 r1 r1 3µ0 mh r −m ❄ ❄ ˆ r. 2 2 5/2 2⇡ (r + h )

Now B = µ0 (K ⇥ ˆ z) ) ˆ z ⇥ B = µ0 ˆ z ⇥ (K ⇥ ˆ z) = µ0 [K ˆ z(K · ˆ z)] = µ0 K. (I used the BAC-CAB rule, and noted that K · ˆ z = 0, because the surface current is in the x y plane.) K=

1 (ˆ z ⇥ B) = µ0

3mh r (ˆ z⇥ˆ r) = 2 2⇡ (r + h2 )5/2

3mh r ˆ. qed 2 2⇡ (r + h2 )5/2


12 CHAPTER 7. ELETRODYNAMICS

159

Problem Problem7.44 7.46 Say the angle between the dipole (m1 ) and the z axis is ✓ (see diagram).

−

B(z) =

µ0 1 [3(m2 · ˆ z) ˆ z 4⇡ (h + z)3

m2 ]

✲x h

for points on the z axis (Eq. 5.89). The torque on m1 is (Eq. 6.1)

−

µ0 [3(m2 · ˆ z)(m1 ⇥ ˆ z) 4⇡(2h)3

But m1 = m(sin ✓ x ˆ + cos ✓ˆ z), m2 = m(sin ✓ x ˆ m1 ⇥ m2 = 2m2 sin ✓ cos ✓ˆ y. N=

✻ θ + ✒ m 1

The field of the image dipole (m2 ) is

N = m1 ⇥ B =

z

cos ✓ˆ z), so m2 · ˆ z=

⇥ 2 µ0 3m sin ✓ cos ✓ y ˆ 3 4⇡(2h)

m cos ✓, m1 ⇥ ˆ z=

m2

❘+ θ

(m1 ⇥ m2 )] .

m sin ✓ y ˆ, and

⇤ µ0 m2 2m2 sin ✓ cos ✓ y ˆ) = sin ✓ cos ✓ y ˆ. 4⇡(2h)3

Evidently the torque is zero for ✓ = 0, ⇡/2, or ⇡. But 0 and ⇡ are clearly unstable, since the nearby ends of the dipoles (minus, in the figure) dominate, and they repel. The stable configuration is ✓ = ⇡/2: parallel to the surface (contrast Prob. 4.6). µ0 m In this orientation, B(z) = 4⇡(h+z) ˆ, and the force on m1 is (Eq. 6.3): 3 x F=r



µ0 m2 4⇡(h + z)3

= z=h

3µ0 m2 ˆ z 4⇡(h + z)4

= z=h

3µ0 m2 ˆ z. 4⇡(2h)4

At equilibrium this force upward balances the weight M g: 3µ0 m2 1 = Mg ) h = 4⇡(2h)4 2

✓

3µ0 m2 4⇡M g

◆1/4

.

1 Incidentally, this is (1/2)1/4 = 0.84 times the height it would adopt in the orientation perpendicular to the plane (Prob. 7.45b).

Problem 7.47 R ˆ f = v⇥B; v = !a sin ✓ ˆ; f = !aB0 sin ✓( ˆ⇥ˆ z). E = f ·dl, and dl = a d✓ ✓. R ⇡/2 ˆ ˆ⇥ˆ ˆ ˆ) = ˆ So E = !a2 B0 0 sin ✓( ˆ⇥ˆ z)·✓ˆ d✓. But ✓·( z) = ˆ z·(✓⇥ z·ˆ r = cos ✓. Z ⇡/2 2 ⇥ ⇤ sin ✓ 1 ⇡/2 E = !a2 B0 sin ✓ cos ✓ d✓ = !a2 B0 = !a2 B0 (same as the rotating disk in Ex. 7.4). 0 2 2 0 z✻ Problem 7.48 Ry p F = IBl; = 2B a a2 x2 dx (a = radius of circle). p p l/2 ✲ E = ddt = 2B a2 y 2 dy = 2Bv a2 y 2 = IR. dt x p p 2Bv 4B 2 v 2 2 2 2 2 2 I= R a y ; l/2 = a y . So F = R (a y ) = mg. mgR ✛✲ vcircle = ; y 4B 2 (a2 y 2 ) Z a Z a a dy 4B 2 4B 2 1 3 4B 2 4 3 16 B 2 a3 tcircle = = (a2 y 2 )dy = (a2 y y ) = ( a )= . v mgR a mgR 3 mgR 3 3 mgR +a a

Contents

}

c Saddle ⃝2005 Pearson Inc., Upper Saddle River, NJ. c 2012 Pearson Education, Inc., Upper River, Education, NJ. All rights reserved. This material is All rights reserved. This material is underexist. all copyright lawsofas they currently protected under all copyright laws as protected they currently No portion this material mayexist. be No portion of this material may be reproduced, any form or any means, without permission in writing from the publisher. reproduced, in any form or by any means, withoutinpermission in by writing from the publisher.

160


Problem 7.49 (a) From Prob. 5.52a, A(r, t) = @ @t (r

[Check: r ⇥ E =

⇥ A) =

@B @t ,

1 4⇡

Z

B(r0 , t) ⇥ rˆ

r

2

d⌧ 0 , so E =

@A . @t

and we recover Faraday’s law.] 2

2

1 Q 1 4⇡R (b) The Coulomb field is zero inside and 4⇡✏ r = 4⇡✏ ˆ r = ✏0Rr2 ˆ r outside. The Faraday field is 2 ˆ r2 0 r 0 @A @t , where A is given (in the quasistatic approximation) by Eq. 5.69, with ! a function of time. Letting !˙ ⌘ d!/dt, 8 µ R!˙ 0 > r sin ✓ ˆ (r < R), > > < 3 E(r, ✓, , t) = > > µ0 R4 !˙ sin ✓ ˆ > R2 : ˆ r (r > R). ✏0 r2 3 r2

Problem 7.50

dB dv dB = m = ma = F = qE, or E = R . But dt dt dt ✓ ◆ I d d 1 d dB 1 1 E·dl = , so E 2⇡R = , so =R , or B = + constant. If at time t = 0 2 dt dt 2⇡R dt dt ✓ 2 ⇡R ◆ 1 1 the field is o↵, then the constant is zero, and B(R) = (in magnitude). Evidently the field at R 2 ⇡R2 must be half the average field over the cross-section of the orbit. qed qBR = mv (Eq. 5.3). If R is to stay fixed, then qR

Problem 7.51 In the quasistatic approximation the magnetic field of the wire is B = (µ0 I/2⇡s) ˆ, or, in Cartesian coordinates, µ0 I µ0 I ⇣ y x ⌘ µ0 I ( y x ˆ + xy ˆ) B= ( sin x ˆ + cos y ˆ) = x ˆ+ y ˆ = , 2⇡s 2⇡s s s 2⇡ x2 + y 2 where x and y are measured from the wire. To convert to the stationary coordinates in the diagram, y ! y B=

vt:

µ0 I [ (y vt) x ˆ + xy ˆ] . 2 2 2⇡ x + (y vt)

Faraday’s law says r⇥E=

@B = @t

µ0 I 2⇡

⇢

x2

[ (y vt) x ˆ + xy ˆ] ( 2v)(y 2 2 2 [x + (y vt) ]

vx ˆ + (y vt)2

vt) .

At t = 0, then, r⇥E =

µ0 Iv 2⇡

⇢

x ˆ [ y2 x ˆ + xy y ˆ] +2 2 2 x +y [x2 + y 2 ]2

=

µ0 Iv 2⇡

⇢

(x2

y2 ) x ˆ + 2xy y ˆ] [x2 + y 2 ]2

=

µ0 Iv ⇣ cos ˆ s + sin 2⇡s2

⌘ ˆ .

Our problem is to find a vector function of s and (it obviously doesn’t depend on z) whose divergence is zero, whose curl is as given above, that goes to zero at large s: E(s, ) = Es (s, ) ˆ s + E (s, ) ˆ + Ez (s, ) ˆ z, c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

161

CHAPTER 7. ELETRODYNAMICS with

1 @ 1 @E (sEs ) + = 0, s @s s @ 1 @Ez µ0 Iv cos (r ⇥ E)s = = s @ 2⇡s2 @Ez µ0 Iv (r ⇥ E) = = sin , 2⇡s2 @s 1 @ @Es (r ⇥ E)z = (sE ) = 0. s @s @ r·E =

The first and last of these are satisfied if Es = E = 0, the middle two are satisfied by Ez = Evidently E =

µ0 Iv sin . 2⇡s

µ0 Iv sin ˆ z. The electric and magnetic fields ride along with the wire. 2⇡s

Problem 7.52 2 1 qQ 1 2 Initially, mv r = 4⇡✏0 r 2 ) T = 2 mv = new orbit, of radius r1 and velocity v1 :

1 1 qQ 2 4⇡✏0 r .

After the magnetic field is on, the electron circles in a

mv12 1 qQ 1 1 1 qQ 1 = + qv1 B ) T1 = mv12 = + qv1 r1 B. 2 r1 4⇡✏0 r1 2 2 4⇡✏0 r1 2 1 ⇠ = r 1 1 + dr = r 1 1 dr r r , while v1 = v + dv, B = dB. To first order, then, ✓ ◆ 1 1 qQ dr 1 qvr 1 1 qQ T1 = 1 + q(vr) dB, and hence dT = T1 T = dB dr. 2 4⇡✏0 r r 2 2 2 4⇡✏0 r2

But r1 = r + dr, so (r1 )

1

qr dB qr dv Now, the induced electric field is E = 2r dB dt (Ex. 7.7), so m dt = qE = 2 dt , or m dv = 2 dB. The increase in qvr 1 2 kinetic energy is therefore dT = d( 2 mv ) = mv dv = 2 dB. Comparing the two expressions, I conclude that dr = 0. qed Problem 7.53 d ↵ E= = ↵. So the current in R1 and R2 is I = ; by Lenz’s law, it flows counterclockwise. Now dt R1 + R 2 ↵R1 the voltage across R1 (which voltmeter #1 measures) is V1 = IR1 = (Vb is the higher potential), R 1 + R2

and V2 =

IR2 =

↵R2 (Vb is lower ). R 1 + R2

Problem 7.54 d dB ⇡↵r2 (a) E = = ⇡r2 = ↵⇡r2 = IR ) (in magnitude) I = . If B is out dt dt R law says the current is clockwise. (b) Inside the shaded region, for a circle of radius s, apply Faraday’s law: I ↵s ˆ ↵s ↵ E · dl = E2⇡s = ⇡s2 ↵ ) E = = ( sin x ˆ + cos y ˆ) = (s sin x ˆ s cos 2 2 2 Z Z p ↵ Along the line from P to Q, dl = dx x ˆ, and y = r/ 2, so V = E · dl = y dx = 2 ↵r2 P is at the higher voltage, and the meter reads . 2 c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

of the page, Lenz’s

↵ (y x ˆ xy ˆ). 2 ↵ r p p (r 2). Thus 2 2

y ˆ) =

162


That’s the simplest way to do it. But you might instead regard the 3/4-circle+chord as a circuit, and use Kirchho↵’s rule (the total emf around a closed loop is zero): V + IR1 ↵A1 = 0, where R1 = 3/4R is the resistance of the curved portion, and A1 = (3/4)⇡r2 + r2 /2 = (r2 /4)(3⇡ + 2) is the area (3/4 of the circle, plus the triangle). Then r2 ⇡r2 ↵ 3 ↵r2 ↵r2 V = ↵ (3⇡ + 2) R= (3⇡ + 2 3⇡) = . X 4 R 4 4 2 Or we could do the same thing, using the small loop at the top: V + IR2 ↵A2 = 0, where R2 = (1/4)R and A2 = (1/4)⇡r2 r2 /2 = (r2 /4)(⇡ 2). V =

⇡↵r2 R R 4

↵

r2 (⇡ 4

2) =

↵r2 (⇡ 4

⇡ + 2) =

↵r2 . X 2

Problem 7.55 E = vBh =

dI dv d2 v hB dI L ; F = IhB = m ; = = dt dt dt2 m dt

Problem 7.56 A point on the upper loop: r2 = (a cos

r

2

= (r2

r1 )2 = (a cos

= a2 cos2

2

= a2 + b2 + z 2

b cos

2

2ab cos

2

cos

2ab(cos

⇥ = (a2 + b2 + z 2 ) 1

2 , a sin

2

2 1)

1

2

1

✓

◆

hB L

v,

d2 v = dt2

hB ! 2 v, with ! = p . mL

a point on the lower loop: r1 = (b cos

+ (a sin

+ b2 cos2

1

cos

2 cos(

2 , z);

hB m

b sin

2

+ a2 sin2

+ sin 2 sin ⇤ ab ⇥ 1 1) =

2 1)

1 , 0).

+ z2

2ab sin

2

1 , b sin

1

sin

2

+ b2 sin2

1)

= a2 + b2 + z 2 2ab cos( ⇤ 2 cos( 2 1) .

2

1

+ z2

1)

dl1 = b d 1 ˆ1 = b d 1 [ sin 1 x ˆ + cos 1 y ˆ]; dl2 = a d 2 ˆ2 = a d 2 [ sin 2 x ˆ + cos 2 y ˆ], so dl1 ·dl2 = ab d 1 d 2 [sin 1 sin 2 + cos 1 cos 2 ] = ab cos( 2 ) d d . 1 1 2 II ZZ µ0 dl1 ·dl2 µ0 ab cos( 2 1) p M= d 2 d 1. r = 4⇡ pab/ 4⇡ 1 2 cos( 2 1) Both integrals run from 0 to 2⇡. Do the 2⇡ Z

1

2

integral first, letting u ⌘

cos u p du = 1 2 cos u

1

Z2⇡ 0

p

2

1:

cos u du 1 2 cos u

(since the integral runs over a complete cycle of cos u, we may as well change the limits to 0 ! 2⇡). Then the 1 integral is just 2⇡, and Z 2⇡ Z 2⇡ µ0 p cos u µ0 p cos u p p M= ab 2⇡ du = ab du. 4⇡ 2 1 2 cos u 1 2 cos u 0 0 (a) If a is small, then p

1

1 ⇠ =1+ 2 cos u

⌧ 1, so (using the binomial theorem) cos u, and

Z

0

2⇡

p

cos u du ⇠ = 1 2 cos u

Z

0

2⇡

cos u du +

Z

2⇡

cos2 u du = 0 + ⇡,

0


163

CHAPTER 7. ELETRODYNAMICS p and hence M = (µ0 ⇡/2) ab

3.

µ0 ⇡a2 b2 (same as in Prob. 7.22). 2(b2 + z 2 )3/2

⇠ = ab/(b2 + z 2 ), so M ⇠ =

Moreover,

(b) More generally,

(1 + ✏)

1/2

1 3 ✏ + ✏2 2 8

=1

5 3 ✏ + · · · =) p 16 1

1 =1+ 2 cos u

cos u +

3 2

2

cos2 u +

5 2

3

cos3 u + · · · ,

so M=

µ0 p ab 2

µ0 p = ab 2

⇢Z

2⇡

Z

cos u du +

0

 3 0 + (⇡) + 2

2⇡

cos2 u du +

0

2

5 (0) + 2

3

3 2

2

Z

2⇡

cos3 u du +

0

3 µ0 ⇡ p ( ⇡) + · · · = ab 4 2

Problem 7.57 Let be the flux of B through a single loop of either coil, so that E1 =

N1

d , dt

E2 =

N2

= I1 L1 + M I2 = N1 ;

In case I1 = 0, we have (b)

E1 =

d 1 dt

=

1 M 2 dI dt

M RI2

Z

0

15 1+ 8

= N1

2⇡

2

and

cos4 u du + · · · +()

2

4

+ ···

◆

qed

.

= N2 . Then

= I2 L2 + M I1 = N2 ,

or

is the flux through one turn): = I1

L1 M L2 M + I2 = I2 + I1 . N1 N1 N2 N2

L2 L1 L2 M M N2 ; if I2 = 0, we have N1 = N2 . Dividing: L1 = M , or L1 L2 = dI1 2 2 + M dI E2 = ddt2 = L2 dI I2 R. qed dt = V1 cos(!t); dt + M dt = dI1 dI2 2 equation by L2 : L1 L2 dt + L2 dt M = L2 V1 cos !t. Plug in L2 dI dt =

M N1 1 L1 dI dt

(c) Multiply the first

2

1

✓

3

d E2 N2 , so = . qed dt E1 N1

Problem 7.58 (a) Suppose current I1 flows in coil 1, and I2 in coil 2. Then (if 1

3

5 2

=

M 2.

qed

I2 R

1 M dI dt .

L2 V1 L2 V1 1 cos !t. L1 dI dt + M M R ! sin !t = V1 cos !t. MR ◆ ✓ ◆ L2 V1 1 L2 ! sin !t ) I1 (t) = sin !t + cos !t . R L1 ! R

1 M 2 dI dt = L2 V1 cos !t ) I2 (t) =

dI1 V1 = dt L1

✓

cos !t

cos !t R L2 N2 N2 = = . The ratio of the amplitudes is . qed V1 cos !t M N1 N1 ✓ ◆ ✓ ◆ V1 1 L2 (V1 )2 1 L2 2 (e) Pin = Vin I1 = (V1 cos !t) sin !t + cos !t = sin !t cos !t + cos !t . L1 ! R L1 ! R (d)

Vout I2 R = = Vin V1 cos !t

L2 V1 MR

(L2 V1 )2 cos2 !t. Average of cos2 !t is 1/2; average of sin !t cos !t is zero. M 2R ✓ ◆   1 L2 1 (L2 )2 1 (L2 )2 (V1 )2 L2 2 So hPin i = (V1 )2 ; hPout i = (V1 )2 = (V ) ; hPin i = hPout i = . 1 2 2 L1 R 2 M R 2 L1 L2 R 2L1 R Pout = Vout I2 = (I2 )2 R =


164


Problem 7.59 (a) The charge flowing into dz in time dt is dq = I(z) dt

I(z + dz) dt =

dI dz dt = (t + dt) dz dt

(t) dz =

d d dt dz ) = dt dt

dI . X dz

Since the left side is a function only of t, and the right side is a function only of z, they must both be constant; call it k: d dI = k ) (t) = kt + C1 ; = k ) I(z) = kz + C2 . dt dz If (0) = 0 and I(0) = 0 the constants C1 and C2 must both be zero: (t) = kt, I(z) = kz. X kt µ0 I ˆ µ0 kz ˆ (b) In the quasistatic approximation, E = ˆ s= ˆ s; B = = . 2⇡✏0 s 2⇡✏0 s 2⇡s 2⇡s ✓ ◆ 1 @ kt r·E = s =0 X s @s 2⇡✏0 s ✓ ◆ 1 @ µ0 kz r·B = =0 X s@ 2⇡s @E ˆ 1 @E @B r⇥E = ˆ z=0= X @z s@ @t @B ˆ 1 @(sB) µ0 k @E r⇥B = + ˆ z= ˆ z = µ0 ✏0 X @z s @s 2⇡s @t For a gaussian cylinder of radius s about the z axis, at height z and with width dz: I I kt kt dz Qenc E · da = (2⇡s) dz = dz = = . X B · da = 0. X 2⇡✏0 s ✏0 ✏0 ✏0

For a circular amperian loop of radius s about the z axis, at height z: I Z I Z d µ0 kz d E · dl = 0 = B · da, X B · dl = (2⇡s) = µ0 I = µ0 Ienc + µ0 ✏0 E · da. X dt 2⇡s dt R R (Note that B · da and E · da are zero through this loop.)

Problem 7.60 (a) The continuity equation says @⇢ r·J. Here the right side is independent of t, so we can integrate: @t = ⇢(t) = ( r·J)t+ constant. The “constant” may be a function of r—it’s only constant with respect to t. So, putting in the r dependence explicitly, and noting that r·J = ⇢(r, ˙ 0), ⇢(r, t) = ⇢(r, ˙ 0)t + ⇢(r, 0). qed R ⇢ rˆ R J⇥ rˆ µ0 1 (b) Suppose E = 4⇡✏0 r 2 d⌧ and B = 4⇡ r 2 d⌧ . We want to show that r·B = 0, r⇥B = 1 @B µ0 J + µ0 ✏0 @E ; r·E = ⇢, and r⇥E = , provided that J is independent of t. @t ✏0 @t ✓ ◆ R ˆ ⇢r 1 We know from Ch. 2 that Coulomb’s law E = 4⇡✏ d⌧ satisfies r·E = ✏10 ⇢ and r⇥E = 0. Since 0 r2

B is constant (in time), the r·E and r⇥E equations are satisfied. From Chapter 5 (specifically, Eqs. 5.475.50) we know that the Biot-Savart law satisfies r·B = 0. It remains only to check r⇥B. The argument in Sect. 5.3.2 carries through until the equation following Eq. 5.54, where I invoked r0 ·J = 0. In its place we now put r0 ·J = ⇢: ˙ Z rˆ µ0 r⇥B = µ0 J (J·r) 2 d⌧ (Eqs. 5.51-5.53) 4⇡ | {z r } ˆ ( J·r0 ) rr 2

(Eq. 5.54)


165 1


Integration by parts yields two terms, one of which becomes a surface integral, and goes to zero. The other is r 0 rˆ r ·J = ( ⇢). ˙ So: 3 2

r

r

µ0 4⇡

r⇥B = µ0 J

Z

rˆ 14 @ ( ⇢)d⌧ ˙ = µ0 J + µ0 ✏0 2 r @t

Contents

Problem Problem 7.61 7.56 1 ( )dz (a) dEz = sin ✓ 4⇡✏0 r 2 p z 2 2 sin ✓ = r ;Z r = z + s  vt z dz 1 p Ez = = 2 + s2 )3/2 4⇡✏0 (z 4⇡✏0 z 2 + s2 vt Problem 7.26 ( ) 1 1 p p Ez = . 4⇡✏0 (vt ✏)2 + s2 (vt)2 + s2

⇢

1 4⇡✏0

Z

⇢ rˆ

r

3

= µ0 J + µ0 ✏0

d⌧

@E . qed @t

y✻ dE θ✼

r ✏

s

$ vt − ϵ ! "# −z

✲z

vt

(b)

E

= =

4⇡✏0 2✏0

(c) Id = ✏0

Z

0

a

(

hp (vt

p (vt

d E = dt 2

1 ✏)2 + s2

✏)2 + a2

(

p

p

1 (vt)2 + s2

(vt)2 + a2

v(vt ✏) p (vt ✏)2 + a2

)

(✏

2⇡s ds =

i vt) + (vt) .

v(vt)

p

2✏0

(vt)2 + a2

+ 2v

)

hp (vt

✏)2 + s2

ia (vt)2 + s2 ˆ s 0 ✕ ✲ˆ z

p

✲ˆ z ❖

ˆ φ

.

w ☛

❄ h ✻

✕

→I

→

As ✏ ! 0, vt < ✏ also ! 0, so Id ! 2 (2v) = v = I. With an infinitesimal gap we attribute the magnetic field to displacement current, instead of real current, but we get the same answer. qed Problem 7.62 ✛ ✲ l B

✛ I 1 1 Q Q ✏0 wl ✏0 w (a) Parallel-plate capacitor: E = ; V = Eh = h)C= = ) C= . ✏0 ✏0 wl V h h

(b) B = µ0 K = µ0

I ; w

= Bhl =

µ0 I µ0 h µ0 h hl = LI ) L = l) L= . w w w

(c) CL = µ0 ✏0 = (4⇡ ⇥ 10 7 )(8.85 ⇥ 10 12 ) = 1.112 ⇥ 10 17 s2 /m2 . p p (Propagation speed 1/ LC = 1/ µ0 ✏0 = 2.999 ⇥ 108 m/s = c.) (d) D = , E = D/✏ = /✏, so just replace ✏0 by ✏; H = K, B = µH = µK, so just replace µ0 by µ.

LC = ✏µ;

p v = 1/ ✏µ.

c 2012 Education, Inc.,Inc., Upper Saddle River, NJ. NJ. All rights reserved. ThisThis material is is c Pearson ⃝2005 Pearson Education, Upper Saddle River, All rights reserved. material protected under all copyright lawslaws as they currently exist. No portion of this material maymay be be protected under all copyright as they currently exist. No portion of this material reproduced, in any formform or by means, without permission in writing fromfrom the publisher. reproduced, in any orany by any means, without permission in writing the publisher.


166


Problem 7.63 (a) J = (E + v⇥B); J finite, = 1 ) E + (v⇥B) = 0. Take the curl: r⇥E + r⇥(v⇥B) = 0. But @B @B Faraday’s law says r⇥E = . So = r⇥(v⇥B). qed @t @t H (b) r·B = 0 ) B·da = 0 for any closed surface. Apply this at time (t + dt) to the surface consisting of S, S 0 , and R: Z Z Z B(t + dt)·da + B(t + dt)·da B(t + dt)·da = 0 S0

R

S

(the sign change in the third term comes from switching outward da to inward da). Z Z Z Z ⇥ ⇤ d = B(t + dt)·da B(t)·da = B(t + dt) B(t) ·da B(t + dt)·da {z } S0 S S | R @B (for infinitesimal dt) @t dt d

=

⇢Z

S

@B ·da dt @t

Z

R

⇥ ⇤ B(t + dt)· (dl⇥v) dt (Figure 7.13).

Since the second term is already first order in dt, we can replace B(t + dt) by B(t) (the distinction would be second order): ⇢Z ✓ ◆ Z I Z @B @B ·da dt B·(dl⇥v) = dt ·da r⇥(v⇥B)·da . d = dt {z } @t S @t C| S S (v⇥B)·dl d = dt

Z  S

@B @t

r⇥(v⇥B) ·da = 0. qed

Problem 7.64 (a) 1 ⇢e cos ↵ + cµ0 ⇢m sin ↵ ✏0 1 1 1 1 (⇢e cos ↵ + cµ0 ✏0 ⇢m sin ↵) = (⇢e cos ↵ + ⇢m sin ↵) = ⇢0e . X ✏0 ✏0 c ✏0 1 1 (r · B) cos ↵ (r · E) sin ↵ = µ0 ⇢m cos ↵ ⇢e sin ↵ c c✏0 1 µ0 (⇢m cos ↵ ⇢e sin ↵) = µ0 (⇢m cos ↵ c⇢e sin ↵) = µ0 ⇢0m . X cµ0 ✏0 ✓ ◆ ✓ ◆ @B @E (r ⇥ E) cos ↵ + c(r ⇥ B) sin ↵ = µ0 Jm cos ↵ + c µ0 Je + µ0 ✏0 sin ↵ @t @t ✓ ◆ @ 1 @B0 µ0 (Jm cos ↵ cJe sin ↵) B cos ↵ E sin ↵ = µ0 J0m .X @t c @t ✓ ◆ ✓ ◆ 1 @E 1 @B (r ⇥ B) cos ↵ (r ⇥ E) sin ↵ = µ0 Je + µ0 ✏0 cos ↵ µ0 Jm sin ↵ c @t c @t 1 @ @E0 µ0 (Je cos ↵ + Jm sin ↵) + µ0 ✏0 (E cos ↵ + cB sin ↵) = µ0 J0e + µ0 ✏0 .X c @t @t

r · E0 = (r · E) cos ↵ + c(r · B) sin ↵ = = r · B0 = = r ⇥ E0 = = r ⇥ B0 = =



167

(b) 1 0 F0 = qe0 (E0 + v ⇥ B0 ) + qm (B0 v ⇥ E0 ) 2 c ✓ ◆ ✓ ◆ 1 1 = qe cos ↵ + qm sin ↵ (E cos ↵ + cB sin ↵) + v ⇥ B cos ↵ E sin ↵ c c ✓ ◆ 1 1 + (qm cos ↵ cqe sin ↵) B cos ↵ E sin ↵ v ⇥ (E cos ↵ + cB sin ↵) c c2 h = qe E cos2 ↵ + cB sin ↵ cos ↵ cB sin ↵ cos ↵ + E sin2 ↵ ✓ ◆i 1 1 +v ⇥ B cos2 ↵ E sin ↵ cos ↵ + E sin ↵ cos ↵ + B sin2 ↵ c c ◆ h✓ 1 1 +qm E sin ↵ cos ↵ + B sin2 ↵ + B cos2 ↵ E sin ↵ cos ↵ c c ✓ ◆i 1 1 1 1 2 2 +v ⇥ B sin ↵ cos ↵ E sin ↵ E cos ↵ B sin ↵ cos ↵ c c2 c2 c ✓ ◆ 1 = qe (E + v ⇥ B) + qm B v ⇥ E = F. qed c2


168

CHAPTER 8. CONSERVATION LAWS

Chapter 8

Conservation Laws Problem 8.1 Example 7.13.

9 1 > E= ˆ s> > 2⇡✏0 s =

1 I 1 S= (E ⇥ B) = ˆ z; 2 ✏ s2 > µ 4⇡ 0 0 µ0 I 1 ˆ > > ; B= 2⇡ s Z Zb Zb I 1 I P = S · da = S2⇡s ds = ds = ln(b/a). 2⇡✏0 s 2⇡✏0 a

But V =

Zb a

Problem 7.62.

E=

E · dl =

✏0

a

2⇡✏0

ˆ z

Zb a

1 ds = ln(b/a), so P = IV. s 2⇡✏0 9 > > =

S=

1 I (E ⇥ B) = y ˆ; µ0 ✏0 w

> µ0 I > x ˆ; w Z Z Ih P = S · da = Swh = , but V = E · dl = h, so P = IV. ✏0 ✏0 B = µ0 K x ˆ=

Problem 8.2 (a) E =

✏0

ˆ z;

=

Q It ; Q(t) = It ) E(t) = ˆ z. 2 ⇡a ⇡✏0 a2

@E 2 I⇡s2 µ0 Is ˆ ⇡s = µ0 ✏0 ) B(s, t) = . @t ⇡✏0 a2 2⇡a2 " # ✓ ◆ ✓ ◆2 ✓ ◆2 ⇤ 1 1 2 1 It 1 µ0 Is µ0 I 2 ⇥ 2 (b) uem = ✏0 E + B = ✏0 + = (ct)2 + (s/2)2 . 2 µ0 2 ⇡✏0 a2 µ0 2⇡a2 2⇡ 2 a4 ✓ ◆✓ ◆ 1 1 It µ0 Is I 2t S= (E ⇥ B) = ( ˆ s) = sˆ s. 2 2 µ0 µ0 ⇡✏0 a 2⇡a 2⇡ 2 ✏0 a4 B 2⇡s = µ0 ✏0


169


@uem µ0 I 2 2 I 2t I 2t I 2t @uem = 2c t = 2 4 ; r·S= r · (s ˆ s) = 2 4 = .X 2 4 2 4 @t 2⇡ a ⇡ ✏0 a 2⇡ ✏0 a ⇡ ✏0 a @t  Z Z b b 2 µ0 I 2 µ0 wI 2 1 s4 2s (c) Uem = uem w2⇡s ds = 2⇡w 2 4 [(ct)2 + (s/2)2 ]s ds = (ct) + 2⇡ a 0 ⇡a4 2 4 4 0  Z µ0 wI 2 b2 b2 I 2t I 2 wtb2 2 = (ct) + . Over a surface at radius b: P = S · da = [b ˆ s · (2⇡bw ˆ s )] = . in 2⇡a4 8 2⇡ 2 ✏0 a4 ⇡✏0 a4 dUem µ0 wI 2 b2 2 I 2 wtb2 = 2c t = = Pin . X (Set b = a for total.) 4 dt 2⇡a ⇡✏0 a4 Problem 8.3 The force is clearly in the z direction, so we need 1 $ ( T · da)z = Tzx dax + Tzy day + Tzz daz = µ0  1 1 2 = Bz (B · da) B daz . µ0 2

✓

Bz Bx dax + Bz By day + Bz Bz daz

1 2 B daz 2

◆

2 µ0 m ˆ (outside), where m = 4 ⇡R3 ( !R). (From µ0 R! ˆ z (inside) and B = (2 cos ✓ ˆ r + sin ✓ ✓) 3 4⇡r3 3 Eq. 5.70, Prob. 5.37, and Eq. 5.88.) We want a surface that encloses the entire upper hemisphere—say a hemispherical cap just outside r = R plus the equatorial circular disk. Now B =

Hemisphere:

Bz = da = B2 = $ ( T · da)z = = =

i ⇥ ⇤ µ0 m h ˆ z = µ0 m 2 cos2 ✓ sin2 ✓ = µ0 m 3 cos2 ✓ 1 . 2 cos ✓ (ˆ r ) + sin ✓ ( ✓) z 3 3 4⇡R 4⇡R 4⇡R3 µ0 m 2 2 R sin ✓ d✓ d ˆ r; B · da = (2 cos ✓)R sin ✓ d✓ d ; daz = R2 sin ✓ d✓ d cos ✓; 4⇡R3 ⇣ µ m ⌘2 ⇣ µ m ⌘2 0 0 4 cos2 ✓ + sin2 ✓ = 3 cos2 ✓ + 1 . 3 3 4⇡R 4⇡R  1 ⇣ µ0 m ⌘2 1 3 cos2 ✓ 1 2 cos ✓R2 sin ✓ d✓ d 3 cos2 ✓ + 1 R2 sin ✓ cos ✓ d✓ d 3 µ0 4⇡R 2 ✓ ◆2  !R 1 2 µ0 R sin ✓ cos ✓ d✓ d 12 cos2 ✓ 4 3 cos2 ✓ 1 3 2 ✓ ◆2 !R2 µ0 9 cos2 ✓ 5 sin ✓ cos ✓ d✓ d . 2 3

µ0 (Fhemi )z = 2

✓

= µ0 ⇡

✓

!R2 3 !R 3

◆2

2

2⇡

◆2 ✓

Z⇡/2

9 cos ✓

0

0+

5 cos ✓ sin ✓ d✓ = µ0 ⇡

3

9 4

5 2

◆

=

µ0 ⇡ 4

✓

wR 3

2

◆2

✓

.


!R2 3

◆2 

9 5 cos4 ✓ + cos2 ✓ 4 2

⇡/2

0

170


Disk:

2 µ0 R!; da = r dr d ˆ = r dr d ˆ z; 3 ✓ ◆2 2 2 B · da = µ0 R!r dr d ; B 2 = µ0 R! ; daz = r dr d . 3 3 ✓ ◆2  ✓ ◆2 1 2 1 1 2 $ µ0 R! r dr d + r dr d = µ0 R! r dr d . ( T · da)z = µ0 3 2 2µ0 3 ✓ ◆2 ZR ✓ ◆ 2 !R !R2 (Fdisk )z = 2µ0 2⇡ r dr = 2⇡µ0 . 3 3 Bz =

0

Total: F=

⇡µ0

✓

!R2 3

◆2 ✓

2+

1 4

◆

ˆ z=

⇡µ0

✓

!R2 2

◆2

ˆ z (agrees with Prob. 5.44).

Alternatively, we could use a surface consisting of the entire equatorial plane, closing it with a hemispherical surface “at infinity,” where (since the field is zero out there) the contribution is zero. We have already done the integral over the disk; it remains to do rest of the integral over the plane, from R to 1. On the plane, µ0 m ˆ µ0 m ✓ = 0, and (for r > R) B = ✓= ˆ z, so Hello 4⇡r3 4⇡r3  1 1 2 1 2 1 ⇣ µ0 m ⌘2 $ ( T · da)z = Bz (Bz daz ) Bz daz ) = Bz daz = ( r dr d ). µ0 2 2µ0 2µ0 4⇡r3 The contribution from the rest of the plane is therefore ✓ ◆2  Z 1 1 ⇣ µ0 m ⌘2 1 µ0 4 4 1 (Frest )z = dr = 2⇡ ⇡R ! 5 4 2µ0 4⇡ r 16⇡ 3 4r R ✓ ◆2 µ0 ⇡ !R2 = . 4 3

1

=

µ0 ⇡

R

✓

R4 ! 3

◆2 

1 4R4

This is the same as (Fhemi )z , so—when added to (Fdisk )z —it will yield the same total force as before. Problem 8.4 (a) $ ( T · da)z = Tzx dax + Tzy day + Tzz daz .

z

✻

r

But for the x y plane dax = day = 0, and daz = a r dr d (I’ll calculate the upper charge). ✓ the force on ◆ r ✼θ 1 $ E 2 ( r dr d ). ( T · da)z = ✏0 Ez Ez ✇ a 2 1 q r Now E = 2 2 cos ✓ ˆ r, and cos ✓ = , so Ez = x❂ r r 4⇡✏ ✓ 0 ◆2 2 q r 0, E 2 = 3 . Therefore 2 2⇡✏0 (r + a2 ) ✓ ◆2 Z1 Z1 1 q r3 dr q2 1 u du (letting u ⌘ r2 ) Fz = ✏0 2⇡ 3 = 4⇡✏ 2 2 2 2 2⇡✏0 (u + a2 )3 0 (r + a ) 0 0 " #1  q2 1 1 a2 q2 1 1 a2 q2 1 = + = 0+ 2 = . X 3 2 4 4⇡✏0 2 (u + a ) 2 (u + a2 ) 4⇡✏0 2 a 2a 4⇡✏0 (2a)2

✲y

0


171

CHAPTER 8. CONSERVATION LAWS 1 q a 2 2 sin ✓ ˆ z, and sin ✓ = r , so 4⇡✏0 r ◆2 ✓ qa 1 $ 2 2 E = Ez = 3 , and hence ( T · da)z = 2 2 2⇡✏0 (r + a )

(b) In this case E =

Fz =

✏0 2

✓

qa 2⇡✏0

◆2

2⇡

Z1 0

r dr 3

(r2 + a2 )

=

q 2 a2 4⇡✏0

"

1 1 2 4 (r + a2 )2

✏0 2

#1

✓

qa 2⇡✏0

◆2

r dr d

Therefore

3.

(r2 + a2 )

 q 2 a2 1 0+ 4 = 4⇡✏0 4a

=

0

q2 1 . X 4⇡✏0 (2a)2

Problem 8.5 (a) E = ˆ z, B = µ0 v x ˆ, g = ✏0 (E ⇥ B) = µ0 2 v y ˆ, p = (dA)g = dAµ0 2 v y ˆ. ✏0 (b) (i) There is a magnetic force, due to the (average) magnetic field at the upper plate: F = q(u ⇥ B) = A[( u ˆ z) ⇥ ( 12 µ0 v x ˆ)] = 12 µ0 2 Avu y ˆ, Z Z I1 = F dt = 12 µ0 2 Av y ˆ u dt = 12 dµ0 2 Av y ˆ.

[The velocity of the patch (of area A) is actually v + u = v y ˆ uˆ z, but the y component produces a magnetic force in the z direction (a repulsion of the plates) which reduces their (electrical) attraction but does not deliver (horizontal) momentum to the plates.] (ii) Meanwhile, in the space immediately above the upper plate the magnetic field drops abruptly to zero (as the plate moves past), inducing an electric field by Faraday’s law. The magnetic field in the vicinity of the top plate (at d(t) = d0 ut) can be written, using Problem 1.46(b), B(z, t) =

µ0 v ✓(d

z) x ˆ,

)

@B = µ0 vu (d @t

z) x ˆ.

In the analogy at the beginning of Section 7.2.2, the Faraday-induced electric field is just like the magnetostatic field of a surface current K = vu x ˆ. Referring to Eq. 5.58, then, ( 1 ˆ, for z < d, 2 µ0 vu y Eind = + 12 µ0 vu y ˆ, for z > d. This induced electric field exerts a force on area A of the bottom plate, F = ( an impulse Z ˆ u dt = 12 µ0 2 Avd y ˆ. I2 = 12 µ0 2 Av y

A)(

1 2 µ0

vu y ˆ), and delivers

(I dropped the subscript on d0 , reverting to the original notation: d is the initial separation of the plates.) The total impulse is thus I = I1 + I2 = dAµ0 2 v y ˆ, matching the momentum initially stored in the fields, from part (a). [I thank Michael Ligare for untangling this surprisingly subtle problem. Incidentally, there is also “hidden momentum” in the original configuration. It is not relevant here; it is (relativistic) mechanical momentum (see Example 12.13), and is delivered to the plates as they come together, so it does not a↵ect the overall conservation of momentum.] Problem 8.6 (a) gem = ✏0 (E ⇥ B) = ✏0 EB y ˆ; pem = ✏0 EBAd y ˆ. Z 1 Z 1 Z 1 Z (b) I = F dt = I(l ⇥ B) dt = IBd(ˆ z⇥x ˆ) dt = (Bd y ˆ) 0

0

0

0

1

✓


dQ dt

◆

dt

172 = (Bdˆ y)[Q(1) I = ✏0 EBAd y ˆ.


Q(0)] = BQd y ˆ.

Problem 8.7 (a) Ex = Ey = 0, Ez =

But the original field was E = /✏0 = Q/✏0 A, so Q = ✏0 EA, and hence

/✏0 . Therefore

Txy = Txz = Tyz = · · · = 0;

$ T =

(b) F =

I

✏0 2 E = 2

Txx = Tyy =

2

2✏0

;

✓ Tzz = ✏0 Ez2

1 2 E 2

◆

=

2 ✏0 2 E = . 2 2✏0

1 1 0 0 @ 0 1 0 A. 2✏0 0 0 +1 2

0

$ T · da (S = 0, since B = 0); integrate over the x y plane: da =

dx dy ˆ z (negative because

outward with respect to a surface enclosing the upper plate). Therefore Fz =

Z

Tzz daz =

2

2✏0

A, and the force per unit area is f =

F = A

2

2✏0

ˆ z.

2 (c) Tzz = /2✏0 is the momentum in the z direction crossing a surface perpendicular to z, per unit area, per unit time. (d) The recoil force is the momentum delivered per unit time, so the force per unit area on the top plate is 2

f=

2✏0

ˆ z

(same as (b)).

Problem 8.8 B = µ0 nI ˆ z (for s < R; outside the solenoid B = 0). The force on a segment ds of spoke is dF = I 0 dl ⇥ B = I 0 µ0 nI ds(ˆ s⇥ˆ z) =

I 0 µ0 nI ds ˆ.

The torque on the spoke is N=

Z

r ⇥ dF = I µ0 nI 0

ZR a

1 s ds( ˆ s ⇥ ˆ) = I 0 µ0 nI R2 2 Z

1 µ0 nI(R2 2

a2 ( ˆ z).

Therefore the angular momentum of the cylinders is L = N dt = a )ˆ z so 1 L= µ0 nIQ(R2 a2 ) ˆ z (in agreement with Eq. 8.34). 2 Problem 8.9 Q 1 (a) Between the shells, E = ˆ r, 4⇡✏0 r2 L=

Z

(r ⇥ g) d⌧ =

QB0 4⇡

Z

g = ✏0 (E ⇥ B) =

2

Z

I 0 dt. But

R

I 0 dt = Q,

QB0 (ˆ r⇥ˆ z). 4⇡r2

1 QB0 [r ⇥ (ˆ r⇥ˆ z)]r2 sin ✓ dr d✓ d = r2 4⇡

Z

r[ˆ r ⇥ (ˆ r⇥ˆ z)] sin ✓ dr d✓ d .


173


Now ˆ r ⇥ (ˆ r⇥ˆ z) = ˆ r(ˆ r·ˆ z) ˆ z(ˆ r·ˆ r) = ˆ r cos ✓ = ˆ z, but L has to be along the z direction, so we pick o↵ the z component of ˆ r: [ˆ r ⇥ (ˆ r⇥ˆ z)]z = cos2 ✓ 1 = sin2 ✓. Lz =

QB0 4⇡

Z

r sin ✓ dr d✓ d = 3

QB0 2⇡ 4⇡

Z

⇡

sin ✓ d✓ 3

0

Z

b

QB0 2

r dr =

a

1 QB0 (b2 3

L=

✓ ◆✓ 2 ◆ 4 b a2 = 3 2

1 QB0 (b2 a2 ). 3

a2 ) ˆ z.

(b) The changing magnetic field induces an electric field (Example 7.7). Assuming symmetry about the z s dB ˆ axis, E = . The force on a patch of area da is dF = E da, and the torque on this patch is dN = s⇥dF. 2 dt The net torque on the sphere at radius a is (using s = a sin ✓ ˆ s and da = a2 sin ✓ d✓ d ): ✓ ◆✓ ◆Z Z Q 1 dB Qa2 dB Qa2 dB 2 Na = ˆ z s da = ˆ z 2⇡ sin3 ✓ d✓ = ˆ z. 2 4⇡a 2 dt 8⇡ dt 3 dt Qb2 dB Q dB 2 Similarly, Nb = ˆ z, so the total torque is N = (b 3 dt 3 dt to the spheres is Z Z 0 Q 2 dB 2 L = N dt = (b a )ˆ z dt = 3 B0 dt Problem 8.10 (a)

(where m =

8 0, > <

a2 ) ˆ z, and the angular momentum delivered Q 2 (b 3

X

a2 )B0 ˆ z.

9 (r < R) > =

82 9 z, (r < R) > > < 3 µ0 M ˆ = ; B= (Ex. 6.1) E= h i > > > ˆ , (r > R) > : 1 Qˆ ; : µ0 m 2 cos ✓ ˆ ; r + sin ✓ ✓ r , (r > R) 4⇡ r3 4⇡✏0 r2 4 3 µ0 Qm ˆ sin ✓, and (ˆ ˆ = ˆ, so ⇡R M ); g = ✏0 (E ⇥ B) = (ˆ r ⇥ ✓) r ⇥ ✓) 3 (4⇡)2 r5 `=r⇥g =

But (ˆ r ⇥ ˆ) = µ0 mQ L= ˆ z (4⇡)2

µ0 mQ sin ✓(ˆ r ⇥ ˆ). (4⇡)2 r4

ˆ and only the z component will survive integration, so (since (✓) ˆz= ✓, Z

sin2 ✓ 2 r sin ✓ dr d✓ d r4

.

Z2⇡ 0

µ0 mQ L= ˆ z (2⇡) (4⇡)2

d = 2⇡;

Z⇡

4 sin ✓ d✓ = ; 3 3

0

✓ ◆✓ ◆ 4 1 = 3 R

Z1

R

2 µ0 M QR2 ˆ z. 9


sin ✓):

1 dr = r2

✓

1 r

◆

1 R

=

1 . R

a

✼θ ✇

r a

✲x

y❂ −q

174

CHAPTER 8. CONSERVATION LAWS z ✻

(b) Apply Faraday’s law to the ring shown: ✓ ◆ I d 2 dM E · dl = E(2⇡r sin ✓) = = ⇡(r sin ✓)2 µ0 dt 3 dt ) E=

θR

µ0 dM (r sin ✓) ˆ. 3 dt

✓ µ0 dM The force on a patch of surface (da) is dF = E da = (r sin ✓) da ˆ z 3 dt ✻ µ0 dM 2 ✶ The torque on the′ patch is✸dN = r ⇥ dF = r sin ✓ da (ˆ r ⇥ ˆ). But r 3 dt only the qmz component (✓ˆz = sin ✓): Z r µ0 dM d N= ˆ z r2 sin2 ✓ r2 sin ✓ d✓ d . θ 3 dt ✲x qe ✓ ◆ Z⇡ Z2⇡ 4 µ0 dM 4 Here r = R; sin3 ✓ d✓ = ; d = 2⇡, so N = ˆ z R4 (2⇡) = 3 3 dt 3 0

◆ Q . 4⇡R2 ˆ and we want (ˆ r ⇥ ˆ) = ✓,

=

2µ0 dM QR2 ˆ z. 9 dt

0

L=

Z

2µ0 QR2 ˆ z 9

N dt =

Z0

dM =

2µ0 M QR2 ˆ z 9

(same as (a)).

M

(c) Let the charge on the sphere at time t be q(t); the charge density is

=

(“south of”) the ring in the figure is qs =

2⇡R

2

Z⇡

sin ✓0 d✓0 =

q(t) . The charge below 4⇡R2

q q ⇡ ( cos ✓0 )|✓ = (1 + cos ✓). 2 2

0

So the total current crossing the ring (flowing “north”) is I(t) =

1 dq (1 + cos ✓), and hence 2 dt

I ˆ = 1 dq (1 + cos ✓) ✓. ˆ The force on a patch of area da is dF = (K ⇥ B) da. ( ✓) 2⇡R sin ✓ 4⇡R dt sin ✓  4 c ⃝2005 Pearson Education, Thisµmaterial is ⇡R3River, M NJ. All rights reserved. 2 Inc., Upper µ0Saddle 0M ˆ this1material ˆ protected under portion may Baveall=copyright µ0 Mlaws ˆ z +as they3 currently (2exist. cos ✓No ˆ r+ sin ✓of✓) = [2 ˆ zbe+ 2 cos ✓ ˆ r + sin ✓ ✓]; 3 3 4⇡ R 2 6 reproduced, in any form or by any means, without permission in writing from the publisher.

K(t) =

K⇥B=

1 dq µ0 M (1 + cos ✓) ˆ [2(✓ ⇥ ˆ z) + 2 cos ✓ (✓ˆ ⇥ ˆ r)]. | {z } 4⇡R dt 6 sin ✓ ˆ

dN = R ˆ r ⇥ dF = =

µ0 M 12⇡

✓

dq dt

◆

µ0 M 24⇡

✓

dq dt

◆

(1 + cos ✓) 2[ ˆ r ⇥ (✓ˆ ⇥ ˆ z) | {z } sin ✓ ˆ r·ˆ ˆ ✓(ˆ z) ˆ z(ˆ r · ✓)

cos ✓ (ˆ r ⇥ ˆ)]R2 sin ✓ d✓ d | {z }

2 ˆ d✓ d = µ0 M R (1 + cos ✓)R2 [cos ✓ ✓ˆ + cos ✓ ✓] 6⇡

ˆ ✓

✓

dq dt

◆

ˆ (1 + cos ✓) cos ✓ d✓ d ✓.


175


ˆz= The x and y components integrate to zero; (✓)

sin ✓, so (using

Z2⇡

d = 2⇡):

0

Nz = =

µ0 M R2 6⇡ µ0 M R2 3

✓ ✓

dq dt

◆

(2⇡)

Z⇡

(1 + cos ✓) cos ✓ sin ✓ d✓ =

µ0 M R2 3

0

◆✓ ◆ dq 2 = dt 3

2µ0 dq M R2 . 9 dt

✓

dq dt

◆✓

sin2 ✓ 2

cos3 ✓ 3

◆

⇡ 0

2µ0 dq M R2 ˆ z. 9 dt

N=

Therefore L=

Z

N dt =

2µ0 M R2 ˆ z 9

Z0

dq =

2µ0 M R2 Q ˆ z (same as (a)). 9

Q

(I used the average field at the discontinuity—which is the correct thing to do—but in this case you’d get the same answer using either the inside field or the outside field.) µ0 Ib b2 ˆ z. (I put the 2 (b2 + z 2 )3/2 origin at the center of the upper loop, with the z axis pointing up, so z = h at the location of the lower loop.) Treat the lower loop as a magnetic dipole, with moment m = Ia ⇡a2 ˆ z. The force on m is given by Eq. 6.3: ✓ ◆ µ0 Ib b2 µ0 ⇡a2 b2 Ia Ib @ 2 3/2 F = r(m · B) = r Ia ⇡a2 = b + z2 ˆ z 2 2 3/2 2 (b + z ) 2 @z ✓ ◆ µ0 ⇡a2 b2 Ia Ib 3 3⇡ a2 b2 h 5/2 = b2 + z 2 (2z) ˆ z= µ0 Ia Ib 2 ˆ z. X 2 2 2 (b + h2 )5/2 Problem 8.11 The magnetic field of the upper loop is given by Eq. 5.41: B =

Problem 8.12 Following the method of Section 7.2.4 (leading up to Eq. 7.30), the power delivered to the two loops is dW = Ea Ia Eb Ib dt (where these are the changing values as the currents are turned on, not the final values). Now Ea = so dW = dt

✓

La Ia

La

dIa dIb + M Ia dt dt

◆

dIa dt

M

dIb , dt

Eb =

Lb

dIb dt

M

dIa , dt

✓ ◆ ✓ ◆ dIb dIa d 1 1 + Lb Ib + M Ib = La Ia2 + Lb Ib2 + M Ia Ib . dt dt dt 2 2

The total work done, then, to increase the currents from zero to their final values, is W = Problem 8.13 d (a) E = ; dt

1 1 La Ia2 + Lb Ib2 + M Ia Ib . qed 2 2

= ⇡a2 B; B = µ0 nIs ; E = Ir R. So Ir =

1 dIs µ0 ⇡a2 n . R dt


176 (b)

I

E · dl =

S=


d ) E(2⇡a) = dt

1 1 (E ⇥ B) = µ0 µ0

✓

1 dIs ˆ µ0 Ir b2 µ0 an . B= ˆ z (Eq. 5.41). 2 dt 2 (b2 + z 2 )3/2 ! µ0 Ir b2 1 dIs ab2 n (ˆ⇥ˆ z) = µ0 Ir ˆ s. 3/2 2 (b2 + z 2 ) 4 dt (b2 + z 2 )3/2

µ0 ⇡a2 n

µ0 an dIs 2 dt

◆

dIs )E= dt

Power: Z

Z1

Z1 1 dIs 1 2 2 P = S · da = (S)(2⇡a) dz = ⇡µ0 a b nIr dz 3/2 2 dt (b2 + z 2 ) 1 1 ✓ ◆ 1 z 1 1 2 The integral is p = 2 = 2. 2 2 2 2 b b b 1 b z +b ✓ ◆ dI s = ⇡µ0 a2 n Ir = (RIr )Ir = Ir 2 R. qed dt Problem 8.14 The fields are zero for s < a and s > b; between the cylinders, E=

1 ˆ s, 2⇡✏0 s

B=

µ0 I ˆ , 2⇡ s

where I = v. (a) From Eq. 8.5: 1 u= 2

✓

1 2 ✏0 E + B µ0 2

◆

" ✓ # ◆2 2 2 1 1 1 ⇣ µ0 ⌘2 2 v 2 1 = ✏0 + = (1 + ✏0 µ0 v 2 ) 2 . 2 2 2 2 2⇡✏0 s µ0 2⇡ s 8⇡ ✏0 s

Writing ✏0 µ0 = 1/c2 , and integrating over the volume between the cylinders: 2 W = ` 8⇡ 2 ✏0

✓

v2 1+ 2 c

◆Z

a

(b) From Eq. 8.29: g = ✏0 (E ⇥ B) = ✏0 p µ0 2 v = ˆ z ` 4⇡ 2 (c) From Eq. 8.10: S=

b

Z

a

b

2 1 2⇡s ds = s2 4⇡✏0

✓

2⇡✏0 s

1 q ˆ r, 4⇡✏0 r2

B=

µ0 v 2⇡s

◆

v2 1+ 2 c

ˆ z=

◆

ln(b/a).

µ0 2 v ˆ z. 4⇡ 2 s2

1 µ0 2 v 2⇡s ds = ln(b/a) ˆ z. 2 s 2⇡

1 1 (E ⇥ B) = g = c2 g. µ0 ✏0 µ0

Problem 8.15 (a) The fields are E =

◆✓

✓

dW = dt

Z

S · da =

µ0

2 2

c v ln(b/a) ˆ z. 2⇡

µ0 N I ˆ (for points inside the toroid—see Eq. 5.60). 2⇡ s

g = ✏0 (E ⇥ B) = ✏0

q µ0 N I µ0 qN I ˆ z= ˆ z. 4⇡✏0 a2 2⇡a 8⇡ 2 a3


177

CHAPTER 8. CONSERVATION LAWS (Note that inside the toroid r ⇡ a ˆ s and s ⇡ a.) p=

✓

◆ µ0 qN I µ0 qN Iwh ˆ z (2⇡awh) = ˆ z. 2 3 8⇡ a 4⇡ a2

(b) The changing magnetic field induces an electric field, as given by Problem 7.19 (with z = 0): µ0 dI E= N hw ˆ z. The impulse delivered to q is therefore 4⇡a2 dt Z Z Z µ0 dI µ0 q I = F dt = qE dt = q N hw ˆ z dt = N hwI ˆ z. X 4⇡a2 dt 4⇡a2 Problem 8.16 According to Eqs. 3.104, 4.14, 5.89, and 6.16, the fields are 8 9 8 9 1 2 > > > > > > P, (r < R), > > > > µ M, (r < R), 0 < 3✏0 = <3 = E= B= > > > > > > > > 1 1 > > : µ0 m [3(m · ˆ : ; r) ˆ r m] , (r > R), ; [3(p · ˆ r ) ˆ r p], (r > R), 3 4⇡ r 4⇡✏0 r3 R where p = (4/3)⇡R3 P, and m = (4/3)⇡R3 M. Now p = ✏0 (E ⇥ B) d⌧ , and there are two contributions, one from inside the sphere and one from outside. Inside: ◆ ✓ ◆ Z ✓ Z 1 2 2 2 4 8 pin = ✏0 µ0 M d⌧ = µ0 (P ⇥ M) d⌧ = µ0 (P ⇥ M) ⇡R3 = µ0 ⇡R3 (M ⇥ P). P ⇥ 3✏0 3 9 9 3 27 Outside: pout = ✏0

1 µ0 4⇡✏0 4⇡

Z

1 {[3(p · ˆ r) ˆ r r6

p] ⇥ [3(m · ˆ r) ˆ r

m]} d⌧.

Now ˆ r ⇥(p⇥m) = p(ˆ r ·m) m(ˆ r ·p), so ˆ r ⇥[ˆ r ⇥(p⇥m)] = (ˆ r ·m)(ˆ r ⇥p) (ˆ r ·p)(ˆ r ⇥m), whereas using the BACCAB rule directly gives ˆ r ⇥[ˆ r ⇥(p⇥m)] = ˆ r[ˆ r ·(p⇥m)] (p⇥m)(ˆ r ·ˆ r). So {[3(p · ˆ r) ˆ r p] ⇥ [3(m · ˆ r) ˆ r m]} = 3(p·ˆ r)(ˆ r ⇥m)+3(m·ˆ r)(ˆ r ⇥p)+(p⇥m) = 3 {ˆ r[ˆ r · (p ⇥ m)] (p ⇥ m)}+(p⇥m) = 2(p⇥m)+3ˆ r[ˆ r ·(p⇥m)]. Z µ0 1 pout = { 2(p ⇥ m) + 3 ˆ r[ˆ r · (p ⇥ m)]} r2 sin ✓ dr d✓ d . 2 16⇡ r6 To evaluate the integral, set the z axis along (p ⇥ m); then ˆ r · (p ⇥ m) = |p ⇥ m| cos ✓. Meanwhile, ˆ r = sin ✓ cos x ˆ + sin ✓ sin y ˆ + cos ✓ ˆ z. But sin and cos integrate to zero, so the x ˆ and y ˆ terms drop out, leaving ✓Z 1 ◆⇢ Z Z µ0 1 pout = dr 2(p ⇥ m) sin ✓ d✓ d + 3|p ⇥ m| ˆ z cos2 ✓ sin ✓ d✓ d 16⇡ 2 r4 0 ✓ ◆1 µ0 1 4⇡ µ0 = 2(p ⇥ m)4⇡ + 3(p ⇥ m) = (p ⇥ m) 16⇡ 2 3r3 R 3 12⇡R3 ✓ ◆ ✓ ◆ µ0 4 3 4 3 4µ0 3 = ⇡R P ⇥ ⇡R M = R (M ⇥ P). 12⇡R3 3 3 27 ✓ ◆ 8 4 4 ptot = + µ0 R3 (M ⇥ P) = µ0 R3 (M ⇥ P). 27 27 9 c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

178


Problem 8.17 (a) From Eq. 5.70 and Prob. 5.37, 8 2 e > < r < R : E = 0, B = µ0 R! ˆ z, with = ; 3 4⇡R2 1 e µ0 m > ˆ with m = 4 ⇡ !R4 . :r > R : E = ˆ r, B = (2 cos ✓ ˆ r + sin ✓ ✓), 4⇡✏0 r2 4⇡ r3 3 The energy stored in the electric field is (Ex. 2.9):

WE =

1 e2 . 8⇡✏0 R

The energy density of the internal magnetic field is: ✓ ◆2 1 2 1 2 e µ0 ! 2 e2 µ0 ! 2 e2 4 3 µ0 e2 ! 2 R uB = B = µ0 R! = , so W = ⇡R = . B in 2µ0 2µ0 3 4⇡R2 72⇡ 2 R2 72⇡ 2 R2 3 54⇡ The energy density in the external magnetic field is: uB =

WBout

1 µ20 m2 e2 ! 2 R4 µ0 1 4 cos2 ✓ + sin2 ✓ = 3 cos2 ✓ + 1 , so 2 6 2µ0 16⇡ r 18(16⇡ 2 ) r6

µ0 e2 ! 2 R4 = (18)(16)⇡ 2

Z1

1 2 r dr r6

Z⇡

3 cos ✓ + 1 sin ✓ d✓ 2

0

R

WB = WBin + Wbout =

Z2⇡

d =

µ0 e2 ! 2 R4 (18)(16)⇡ 2

0

✓

1 3R3

◆

(4)(2⇡) =

µ0 e2 ! 2 R . 108⇡

µ0 e ! R µ0 e ! R 1 e2 µ0 e2 ! 2 R (2 + 1) = ; W = WE + WB = + . 108⇡ 36⇡ 8⇡✏0 R 36⇡ 2

2

2

(b) Same as Prob. 8.10(a), with Q ! e and m !

2

1 µ0 e2 !R e!R2 : L = ˆ z. 3 18⇡

µ0 e2 ~ 9⇡~ (9)(⇡)(1.05 ⇥ 10 34 ) !R = ) !R = = = 9.23 ⇥ 1010 m/s. 2 18⇡ 2 µ0 e (4⇡ ⇥ 10 7 )(1.60 ⇥ 10 19 )2 " " ✓ ◆2 # ✓ ◆2 # ✓ ◆2 1 e2 2 !R 2 !R 2 9.23 ⇥ 1010 2 1+ = mc ; 1 + =1+ = 2.10 ⇥ 104 ; 8⇡✏0 R 9 c 9 c 9 3 ⇥ 108 (c)

(2.10 ⇥ 104 )(1.6 ⇥ 10 19 )2 9.23 ⇥ 10 10 11 = 2.95 ⇥ 10 m; ! = = 3.13 ⇥ 1021 rad/s. 8⇡(8.85 ⇥ 10 12 )(9.11 ⇥ 10 31 )(3 ⇥ 108 )2 2.95 ⇥ 10 11 Since !R, the speed of a point on the equator, is 300 times the speed of light, this “classical” model is clearly unrealistic. Problem 8.18 Maxwell’s equations with magnetic charge (Eq. 7.44): R=

(i) r · E =

1 ⇢e , ✏0

(ii) r · B = µ0 ⇢m ,

(iii) r ⇥ E =

µ0 Jm

@B , @t

(iv) r ⇥ B = µ0 Je + µ0 ✏0

@E . @t


179

CHAPTER 8. CONSERVATION LAWS The Lorentz force law becomes (Eq. 8.44) ✓ F = qe (E + v ⇥ B) + qm B Following the argument in Section 8.1.2:  ✓ F · dl = qe (E + v ⇥ B) + qm B dW = dt

Z

◆ 1 v ⇥ E . c2

◆ 1 v ⇥ E · v dt = (qe E + qm B) · v dt, c2

(E · Je + B · Jm ) d⌧

(which generalizes Eq. 8.6). Use (iii) and (iv) to eliminate Je and Jm : (E · Je + B · Jm ) = but r · (E ⇥ B) = B · (r ⇥ E)

1 E · (r ⇥ B) µ0

✏0 E ·

@E @t

1 B · (r ⇥ E) µ0

E · (r ⇥ B), so 1 r · (E ⇥ B) µ0

(E · Je + B · Jm ) =

1 @B B· , µ0 @t

1 @ 2 @t

✓

1 2 ✏0 E + B µ0

1 µ0

I

(E ⇥ B) · da,

2

◆

(generalizing Eq. 8.8). Thus dW = dt

d dt

Z

1 2

✓

✏0 E 2 +

1 2 B µ0

◆

d⌧

which is identical to Eq. 8.9. Evidently Poynting’s theorem is unchanged(!), and u=

1 2

✓

✏0 E 2 +

◆ 1 2 B , µ0

S=

1 (E ⇥ B), µ0

the same as before. To construct the stress tensor we begin with the generalization of Eq. 8.13: ✓ ◆ Z  1 F= (E + v ⇥ B) ⇢e + B v ⇥ E ⇢m d⌧. c2 The force per unit volume (Eq. 8.14) ✓ f = (⇢e E + Je ⇥ B) + ⇢m B ✓ 1 = ✏0 (r · E)E + r⇥B µ0

becomes

◆ 1 Jm ⇥ E c2 ◆ ✓ ◆ @E 1 1 1 @B ✏0 ⇥B+ (r · B)B µ0 ✏0 r⇥E ⇥E @t µ0 µ0 µ0 @t 1 @ = ✏0 [(r · E)E E ⇥ (r ⇥ E)] + [(r · B)B B ⇥ (r ⇥ B)] ✏0 (E ⇥ B) µ0 @t   1 1 1 @ = ✏0 (r · E)E r(E 2 ) + (E · r)E + (r · B)B r(B 2 ) + (B · r)B ✏0 (E ⇥ B). 2 µ0 2 @t


z

✻

+q a r a

✼θ ✇

y❂ CHAPTER 8. CONSERVATION LAWS −q

180

This is identical to Eq. 8.16, so the stress tensor is the same as before: ✓

Tij ⌘ ✏0 Ei Ej

1 2

ij E

2

◆

1 + µ0

✓

1 2

Bi Bj

ij B

2

◆

.

Likewise, Eq. 8.20 is still valid. In fact, this argument is more straightforward when you include magnetic charge, since you don’t need artificially to insert the (r · B)B term (after Eq. 8.15). The electromagnetic momentum density (Eq. 8.29) also stays the same, since the argument in Section 8.2.3 is formulated entirely in terms of the fields: g = ✏0 (E ⇥ B). Problem 8.19

z

qe r E= ; 4⇡✏0 r3 µ0 qm r0 µ0 qm (r d ˆ z) B= = . 3 2 2 0 4⇡ r 4⇡ (r + d 2rd cos ✓)3/2

✻

✶ ✸

r′ qm d

r θ

✲x

qe

Momentum density (Eq. 8.32): g = ✏0 (E ⇥ B) =

µ0 qe qm ( d)(r ⇥ ˆ z) . 2 3/2 3 2 2 (4⇡) r (r + d 2rd cos ✓)

Angular momentum density (Eq. 8.33): ` = (r ⇥ g) =

µ0 qe qm d r ⇥ (r ⇥ ˆ z) . But r ⇥ (r ⇥ ˆ z) = r(r · ˆ z) (4⇡)2 r3 (r2 + d2 2rd cos ✓)3/2

r2 ˆ z = r2 cos ✓ ˆ r

r2 ˆ z.

The x and y components will integrate to zero; using (ˆ r)z = cos ✓, we have: Z µ0 qe qm d r2 (cos2 ✓ 1) L= ˆ z r2 sin ✓ dr d✓ d . Let u ⌘ cos ✓ : 2 3/2 3 2 2 (4⇡) r (r + d 2rd cos ✓) Z1 Z1 r 1 u2 µ0 qe qm d = ˆ z (2⇡) du dr. 3/2 (4⇡)2 (r2 + d2 2rdu) 1 0

Do the r integral first: Z1 0

r dr (r2 + d2

3/2

2rdu)

=

d(1

(ru d) p 2 u ) r 2 + d2

1

2rdu

0

=

u d (1

u2 )

+

c ⃝2005 Pearson Education, Inc., Upper Saddle River, NJ. A protected under all copyright laws as they currently exist. N d u +or1 by any means, 1 without permissio reproduced, in any form = = . 2 2 d (1 u ) d d (1 u ) d(1 u)

Then µ0 qe qm d 1 L= ˆ z 8⇡ d

Z1 1

(1 u2 ) µ0 qe qm du = ˆ z (1 u) 8⇡

Z1 1

(1 + u)du =

✓ ◆ µ0 qe qm u2 ˆ z u+ 8⇡ 2

1

= 1

µ0 qe qm ˆ z. 4⇡


181


Problem 8.20 Z Z Z p = ✏0 (E ⇥ B) d⌧ = ✏0 (rV ) ⇥ B d⌧ = ✏0 [r ⇥ (V B) V V V I Z Z 1 = ✏0 V B ⇥ da + ✏0 µ0 V J d⌧ = 2 V J d⌧. c V S V

V r ⇥ B] d⌧

(I used Problem 1.61(b) in the penultimate step. Here V is all of space, and S is its surface at infinity, where B = 0, so the surface integral vanishes.) Using V (r) ⇡ V (0) + (rV ) · r = V (0) E(0) · r, Z Z 1 1 p = 2 V (0) J d⌧ [E(0) · r]J d⌧. c c2 R R R For a current loop, J d⌧ ! I dl = I dl = 0, and (Eq. 1.108): Z Z Z [E(0) · r]J d⌧ ! [E(0) · r]I dl = I [E(0) · r] dl = Ia ⇥ E(0) = m ⇥ E. So p=

1 (m ⇥ E). X c2

Problem 8.21 (a) The rotating shell at radius b produces a solenoidal magnetic field: B = µ0 K ˆ z, where K =

b !b b,

and

b

Q . So B = 2⇡bl

=

µ0 !b Q ˆ z (a < s < b). 2⇡l

(Note that if angular velocity is defined with respect to the z axis, then !b is a negative number.) The shell at a also produces a magnetic field (µ0 !a Q/2⇡l) ˆ z, in the region s < a, so the total field inside the inner shell is B=

µ0 Q (!a 2⇡l

!b ) ˆ z, (s < a).

Meanwhile, the electric field is

1 Q ˆ s= ˆ s, (a < s < b). 2⇡✏0 s 2⇡✏0 ls ✓ ◆✓ ◆ Q µ0 !b Q µ0 !b Q2 ˆ µ0 !b Q2 g = ✏0 (E ⇥ B) = ✏0 (ˆ s⇥ˆ z) = ; ` = r ⇥ g = (r ⇥ ˆ). 2⇡✏0 ls 2⇡l 4⇡ 2 l2 s 4⇡ 2 l2 s E=

Now r ⇥ ˆ = (s ˆ s+zˆ z) ⇥ ˆ = s ˆ z L=

µ0 !b Q2 ˆ z 4⇡ 2 l2

zˆ s, and the ˆ s term integrates to zero, so Z

d⌧ =

µ0 !b Q2 ⇡(b2 4⇡ 2 l2

a2 )l ˆ z=

µ0 !b Q2 (b2 4⇡l

a2 )

ˆ z.

(b) The extra electric field induced by the changing magnetic field due to the rotating shells is given by d 1 d ˆ E 2⇡s = )E= , and in the region a < s < b dt 2⇡s dt ✓ ◆ µ0 Q µ0 Q!b µ0 Q 1 µ0 Q d!a d!b ˆ = (!a !b ) ⇡a2 ⇡ s2 a2 = !a a2 !b s2 ; E(s) = a2 s2 . 2⇡l 2⇡l 2l 2⇡s 2l dt dt c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

182


In particular, µ0 Qa 4⇡l

E(a) =

✓

d!a dt

d!b dt

◆

ˆ,

and E(b) =

µ0 Q 4⇡lb

✓

2 d!a

a

dt

b

2 d!b

dt

◆

ˆ.

The torque on a shell is N = r ⇥ qE = qsE ˆ z, so ✓

◆✓ ◆ Z 1 µ0 Qa d!a d!b µ0 Q2 a2 Na = Qa ˆ z; La = Na dt = (!a !b ) ˆ z. 4⇡l dt dt 4⇡l 0 ✓ ◆✓ ◆ Z 1 µ0 Q d!a d!b µ0 Q2 2 Nb = Qb a2 b2 ˆ z; Lb = Nb dt = a ! a b2 ! b ˆ z. 4⇡lb dt dt 4⇡l 0 Ltot = La + Lb =

µ0 Q2 2 a !a 4⇡l

a2 !a + a2 !b ˆ z=

b2 ! b

µ0 Q2 !b 2 (b 4⇡l

a2 ) ˆ z.

Thus the reduction in the final mechanical angular momentum (b) is equal to the residual angular momentum in the fields (a). X Problem 8.22 B = µ0 nI ˆ z, (s < R);

E=

q 4⇡✏0

r r 3 , where r

g = ✏0 (E ⇥ B) = ✏0 (µ0 nI)

✓

q 4⇡✏0

◆

= (x 1

r

3

a, y, z).

(r ⇥ ˆ z) =

µ0 qnI [y x ˆ 4⇡ r 3

(x

a) y ˆ].

Linear Momentum. Z

Z

yx ˆ (x a) y ˆ dx dy dz. The x ˆ term is odd in y; it integrates to zero. [(x a)2 + y 2 + z 2 ]3/2 µ0 qnI (x a) = y ˆ dx dy dz. Do the z integral first : 4⇡ [(x a)2 + y 2 + z 2 ]3/2 1 z 2 p = . [(x a)2 + y 2 ] [(x a)2 + y 2 ] (x a)2 + y 2 + z 2 1 Z µ0 qnI (x a) = y ˆ dx dy. Switch to polar coordinates : 2⇡ [(x a)2 + y 2 ]

p=

g d⌧ =

µ0 qnI 4⇡ Z

x = s cos , y = s sin , dx dy ) s ds d ; [(x a)2 + y 2 ] = s2 + a2 2sa cos . Z µ0 qnI (s cos a) = y ˆ s ds d 2⇡ (s2 + a2 2sa cos ) ✓ ◆ Z 2⇡ Z 2⇡ cos d 2⇡ A d 2⇡ Now = 1 p ; =p . 2 2 2 (A + B cos ) B (A + B cos ) A B A B2 0 0 p Here A2 B 2 = (s2 + a2 )2 4s2 a2 = s4 + 2s2 a2 + a4 4s2 a2 = (s2 a2 )2 ; A2 B 2 = a2 ✓ 2 ◆ Z  Z R µ0 qnI a + s2 2a2 µ0 qnI µ0 qnIR2 = y ˆ 1 + s ds = y ˆ s ds = y ˆ. 2a a2 s2 (a2 s2 ) a 2a 0

s2 .


183

CHAPTER 8. CONSERVATION LAWS Angular Momentum.

µ0 qnI µ0 qnI r ⇥ [y x ˆ (x a) y ˆ] = z(x a) x ˆ + zy y ˆ [x(x a) + y 2 ] ˆ z . 4⇡ r 3 4⇡ r 3 The x ˆ and y ˆ terms are odd in z, and integrate to zero, so Z µ0 qnI x2 + y 2 xa L= ˆ z dx dy dz. The z integral is the same as before. 4⇡ [(x a)2 + y 2 + z 2 ]3/2 Z Z µ0 qnI x2 + y 2 xa µ0 qnI s a cos = ˆ z dx dy = ˆ z s2 ds d 2 2 2 2⇡ [(x a) + y ] 2⇡ (s + a2 2sa cos ) ✓ ◆ Z  Z R 2 s2 a2 + s2 s s2 + 1 s ds = µ qnI ˆ z s ds = zero. = µ0 qnI ˆ z 0 2 2 2 2 2 a s a s s2 0 a ` = r⇥g =

Problem 8.23 (a) If Z we’re only interested in the work done on free charges and currents, Eq. 8.6 becomes dW @D @D = (E · Jf ) d⌧ . But Jf = r ⇥ H (Eq. 7.56), so E · Jf = E · (r ⇥ H) E · . From product dt @t @t V @B rule #6, r · (E ⇥ H) = H · (r ⇥ E) E · (r ⇥ H), while r ⇥ E = , so @t @B @B @D E · (r ⇥ H) = H · r · (E ⇥ H). Therefore E · Jf = H · E· r · (E ⇥ H), and hence @t @t @t ◆ Z ✓ I dW @D @B = E· +H· d⌧ (E ⇥ H) · da. dt @t @t V S This is Poynting’s theorem for the fields in matter. Evidently the Poynting vector, representing the power per unit area transported by the fields, is S = E ⇥ H, and the rate of change of the electromagnetic energy density @uem @D @B is =E· +H· . @t @t @t 1 For linear media, D = ✏E and H = B, with ✏ and µ constant (in time); then µ @uem @E 1 @B 1 @ 1 @ 1 @ = ✏E · + B· = ✏ (E · E) + (B · B) = (E · D + B · H) , @t @t µ @t 2 @t 2µ @t 2 @t so uem = 12 (E · D + B · H). qed (b) If we’re only interested in the force on free charges and currents, Eq. 8.13 f = ⇢f E + Jf ⇥ B. ✓ becomes ◆ @D @D But ⇢f = r · D, and Jf = r ⇥ H , so f = E(r · D) + (r ⇥ H) ⇥ B ⇥ B. Now @t ✓ ◆@t @ @D @B @B @D @ (D ⇥ B) = ⇥B+D⇥ , and = r ⇥ E, so ⇥B = (D ⇥ B) + D ⇥ (r ⇥ E), and @t @t @t @t @t @t @ hence f = E(r · D) D ⇥ (r ⇥ E) B ⇥ (r ⇥ H) (D ⇥ B). As before, we can with impunity add the @t term H(r · B), so f = {[E(r · D)

D ⇥ (r ⇥ E)] + [H(r · B)

B ⇥ (r ⇥ H)]}

@ (D ⇥ B). @t

The term in curly brackets can be written as the divergence of a stress tensor (as in Eq. 8.19), and the last term is (minus) the rate of change of the momentum density, g = D ⇥ B. c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

184


Problem 8.24 (a) Initially, the disk will rise like a helicopter. The force on one charge (velocity v = !R ˆ + vz ˆ z) is Fi = q(v ⇥ B) = qk

ˆ s ˆ ˆ z ⇣ s 0 !R vz = qk 2!Rz ˆ R 0 2z

⌘ Rvz ˆ + !R2 ˆ z .

The net force on all the charges is n X

d2 z = dt2

d2 z F= Fi = nqkR ! ˆ z=M 2 ˆ z; dt i=1 2

✓

nqkR2 M

◆

!.

[1]

The net torque on the disk is N=

n X i=1

(ri ⇥ Fi ) = n(R ˆ s) ⇥ ( qkRvz ˆ) =

nqkR2 vz ˆ z=I

d! ˆ z dt

where I is the moment of inertia of the disk. So ✓

d! = dt

nqkR2 I

◆

dz . dt

[2]

Di↵erentiate [2], and combine with [1]: d2 z = dt2

✓

I nqkR2

◆

d2 ! = dt2

✓

nqkR2 M

◆

!

)

d2 ! = ↵!, dt2

where

nqkR2 ↵⌘ p . MI

The solution (with initial angular velocity !0 and initial angular acceleration 0) is !(t) = !0 cos ↵t. Meanwhile, dz = dt

r ◆ ✓ ◆ I d! I I = !0 ↵ sin ↵t = !0 sin ↵t. nqkR2 dt nqkR2 M r Z t r I !0 I z(t) = !0 sin ↵t dt = (1 cos ↵t) . M 0 ↵ M ✓

[The problem is a little cleaner if you make the disk massless, and assign a mass m to each of the charges. Then M ! nm and I ! nmR2 , so ↵ = qkR/m and z(t) ! (!0 R/↵)(1 cos ↵t).] (b) The disk rises and falls harmonically, as its rotation slows down and speeds up. The total energy is E=

1 1 1 I 1 1 1 M vz2 + I! 2 = M !02 sin2 ↵t + I!02 cos2 ↵t = I!02 (sin2 ↵t + cos2 ↵t) = I!02 , 2 2 2 M 2 2 2

which is constant (and equal to the initial energy). [Of course, if you didn’t notice that the rotation rate is changing, you might think the magnetic force is doing work, as the disk oscillates up and down.]


185

CHAPTER 9. ELECTROMAGNETIC WAVES

Chapter 9

Electromagnetic Waves Problem 9.1 h i 2 @f1 @ 2 f1 b(z vt)2 2 b(z vt)2 = 2Ab(z vt)e b(z vt) ; = 2Ab e 2b(z vt) e ; @z @z 2 h i 2 2 2 @f1 @ 2 f1 @ 2 f1 = 2Abv(z vt)e b(z vt) ; = 2Abv ve b(z vt) + 2bv(z vt)2 e b(z vt) = v 2 2 . X 2 @t @t @z @f2 @ 2 f2 = Ab cos[b(z vt)]; = Ab2 sin[b(z vt)]; @z @z 2 @f2 @ 2 f2 @ 2 f2 = Abv cos[b(z vt)]; = Ab2 v 2 sin[b(z vt)] = v 2 2 . X 2 @t @t @z 2Ab(z vt) @ 2 f3 2Ab 8Ab2 (z vt)2 @f3 = ; = + ; @z [b(z vt)2 + 1]2 @z 2 [b(z vt)2 + 1]2 [b(z vt)2 + 1]3 2 @f3 @ 2 f3 2Abv 2 8Ab2 v 2 (z vt)2 2Abv(z vt) 2 @ f3 = ; = + = v .X @t [b(z vt)2 + 1]2 @t2 [b(z vt)2 + 1]2 [b(z vt)2 + 1]3 @z 2 h i 2 @f4 @ 2 f4 2 b(bz 2 +vt) 2 2 b(bz 2 +vt) = 2Ab2 ze b(bz +vt) ; = 2Ab e 2b z e ; @z @z 2 2 2 2 2 @f4 @ f4 @ f4 = Abve b(bz +vt) ; = Ab2 v 2 e b(bz +vt) 6= v 2 2 . 2 @t @t @z 2 @f5 @f5 3 @ f5 2 = Ab cos(bz) cos(bvt) ; = Ab sin(bz) cos(bvt)3 ; = 3Ab3 v 3 t2 sin(bz) sin(bvt)3 ; @z @z 2 @t 2 @ 2 f5 3 3 3 6 6 4 3 2 @ f5 = 6Ab v t sin(bz) sin(bvt) 9Ab v t sin(bz) cos(bvt) = 6 v . @t2 @z 2 Problem 9.2 @f @2f = Ak cos(kz) cos(kvt); = Ak 2 sin(kz) cos(kvt); @z @z 2 2 @f @2f 2 2 2@ f = Akv sin(kz) sin(kvt); = Ak v sin(kz) cos(kvt) = v .X @t @t2 @z 2 Use the trig identity sin ↵ cos = 12 [sin(↵ + ) + sin(↵ )] to write f=

A {sin[k(z + vt)] + sin[k(z 2

vt)]} ,


186


which is of the form 9.6, with g = (A/2) sin[k(z

vt)] and h = (A/2) sin[k(z + vt)].

Problem 9.3 (A3 )2 = A3 ei

3

A3 e

i

3

= A1 ei

1

+ A2 ei

2

A1 e

= (A1 )2 + (A2 )2 + A1 A2 ei 1 e i 2 + e i 1 ei p A3 = (A1 )2 + (A2 )2 + 2A1 A2 cos( 1 2 ).

A3 ei

3

3

= A3 (cos

3

+ i sin

= (A1 cos

1

+ A2 cos

= tan

1

✓

A1 sin A1 cos

3)

= A1 (cos 2)

1

+ i sin

+ i(A1 sin

+ A2 sin 1 + A2 cos 1

2 2

◆

1

1)

2

i

1

+ A2 e

2

= (A1 )2 + (A2 )2 + A1 A2 2 cos(

+ A2 (cos

+ A2 sin

i

2 ).

+ i sin 2 ) A3 sin tan 3 = A3 cos

2 );

1

2

3 3

=

A1 sin A1 cos

+ A2 sin 1 + A2 cos 1

2

;

2

.

Problem 9.4

@2f 1 @2f = 2 2 . Look for solutions of the form f (z, t) = Z(z)T (t). Plug 2 @z v @t d2 Z 1 d2 T 1 d2 Z 1 d2 T this in: T 2 = 2 Z 2 . Divide by ZT : = . The left side depends only on z, and the dz v dt Z dz 2 v 2 T dt2 right only on t, so both must be constant. Call 9 the constant k 2 . 8 side 2 d Z > > 2 > ) Z(z) = Aeikz + Be ikz , > > > < dz 2 = k Z = > 2 > > > > : d T = (kv)2 T ) T (t) = Ceikvt + De ikvt . > ; dt2 (Note that k must be real, else Z and T blow up; with no loss of generality we can assume k is positive.) f (z, t) = Aeikz + Be ikz Ceikvt + De ikvt = A1 ei(kz+kvt) + A2 ei(kz kvt) + A3 ei( kz+kvt) + A4 ei( kz kvt) . The general linear combination of separable solutions is therefore Z 1h i f (z, t) = A1 (k)ei(kz+!t) + A2 (k)ei(kz !t) + A3 (k)ei( kz+!t) + A4 (k)ei( kz !t) dk, The wave equation (Eq. 9.2) says

0

where ! ⌘ kv. But we can combine the third term with the first, by allowing k to run negative (! = |k|v remains positive); likewise the second and the fourth: Z 1h i f (z, t) = A1 (k)ei(kz+!t) + A2 (k)ei(kz !t) dk. 1

Because (in the end) we shall only want the the real part of f , it suffices to keep only one of these terms (since k goes negative, both terms include waves traveling in both directions); the second is traditional (though either would do). Specifically, Z 1 Re(f ) = [Re(A1 ) cos(kz + !t) Im(A1 ) sin(kz + !t) + Re(A2 ) cos(kz !t) Im(A2 ) sin(kz !t)] dk. 1

The first term, cos(kz + !t) = cos( kz !t), combines with the third, cos(kz !t), since the negative k is picked up in the other half of the range of integration, and the second, sin(kz +!t) = sin( kz !t), combines with the fourth for the same reason. So the general solution, for our purposes, can be written in the form Z 1 i(kz !t) ˜ f˜(z, t) = A(k)e dk qed (the tilde0 s remind us that we want the real part). 1


187


Problem 9.5

@gI 1 @gI @hR 1 @hR @gT 1 @gT = ; = ; = . @z v1 @t @z v1 @t @z v2 @t 1 @gI ( v1 t) 1 @hR (v1 t) 1 @gT ( v2 t) v1 Equation 9.27 ) + = ) gI ( v1 t) hR (v1 t) = gT ( v2 t) +  v1 @t v1 @t v2 @t v2 (where  is a constant). ✓ ◆ ✓ ◆ v1 2v2 Adding these equations, we get 2gI ( v1 t) = 1 + gT ( v2 t)+, or gT ( v2 t) = gI ( v1 t)+0 v2 v1 + v2 v2 (where 0 ⌘  ). Now gI (z, t), gT (z, t), and hR (z, t) are each functions of a single variable u (in the v1 + v2 first case u = z v1 t, in the second u = z v2 t, and in the third u = z + v1 t). Thus ✓ ◆ 2v2 gT (u) = gI (v1 u/v2 ) + 0 . v1 + v2 ✓ ◆ ✓ ◆ v1 v1 Multiplying the first equation by v1 /v2 and subtracting, 1 gI ( v1 t) 1+ hR (v1 t) =  ) v2 v2 ✓ ◆ ✓ ◆ ✓ ◆ v2 v1 v2 v2 v1 hR (v1 t) = gI ( v1 t)  , or hR (u) = gI ( u) + 0 . v1 + v2 v1 + v2 v1 + v2 Equation 9.26 ) gI ( v1 t) + hR (v1 t) = gT ( v2 t). Now

[The 8notation is tricky, so here’s an example: for a sinusoidal wave, < gI = AI cos(k1 z !t) = AI cos[k1 (z v1 t)] ) gI (u) = AI cos(k1 u). gT = AT cos(k2 z !t) = AT cos[k2 (z v2 t)] ) gT (u) = AT cos(k2 u). : hR = AR cos( k1 z !t) = AR cos[ k1 (z + v1 t)] ) hR (u) = AR cos( k1 u). AT 2v2 AR v2 v1 v1 Here 0 = 0, and the boundary conditions say = , = (same as Eq. 9.32), and 2 k1 = k2 AI v1 + v2 AI v1 + v2 v (consistent with Eq. 9.24).] Problem 9.6 (a) T sin ✓+

T sin ✓ = ma ) T

✓

@f @z

0+

@f @z

0

◆

=m

@2f @t2

. 0

◆ ✓ im! 2 ˜ A˜R )] = m( ! 2 A˜T ), or k1 (A˜I A˜R ) = k2 AT . T ✓ ◆ ✓ ◆ m! 2 ˜ 2k1 ˜ ˜ Multiply first equation by k1 and add: 2k1 AI = k1 + k2 i AT , or AT = A˜I . T k1 + k2 im! 2 /T ✓ ◆ 2k1 (k1 + k2 im! 2 /T ) ˜ k1 k2 + im! 2 /T ˜ A˜R = A˜T A˜I = A = AI . I k1 + k2 im! 2 /T k1 + k2 im! 2 /T ✓ ◆ p 2 ˜ If the second string is massless, so v2 = T /µ2 = 1, then k2 /k1 = 0, and we have AT = A˜I , 1 i ✓ ◆ ✓ ◆ 1+i m! 2 m(k1 v1 )2 mk1 T k1 1+i ˜ ˜ AR = AI , where ⌘ = = , or = m . Now = Aei , with 1 i k1 T k1 T T µ1 µ1 1 i ✓ ◆✓ ◆ 2 1+i 1 i (1 + i )2 1 + 2i A2 = = 1 ) A = 1, and ei = = ) 1 i 1+i (1 i )(1 + i ) 1+ 2 ✓ ◆ 2 2 i R i i I 1 tan = . Thus A e = e A e ) A = A , = + tan . R I R I R I 2 2 1 1 ✓ ◆ ✓ ◆✓ ◆ 2 2 2 4 2 i 2 Similarly, = Ae ) A = = )A= p . 1 i 1 i 1+i 1+ 2 1+ 2 (b) A˜I + A˜R = A˜T ; T [ik2 A˜T

ik1 (A˜I


188 Aei =

(1

2(1 + i ) 2(1 + i ) = ) tan i )(1 + i ) (1 + 2 ) 2

AT = p


1+

2

AI ;

Problem 9.7 @2f (a) F = T 2 @z

T

=

I

+ tan

1

.

= . So AT ei

T

2 =p 1+

2

ei AI ei I ;

@2f @2f @2f @f , or T 2 = µ 2 + . 2 @t @z @t @t d2 F˜ (b) Let f˜(z, t) = F˜ (z)e i!t ; then T e i!t 2 = µ( ! 2 )F˜ e i!t + ( i!)F˜ e i!t ) dz d2 F˜ d2 F˜ ! ˜ ˜ 2˜ ˜ ˜ ˜ ikz ˜ ikz T 2 = !(µ! + i )F , = k F , where k˜2 ⌘ (µ! + i ). Solution : F˜ (z) = Ae + Be . 2 dz dz T ! Resolve k˜ into its real and imaginary parts: k˜ = k + i ) k˜2 = k 2 2 + 2ik = (µ! + i ). T ⇣ ! ⌘2 1 ! ! µ! 2 2 2 2 4 2 2 2k = )= ; k  =k = ; or k k (µ! /T ) (! /2T )2 = 0 ) 2 T 2kT 2T k T i µ! 2 h i p p 1h k2 = (µ! 2 /T ) ± (µ! 2 /T )2 + 4(! /2T )2 = 1 ± 1 + ( /µ!)2 . But k is real, so k 2 is positive, so 2 2T r q h i 1/2 p p µ ! we need the plus sign: k = ! 1 + 1 + ( /µ!)2 .  = =p 1 + 1 + ( /µ!)2 . 2T 2kT 2T µ Plugging this in, F˜ = Aei(k+i)z + Be i(k+i)z = Ae z eikz + Bez e ikz . But the B term gives an exponentially increasing function, which we don’t want (I assume the waves are propagating in the +z direction), ˜ z ei(kz !t) . (The actual displacement of the string is the real part so B = 0, and the solution is f˜(z, t) = Ae z

@f @t

z=µ z

of this, of course.) (c) The wave is attenuated by the factor e z , which becomes 1/e when p q p 1 2T µ z= = 1 + 1 + ( /µ!)2 ; this is the characteristic penetration depth.  ✓ ◆ k1 k i ˜ ˜ (d) This is the same as before, except that k2 ! k + i. From Eq. 9.29, AR = AI ; k1 + k + i s ✓ ◆2 ✓ ◆✓ ◆ AR k1 k i k1 k + i (k1 k)2 + 2 (k1 k)2 + 2 = = . A = AI R AI k1 + k + i k1 + k i (k1 + k)2 + 2 (k1 + k)2 + 2 p (where k1 = !/v1 = ! µ1 /T , while k and  are defined in part b). Meanwhile ✓ ◆ ✓ ◆ k1 k i (k1 k i)(k1 + k + i) (k1 )2 k 2 2 2ik1 2k1  1 = = ) R = tan . k1 + k + i (k1 + k)2 + 2 (k1 + k)2 + 2 (k1 )2 k 2 2 Problem 9.8 (a) fv (z, t) = A cos(kz !t) x ˆ; fh (z, t) = A cos(kz !t + 90 ) y ˆ= A sin(kz !t) y ˆ. Since fv2 +fh2 = A2 , the vector sum f = fv +fh lies on a circle of radius A. At time t = 0, f = A cos(kz) x ˆ A sin(kz) y ˆ. At time t = ⇡/2!, f = A cos(kz 90 ) x ˆ A sin(kz 90 ) y ˆ = A sin(kz) x ˆ + A cos(kz) y ˆ. Evidently it circles counterclockwise . To make a wave circling the other way, use h = 90 .



189

(b)

(c) Shake it around in a circle, instead of up and down. Problem 9.9 ⇣ ! ⌘ ! ! x ˆ; n ˆ=ˆ z. k · r = x ˆ · (x x ˆ+yy ˆ +zˆ z) = x; k ⇥ n ˆ= c c c ⇣! ⌘ ⇣! ⌘ E0 E(x, t) = E0 cos x + !t ˆ z; B(x, t) = cos x + !t y ˆ. c c c

(a) k =

x ˆ⇥ˆ z=y ˆ.

◆ x ˆ+y ˆ+ˆ z x ˆ ˆ z p ; n ˆ = p . (Since n ˆ is parallel to the x z plane, it must have the form ↵ x ˆ+ ˆ z; 3 2 p since n ˆ · k = 0, = ↵; and since it is a unit vector, ↵ = 1/ 2.) (b) k =

! c

✓

x ˆ y ˆ ˆ z ! ! 1 1 ˆ⇥n k · r = p (ˆ x+y ˆ+ˆ z) · (x x ˆ+yy ˆ +zˆ z) = p (x + y + z); k ˆ=p 1 1 1 =p ( x ˆ + 2y ˆ 3c 3c 6 1 0 1 6  ✓ ◆ ! x ˆ ˆ z p E(x, y, z, t) = E0 cos p (x + y + z) !t ; 3c 2  ✓ E0 ! x ˆ + 2ˆ y p B(x, y, z, t) = cos p (x + y + z) !t c 3c 6


ˆ z

◆

.

ˆ z).

190


Problem 9.10 I 1.3 ⇥ 103 P = = = 4.3 ⇥ 10 6 N/m2 . For a perfect reflector the pressure is twice as great: c 3.0 ⇥ 108 8.6 ⇥ 10 6 N/m2 . Atmospheric pressure is 1.03 ⇥ 105 N/m2 , so the pressure of light on a reflector is (8.6 ⇥ 10

6

)/(1.03 ⇥ 105 ) = 8.3 ⇥ 10

11

atmospheres.

Problem 9.11 The fields are E(z, t) = E0 cos(kz (a) The electric force is Fe = qE = qE0 cos(kz qE0 v= x ˆ m But vave = C = 0, so v = (b) The magnetic force is Fm = q(v ⇥ B) = q

✓

qE0 m!

Z

cos(kz

qE0 sin(kz m!

◆✓

E0 c

◆

1 E0 cos(kz c dv !t) x ˆ = ma = m , so dt

!t) x ˆ, B(z, t) =

qE0 sin(kz m!

!t) dt =

!t) y ˆ, with ! = ck.

!t) x ˆ + C.

!t) x ˆ.

sin(kz

!t) cos(kz

q 2 E02 sin(kz m!c

!t)(ˆ x⇥y ˆ) =

!t) cos(kz

!t) ˆ z.

Z T q 2 E02 ˆ z sin(kz !t) cos(kz !t) dt, where T = 2⇡/! is the m!c 0 ⇤ 1 ⇥ 2 = sin (kz 2⇡) sin2 (kz) = 0, so (Fm )ave = 0. 2!

(c) The (time) average force is (Fm )ave = 1 sin2 (kz !t) 2! (d) Adding in the damping term,

period. The integral is

F = qE

mv = qE0 cos(kz

T 0

mv = m

!t)ˆ x

dv dv qE0 ) + v= cos(kz dt dt m

dv = A! sin(kz dt in, and using the trig identity cos u = cos ✓ cos(u + ✓) + sin ✓ sin(u + ✓), The steady state solution has the form v = A cos(kz

A! sin(kz

!t + ✓) + A cos(kz

!t + ✓) =

!t + ✓) x ˆ,

qE0 [cos ✓ cos(kz m

!t) x ˆ.

!t + ✓) x ˆ. Putting this

!t + ✓) + sin ✓ sin(kz

!t + ✓)] .

Equating like terms: A! =

qE0 qE0 ! sin ✓, A = cos ✓ ) tan ✓ = , A2 (! 2 + m m

2

)=

✓

qE0 m

◆2

)A=

So v=

qE p 0 m !2 +

2

cos(kz

!t + ✓)ˆ x, ✓ ⌘ tan

1

(!/ ); Fm =

q2 E 2 p 0 mc ! 2 +

2

cos(kz

qE p 0 m !2 +

2

!t + ✓) cos(kz

To calculate the time average, write cos(kz !t + ✓) = cos ✓ cos(kz !t) sin ✓ sin(kz know that the average of cos(kz !t) sin(kz !t) is zero, so Z T q 2 E02 p (Fm )ave = ˆ z cos ✓ cos2 (kz !t) dt. mc ! 2 + 2 0

.

!t) ˆ z.

!t). We already


191


The integral is T /2 = ⇡/!, and cos ✓ = /

p !2 +

2

(see figure), so (Fm )ave =

⇡ q 2 E02 m!c(! 2 +

2)

ˆ z.

w q g Problem 9.12 hf gi =

1 T

Z

T

0

ab = 2T

Z

0

a cos(k · r T

[cos(2k · r

!t +

a )b cos(k

2!t +

a

+

b)

·r

!t +

+ cos(

b ) dt

a

b )]

dt =

ab cos( 2T

b )T

a

=

1 ab cos( 2

a

b ).

Meanwhile, in the complex notation: f˜ = a ˜eik·r !t) , g˜ = ˜beik·r✓ !t) , where a ˜ = aei a , ˜b = bei b . So ◆ 1˜⇤ 1 i(k·r !t)˜⇤ i(k·r !t) 1 ˜⇤ 1 1˜⇤ 1 f g˜ = a ˜e b e = a ˜b = abei( a b ) , Re f g˜ = ab cos( a qed b ) = hf gi. 2 2 2 2 2 2 Problem 9.13

✓

Tij = ✏0 Ei Ej

1 2

ij E

2

◆

1 + µ0

✓

Bi Bj

1 2

ij B

2

◆

.

With the fields in Eq. 9.48, E has only an x component, and B only a y component. So all the “o↵-diagonal” (i 6= j) terms are zero. As for the “diagonal” elements: ✓ ◆ ✓ ◆ ✓ ◆ 1 2 1 1 2 1 1 2 2 Txx = ✏0 Ex Ex E + B = ✏0 E B = 0. 2 µ0 2 2 µ0 ✓ ◆ ✓ ◆ ✓ ◆ 1 2 1 1 2 1 1 2 Tyy = ✏0 E + By By B = ✏0 E 2 + B = 0. 2 µ0 2 2 µ0 ✓ ◆ ✓ ◆ 1 2 1 1 2 Tzz = ✏0 E + B = u. 2 µ0 2 So Tzz =

✏0 E02 cos2 (kz

!t + ) (all other elements zero).

The momentum of these fields is in the z direction, and it is being transported in the z direction, so yes, it does make sense that Tzz should be the only nonzero element in Tij . According $ to Sect. 8.2.3, T · da is the rate at which momentum crosses an area da. Here we have no momentum crossing areas oriented in the x or y direction; the momentum per unit time per unit area flowing across a surface oriented in the z direction is Tzz = u = gc (Eq. 9.59), so p = gcA t, and hence p/ t = gcA = momentum per unit time crossing area A. Evidently momentum flux density = energy density. X Problem 9.14 R=

✓

E0 R E0 I

◆2

(Eq. 9.86) ) R =

✓

1 1+

◆2

(Eq. 9.82), where

⌘

µ1 v1 ✏2 v2 . T = µ2 v2 ✏1 v1


✓

E0T E0I

◆2

(Eq. 9.87)

192

) T = T +R=

✓

2 1+

◆2


✏2 v2 µ1 ✏2 µ2 v2 µ1 (Eq. 9.82). [Note that = = ✏1 v1 µ2 ✏1 µ1 v1 µ2

⇥ 1 4 + (1 2 (1 + )

⇤ )2 =

1 (4 + 1 (1 + )2

2 +

2

)=

✓

v1 v2

◆2

v2 µ1 v1 = = .] v1 µ2 v2

1 (1 + 2 + (1 + )2

2

) = 1. X

Problem 9.15 ˜0 x ˜0 n ˜0 n ˜0 y ˜0 (ˆ Equation 9.78 is replaced by E ˆ+E ˆ = E ˆ T , and Eq. 9.80 becomes E ˆ E z⇥n ˆR ) = I R R T I R ˜ ˜ ˜ E0T (ˆ z⇥n ˆ T ). The y component of the first equation is E0R sin ✓R = E0T sin ✓T ; the x component of the ˜0 sin ✓R = ˜0 sin ✓T . Comparing these two, we conclude that sin ✓R = sin ✓T = 0, and hence second is E E R T ✓R = ✓T = 0. qed Problem 9.16 Aeiax + Beibx = Ceicx for all x, so (using x = 0), A + B = C. Di↵erentiate: iaAeiax + ibBeibx = icCeicx , so (using x = 0), aA + bB = cC. Di↵erentiate again: a2 Aeiax b2 Beibx = c2 Ceicx , so (using x = 0), a2 A + b2 B = c2 C. 2 a A + b2 B = c(cC) = c(aA + bB); (A + B)(a2 A + b2 B) = (A + B)c(aA + bB) = cC(aA + bB); a2 A2 + b2 AB + a2 AB + b2 B 2 = (aA + bB)2 = a2 A2 + 2abAB + b2 B 2 , or (a2 + b2 2ab)AB = 0, or (a b)2 AB = 0. But A and B are nonzero, so a = b. Therefore (A + B)eiax = Ceicx . a(A + B) = cC, or aC = cC, so (since C 6= 0) a = c. Conclusion: a = b = c. qed Problem 9.17 8 9 ˜I = E ˜0 ei(kI ·r !t) y < (i) ✏1 E1? = ✏2 E2? , (iii) E1 = E2 , Boundary conditions: > k k : (ii) B ? = B ? , (iv) µ11 B1 = µ12 B2 . 1 2 sin ✓2 v2 = . [Note: kI · r !t = kR · r sin ✓1 v1 exponential factors in applying the boundary conditions.] Law of refraction:

!t = kT · r

!t, at z = 0, so we can drop all

˜0 + E ˜0 = E ˜0 . Boundary condition (i): 0 = 0 (trivial). Boundary condition (iii): E I R T ✓ ◆ 1 ˜ 1 ˜ 1 ˜ ˜0 + E ˜0 = v1 sin ✓2 E ˜0 . Boundary condition (ii): E0I sin ✓1 + E0R sin ✓1 = E0T sin ✓2 ) E I R T v1 v1 v2 v2 sin ✓1 But the term in parentheses is 1,by the law of refraction, so this is the same as (iii). 1 1 ˜ 1 ˜ 1 ˜ Boundary condition (iv): E0I ( cos ✓1 ) + E E0 ( cos ✓2 ) ) 0R cos ✓1 = µ1 v1 v1 µ2 v2 T ✓ ◆ ˜0 ˜0 = µ1 v1 cos ✓2 E ˜0 . Let ↵ ⌘ cos ✓2 ; ⌘ µ1 v1 . Then E ˜0 ˜0 = ↵ E ˜0 . E E E I R T I R T µ2 v2 cos ✓1 cos ✓1 µ2 v2 ✓ ◆ 2 ˜0 and E ˜ 0 : 2E ˜0 = (1 + ↵ )E ˜0 ) E ˜0 = ˜0 ; Solving for E E R T I T T I 1+↵ c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

193


◆ ✓ ◆ 2 1+↵ ˜0 ) E ˜0 = 1 ↵ ˜0 . E E I R I 1+↵ 1+↵ 1+↵ Since ↵ and are positive, it follows that 2/(1 + ↵ ) is positive, and hence the transmitted wave is in phase ✓ ◆ 2 with the incident wave, and the (real) amplitudes are related by E0T = E0I . The reflected wave is 1+↵ ˜0 = E ˜0 E R T

˜0 = E I

✓

in phase if ↵ < 1 and 180 out of phase if ↵ > 1; the (real) amplitudes are related by E0R =

1 ↵ 1+↵

E0 I .

These are the Fresnel equations for polarization perpendicular to the plane of incidence. q p 2 1 sin2 ✓/ 2 sin2 ✓ To construct the graphs, note that ↵ = = , where ✓ is the angle of incidence, cos ✓ cos ✓ p 2 2.25 sin ✓ so, for = 1.5, ↵ = . [In the figure, the minus signs on the vertical axis should be decimal cos ✓ points.]

Isqthere a Brewster’s angle? Well, E0R = 0 would mean that ↵ = 1, and hence that ✓ ◆2 ✓ ◆2 1 (v2 /v1 )2 sin2 ✓ 1 µ2 v2 v2 µ2 v2 ↵= = = , or 1 sin2 ✓ = cos2 ✓, so cos ✓ µ1 v1 v1 µ1 v1 ✓ ◆2 ⇥ 2 ⇤ v2 1= sin ✓ + (µ2 /µ1 )2 cos2 ✓ . Since µ1 ⇡ µ2 , this means 1 ⇡ (v2 /v1 )2 , which is only true for optically v1 indistinguishable media, in which case there is of course no reflection—but that would be true at any angle, not just at a special “Brewster’s angle”. [If µ2 were substantially di↵erent from µ1 , and the relative velocities were just right, it would be possible to get a Brewster’s angle for this case, at ✓

v1 v2

◆2

=1

cos ✓ + 2

✓

µ2 µ1

◆2

cos2 ✓ ) cos2 ✓ =

(v1 /v2 )2 (µ2 /µ1 )2

1 (µ2 ✏2 /µ1 ✏1 ) 1 (✏2 /✏1 ) = = 2 1 (µ2 /µ1 ) 1 (µ2 /µ1 )

(µ1 /µ2 ) . (µ1 /µ2 )

But the media would be very peculiar.] By the same token, R is either always 0, or always ⇡, for a given interface—it does not switch over as you change ✓, the way it does for polarization in the plane of incidence. In particular, if = 3/2, then ↵ > 1, for ↵ = In general, for

p

2.25 sin2 ✓ > 1 if 2.25 cos ✓

> 1, ↵ > 1, and hence

sin2 ✓ > cos2 ✓, or 2.25 > sin2 ✓ + cos2 ✓ = 1. X

< 1, ↵ < 1, and ✓R = 0.◆ 2 1 At normal incidence, ↵ = 1, so Fresnel’s equations reduce to E0T = E0 I ; E 0 R = 1+ 1+ consistent with Eq. 9.82. R

= ⇡. For


E0I ,

194


Reflection and Transmission coefficients: R = T =

✏2 v2 ↵ ✏1 v1

✓

E0T E0 I

◆2

✓

= ↵

R+T =

◆2

2 1+↵

(1

✓

E0 R E0I

◆2

=

✓

1 ↵ 1+↵

◆2

.

Referring to Eq. 9.116,

.

↵ )2 + 4↵ (1 + ↵ )2

=

1

2↵ + ↵2 2 + 4↵ (1 + ↵ )2

=

(1 + ↵ )2 = 1. X (1 + ↵ )2

Problem 9.18 Equation p 9.106 ) = 2.42; Eq. 9.110 ) 1 (sin ✓/2.42)2 ↵= . cos ✓ ✓ ◆ E0 R ↵ (a) ✓ = 0 ) ↵ = 1. Eq. 9.109 ) = E0I ↵+ 1 2.42 1.42 = = = 0.415; 3.42 ✓ ◆1 + 2.42 E0T 2 2 = = = 0.585. E0I ↵+ 3.42 (b) Equation 9.112 ) ✓B = tan 1 (2.42) = 67.5 . (c) E0R = E0T ) ↵ = 2 ) ↵ = + 2 = 4.42; (4.42)2 cos2 ✓ = 1 sin2 ✓/(2.42)2 ; (4.42)2 (1 sin2 ✓) = (4.42)2 (4.42)2 sin2 ✓ = 1 0.171 sin2 ✓; 19.5 1 = (19.5 0.17) sin2 ✓; 18.5 = 19.3 sin2 ✓; sin2 ✓ = 18.5/19.3 = 0.959; sin ✓ = 0.979; ✓ = 78.3 . Problem 9.19 (a) Equation 9.120 ) ⌧ = ✏/ . Now ✏ = ✏0 ✏r (Eq. 4.34), ✏r ⇠ = n2 (Eq. 9.70), and for glass the index 2 12 of refraction is typically around 1.5, so ✏ ⇡ (1.5) ⇥ 8.85 ⇥ 10 = 2 ⇥ 10 11 C2 /N m2 , while = 1/⇢ ⇡ 12 1 11 12 10 (⌦ m) (Table 7.1). Then ⌧ = (2⇥10 )/10 = 20 s. (But the resistivity of glass varies enormously from one type to another, so this answer could be o↵ by a factor of 100 in either direction.) (b) For silver, ⇢ = 1.59 ⇥ 10 8 (Table 7.1), and ✏ ⇡ ✏0 , so !✏ = 2⇡ ⇥ 1010 ⇥ 8.85 ⇥ 10 12 = 0.56. Since = 1/⇢ = 6.25 ⇥ 107 !✏, the skin depth (Eq. 9.128) is 1 d= ⇠ = 

r

2 ! µ

=

r

2⇡ ⇥

1010

2 ⇥ 6.25 ⇥ 107 ⇥ 4⇡ ⇥ 10

7

= 6.4 ⇥ 10

7

m = 6.4 ⇥ 10

4

mm.

I’d plate silver to a depth of about 0.001 mm; there’s no point in making it any thicker, since the fields don’t penetrate much beyond this anyway. 8 7 6 12 5 (c) For copper, Table 7.1 gives r = 1/(1.68 ⇥ 10 ) = 6 ⇥ 10 , !✏0 = (2⇡ ⇥ 10 ) ⇥ (8.85 ⇥ 10 ) = 6 ⇥ 10 . ! µ Since !✏, Eq. 9.126 ) k ⇡ , so (Eq. 9.129) 2 = 2⇡

r

2 = 2⇡ ! µ0

r

2 2⇡ ⇥ 106 ⇥ 6 ⇥ 107 ⇥ 4⇡ ⇥ 10

7

= 4 ⇥ 10

4

m = 0.4 mm.


195

CHAPTER 9. ELECTROMAGNETIC WAVES ! ! = k 2⇡

From Eq. 9.129, the propagation speed is v =

= ⌫ = (4 ⇥ 10

4

) ⇥ 106 = 400 m/s. In vacuum,

c 3 ⇥ 108 = = 300 m; v = c = 3 ⇥ 108 m/s. (But really, in a good conductor the skin depth is so small, ⌫ 106 compared to the wavelength, that the notions of “wavelength” and “propagation speed” lose their meaning.) =

Problem 9.20 (a) Use the binomial expansion for the square root in Eq. 9.126: ⇠ =!

r

 ✏µ 1 ⇣ ⌘2 1+ 2 2 ✏!

1/2

1

=!

r

✏µ 1 p = 2 2 ✏! 2

r

µ . ✏

r 1 ⇠ 2 ✏ . qed =  µ 8 < ✏ = ✏r ✏0 = 80.1 ✏0 (Table 4.2), For pure water, µ = µ0 (1 + m ) = µ0 (1 9.0 ⇥ 10 6 ) ⇠ = µ0 (Table 6.1), : = 1/(2.5 ⇥ 105 ) (Table 7.1). r (80.1)(8.85 ⇥ 10 12 ) 5 So d = (2)(2.5 ⇥ 10 ) = 1.19 ⇥ 104 m. 4⇡ ⇥ 10 7 (b) In this case ( /✏!)2 dominates, so (Eq. 9.126) k ⇠ = , and hence (Eqs. 9.128 and 9.129) 2⇡ ⇠ 2⇡ = = 2⇡d, or d = . qed = k  2⇡ r r r r ✏µ !µ (1015 )(4⇡ ⇥ 10 7 )(107 ) 1 1 ⇠ Meanwhile  = ! = = = 8 ⇥ 107 ; d = = = 2 ✏! 2 2  8 ⇥ 107 1.3 ⇥ 10 8 = 13 nm. So the fields do not penetrate far into a metal—which is what accounts for their opacity. So (Eq. 9.128) d =

(c) Since k ⇠ = , as we found in (b), Eq. 9.134 says = tan 1 (1) = 45 . qed r r r B0 ⇠ µ B0 (107 )(4⇡ ⇥ 10 Meanwhile, Eq. 9.137 says = . For a typical metal, then, = = ✏µ E0 ✏! ! E0 1015 10 7 s/m. (In vacuum, the ratio is 1/c = 1/(3 ⇥ 108 ) = 3 ⇥ 10 about 100 times larger in a metal.) Problem 9.21✓ ◆  1 1 1 (a) u = ✏E 2 + B 2 = e 2z ✏E02 cos2 (kz 2 µ 2 over a full cycle, using hcos2 i = 12 and Eq. 9.137: 1 hui = e 2

2z



✏ 2 1 2 1 E + B = e 2 0 2µ 0 4 r

⇣

⌘2

2z

"

✏E02

!t +

1+

=

humag i B 2 /µ 1 = 02 = µ✏ huelec i E0 ✏ µ✏

E)

r

1+

⇣

✏!

2z

⌘2

✏E02

=

r

=

s/m, so the magnetic field is comparatively

+

1 2 B cos2 (kz µ 0

# r ⇣ ⌘2 1 2 1 + E0 ✏µ 1 + = e µ ✏! 4

2 k2 1 , so hui = e ✏! ✏µ ! 2 4 magnetic contribution to the electric contribution is But Eq. 9.126 ) 1 +

9

7)

2z

!t +

✏E02

2 k2 k2 = E2e 2 ✏µ ! 2µ! 2 0

1+

⇣

✏!

⌘2


"

1+

2z

> 1. qed

E

+ ) . Averaging

r

1+

⇣

✏!

⌘2

#

.

. So the ratio of the

196

1 cos(kz !t+ E ) cos(kz !t+ E + ) ˆ z; hSi = E0 B0 e 2z cos ˆ z. [The 2µ R 2⇡ 1 average of the product of the cosines is (1/2⇡) 0 cos ✓ cos(✓+ ) d✓ = (1/2) cos .] So I = E0 B0 e 2z cos = 2µ ✓ ◆ 1 2 2z K k E0 e cos , while, from Eqs. 9.133 and 9.134, K cos = k, so I = E 2 e 2z . qed 2µ ! 2µ! 0 (b) S =

1 1 (E⇥B) = E0 B0 e µ µ


2z

Problem 9.22

! ! 2 2 ˜0 E 1 ˜ 1 ˜ 1 ˜⇤ µ1 v1 ˜ R According to Eq. 9.147, R = = = , where ˜ = k2 ⇤ ˜ ˜ ˜ ˜ µ2 ! E0 I 1+ 1+ 1+ µ1 v1 = (k2 + i2 ) (Eqs. 9.125 and 9.146). Since silver is a good conductor ( ✏!), Eq. 9.126 reduces to µ2 ! r r r r r !µ2 µ1 v1 !µ2 ⇠ k2 = ⇠ ! ✏2 µ2 2 = = , so ˜ = (1 + i) = µ1 v1 (1 + i). 2 ✏2 ! 2 µ2 ! 2 s 2µ2 ! r r r µ0 (6 ⇥ 107 )(4⇡ ⇥ 10 7 ) Let ⌘ µ1 v1 = µ0 c =c = (3 ⇥ 108 ) = 29. Then 2µ2 ! 2µ0 ! 2! (2)(4 ⇥ 1015 ) ✓ ◆✓ ◆ 1 i 1 +i (1 )2 + 2 R= = = 0.93. Evidently 93% of the light is reflected. 1+ +i 1+ i (1 + )2 + 2 Problem 9.23 p (a) We are told that v = ↵ , where ↵ is a constant. But = 2⇡/krand v = !/k, so p p p d! 1 1 2⇡ 1 p 1 ! = ↵k 2⇡/k = ↵ 2⇡k. From Eq. 9.150, vg = = ↵ 2⇡ p = ↵ = ↵ = v, or v = 2vg . dk 2 k 2 2 2 k i(px Et) p E p2 ~k 2 ! E p ~k (b) = i(kx !t) ) k = , ! = = = . Therefore v = = = = ; ~ ~ ~ 2m~ 2m k p 2m 2m vg =

d! 2~k ~k p 1 = = = . So v = vg . Since p = mvc (where vc is the classical speed of the particle), it dk 2m m m 2

follows that vg (not v) corresponds to the classical veloctity. Problem 9.24

s ✓ ◆ 1 qd 1 q2 q2 2 E = ) F = qE = x = kspring x = m!0 x (Eq. 9.151). So !0 = . 3 3 4⇡✏0 a 4⇡✏0 a 4⇡✏0 ma3 s !0 1 (1.6 ⇥ 10 19 )2 ⌫0 = = = 7.16 ⇥ 1015 Hz. This is ultraviolet. 2⇡ 2⇡ 4⇡(8.85 ⇥ 10 12 )(9.11 ⇥ 10 31 )(0.5 ⇥ 10 10 )3 From Eqs. 9.173 and 9.174, ⇢ 0 s # nq 2 f N = # of molecules per unit volume = Avogadro = 22.4 liters A= , 2 2m✏0 !0 f = # of electrons per molecule = 2 (for H2 ).

6.02⇥1023 22.4⇥10 3

= 2.69 ⇥ 1025 ,

(2.69 ⇥ 1025 )(1.6 ⇥ 10 19 )2 = 4.2 ⇥ 10 5 (which is about 1/3 the actual value); (9.11 ⇥ 10 31 )(8.85 ⇥ 10 12 )(4.5 ⇥ 1016 )2 ✓ ◆2 ✓ ◆2 2⇡c 2⇡ ⇥ 3 ⇥ 108 B = = = 1.8 ⇥ 10 15 m2 (which is about 1/4 the actual value). !0 4.5 ⇥ 1016 =

So even this extremely crude model is in the right ball park.


197

CHAPTER 9. ELECTROMAGNETIC WAVES Problem 9.25

N q2 (!02 ! 2 ) Equation 9.170 ) n = 1 + . Let the denominator ⌘ D. Then 2m✏0 [(!02 ! 2 )2 + 2 ! 2 ] ⇢ ⇤ ⇥ ⇤ dn N q2 2! (!02 ! 2 ) ⇥ 2 = 2(!02 ! 2 )( 2!) + 2 2! = 0 ) 2!D = (!02 ! 2 ) 2(!02 ! 2 ) 2!; d! 2m✏0 D D2 2 (!02 ! 2 )2 + 2 ! 2 = 2(!02 ! 2 )2 (!02 ! 2 ), or (!02 ! 2 )2 = 2 (! 2 + !02 ! 2 ) = 2 !02 ) (!02 ! 2 ) = ±!0 ; p ! 2 = ! 2 ⌥ !0 , ! = !0 1 ⌥ /!0 ⇠ /2, and the = !0 (1 ⌥ /2!0 ) = !0 ⌥ /2. So !2 = !0 + /2, !1 = !0 0

width of the anomalous region is

! = !2

!1 = .

N q2 , so at the maximum (! = ! ), ↵ = . 0 max ! 2 )2 + 2 ! 2 m✏0 c ✓ ◆ N q2 !2 !2 At !1 and !2 , ! 2 = !02 ⌥ !0 , so ↵ = = ↵ . But max m✏0 c 2 !02 + 2 ! 2 ! 2 + !02 ✓ ◆ ✓ ◆ ✓ ◆ !2 !02 ⌥ !0 1 (1 ⌥ /!0 ) ⇠ 1 1 1 ⇠ ⇠ = = 1⌥ 1± 1⌥ = = = . ! 2 + !02 2!02 ⌥ !0 2 (1 ⌥ /2!0 ) 2 !0 2!0 2 2!0 2 1 ⇠ So ↵ = 2 ↵max at !1 and !2 . qed From Eq. 9.171, ↵ =

2

2

Nq ! m✏0 c (!02

Problem 9.26 Equation 9.170 for a single resonance: ck =1+ !

✓

N q2 2m✏0

◆

 (!02 ! 2 ) a! 2 (! 2 ! 2 ) ) ck = ! 1 + 2 0 20 2 2 2 2 2 2 (!0 ! ) + ! (!0 ! ) + 2 ! 2

where a ⌘

N q2 . 2m✏0 !02

From Eq. 9.150, the group velocity is vg = d!/dk, so c dk =c vg d!  = 1+

a!02 (!02 ! 2 ) [(! 2 + !a!02 0 2 2 2 2 ! ) + ! 2 ! 2 )[(!02 ! 2 )2 + 2 ! 2 ] 2 (!0

! 2 )2 +

2! 2 [(!02 [(!02 2 2 (! 2 ! 2 )3 + 2! 2 (!02 ! 2 )2 ! (!02 = 1 + a!02 0 2 2 2 2 [(!0 ! ) + ! 2 ]2 2 2 (! 2 ! 2 )2 ! = 1 + a!02 (!02 + ! 2 ) 20 . [(!0 ! 2 )2 + 2 ! 2 ]2 = 1 + a!0

⇢ (! 2 vg = c 1 + a!02 (!02 + ! 2 ) 20 [(!0 (a) For

! 2 ]( 2!) (!02 ! 2 )[2(!02 ! 2 )( 2!) + 2 (2!)] [(!02 ! 2 )2 + 2 ! 2 ]2 ! 2 )2 + 2 ! 2 ] + 4! 2 (!02 ! 2 )2 2 2 ! 2 (!02 ! 2 ) ! 2 )2 + 2 ! 2 ]2 !2 ) 2 2 !4

2

(!02

! 2 )2 ! 2 )2 +

2

!2 2 ! 2 ]2

1

.

= 0 and a = 0.003:  vg (! 2 + ! 2 ) = 1 + a!02 20 c (!0 ! 2 )2

1

 1+x ) y = 1 + 0.003 (1 x)2


1

.

198

CHAPTER 9. ELECTROMAGNETIC WAVES 1.0

0.8

0.6

0.4

0.2

0.0

(b) For

0.5

1.0

1.5

2.0

= (0.1)!0 and a = 0.003: ⇢ (1 x)2 0.01x y = 1 + 0.003(1 + x) [(1 x)2 + 0.01x]2

1

.

2.5

2.0

1.5

1.0

0.0

0.5

Problem 9.27 ˜ = (a) From Eqs. 9.176 and 9.177, r ⇥ E

In the terminology of Eq. 9.178: ˜ ˜y ˜0 @E @E z ˜ x = @ Ez (r ⇥ E) = @y @z @y

˜0 ik E y

1.0

1.5

˜ @B ˜ 0 ei(kz = i! B @t !

˜z @E = @x

˜0 ik E x

˜0 @E z @x

˜ ˜ z = @ Ey (r ⇥ E) @x

˜x @E = @y

˜0 @E y @x

˜0 @E x @y

˜ ˜ x = @ Bz (r ⇥ B) @y

˜y @B = @z

˜0 @B z @y

˜0 ik B y

˜ ˜ y = @ Bx (r ⇥ B) @z

˜z @B = @x

˜0 ik B x

˜0 @B z @x

˜ ˜ z = @ By (r ⇥ B) @x

˜x @B = @y

˜0 @B y @x

˜0 @B x @y

!t)

˜ ˜ = 1 @E = ; r⇥B c2 @t @Ez @y

ei(kz

!t)

.

So (ii)

ei(kz

!t)

.

So (iii) ikEx

ei(kz !

!t)

.

So (i)

ei(kz

!t)

.

So (v)

ei(kz

!t)

.

So (vi) ikBx

@Bz = @x

@By @x

@Bx = @y

!

˜ ˜ y = @ Ex (r ⇥ E) @z

2.0

!

!

!

@Ey @x @Bz @y

i! ˜ i(kz E0 e c2

!t)

.

ikEy = i!Bx . @Ez = i!By . @x @Ex = i!Bz . @y ikBy =

i! Ex . c2 i! Ey . c2

i! Ez . c2 @Ez @Bz This confirms Eq. 9.179. Now multiply (iii) by k, (v) by !, and subtract: ik 2 Ex k ! + i!kBy = @x @y ✓ ◆ ✓ ◆ i! 2 !2 @Ez @Bz i @Ez @Bz ik!By + 2 Ex ) i k 2 Ex = k +! , or (i) Ex = k +! . c c2 @x @y (!/c)2 k 2 @x @y ✓ 2 ◆ 2 @Ez @Bz i! ! 2 Multiply (ii) by k, (vi) by !, and add: k ik 2 Ey +i!kBx ! = i!kBx E ) i k Ey = y @y @x c2 c2 ei(kz

!t)

.

So (iv)



◆ @Ez @Bz ! . k2 @y @x ! @Ez !k @Bz !2 !k Multiply (ii) by !/c2 , (vi) by k, and add: 2 i 2 Ey + ik 2 Bx k = i 2 Bx i 2 Ey ) c @y c @x c c ✓ ◆ ✓ ◆ 2 ! @B ! @E i @B ! @E z z z z i k2 Bx = k , or (iii) Bx = k . c2 @x c2 @y (!/c)2 k 2 @x c2 @y !k ! @Ez @Bz !2 i!k Multiply (iii) by !/c2 , (v) by k, and subtract: i 2 Ex k + ik 2 By = i 2 By + 2 Ex ) 2 c c @x @y c c ✓ ◆ ✓ ◆ !2 ! @Ez @Bz i @Bz ! @Ez 2 i k By = 2 +k , or (iv) By = k + 2 . c2 c @x @y (!/c)2 k 2 @y c @x This completes the confirmation of Eq. 9.180. ! ˜0 ˜x ˜y ˜z ˜0 @E @E @E @E @E @Ex @Ey y x ˜ ˜ (b) r · E = + + = + + ik E0z ei(kz !t) = 0 ) + + ikEz = 0. @x @y @z @x @y @x @y ✓ 2 ◆ ✓ ◆ i @ Ez @ 2 Bz i @ 2 Ez @ 2 Bz k + ! + k ! + ikEz = 0, Using Eq. 9.180, (!/c)2 k 2 @x2 @x@y (!/c)2 k 2 @2y @x@y ⇤ @ 2 Ez @ 2 Ez ⇥ or + 2 + (!/c)2 k 2 Ez = 0. 2 @x @ y ˜ = 0 ) @Bx + @By + ikBz = 0 ) Likewise, r · B @x ◆ @y ✓ 2 ✓ 2 ◆ i @ Bz ! @ 2 Ez i @ Bz ! @Ez k + k + 2 + ikBz = 0 ) (!/c)2 k 2 @x2 c2 @x@y (!/c)2 k 2 @y 2 c @x@y 2 2 ⇤ @ Bz @ Bz ⇥ + 2 + (!/c)2 k 2 Bz = 0. 2 @x @ y This confirms Eqs. 9.181. [You can also do it by putting Eq. 9.180 into Eq. 9.179 (i) and (iv).] k

@Ez @Bz i +! , or (ii) Ey = @y @x (!/c)2

✓

199

k

Problem 9.28 Here Ez = 0 (TE) and !/c = k (n = m = 0), so Eq. 9.179(ii) ) Ey = cBx , Eq. 9.179(iii) ) Ex = cBy , ⇣ ⌘ ⇣ ⇣ @Bz ! ! ⌘ @Bz ! ⌘ Eq. 9.179(v) ) = i kBy E = i kB B = 0, Eq. 9.179(vi) ) = i kB + Ey = x y y x @y c2 c @x c2 ⇣ ⌘ ! @Bz @Bz i kBx Bx = 0. So = = 0, and since Bz is a function only of x and y, this says Bz is in fact c @x @y I Z @B a constant (as Eq. 9.186 also suggests). Now Faraday’s law (in integral form) says E · dl = · da, @t H R @B and Eq. 9.176 ) = i!B, so E · dl = i! B · da. Applied to a cross-section of the waveguide this gives @t Z I i(kz !t) E · dl = i!e Bz da = i!Bz ei(kz !t) (ab) (since Bz is constant, it comes outside the integral). But

if the boundary is just inside the metal, where E = 0, it follows that Bz = 0. So this would be a TEM mode, which we already know cannot exist for this guide. Problem 9.29 1 c c Here a = 2.28 cm and b = 1.01 cm, so ⌫10 = !10 = = 0.66 ⇥ 1010 Hz; ⌫20 = 2 = 1.32 ⇥ 1010 Hz; 2⇡ 2a 2a r c c c c 1 1 10 10 10 ⌫30 = 3 = 1.97 ⇥ 10 Hz; ⌫01 = = 1.49 ⇥ 10 Hz; ⌫02 = 2 = 2.97 ⇥ 10 Hz; ⌫11 = + 2 = 2 2a 2b 2b 2 a b 1.62 ⇥ 1010 Hz. Evidently just four modes occur: 10, 20, 01, and 11. To get only one mode you must drive the waveguide at a frequency between ⌫10 and ⌫20 : c 0.66 ⇥ 1010 < ⌫ < 1.32 ⇥ 1010 Hz. = , so 10 = 2a; 20 = a. 2.28 cm < < 4.56 cm. ⌫ c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

200


Problem 9.30 1 ˜ ⇥B ˜ ⇤ ). Here (Eq. 9.176) E ˜ =E ˜ 0 ei(kz From Prob. 9.11, hSi = Re(E 2µ0 the TEmn mode (Eqs. 9.180 and 9.186) ✓ ◆ ⇣ m⇡x ⌘ ⇣ n⇡y ⌘ ik m⇡ ⇤ Bx = B sin cos ; 0 (!/c)2 k 2 a a b ✓ ◆ ⇣ m⇡x ⌘ ⇣ n⇡y ⌘ ik n⇡ By⇤ = B cos sin ; 0 (!/c)2 k 2 b a b ⇣ m⇡x ⌘ ⇣ n⇡y ⌘ Bz⇤ = B0 cos cos ; a ✓ ◆b ⇣ m⇡x ⌘ ⇣ n⇡y ⌘ i! n⇡ Ex = B cos sin ; 0 (!/c)2 k 2 b a b ✓ ◆ ⇣ m⇡x ⌘ ⇣ n⇡y ⌘ i! m⇡ B sin Ey = cos ; 0 (!/c)2 k 2 a a b Ez = 0.

!t)

˜⇤ = B ˜ ⇤e , B 0

i(kz !t)

⇣ m⇡x ⌘

⇣ n⇡y ⌘

, and, for

So hSi = Z Ra 0

hSi · da =

!k⇡ 2 B02 [(!/c)2

2

k2 ]

⇣ ⌘ n 2 b

1 !k⇡ 2 B02 ab 8µ0 [(!/c)2 k 2 ]2

sin2 (m⇡x/a) dx = Similarly,

Ra 0

cos2

⇣ m⇡x ⌘

⇣ ⌘ m 2 a

cos2 (m⇡x/a) dx = a/2;

a

+ Rb 0

sin2

⇣ n ⌘2

⇣ n⇡y ⌘ b

⇣ m ⌘2

+

a

a

cos2

b

sin2 (n⇡y/b) dy =

Rb 0

cos2 (n⇡y/b) dy = b/2.]

✓ ◆ 1 1 ˜ ˜⇤ ⇤ ˜ ˜ hui = ✏0 E · E + B·B 4 µ0  ⇣ n ⌘2 ⇣ ⌘ ⇣ ⌘ ⇣ m ⌘2 ⇣ ⌘ ⇣ ⌘ ✏0 ! 2 ⇡ 2 B02 2 n⇡y 2 m⇡x 2 m⇡x 2 n⇡y = cos sin + sin cos 4 [(!/c)2 k 2 ]2 b a b a a b ⇢ ⇣ ⌘ ⇣ ⌘ 1 m⇡x n⇡y + B02 cos2 cos2 4µ0 a b ⇣ ⌘ ⇣ m⇡x ⌘ ⇣ ⌘ ⇣ m ⌘2 ⇣ ⌘ ⇣ ⌘ k 2 ⇡ 2 B02 n 2 2 n⇡y 2 m⇡x 2 2 n⇡y + cos sin + sin cos 2 b a b a a b [(!/c)2 k 2 ] Z

ab hui da = 4

(

o ˆ z .

[In the last step I used

.

b

sin2

.

) ⇣ ⌘ ⇣ ⌘ ✏0 ! 2 ⇡ 2 B02 n 2 ⇣ m ⌘2 B02 1 k 2 ⇡ 2 B02 n 2 ⇣ m ⌘2 + + + + . 4 [(!/c)2 k 2 ]2 b a 4µ0 4µ0 [(!/c)2 k 2 ]2 b a

⇥ These results can be ⇥simplified, using Eq. 9.190 to write (!/c)2 ⇤ and Eq. 9.188 to write (m/a)2 + (n/b)2 = (!mn /⇡c)2 : Z

hSi · da =

!kabc2 2 B0 ; 2 8µ0 !mn

Z

hui da =

⇤ k 2 = (!mn /c)2 , ✏0 µ0 = 1/c2 to eliminate ✏0 , ! 2 ab B02 . 2 8µ0 !mn


201

CHAPTER 9. ELECTROMAGNETIC WAVES Evidently R hSi · da energy per unit time kc2 cp 2 = R = = ! energy per unit length ! ! hui da

2 !mn = vg (Eq. 9.192). qed

Problem 9.31 Following Sect. 9.5.2, the problem is to solve Eq. 9.181 with Ez 6= 0, Bz = 0, subject to the boundary conditions 9.175. Let Ez (x, y) = X(x)Y (y); as before, we obtain X(x) = A sin(kx x) + B cos(kx x). But the boundary condition requires Ez = 0 (and hence X = 0) when x = 0 and x = a, so B = 0 and kx = m⇡/a. But this time m = 1, 2, 3, . . . , but not zero, since m = 0 would kill X entirely. The same goes for Y (y). Thus ⇣ m⇡x ⌘ ⇣ n⇡y ⌘ Ez = E0 sin sin with n, m = 1, 2, 3, . . . . a b p The rest is the same as for TE waves: !mn = c⇡ (m/a)2 + (n/b)2 is the cuto↵ frequency, the wave p p velocity is v = c/ 1 (!mn /!)2 , and the group velocity is v = c 1 (!mn /!)2 . The lowest TM mode is g p 2 2 11, with cuto↵ frequency !11 = c⇡ (1/a) + (1/b) . So the ratio of the lowest TM frequency to the lowest p p c⇡ (1/a)2 + (1/b)2 TE frequency is = 1 + (a/b)2 . (c⇡/a)

Problem 9.32 1 @ 1 @ @Es ˆ 1 @Es E0 k sin(kz !t) ˆ ? (a) r·E = (sEs ) = 0 X; r·B = (B ) = 0 X; r⇥E = ˆ z= = s @s s@ @z s @ s @B E0 ! sin(kz !t) ˆ @B 1 @ E0 k sin(kz !t) ? = X (since k = !/c); r ⇥ B = ˆ s+ (sB ) ˆ z= ˆ s= @t c s @z s @s c s 1 @E E0 ! sin(kz !t) = 2 ˆ s X. Boundary conditions: E k = Ez = 0 X; B ? = Bs = 0 X. c2 @t c s (b) To determine , use Gauss’s law for a cylinder of radius s and length dz: cos(kz !t) 1 1 E · da = E0 (2⇡s) dz = Qenc = dz ) = 2⇡✏0 E0 cos(kz !t). s ✏0 ✏0 To determine I, use Ampére’s law for a circle of radius s (note that the displacement current through this I E0 cos(kz !t) 2⇡E0 loop is zero, since E is in the ˆ s direction): B·dl = (2⇡s) = µ0 Ienc ) I = cos(kz !t). c s µ0 c I

The charge and current on the outer conductor are precisely the opposite of these, since E = B = 0 inside the metal, and hence the total enclosed charge and current must be zero. Problem 9.33 Z 1 Z 1 ikz ˜ ˜ ⇤ e ikz dk. Let l ⌘ k; then f˜(z, 0)⇤ = f˜(z, 0) = A(k)e dk ) f˜(z, 0)⇤ = A(k) 1 Z 1 Z 1 Z11 ⇤ ilz ⇤ ilz ˜ l) e ( dl) = ˜ l) e dl = ˜ k)⇤ eikz dk (renaming the dummy variable l ! k). A( A( A( 1 1 1 h i 1h i Z 1 1h i ⇤ ˜ ˜ ˜ ˜ ˜ k)⇤ eikz dk. Therefore f (z, 0) = Re f (z, 0) = f (z, 0) + f (z, 0) = A(k) + A( 2 1 2 Z 1 i 1h˜ 1 ⇤ ikz ˜ k) = A(k) + A( f (z, 0)e dz. 2 2⇡ 1 Z 1 Z 1 ˙ ˙ ikz ˜ ˜ Meanwhile, f˜(z, t) = A(k)( i!)ei(kz !t) dk ) f˜(z, 0) = [ i! A(k)]e dk. 1

1


202


(Note that ! = |k|v, here, so it does not come outside the integral.) ˙ f˜(z, 0)⇤ =

Z Z

1 1 1

˜ ⇤ ]e [i! A(k)

ikz

dk =

Z

˜ ⇤ ]e [i|k|v A(k)

1

Z

ikz

dk =

Z

1

˜ l)⇤ ]eilz ( dl) [i|l|v A(

1

1

˜ k)⇤ ]eikz dk. [i! A( h i 1h i Z 1 1 ˙ ˙ ˙ ˜ ˜ k)⇤ ]eikz dk. f˙(z, 0) = Re f˜(z, 0) = f˜(z, 0) + f˜(z, 0)⇤ = [ i! A(k) + i! A( 2 1 2 =

˜ k)⇤ ]eikz dk = [i|k|v A(

1

1

1

Z 1 i 1h˜ i ˙ ˜ k)⇤ = 1 A(k) A( f (z, 0) e 2 2⇡ ! 1 1 Z 1 1 i ˜ Adding these two results, we get A(k) = f (z, 0) + f˙(z, 0) e ikz dz. qed 2⇡ ! 1

i! h ˜ A(k) 2

Z i ˜ k)⇤ = 1 A( 2⇡

1

f˙(z, 0)e

ikz

dz, or

ikz

dz.

Problem 9.34 (a) Since (E ⇥ B) points in the direction of propagation, B = (b) From Eq. 7.63, K ⇥ ( ˆ z) =

E0 [cos(kz c

!t) + cos(kz + !t)] y ˆ.

1 E0 2E0 B= [2 cos(!t)] y ˆ, K = cos(!t) x ˆ. µ0 µ0 c µ0 c

(c) The force per unit area is f = K ⇥ Bave =

average of cos2 (!t) is 1/2, so

2E02 [cos(!t) x ˆ] ⇥ [cos(!t) y ˆ] = 2✏0 E02 cos2 (!t) ˆ z. The time µ0 c2

fave = ✏0 E02 . This is twice the pressure in Eq. 9.64, but that was for a perfect absorber, whereas this is a perfect reflector. Problem 9.35 (a) (i) Gauss’s law: r · E = (ii) Faraday’s law:

1 @E = 0. X r sin ✓ @

@B 1 @ 1 @ = r⇥E= (sin ✓E ) ˆ r (rE ) ✓ˆ @t r sin ✓ @✓ r @r  ✓ ◆  ✓ ◆ 1 @ sin2 ✓ 1 1 @ 1 ˆ = E0 cos u sin u ˆ r E0 sin ✓ cos u sin u ✓. r sin ✓ @✓ r kr r @r kr @ @ But cos u = k sin u; sin u = k cos u. @r ✓ @r ◆ ✓ ◆ 1 E0 1 1 1 1 ˆ = 2 sin ✓ cos ✓ cos u sin u ˆ r E0 sin ✓ k sin u + 2 sin u cos u ✓. r sin ✓ r kr r kr r Integrating with respect to t, and noting that

B=

2E0 cos ✓ !r2

✓

sin u +

Z

cos u dt =

1 sin u and !

Z

sin u dt =

1 cos u, we obtain !

◆ ✓ ◆ 1 E0 sin ✓ 1 1 ˆ cos u ˆ r+ k cos u + 2 cos u + sin u ✓. kr !r kr r


203

CHAPTER 9. ELECTROMAGNETIC WAVES (iii) Divergence of B:

1 @ 1 @ r2 Br + (sin ✓B✓ ) 2 r @r r sin ✓ @✓  ✓ ◆  ✓ ◆ 1 @ 2E0 cos ✓ 1 1 @ E0 sin2 ✓ 1 1 = 2 sin u + cos u + k cos u + 2 cos u + sin u r @r ! kr r sin ✓ @✓ !r kr r ✓ ◆ 1 2E0 cos ✓ 1 1 = 2 k cos u cos u sin u r ! kr2 r ✓ ◆ 1 2E0 sin ✓ cos ✓ 1 1 + k cos u + 2 cos u + sin u r sin ✓ !r kr r

r·B =

2E0 cos ✓ = !r2

✓

k cos u

1 cos u kr2

1 sin u r

1 1 k cos u + 2 cos u + sin u kr r

◆

= 0. X

(iv) Ampére/Maxwell:  1 @ @Br ˆ r⇥B = (rB✓ ) r @r @✓ ⇢  ✓ ◆  ✓ ◆ 1 @ E0 sin ✓ 1 1 @ 2E0 cos ✓ 1 = k cos u + 2 cos u + sin u sin u + cos u r @r ! kr r @✓ !r2 kr ✓ ◆ E0 sin ✓ 2 1 1 k 2 2 2 = k sin u cos u sin u sin u + cos u + 2 sin u + 3 cos u ˆ !r kr3 r2 r2 r r kr ✓ ◆ ✓ ◆ k E0 sin ✓ 1 1 E sin ✓ 1 0 = k sin u + cos u ˆ = k sin u + cos u ˆ. ! r r c r r ✓ ◆ ⌘ 1 @E 1 E0 sin ✓ ⇣ ! 1 ! E sin ✓ 1 0 ˆ = 2 ! sin u + cos u = 2 k sin u + cos u ˆ c2 @t c r kr c k r r ✓ ◆ 1 E0 sin ✓ 1 = k sin u + cos u ˆ = r ⇥ B. X c r r (b) Poynting Vector: ✓ ◆ ✓ ◆ 1 E0 sin ✓ 1 2E0 cos ✓ 1 (E ⇥ B) = cos u sin u sin u + cos u ✓ˆ µ0 µ0 r kr !r2 kr ✓ ◆ E0 sin ✓ 1 1 + k cos u + 2 cos u + sin u ( ˆ r) !r kr r ⇢  E 2 sin ✓ 2 cos ✓ 1 1 = 0 2 sin u cos u + (cos2 u sin2 u) sin u cos u ✓ˆ µ0 !r r kr k2 r2 ✓ ◆ 1 1 1 1 1 2 sin ✓ k cos2 u + 2 cos2 u + sin u cos u + sin u cos u sin u cos u sin u ˆ r kr r r k2 r3 kr2 ⇢ ✓ ◆ E02 sin ✓ 2 cos ✓ 1 1 1 sin u cos u + (cos2 u sin2 u) ✓ˆ µ0 !r2 r k2 r2 kr = ✓ ◆ 2 1 1 + sin ✓ + 2 3 sin u cos u + k cos2 u + 2 (sin2 u cos2 u) ˆ r . r k r kr

S=


ˆ

204


Averaging over a full cycle, using hsin u cos ui = 0, hsin2 ui = hcos2 ui = 12 , we get the intensity: I = hSi =

E02 sin ✓ µ0 !r2

✓

◆ k E 2 sin2 ✓ sin ✓ ˆ r= 0 ˆ r. 2 2µ0 cr2

It points in the ˆ r direction, and falls o↵ as 1/r2 , as we would expect for a spherical wave. Z Z Z ⇡ E02 sin2 ✓ 2 E02 4⇡ E02 3 (c) P = I · da = r sin ✓ d✓ d = 2⇡ sin ✓ d✓ = . 2µ0 c r2 2µ0 c 3 µ0 c 0 Problem 9.36

z<0:

(

˜ I (z, t) = E ˜ I (z, t) = 1 E ˜I ei(k1 z !t) x ˜ i(k1 z E ˆ, B v1 I e i( k1 z !t) ˜ ˜ ˜ ˜R ei( ER (z, t) = ER e x ˆ, BR (z, t) = v11 E

0
(

˜ r (z, t) = E ˜r ei(k2 z E ˜ l (z, t) = E ˜l ei( k2 z E

z>d:

n

˜ T (z, t) = E ˜T ei(k3 z E k

k

!t)

˜ r (z, t) = x ˆ, B ˜ l (z, t) = x ˆ, B

!t)

!t) k

˜ T (z, t) = x ˆ, B

!t)

y ˆ

k1 z !t)

1 ˜ i(k2 z !t) y ˆ v2 Er e 1 ˜ i( k2 z !t) E e v2 l 1 ˜ i(k3 z !t) v 3 ET e

y ˆ.

y ˆ.

y ˆ.

k

Boundary conditions: E1 = E2 , B1 = B2 , at each boundary (assuming µ1 = µ2 = µ3 = µ0 ):

z=0:

z=d:

8 ˜ ˜ ˜ ˜ > < EI + ER = Er + El ; > : 1E ˜I v1

1 ˜ 1 ˜ ER = E r v1 v2

8 ˜ ik2 d ˜l e +E > < Er e > : 1E ˜r eik2 d v2

ik2 d

1 ˜ El e v2

1 ˜ ˜I El ) E v2

˜R = (E ˜r E

˜l ), where E

⌘ v1 /v2 .

˜T eik3 d ; =E ik2 d

=

1 ˜ ik3 d ˜r eik2 d ET e )E v3

˜l e E

ik2 d

˜T eik3 d , where ↵ ⌘ v2 /v3 . = ↵E

˜R , E ˜r , and E ˜l , to obtain a single equation for We have here four equations; the problem is to eliminate E ˜T in terms of E ˜I . E ˜R : ˜I = (1 + )E ˜r + (1 ˜l ; Add the first two to eliminate E 2E )E ˜l : ˜r eik2 d = (1 + ↵)E ˜T eik3 d ; Add the last two to eliminate E 2E ˜r : 2E ˜l e ik2 d = (1 ↵)E ˜ T eik3 d . Subtract the last two to eliminate E c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

205

CHAPTER 9. ELECTROMAGNETIC WAVES Plug the last two of these into the first: 1 ˜I = (1 + ) 1 e ik2 d (1 + ↵)E ˜T eik3 d + (1 ˜T eik3 d 2E ) eik2 d (1 ↵)E 2 2 ⇥ ⇤ ˜I = (1 + ↵)(1 + )e ik2 d + (1 ↵)(1 ˜T eik3 d 4E )eik2 d E ⇥ ⇤ ˜T eik3 d = (1 + ↵ ) e ik2 d + eik2 d + (↵ + ) e ik2 d eik2 d E ˜T eik3 d . = 2 [(1 + ↵ ) cos(k2 d) i(↵ + ) sin(k2 d)] E Now the transmission coefficient is T =

T

1

1 ↵ 1 = 4↵ 1 = 4↵ 1 = 4↵ =

v3 ✏3 ET20 v3 = v1 ✏1 EI20 v1

✓

µ0 ✏3 µ0 ✏1

◆ ˜ 2 ˜T |2 ˜T |2 |ET | v1 |E |E = =↵ , so ˜I |2 ˜I |2 ˜I |2 v3 |E |E |E

˜I |2 |E 1 1 = [(1 + ↵ ) cos(k2 d) i(↵ + ) sin(k2 d)] eik3 d ˜T |2 ↵ 2 |E ⇥ ⇤ (1 + ↵ )2 cos2 (k2 d) + (↵ + )2 sin2 (k2 d) . But cos2 (k2 d) = 1 2

⇥ (1 + ↵ )2 + (↵2 + 2↵ + ⇥ (1 + ↵ )2

But n1 =

(1

↵2 )(1

1

2

2

2↵

⇤ ) sin2 (k2 d) .

c c c n3 , n2 = , n3 = , so ↵ = , v1 v2 v3 n2

 1 (n2 = (n1 + n3 )2 + 1 4n1 n3

n22 )(n23 n22

n22 )

↵2

=

2

sin2 (k2 d).

⇤ ) sin2 (k2 d)

n2 . n1

sin2 (k2 d) .

Problem 9.37 T = 1 ) sin kd = 0 )p kd = 0, ⇡, 2⇡ . . .. The minimum (nonzero) thickness is d p = ⇡/k. But k = !/v = p 2⇡⌫/v = 2⇡⌫n/c, and n = ✏µ/✏0 µ0 (Eq. 9.69), where (presumably) µ ⇡ µ0 . So n = ✏/✏0 = ✏r , and hence 8 ⇡c c 3 ⇥ 10 p d= = 9.49 ⇥ 10 3 m, or 9.5 mm. p = p = 2⇡⌫ ✏r 2⌫ ✏r 2(10 ⇥ 109 ) 2.5 Problem 9.38 From Eq. 9.199, T

⇢ 1 [(16/9) (9/4)][1 (9/4)] [(4/3) + 1]2 + sin2 (3!d/2c) 4(4/3)(1) (9/4)  3 49 ( 17/36)( 5/4) 49 85 = + sin2 (3!d/2c) = + sin2 (3!d/2c). 16 9 (9/4) 48 (48)(36) 48 T = . 49 + (85/36) sin2 (3!d/2c) 1

=

48 48 = 0.935; Tmax = = 0.980. Not much 49 + (85/36) 49 variation, and the transmission is good (over 90%) for all frequencies. Since Eq. 9.199 is unchanged when you switch 1 and 3, the transmission is the same either direction, and the fish sees you just as well as you see it. Since sin2 (3!d/2c) ranges from 0 to 1, Tmin =


206


Problem 9.39 ˜ T (r, t) = E ˜ 0 ei(kT ·r (a) Equation 9.91 ) E T q

kT (x sin ✓T + z cos ✓T ) = xkT sin ✓T + izkT

!t)

sin ✓T 2

; kT · r = kT (sin ✓T x ˆ + cos ✓T ˆ z) · (x x ˆ + yy ˆ + zˆ z) = 1 = kx + iz, where

⇣ !n ⌘ n !n1 2 1 sin ✓I = sin ✓I , c n2 c q q !n2 ! 1= (n1 /n2 )2 sin2 ✓I 1 = n21 sin2 ✓I c c ˜ T (r, t) = E ˜ 0 e z ei(kx !t) . qed E T k ⌘ kT sin ✓T = q  ⌘ kT sin2 ✓T

˜0 E R ˜0 E

2

n22 . So

2

↵ . Here is real (Eq. 9.106) and ↵ is purely imaginary (Eq. 9.108); write ↵ = ia, ↵+ I ✓ ◆✓ ◆ ia ia a2 + 2 with a real: R = = 2 = 1. ia + ia + a + 2 (b) R =

=

(c) From Prob. 9.17, E0R =

1 ↵ 1+↵

E0I , so R =

1 ↵ 1+↵

2

=

2

1 ia 1 + ia

=

(1 ia )(1 + ia ) = 1. (1 + ia )(1 ia )

(d) From the solution to Prob. 9.17, the transmitted wave is ˜ t) = E ˜0 ei(kT ·r E(r, T

!t)

y ˆ,

˜ t) = 1 E ˜0 ei(kT ·r B(r, v2 T

Using the results in (a): kT · r = kx + iz, sin ✓T = ˜ t) = E ˜0 e E(r, T

z i(kx !t)

e

y ˆ,

!t)

( cos ✓T x ˆ + sin ✓T ˆ z).

ck c , cos ✓T = i : !n2 !n2

˜ t) = 1 E ˜0 e B(r, v2 T

z i(kx !t)

e

✓

i

◆ c ck x ˆ+ ˆ z . !n2 !n2

˜0 is real. Then We may as well choose the phase constant so that E T

E(r, t) = E0 e z cos(kx !t) y ˆ; 1 c B(r, t) = E0 e z Re {[cos(kx !t) + i sin(kx !t)] [ i x ˆ + kˆ z]} v2 !n2 1 = E0 e z [ sin(kx !t) x ˆ + k cos(kx !t) ˆ z] . qed ! (I used v2 = c/n2 to simplfy B.) c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

207


⇤ @ ⇥ E0 e z cos(kx !t) = 0. X @y   @ E0 z @ E0 z (ii) r · B = e  sin(kx !t) + e k cos(kx !t) @x ! @z ! ⇤ E0 ⇥ z = e k cos(kx !t) e z k cos(kx !t) = 0. X ! x ˆ y ˆ ˆ z @Ey @Ey (iii) r ⇥ E = @/@x @/@y @/@z = x ˆ+ ˆ z @z @x 0 Ey 0

(e)

(i) r · E =

= E0 e z cos(kx !t) x ˆ E0 e z k sin(kx !t) ˆ z. @B E0 z = e [ ! cos(kx !t) x ˆ + k! sin(kx !t) ˆ z] @t ! = E0 e z cos(kx !t) x ˆ kE0 e z sin(kx !t) ˆ z = r ⇥ E. X ◆ ✓ x ˆ y ˆ ˆ z @Bx @Bz (iv) r ⇥ B = @/@x @/@y @/@z = y ˆ @z @x Bx 0 Bz  E0 2 z E0 z 2 E0 =  e sin(kx !t) + e k sin(kx !t) y ˆ = (k 2 2 ) e z sin(kx !t) y ˆ. ! ! ! ⇣ ! ⌘2 ⇥ ⇤ ⇣ n2 ! ⌘2 Eq. 9.202 ) k 2 2 = n21 sin2 ✓I (n1 sin ✓I )2 + (n2 )2 = = ! 2 ✏2 µ2 . c c = ✏2 µ2 !E0 e @E µ2 ✏2 = µ2 ✏2 E0 e @t

z

z

sin(kx

!t) y ˆ. !t) y ˆ = r ⇥ B X.

! sin(kx

(f) S=

=

1 1 E02 (E ⇥ B) = e µ2 µ2 ! E02 e µ2 !

2z

⇥ k cos2 (kx

2z

x ˆ 0  sin(kx

!t) x ˆ

y ˆ cos(kx !t) 0

 sin(kx

!t)

!t) cos(kx

ˆ z 0 k cos(kx

!t)

⇤ !t) ˆ z .

E02 k 2z e x ˆ. On average, 2µ2 ! then, no energy is transmitted in the z direction, only in the x direction (parallel to the interface). qed Averaging over a complete cycle, using hcos2 i = 1/2 and hsin cosi = 0, hSi =

Problem 9.40 Look for solutions of the form E = E0 (x, y, z)e i!t , B = B0 (x, y, z)e i!t , subject to the boundary condik ? tions equations, in the form of Eq. 9.177, give ⇢ E = 0, B = 0 at all surfaces. Maxwell’s r · E = 0 ) r · E0 = 0; r ⇥ E = @B ) r ⇥ E0 = i!B0 ; @t 1 @E r · B = 0 ) r · B0 = 0; r ⇥ B = c2 @t ) r ⇥ B0 = i! c2 E0 . From now on I’ll leave o↵ the subscript (0). The problem is to solve the (time independent) equations ⇢ r · E = 0; r ⇥ E = i!B; r · B = 0; r ⇥ B = i! c2 E. c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

208


From r ⇥ E = i!B it follows that I can get B once I know E, so I’ll on the latter for the moment. ✓ concentrate ◆ 2 i! ! r ⇥ (r ⇥ E) = r(r · E) r2 E = r2 E = r ⇥ (i!B) = i! E = E. So c2 c2 ⇣ ! ⌘2 ⇣ ! ⌘2 ⇣ ! ⌘2 r 2 Ex = Ex ; r 2 Ey = Ey ; r 2 E z = Ez . Solve each of these by separation of variables: c c c ⇣ ! ⌘2 2 2 2 d X d Y d Z 1 d2 X 1 d2 Y 1 d2 Z Ex (x, y, z) = X(x)Y (y)Z(z) ) Y Z 2 +ZX 2 +XY 2 = XY Z, or + + = dx dy dz c X dx2 Y dy 2 Z dz 2 d2 X d2 Y d2 Z 2 (!/c) . Each term must be a constant, so = kx2 X, = ky2 Y, = kz2 Z, with 2 2 dx dy dz 2 2 kx2 + ky2 + kz2 = (!/c) . The solution is Ex (x, y, z) = [A sin(kx x) + B cos(kx x)][C sin(ky y) + D cos(ky y)][E sin(kz z) + F cos(kz z)]. But Ek = 0 at the boundaries ) Ex = 0 at y = 0 and z = 0, so D = F = 0, and Ex = 0 at y = b and z = d, so ky = n⇡/b and kz = l⇡/d, where n and l are integers. A similar argument applies to Ey and Ez . Conclusion: Ex (x, y, z) = [A sin(kx x) + B cos(kx x)] sin(ky y) sin(kz z), Ey (x, y, z) = sin(kx x)[C sin(ky y) + D cos(ky y)] sin(kz z), Ez (x, y, z) = sin(kx x) sin(ky y)[E sin(kz z) + F cos(kz z)], where kx = m⇡/a. (Actually, there is no reason at this stage to assume that kx , ky , and kz are the same for all three components, and I should really affix a second subscript (x for Ex , y for Ey , and z for Ez ), but in a moment we shall see that in fact they do have to be the same, so to avoid cumbersome notation I’ll assume they are from the start.) Now r·E = 0 ) kx [A cos(kx x) B sin(kx x)] sin(ky y) sin(kz z)+ky sin(kx x)[C cos(ky y) D sin(ky y)] sin(kz z)+ kz sin(kx x) sin(ky y)[E cos(kz z) F sin(kz z)] = 0. In particular, putting in x = 0, kx A sin(ky y) sin(kz z) = 0, and hence A = 0. Likewise y = 0 ) C = 0 and z = 0 ) E = 0. (Moreover, if the k’s were not equal for di↵erent components, then by Fourier analysis this equation could not be satisfied (for all x, y, and z) unless the other three constants were also zero, and we’d be left with no field at all.) It follows that (Bkx + Dky + F kz ) = 0 (in order that r · E = 0), and we are left with E = B cos(kx x) sin(ky y) sin(kz z) x ˆ + D sin(kx x) cos(ky y) sin(kz z) y ˆ + F sin(kx x) sin(ky y) cos(kz z) ˆ z, with kx = (m⇡/a), ky = (n⇡/b), kz = (l⇡/d) (l, m, n all integers), and Bkx + Dky + F kz = 0.

Or:

The corresponding magnetic field is given by B = (i/!)r ⇥ E: ✓ ◆ i @Ez @Ey i Bx = = [F ky sin(kx x) cos(ky y) cos(kz z) ! @y @z ! ✓ ◆ i @Ex @Ez i By = = [Bkz cos(kx x) sin(ky y) cos(kz z) ! @z @x ! ✓ ◆ i @Ey @Ex i Bz = = [Dkx cos(kx x) cos(ky y) sin(kz z) ! @x @y !

B =

i (F ky ! i (Dkx !

Dkz ) sin(kx x) cos(ky y) cos(kz z) x ˆ

i (Bkz !

Dkz sin(kx x) cos(ky y) cos(kz z)] , F kx cos(kx x) sin(ky y) cos(kz z)] , Bky cos(kx x) cos(ky y) sin(kz z)] .

F kx ) cos(kx x) sin(ky y) cos(kz z) y ˆ

Bky ) cos(kx x) cos(ky y) sin(kz z) ˆ z. c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


209

These automatically satisfy the boundary condition B ? = 0 (Bx = 0 at x = 0 and x = a, By = 0 at y = 0 and y = b, and Bz = 0 at z = 0 and z = d). As a check, let’s see if r · B = 0 : r·B =

=

i i (F ky Dkz )kx cos(kx x) cos(ky y) cos(kz z) (Bkz F kx )ky cos(kx x) cos(ky y) cos(kz z) ! ! i (Dkx Bky )kz cos(kx x) cos(ky y) cos(kz z) ! i (F kx ky Dkx kz + Bkz ky F kx ky + Dkx kz Bky kz ) cos(kx x) cos(ky y) cos(kz z) = 0. X !

The boxed equations satisfy all of Maxwell’s equations, and they meet the boundary conditions. For TE modes, we pick Ez = 0, so F = 0 (and hence Bkx + Dky = 0, leaving only the overall amplitude undetermined, for given l, m, and n); for TM modes we want Bz = 0 (so Dkx Bky = 0, again leaving only one amplitude undetermined, since Bkx + Dky + F kz = 0). In either case (TElmn or TMlmn ), the frequency is given by p ⇥ ⇤ ! 2 = c2 (kx2 + ky2 + kz2 ) = c2 (m⇡/a)2 + (n⇡/b)2 + (l⇡/d)2 , or ! = c⇡ (m/a)2 + (n/b)2 + (l/d)2 .


210

CHAPTER 10. POTENTIALS AND FIELDS

Chapter 10

Potentials and Fields Problem 10.1 22 V +

@L = r2 V @t

22 A

rL = r2 A

@2V @ @2V @ + (r · A) + µ ✏ = r2 V + (r · A) = 0 0 2 2 @t @t @t @t ✓ ◆ @2A @V µ0 ✏0 2 r r · A + µ0 ✏0 = µ0 J. X @t @t

µ0 ✏0

Problem 10.2 ◆ Z ✓ 1 1 2 (a) W = ✏0 E 2 + B d⌧. At t1 = d/c, x 2 µ0 At T2 = (d + h)/c, ct2 = d + h: E= so B 2 =

1 2 E , and c2

Therefore 1 µ2 k 2 W (t2 ) = (2✏0 ) 0 2 4

✓

µ0 k (d + h 2

✏0 E 2 +

(d+h) Z

1 2 B µ0

(d + h

◆

d = ct1 , so E = 0, B = 0, and hence W (t1 ) = 0.

x) ˆ z,

B=

✓ = ✏0 E 2 +

1 µ0 k (d + h c 2

1 1 2 E µ0 ✏0 c2

✏0 µ20 k 2 lw x) dx (lw) = 4 2

d

(b) S(x) =

1 ⇢. X ✏0



◆

x) y ˆ,

= 2✏0 E 2 .

(d + h 3

1 1 2 1 2 µ0 k 2 (E ⇥ B) = E [ ˆ z ⇥ (±ˆ y)] = ± E x ˆ= ± (ct 4c µ0 µ0 c µ0 c

x)3

d+h

= d

✏0 µ20 k 2 lwh3 . 12

|x|)2 x ˆ

(plus sign for x > 0, as here). For |x| > ct, S = 0.

So the energy per unit time entering the box in this time interval is Z dW µ0 k 2 lw = P = S(d) · da = (ct d)2 . dt 4c Note that no energy flows out the top, since S(d + h) = 0.


211

CHAPTER 10. POTENTIALS AND FIELDS Zt2

(c) W =

µ0 k 2 lw P dt = 4c

t1

(d+h)/c Z

(ct

d)2 dt =

d/c

 µ0 k 2 lw (ct d)3 4c 3c

(d+h)/c

µ0 k 2 lwh3 . 12c2

= d/c

Since 1/c2 = µ0 ✏0 , this agrees with the answer to (a). Problem 10.3 (a) E=

rV

@A 1 q = ˆ r. @t 4⇡✏0 r2

B = r⇥A = 0.

This is a funny set of potentials for a stationary point charge q at the origin. (V =

1 q , A = 0 would, of 4⇡✏0 r

course, be the customary choice.) Evidently ⇢ = q 3 (r); J = 0. (b) @ =0 @t

V =V 0

✓

1 q 4⇡✏0 r

◆

1 q = ; A0 = A + r = 4⇡✏0 r

1 qt ˆ r+ 4⇡✏0 r2

✓

1 qt 4⇡✏0

◆✓

◆ 1 ˆ r = 0. r2

This gauge function transforms the “funny” potentials into the “ordinary” potentials of a stationary point charge. Problem 10.4 @A = A0 cos(kx !t) y ˆ( !) = A0 ! cos(kx @t @ B = r⇥A = ˆ z [A0 sin(kx !t)] = A0 k cos(kx !t) ˆ z. @x E=

rV

!t) y ˆ,

Hence r·E = 0 X, r·B = 0 X. r⇥E = ˆ z so r⇥E =

@ [A0 ! cos(kx @x

!t)] =

A0 !k sin(kx

!t) ˆ z,

@B = @t

A0 !k sin(kx

!t) ˆ z,

@B X. @t

r⇥B = So r⇥B = µ0 ✏0

y ˆ

@ [A0 k cos(kx @x

!t)] = A0 k 2 sin(kx

!t) y ˆ,

@E = A0 ! 2 sin(kx @t

@E provided k 2 = µ0 ✏0 ! 2 , or, since c2 = 1/µ0 ✏0 , ! = ck. @t

Problem 10.5 @V Ex. 10.1: r·A = 0; = 0. Both Coulomb and Lorentz. @t ✓ ◆ qt ˆ r qt 3 @V Prob. 10.3: r·A = r· = (r); = 0. Neither. 4⇡✏0 r2 ✏0 @t @V Prob. 10.4: r·A = 0; = 0. Both. @t


!t) y ˆ.

212 Problem 10.6 Suppose r·A 6=


@V @V . (Let r·A + µ0 ✏0 = @t @t 0 @V that A0 and V 0 (Eq. 10.7) do obey r·A0 = µ0 ✏0 . @t µ0 ✏0

r·A0 + µ0 ✏0

—some known function.) We want to pick

@V 0 @V = r·A + r2 + µ0 ✏0 @t @t

µ0 ✏0

@2 = @t2

such

+ 22 .

This will be zero provided we pick for the solution to 22 = , which by hypothesis (and in fact) we know how to solve. Rt We could always find a gauge in which V 0 = 0, simply by picking = 0 V dt0 . We cannot in general pick A = 0—this would make B = 0. [Finding such a gauge function would amount to expressing A as r , and we know that vector functions cannot in general be written as gradients—only if they happen to have curl zero, which A (ordinarily) does not.] Problem 10.7 (a) Using Eq. 1.99, r·J= (b) From Eq. 10.10,

q˙ r· 4⇡

✓

ˆ r r2

V (r, t) =

◆

=

q 4⇡✏0

Z

q˙ 4⇡ 3 (r) = 4⇡ 3

(r0 )

r

d⌧ 0 =

q˙ 3 (r) =

@⇢ . X @t

1 q(t) . 4⇡✏0 r

By symmetry, B = 0 (what direction could it point?), so r ⇥ A = 0, r · A = 0, and A ! 0 at infinity. Evidently A = 0. (c) E =

@A 1 q(t) = ˆ r. B = 0. Checking Maxwell’s equations: @t 4⇡✏0 r2 ✓ ◆ q ˆ r q q 3 (r) ⇢ r· = 4⇡ 3 (r) = = . X 2 4⇡✏0 r 4⇡✏0 ✏0 ✏0 0. X @B 0= . X @t ✓ ◆ ✓ ◆ @E q˙ 1 1 q˙ 0; µ0 J + µ0 ✏0 = µ0 ˆ r + µ ✏ ˆ r = 0. X 0 0 @t 4⇡ r2 4⇡✏0 r2

rV

r·E = r·B =

r⇥E = r⇥B =

[Note that the displacement current exactly cancels the conduction current. Physically, this configuration is a point charge at the origin that is changing with time as current flows in symmetrically (from infinity).] Problem 10.8 Noting the A is independent of t and B is independent of r, use Eq. 10.19:  dA @A 1 = + (v · r)A = (v · r) (r ⇥ B) dt @t 2  1 @ @ @ = vx + vy + vz [(yBz zBy ) x ˆ + (zBx xBz ) y ˆ + (xBy yBx ) ˆ z] 2 @x @y @z 1 = [vx ( Bz y ˆ + By ˆ z) + vy (Bz x ˆ Bx ˆ z) + vz ( By x ˆ + Bx y ˆ)] 2 1 1 = [(vy Bz vz By ) x ˆ + (vz Bx vx Bz ) y ˆ + (vx By vy Bx ) ˆ z] = (v ⇥ B). X 2 2 c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

213

CHAPTER 10. POTENTIALS AND FIELDS Equation 10.20 says d dp (p + qA) = dt dt

q (v ⇥ B) = qr(v · A) = 2

q r[v · (r ⇥ B)], 2

or

dp q q = (v ⇥ B) r[r · (B ⇥ v)]. dt 2 2 Now, for a vector c that is independent of position, ✓ ◆ @ @ @ r(r · c) = x ˆ +y ˆ +ˆ z (xcx + ycy + zcz ) = cx x ˆ + cy y ˆ + cz ˆ z) = c. @x @y @z In this case c = (B ⇥ v) =

(v ⇥ B), so dp q q = (v ⇥ B) + (v ⇥ B) = q(v ⇥ B). X dt 2 2

Problem 10.9 d d (T + qV ) = dt dt

✓

1 mv 2 2

◆

+q

 dV dv @V = mv · +q + (v · r)V dt dt @t

=v·F+q



@V + (v · r)V . @t

But Eq. 10.17 says v · F = qv · so



rV

  @A @V q (v · r)V + v · +q + (v · r)V @t @t

d (T + qV ) = dt

Problem 10.10 From the product rule: ✓ ◆ ✓ ◆ J 1 1 r· = (r·J) + J · r , But r

1

r

=

r0

1

r

r·

r

r

, since

r

✓

J

r

◆

=

r

=r 1

r

But r·J = and

 @A q v · rV + v · , @t

@A + v ⇥ (r ⇥ A) = @t

0

r·

✓

J

r

=q

◆

=

✓

@V @t

1

r

@A v· @t

◆

✓

(r ·J) + J · r 0

0

1

r

◆

r0 . So

(r·J)

✓ ◆ 1 1 1 J · r0 = (r·J) + (r0 ·J)

r

r

r

r0 ·

@Jx @Jy @Jz @Jx @tr @Jy @tr @Jz @tr + + = + + , @x @y @z @tr @x @tr @y @tr @z

@tr = @x

@ [q(V @t

=

1@r , c @x

@tr = @y

1@r , c @y

@tr = @z

1@r , c @z


✓

J

r

◆

.

.

v · A)] .

214 so r·J = Similarly,


 1 @Jx @ r @Jy @ r @Jz @ r + + c @tr @x @tr @y @tr @z

=

1 @J · (r r ). c @tr

@⇢ 1 @J · (r0 r ). @t c @tr [The first term arises when we di↵erentiate with respect to the explicit r0 , and use the continuity equation.] thus ✓ ◆   ✓ ◆ ✓ ◆ J 1 1 @J 1 @⇢ 1 @J J 1 @⇢ J r· = · (r0 r ) + · (r0 r ) r· = r0 · r r r r r @t r c @tr @t c @tr r0 ·J =

(the other two terms cancel, since r r = r0 r ). Therefore:  ✓ ◆  Z Z Z I µ0 @ ⇢ J @ 1 ⇢ µ0 J 0 r·A = r· r d⌧ r d⌧ = µ0 ✏0 @t 4⇡✏0 r d⌧ 4⇡ r · da. 4⇡ @t @V The last term is over the suface at “infinity”, where J = 0, so it’s zero. Therefore r·A = µ0 ✏0 .X @t Problem 10.11 (a) As in Ex. 10.2, for t < s/c, A = 0; for t > s/c, 8 p p 2 2 p 2 2 9 (ct) s (ct)2 s2 (ct) s > > p > > Z Z Z = ⇣µ ⌘ k(t s2 + z 2 /c) µ0 k < dz 1 0 p p A(s, t) = ˆ z 2 dz = ˆ z t dz > 4⇡ 2⇡ > c s2 + z 2 s2 + z 2 > > : ; 0 0 0 " ! # p ✓ ◆ ct + (ct)2 s2 µ0 k 1p = ˆ z t ln (ct)2 s2 . Accordingly, 2⇡ s c ( ! p ct + (ct)2 s2 µ0 k E(s, t) = ˆ z ln + 2⇡ s !✓ ◆ ! ) s 1 1 2c2 t 1 2c2 t p p t c+ p s 2 (ct)2 s2 2c (ct)2 s2 ct + (ct)2 s2 ( ! ) p ct + (ct)2 s2 µ0 k ct ct p = ˆ z ln +p 2⇡ s (ct)2 s2 (ct)2 s2 ! p ct + (ct)2 s2 µ0 k = ln ˆ z (or zero, for t < s/c). 2⇡ s @A = @t

B(s, t) =

=

=

@Az ˆ @s  8 > ! s 1 p ( 2s) > 2 (ct)2 s2 µ0 k < s p t 2 2⇡ > s2 > : ct + (ct) ( ) µ0 k ct2 s ˆ= p + p 2⇡ s (ct)2 s2 c (ct)2 s2

ct s2

p

(ct)2

s2

9 > > =

1 ( 2s) ˆ p 2c (ct)2 s2 > > ;

µ0 k ( c2 t2 + s2 ) ˆ µ0 k p p = (ct)2 2⇡ sc (ct)2 s2 2⇡sc

s2 ˆ.


215

CHAPTER 10. POTENTIALS AND FIELDS (b) A(s, t) =

µ0 ˆ z 4⇡

Z

1

q0 (t

1

r

r

/c)

dz. But

A(s, t) = Now z =

Now (t

p

r

r

/c) = c ( r

E(s, t) =

=

p

s2 + z 2 , so the integrand is even in z:

⇣ µ q ⌘ Z 1 (t 0 0 ˆ z 2 4⇡ 0

r

r

/c)

dz.

r d r , and z = 0 ) r = s, z = 1 ) r = 1. So: 1 2r dr p =p 2 2 r 2 s2 r s2 ✓ ◆ Z 1 r pr d r . µ0 q0 1 A(s, t) = ˆ z t r 2⇡ c r 2 s2 s Z 1 µ0 q0 (r ct) p ct) (Ex. 1.15); therefore A = ˆ zc d r , so 2 2⇡ r s2 s

s2 ) dz =

2

A(s, t) =

r

µ0 q0 c 1 p 2⇡ (ct)2

B(s, t) =

ˆ z (or zero, if ct < s); s2 ✓ ◆ µ0 q0 c 1 2c2 t µ0 q0 c3 t @A = ˆ z = ˆ z (or zero, for t < s/c); @t 2⇡ 2 [(ct)2 s2 ]3/2 2⇡[(ct)2 s2 ]3/2 ✓ ◆ @Az ˆ µ0 q0 c 1 2s µ0 q0 cs ˆ= ˆ (or zero, for t < s/c). = @t 2⇡ 2 [(ct)2 s2 ]3/2 2⇡[(ct)2 s2 ]3/2

Problem 10.12 µ0 A= 4⇡

Z

I(tr )

r

µ0 k dl = 4⇡

Z

(t

r r

/c)

µ0 k dl = 4⇡

⇢ Z dl t

r

( Z Z Z b R µ0 kt 1 1 But for the complete loop, dl = 0, so A = dl + dl + 2 x ˆ 4⇡ a 1 b 2 a R circle), 2 dl = 2b x ˆ (outer circle), so  µ0 kt 1 1 µ0 kt A= (2a) + ( 2b) + 2 ln(b/a) x ˆ) A= ln(b/a) x ˆ, 4⇡ a b 2⇡

E=

Z 1 dl . c ) R dx . Here 1 dl = 2a x ˆ (inner x @A = @t

µ0 k ln(b/a) x ˆ. 2⇡

The changing magnetic field induces the electric field. Since we only know A at one point (the center), we can’t compute r ⇥ A to get B. Problem 10.13 ˙ t) = 0, so Eq. 10.36 ) In this case ⇢(r, ˙ t) = ⇢(r, ˙ 0) and J(r,  Z r (Eq. 10.18), so 1 ⇢(r0 , 0) + ⇢(r ˙ 0 , 0)tr ⇢(r ˙ 0 , 0) ˆ d⌧ 0 , but tr = t r E(r, t) = + 2 r 4⇡✏0 cr c Z  0 Z 1 ⇢(r , 0) + ⇢(r ˙ 0 , 0)t ⇢(r ˙ 0 , 0)( r /c) ⇢(r ˙ 0 , 0) 1 ⇢(r0 , t) 0 ˆ d⌧ 0 = ˆ r = + r2 r2 r 2 r d⌧ . qed 4⇡✏0 cr 4⇡✏0


216


Problem 10.14 ˙ r ) = J(t), ˙ In this approximation we’re dropping the higher derivatives of J, so J(t and Eq. 10.38 )  Z r ˙ 0 r µ0 1 0 0 ˙ 0 ˆ B(r, t) = r 2 J(r , t) + (tr t)J(r , t) + c J(r , t) ⇥ r d⌧ , but tr t = c (Eq. 10.25), so 4⇡ Z µ0 J(r0 , t) ⇥ rˆ = d⌧ 0 . qed r2 4⇡ Problem 10.15 At time t the charge is at r(t) = a[cos(!t) x ˆ + sin(!t) y ˆ], so v(t) = !a[ sin(!t) x ˆp + cos(!t) y ˆ]. Therefore r = z ˆz a[cos(!tr ) xˆ + sin(!tr ) yˆ], and hence r 2 = z 2 + a2 (of course), and r = z 2 + a2 .

rˆ

·v =

1

r (r

· v) =

1

r

!a [ sin(!tr ) cos(!tr ) + sin(!tr ) cos(!tr )] = 0, so 2

✓

1

rˆ

·v c

◆

= 1.

Therefore V (z, t) =

1 q q!a p p ; A(z, t) = [ sin(!tr ) x ˆ + cos(!tr ) y ˆ), where tr = t 4⇡✏0 z 2 + a2 4⇡✏0 c2 z 2 + a2

p

z 2 + a2 . c

Problem 10.16 Term under square root in (Eq. 10.49) is: I = c4 t2 2c2 t(r · v) + (r · v)2 + c2 r2 c4 t2 v 2 r2 + v 2 c2 t2 = (r · v)2 + (c2 v 2 )r2 + c2 (vt)2 2c2 (r · vt). put in vt = r R2 . = (r · v)2 + (c2 v 2 )r2 + c2 (r2 + R2 2r · R) 2c2 (r2 r · R) = (r · v)2

r2 v 2 + c2 R2 .

but (r · v)2

r2 v 2 = ((R + vt) · v)2 (R + vt)2 v 2 = (R · v)2 + v 4 t2 + 2(R · v)v 2 t R2 v 2 2(R · v)tv 2 = (R · v)2 R2 v 2 = R2 v 2 cos2 ✓ R2 v 2 = R2 v 2 1 =

v 2 t2 v 2 cos2 ✓

R2 v 2 sin2 ✓.

Therefore I=

✓ R2 v 2 sin2 ✓ + c2 R2 = c2 R2 1

Hence V (r, t) =

1 q 4⇡✏0 R 1

q v2 c2

sin2 ✓

◆ v2 2 sin ✓ . c2 . qed

Problem 10.17 Once seen, from a given point x, the particle will forever remain in view—to disappear it would have to travel faster than light. c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


217

Problem 10.18 First calculateptr : tr = t |r w(tr )|/c ) p c(tr t) = x b2 + c2 t2r ) c(tr t) + x = b2 + c2 t2r ; 2 2 2 2 2 c tr 2c tr t + c t + 2xctr 2xct + x2 = b2 + c2 t2r ; 2ctr (x ct) + (x2 2xct + c2 t2 ) = b2 ; b2 (x ct)2 2ctr (x ct) = b2 (x ct)2 , or tr = . 2c(x ct) 1 qc Now V (x, t) = , and r c r · v = r (c v); r = c(t tr ). 4⇡✏0 ( r c r · v) 1 1 c2 tr c2 tr c2 tr + c(x ct) c2 tr c(x ct) v= p 2c2 tr = = ; (c v) = = ; 2 b2 + c2 t2r c(tr t) + x ctr + (x ct) ctr + (x ct) ctr + (x ct) ct) c2 (t tr )(x ct) b2 (x ct)2 b2 + (x ct)2 r c r ·v = c(tct +tr )c(x = ; ctr +(x ct) = +(x ct) = ; (x ct) ctr + (x ct) 2(x ct) 2(x ct) r 2 2 2 2 2 2 2 2ct(x ct) b + (x ct) (x ct)(x + ct) b (x c t b ) t tr = = = . Therefore ct) 2c(x ct) 2c(x ct)  2c(x 1 b2 + (x ct)2 1 2c(x ct) b2 + (x ct)2 = = . 2 2 2 r c r ·v 2(x ct) c (x ct) [2ct(x ct) b + (x ct) ] c(x ct) [2ct(x ct) b2 + (x ct)2 ] The term in square brackets simplifies to (2ct + x ct)(x ct) b2 = (x + ct)(x ct) b2 = x2 c2 t2 b2 . q 4⇡✏0 (x Meanwhile

So V (x, t) =

A = =

b2 + (x ct)2 . ct)(x2 c2 t2 b2 )

 2 V c2 tr V b (x ct)2 2(x ct) q v = x ˆ = 2 2 c ctr + (x ct) c 2c(x ct) b2 + (x ct)2 4⇡✏0 (x q 4⇡✏0 c (x

b2 (x ct)2 x ˆ. ct)(x2 c2 t2 b2 )


b2 + (x ct)2 x ˆ ct)(x2 c2 t2 b2 )

218


Problem 10.19 From Eq. c(t tr ) = r ) c2 (t ✓tr )2 = r ◆2 = r · r . Di↵erentiate with respect to t: ✓ 10.44, ◆ @tr @r @tr @r 2c2 (t tr ) 1 = 2r · , or c r 1 = r · . Now r = r w(tr ), so @t @t @t @t ✓ ◆ @r @w @w @tr @tr @tr @tr @tr r · v) = = = = v ; cr 1 = r ·v ; cr = (c r @t @t @tr @t @t @t @t @t @tr @tr cr ( r · u) (Eq. 10.71), and hence = . qed r @t @t ·u v Now Eq. 10.47 says A(r, t) = 2 V (r, t), so c ✓ ◆ ✓ ◆ @A 1 @v @V 1 @v @tr @V = 2 V +v = 2 V +v @t c @t @t c @tr @t @t  1 @tr 1 qc 1 qc @ = 2 a +v ( r c r · v) c @t 4⇡✏0 r · u 4⇡✏0 ( r · u)2 @t  ✓ ◆ 1 qc a @tr v @r @r @v = 2 c ·v r · . c 4⇡✏0 r · u @t ( r · u)2 @t @t @t ✓ ◆ @r @tr @r @tr But r = c(t tr ) ) =c 1 , r = r w(tr ) ) = v (as above), and @t @t @t @t @v @v @tr @tr = =a . @t @tr @t @t ⇢  ✓ ◆ q @tr @tr @tr @tr 2 r r = a( · u) v c 1 + v2 ·a 2 4⇡✏0 c( r · u) @t @t @t @t ⇢ ⇥ ⇤ q @t r = c2 v + ( r · u)a + (c2 v 2 + r · a)v 4⇡✏0 c( r · u)2 @t ⇢ ⇥ ⇤ cr q = c2 v + ( r · u)a + (c2 v 2 + r · a)v 2 r ·u 4⇡✏0 c( r · u) ⇥ 2 ⇤ q 2 2 = c v( r · u) + c r ( r · u)a + c r (c v + r · a)v 3 4⇡✏0 c( r · u)  ✓ ◆ r a + r (c2 v2 + r · a)v . qed qc 1 r r = ( c · v) v + 4⇡✏0 ( r c r · v)3 c c Problem 10.20 r ⇥(c2 v2 )u + r ⇥ (u ⇥ a)⇤. Here q E= 4⇡✏0 ( r · u)3 v = vx ˆ, a = a x ˆ, and, for points to the right, rˆ = x ˆ. So u = (c v) x ˆ, u ⇥ a = 0, and r · u = r (c v).

✓ ◆ r q q q 1 (c + v)(c v)2 1 c+v 2 2 E= (c v )(c v) x ˆ= x ˆ= x ˆ; 4⇡✏0 r 3 (c v)3 4⇡✏0 r 2 (c v)3 4⇡✏0 r 2 c v 1 B = rˆ ⇥ E = 0. qed c For field points to the left, rˆ = x ˆ and u = (c + v) x ˆ, so r · u = r (c + v), and ✓ ◆ r q q 1 c v 2 2 E= (c v )(c + v) x ˆ = x ˆ; B = 0. 4⇡✏0 r 3 (c + v)3 4⇡✏0 r 2 c + v c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

219


Problem 10.21 By Gauss’s law (in integral form) the answer has to be q/✏0 . b 1 q(1 v 2 /c2 ) R E= (Eq. 10.75), so 2 4⇡✏0 (1 vc2 sin2 ✓)3/2 R2 I Z Z ⇡ q(1 v 2 /c2 ) R2 sin ✓ d✓ d q(1 v 2 /c2 ) E·da = = 2⇡ 2 4⇡✏0 4⇡✏0 R2 (1 vc2 sin2 ✓)3/2 0 (1 Let u ⌘ cos ✓, so du = sin ✓d✓, sin2 ✓ = 1 u2 . I Z Z q(1 v 2 /c2 ) 1 du q(1 v 2 /c2 ) ⇣ c ⌘3 E·da = = v2 v 2 2 3/2 2✏0 2✏0 v 1 [1 c2 + c2 u ] The integral is:

c2 v2

1

u q

c2 v2

I

+1

1 + u2

E·da =

q(1

= 1

2 c2 v2

1

c v

=

⇣ v ⌘3 c

v 2 /c2 ) ⇣ c ⌘3 ⇣ v ⌘3 2✏0 v c (1

(1

sin ✓ d✓ . 2 v2 3/2 c2 sin ✓)

1

du c2 v2

1

1 + u2

3/2

.

2 . So v 2 /c2 )

2 q = . v 2 /c2 ) ✏0

X

Problem 10.22 Z

ˆ R dx . ⇥ 2 4⇡✏0 R 1 (v/c)2 sin2 ✓⇤3/2 The horizontal components cancel; the vertical comˆ is sin ✓ (see diagram). Here d = R sin ✓, so ponent of R 1 sin2 ✓ x d = ; = cot ✓, so dx = d( csc2 ✓) d✓ = d✓; 2 2 R d d sin2 ✓ (a) E =

(1

v /c ) 2

2

1 d sin2 ✓ d✓ dx = d✓ = . Thus 2 2 2 R d sin ✓ d ✓ ◆Z ⇡ y ˆ sin ✓ 2 E= (1 v 2 /c2 ) ⇥ ⇤3/2 d✓. Let z ⌘ cos ✓, so sin ✓ = 1 4⇡✏0 d 0 1 (v/c)2 sin2 ✓ Z (1 v 2 /c2 ) y ˆ 1 1 = dz 3/2 2 4⇡✏0 d 1 [1 (v/c) + (v/c)2 z 2 ] " # +1 (1 v 2 /c2 ) y ˆ 1 z p = 4⇡✏0 d (v/c)3 (c2 /v 2 1) (c/v)2 1 + z 2 1 =

(1

v 2 /c2 ) c 4⇡✏0 d v (1

2 1 p v 2 /c2 ) (c/v)2

1 (b) B = 2 (v ⇥ E) for each segment dq = c same formula holds for the total field: B=

1+1

y ˆ=

z2.

1 2 y ˆ (same as for a line charge at rest). 4⇡✏0 d

dx. Since v is constant, it comes outside the integral, and the

1 1 1 2 1 2 µ0 2 v (v ⇥ E) = 2 v (ˆ x⇥y ˆ) = µ0 ✏0 v ˆ z= ˆ z. c2 c 4⇡✏0 d 4⇡✏0 d 4⇡ d


220

But v = I, so B =


µ0 2I ˆ (the same as we got in magnetostatics, Eq. 5.39 and Ex. 5.7). 4⇡ d

Problem 10.23 w(t) = R[cos(!t) x ˆ + sin(!t) y ˆ]; v(t) = R![ sin(!t) x ˆ + cos(!t) y ˆ]; a(t) = R! 2 [cos(!t) x ˆ + sin(!t) y ˆ] = r = w(tr ); r = R; tr = t R/c; rˆ = [cos(!tr ) xˆ + sin(!tr ) yˆ];

r

! 2 w(t);

u = c rˆ v(tr ) = c[cos(!tr ) x ˆ + sin(!tr ) y ˆ] !R[ sin(!tr ) x ˆ + cos(!tr ) y ˆ] = {[c cos(!tr ) !R sin(!tr )] x ˆ + [c sin(!tr ) + !R cos(!tr )] y ˆ} ; ⇥ (u ⇥ a) = ( r · a)u ( r · u)a; r · a = w · ( ! 2 w) = ! 2 R2 ; ⇥ ⇤ r · u = R c cos2 (!tr ) !R sin(!tr ) cos(!tr ) + c sin2 (!tr ) + !R sin(!tr ) cos(!tr ) = Rc;

v 2 = (!R)2 . So (Eq. 10.72):

⇤ q R ⇥ q cu Ra u c2 ! 2 R2 + u(!R)2 a(Rc) = 4⇡✏0 (Rc)3 4⇡✏0 (Rc)2 q 1 = [c2 cos(!tr ) !Rc sin(!tr )] x ˆ [c2 sin(!tr ) + !Rc cos(!tr )] y ˆ 4⇡✏0 (Rc)2

E=

+ R2 ! 2 cos(!tr ) x ˆ + R2 ! 2 sin(!tr ) y ˆ =

⇥ 2 2 q 1 ! R 2 4⇡✏0 (Rc)

⇤ ⇥ c2 cos(!tr ) + !Rc sin(!tr ) x ˆ + ! 2 R2

c2 sin(!tr )

⇤ !Rc cos(!tr ) y ˆ .

1 rˆ ⇥ E = 1 rˆ x Ey rˆ y Ex ˆz c c ⇥ ⇤ 1 q 1 = cos(!tr ) ! 2 R2 c2 sin(!tr ) !Rc cos(!tr ) c 4⇡✏0 (Rc)2 ⇥ ⇤ sin(!tr ) ! 2 R2 c2 cos(!tr ) + !Rc sin(!tr ) ˆ z ⇥ ⇤ q 1 q 1 q ! = !Rc cos2 (!tr ) !Rc sin2 (!tr ) ˆ z= !Rc ˆ z= ˆ z. 4⇡✏0 R2 c3 4⇡✏0 R2 c3 4⇡✏0 Rc2

B =

Notice that B is constant in time. To obtain the field at the center of a circular ring of charge, let q ! (2⇡R); for this ring to carry current 2⇡I 1 I, we need I = v = !R, so = I/!R, and hence q ! (I/!R)(2⇡R) = 2⇡I/!. Thus B = ˆ z, or, 4⇡✏0 Rc2 µ0 I since 1/c2 = ✏0 µ0 , B = ˆ z, the same as Eq. 5.41, in the case z = 0. 2R


221


Problem 10.24 ( , t) = 0 | sin(✓/2)|, where ✓ = !t. So the (retarded) scalar potential at the center is (Eq. 10.26) Z Z 2⇡ 1 1 !tr )/2]| 0 |sin[( V (t) = dl0 = ad r 4⇡✏0 4⇡✏0 0 a Z 2⇡ 2⇡ 0 0 = sin(✓/2) d✓ = [ 2 cos(✓/2)] 4⇡✏0 0 4⇡✏0 0 =

0

4⇡✏0

[2

( 2)] =

0

⇡✏0

.

(Note: at fixed tr , d = d✓, and it goes through one full cycle of or ✓.) Meanwhile I( , t) = v = 0 !a |sin[( !t)/2]| ˆ. From Eq. 10.26 (again) Z Z µ0 I 0 µ0 2⇡ 0 !a |sin[( !tr )/2]| ˆ A(t) = dl = ad . r 4⇡ 4⇡ 0 a But tr = t a/c is again constant, for the integration, and ˆ = sin x ˆ + cos y ˆ. Z 2⇡ µ0 0 !a = |sin[( !tr )/2]| ( sin x ˆ + cos y ˆ) d . Again, switch variables to ✓ = !tr , 4⇡ 0 and integrate from ✓ = 0 to ✓ = 2⇡ (so we don0 t have to worry about the absolute value). Z µ0 0 !a 2⇡ = sin(✓/2) [ sin(✓ + !tr ) x ˆ + cos(✓ + !tr ) y ˆ] d✓. Now 4⇡ 0 Z

2⇡

0

1 sin (✓/2) sin(✓ + !tr ) d✓ = 2

= 2⇡

2⇡

[cos (✓/2 + !tr )

cos (3✓/2 + !tr )] d✓

0

 1 = 2 sin (✓/2 + !tr ) 2 =

Z

Z

sin (✓/2) cos(✓ + !tr ) d✓ =

0

= = = A(t) =

µ0

0 !a

4⇡

2⇡

2 sin (3✓/2 + !tr ) 3 0 1 1 sin(⇡ + !tr ) sin(!tr ) sin(3⇡ + !tr ) + sin(!tr ) 3 3 2 4 2 sin(!tr ) + sin(!tr ) = sin(!tr ). 3 3 Z 2⇡ 1 [ sin (✓/2 + !tr ) + sin (3✓/2 + !tr )] d✓ 2 0  2⇡ 1 2 2 cos (✓/2 + !tr ) cos (3✓/2 + !tr ) 2 3 0 1 1 cos(⇡ + !tr ) cos(!tr ) cos(3⇡ + !tr ) + cos(!tr ) 3 3 2 4 2 cos(!tr ) + cos(!tr ) = cos(!tr ). So 3 3

✓ ◆ 4 [sin(!tr ) x ˆ 3

cos(!tr ) y ˆ] =

µ0

0 !a {sin[!(t 3⇡

a/c)] x ˆ


cos[!(t

a/c)] y ˆ} .

222


Problem 10.25

1

r·A=

1 @V c2 @t

Problem 10.26 ⇢(r, t) =

8 <

Q 3Q = , (r < R = vt), (4/3)⇡R3 4⇡v 3 t3 : 0, (otherwise).

For a point at the center, tr = t r/c, so 8 3Q 3Q < = 4⇡(v/c)3 (ct ⇢(r, tr ) = 4⇡v 3 (t r/c)3 : 0,

✓

vt , r < v(t r/c) ) r < r)3 1 + v/c (otherwise).

◆

,

Let a ⌘ vt/(1 + v/c); then Z a Z a 3Q 1 3Q r2 2 4⇡r dr = dr Qe↵ = 3 3 3 4⇡(v/c) 0 (ct r) (v/c) 0 (ct r)3  a 3Q 2ct (ct)2 = ln(ct r) + (v/c)3 (ct r) 2(ct r)2 0  3Q 2ct (ct)2 1 = ln(ct a) + ln(ct) 2 + (v/c)3 (ct a) 2(ct a)2 2 " ✓ ◆ ✓ ◆ ✓ ◆2 # 3Q 3 ct ct 1 ct = + ln 2 + . (v/c)3 2 ct a ct a 2 ct a

cation, Inc., Upper Saddle River, NJ. All rights reserved. This material is opyright laws as they currently exist. No portion of this material may be orm or by any means, without permission in writing from the publisher.


CHAPTER 10. POTENTIALS AND FIELDS Now, ct

223

vt ct ⇣ v v⌘ ct ct v = 1+ = , so = 1 + , and hence 1 + v/c 1 + v/c c c 1 + v/c ct a c  ⇣ ⌘ ⇣ ⌘ ⇣ ⌘ 3Q 3 v v 1 v 2 = + ln 1 + 2 1+ + 1+ 3 (v/c) 2 c c 2 c   ⇣ ⇣ ⌘ 3Q 3 v v 1 v 1 ⇣ v ⌘2 3Q v ⌘ v 1 ⇣ v ⌘2 = + ln 1 + 2 2 + + + = ln 1 + + . (v/c)3 2 c c 2 c 2 c (v/c)3 c c 2 c

a = ct

Qe↵

If ✏ ⌧ 1, then ln(1 + ✏) = ✏ 12 ✏2 + 13 ✏3 14 ✏4 + . . ., so for v ⌧ c,  ✓ 3Q v 1 ⇣ v ⌘2 1 ⇣ v ⌘3 1 ⇣ v ⌘4 v 1 ⇣ v ⌘2 Qe↵ ⇡ + + = Q 1 (v/c)3 c 2 c 3 c 4 c c 2 c

3v 4c

◆

. X

Problem 10.27 Using Product Rule #5, Eq. 10.50 ) ⇥ ⇤ 1/2 µ0 r·A = qcv · r (c2 t r · v)2 + (c2 v 2 )(r2 c2 t2 ) 4⇡ ⇢ ⇤ 3/2 ⇥ 2 ⇤ µ0 qc 1⇥ 2 = v· (c t r · v)2 + (c2 v 2 )(r2 c2 t2 ) r (c t r · v)2 + (c2 v 2 )(r2 c2 t2 ) 4⇡ 2 ⇤ 3/2 µ0 qc ⇥ 2 = (c t r · v)2 + (c2 v 2 )(r2 c2 t2 ) v· 2(c2 t r · v)r(r · v) + (c2 v 2 )r(r2 ) . 8⇡ Product Rule #4 )

r(r · v) = v ⇥ (r ⇥ r) + (v · r)r, but r ⇥ r = 0, ✓ ◆ @ @ @ (v · r)r = vx + vy + vz (x x ˆ+yy ˆ +zˆ z) = vx x ˆ + vy y ˆ + vz ˆ z = v, and @x @y @z r(r2 ) = r(r · r) = 2r ⇥ (r ⇥ r) + 2(r · r)r = 2r. So

⇤ 3/2 ⇥ ⇤ µ0 qc ⇥ 2 (c t r · v)2 + (c2 v 2 )(r2 c2 t2 ) v · 2(c2 t r · v)v + (c2 v 2 )2r 8⇡ ⇤ 3/2 µ0 qc ⇥ 2 = (c t r · v)2 + (c2 v 2 )(r2 c2 t2 ) (c2 t r · v)v 2 (c2 v 2 )(r · v) . 4⇡ But the term in curly brackets is : c2 tv 2 v 2 (r · v) c2 (r · v) + v 2 (r · v) = c2 (v 2 t r · v). µ0 qc3 (v 2 t r · v) = . 4⇡ [(c2 t r · v)2 + (c2 v 2 )(r2 c2 t2 )]3/2

r·A =

Meanwhile, from Eq. 10.49,

✓ ◆ ⇤ 3/2 1 1 ⇥ 2 µ0 ✏0 qc (c t r · v)2 + (c2 v 2 )(r2 c2 t2 ) ⇥ 4⇡✏0 2 ⇤ @ ⇥ 2 (c t r · v)2 + (c2 v 2 )(r2 c2 t2 ) @t ⇤ 3/2 ⇥ 2 µ0 qc ⇥ 2 = (c t r · v)2 + (c2 v 2 )(r2 c2 t2 ) 2(c t r · v)c2 + (c2 8⇡ µ0 qc3 (c2 t r · v c2 t + v 2 t) = = r · A. X 4⇡ [(c2 t r · v)2 + (c2 v 2 )(r2 c2 t2 )]3/2

@V µ0 ✏0 = @t


⇤ v 2 )( 2c2 t)

224


Problem 10.28 (a) F2 =

q1 q2 1 x ˆ. 4⇡✏0 (b2 + c2 t2 )

(This is just Coulomb’s law, since q1 is at rest.)  Z 1 q1 q2 1 q1 q2 ⇥ q1 q2 1 1 1 (b) I2 = dt = tan (ct/b) = tan 2 + c2 t2 ) 4⇡✏0 (b 4⇡✏ bc 4⇡✏ 0 0 bc 1 1 q1 q2 h ⇡ ⇣ ⇡ ⌘i q1 q2 ⇡ = = . 4⇡✏0 bc 2 2 4⇡✏0 bc ✓ ◆ q2 1 c v (c) From Prob. 10.20, E = x ˆ. Here x 4⇡✏0 x2 c + v and v are to be evaluated at the p retarded time tr , which is given by c(t tr ) = x(tr ) = b2 + c2 t2r ) c2 t2 2cttr + c2 t2 b2 c2 t2r = b2 + c2 t2r ) tr = . Note: As we found 2c2 t in Prob. 10.17, q2 first “comes into view” (for q1 ) at time t = 0. Before that it can exert no force on q1 , and there is no retarded time. From the graph of tr versus t we see that tr ranges all the way from 1 to 1 while t > 0.

1

(1)

tan

1

⇤ ( 1)

2c2 t2 c2 t2 + b2 b2 + c2 t2 1 2c2 t c2 t x(tr ) = c(t tr ) = = (for t > 0). v(t) = p = , so 2ct 2ct 2 b2 + c2 t2 x ✓ 2 2 ◆ ✓ ◆ ✓ ◆ c t b2 2ct c2 t2 b2 v(tr ) = = c (for t > 0). Therefore 2t b2 + c2 t2 c2 t2 + b2 c v (c2 t2 + b2 ) (c2 t2 b2 ) 2b2 b2 q2 4c2 t2 b2 = 2 2 = 2 2 = 2 2 (for t > 0). E = x ˆ) 2 2 2 2 2 2 2 2 c+v (c t + b ) + (c t b ) 2c t c t 4⇡✏0 (b + c t ) c2 t2 8 t < 0; < 0, q1 q2 4b2 F1 = x ˆ, t > 0. : 4⇡✏0 (b2 + c2 t2 )2

Z q1 q2 2 1 1 (d) I1 = 4b dt. The integral is 2 + c2 t2 )2 4⇡✏0 (b 0 ✓ 2 ◆ Z 1 Z 1 1 1 1 1 c t 1 1 ⇣ ⇡c ⌘ ⇡ dt = + dt = = . 4 2 2 2 4 2 2 2 2 2 2 2 c 0 [(b/c) + t ] c 2b (b/c) + t 0 [(b/c) + t )] 2c b 2b 4cb3 0 q1 q2 ⇡ So I1 = . 4⇡✏0 bc (e) F1 6= F2 , so Newton’s third law is not obeyed. On the other hand, I1 = I2 in this instance, which suggests that the net momentum delivered from (1) to (2) is equal and opposite to the net momentum delivered from (2) to (1), and hence that the total mechanical momentum is conserved. (In general the fields might carry o↵ some momentum, leaving the mechanical momentum altered; but that doesn’t happen in the present case.) Problem 10.29 The electric field of q1 at q2 [Eq. 10.75, with ✓ = 45 and R = ( vt x ˆ + vt y ˆ)] is E1 =

q1 1 v 2 /c2 1 p ( x ˆ+y ˆ). 2 2 3/2 4⇡✏0 (1 v /2c ) 2 2(vt)2


225

CHAPTER 10. POTENTIALS AND FIELDS The magnetic field [Eq. 10.76, with v1 = B1 =

1 (v1 ⇥ E) = c2

vx ˆ] is v (ˆ x ⇥ E) = c2

v q1 1 v 2 /c2 1 p ˆ z. c2 4⇡✏0 (1 v 2 /2c2 )3/2 2 2(vt)2

The force on q2 is therefore [Lorentz force law with v2 =

vy ˆ] ✓

q1 q2 1 v 2 /c2 1 p F2 = q2 (E1 + v2 ⇥ B1 ) = 4⇡✏0 (1 v 2 /2c2 )3/2 2 2(vt)2

◆ v2 x ˆ+y ˆ+ 2 x ˆ . c

The electric field of q2 at q1 is reversed, E2 = E1 ; so is the magnetic field B2 = B1 . The electric force is also reversed, but the magnetic force now points in the y direction instead of the x direction. So the force on q1 is F1 =

q1 q2 1 v 2 /c2 1 p 4⇡✏0 (1 v 2 /2c2 )3/2 2 2(vt)2

✓

x ˆ

y ˆ+

◆ v2 y ˆ . c2

The forces are equal in magnitude, but not opposite in direction. No, Newton’s third law is not obeyed. Problem 10.30 c(t

t1 ) = vt1 , t = t1 (1 + v/c), t1 =

x1 = vt1 =

vt . 1 + v/c

x2 = vt2 + L =

V (0, t) =

If L ⌧ vt, V = V =

4⇡✏0 ( r

q

r

1 4⇡✏0

Z

t . 1 + v/c

t2 ) = vt2 + L, t

c(t

v(t L/c) vt +L= 1 + v/c x2

x1

x

dx =

4⇡✏0

ln

✓

x2 x1

· v/c)

(In the limit L ! 0, tr = t1 = t2 .)

t L/c . 1 + v/c

vL/c + L + vL/c vt + L = . 1 + v/c 1 + v/c ◆

✓ ◆ ✓ ◆ L L q 1 ln 1 + ⇡ = . 4⇡✏0 vt 4⇡✏0 vt 4⇡✏0 vt . Here ( r

L/c = t2 (1 + v/c), t2 =

=

4⇡✏0

ln

✓

vt + L vt

◆

.

The Liénard-Wiechert potential is

t · v/c) = vtr + v 2 tr /c = v(1 + v/c)tr = v(1 + v/c) = vt. 1 + v/c ✓ ◆ q 1 So V = . X 4⇡✏0 vt

r

Problem 10.31 1 1 (E ⇥ B); B = 2 (v ⇥ E) (Eq. 10.76). µ0 c ⇥ ⇤ 1 So S = [E ⇥ (v ⇥ E)] = ✏0 E 2 v (v · E)E . 2 µ0 c R The power crossing the plane is P = S · da, S=


226


and da = 2⇡r dr x ˆ (see diagram). So P = ✏0

Z

(E 2 v

= 2⇡✏0 v

Z

Ex2 v)2⇡r dr; Ex = E cos ✓, so E 2

E 2 sin2 ✓ r dr. From Eq. 10.75, E =

✓

Ex2 = E 2 sin2 ✓.

q 1 4⇡✏0 2 R2 ⇥1

ˆ R (v/c)2

sin ✓ 2

⇤3/2 where

1

⌘p

1

v 2 /c2

.

◆2 Z 1 q 1 r sin2 ✓ 1 1 cos ✓ = 2⇡✏0 v = . ⇥ ⇤3 dr. Now r = a tan ✓ ) dr = a 2 d✓; 2 2 4⇡✏0 cos ✓ R a 0 R4 1 (v/c)2 sin ✓ Z ⇡/2 v q2 1 sin3 ✓ cos ✓ 2 = ⇥ ⇤3 d✓. Let u ⌘ sin ✓, so du = 2 sin ✓ cos ✓ d✓. 4 2 2 2 2 4⇡✏0 a 0 1 (v/c) sin ✓ ✓ 4◆ Z 1 vq 2 u vq 2 vq 2 = du = = . 3 2 4 2 4 2 16⇡✏0 a 16⇡✏0 a 2 32⇡✏0 a2 (v/c) u] 0 [1 Problem 10.32 (a) F12 (t) =

1 q1 q2 ˆ z. 4⇡✏0 (vt)2

ˆ = (b) From Eq. 10.75, with ✓ = 180 , R = vt, and R F21 (t) =

ˆ z:

1 q1 q2 (1 v 2 /c2 ) ˆ z. 4⇡✏0 (vt)2

No, Newton’s third law does not hold: F12 6= F21 ,

because of the extra factor (1

v 2 /c2 ).

R (c) From Eq. 8.28, p = ✏0 (E⇥B) d⌧ . Here E = E1 +E2 , whereas B = B2 , so E⇥B = (E1 ⇥B2 )+(E2 ⇥B2 ). But the latter, when integrated over all space, is independent of time. We want only the time-dependent part: Z 1 q1 1 p(t) = ✏0 (E1 ⇥ B2 ) d⌧. Now E1 = ˆ r, while, from Eq. 10.76, B2 = 2 (v ⇥ E2 ), and (Eq. 10.75) 4⇡✏0 r2 c ˆ q2 (1 v 2 /c2 ) R r sin ✓ E2 = . But R = r vt; R2 = r2 + v 2 t2 2rvt cos ✓; sin ✓0 = . So 2 4⇡✏0 (1 v 2 sin ✓0 /c2 )3/2 R2 R q2 (1 v 2 /c2 ) (r vt) E2 = . Finally, noting that v ⇥ (r vt) = v ⇥ r = vr sin ✓ ˆ, we get 4⇡✏0 [1 (vr sin ✓/Rc)2 ]3/2 R3 Z 2 2 q2 (1 v 2 /c2 ) vr sin ✓ 1 r sin ✓ (ˆ r ⇥ ˆ) ˆ. So p(t) = ✏0 q1 q2 (1 v /c )v B2 = . 2 2 2 3/2 4⇡✏0 c 4⇡✏0 4⇡✏0 c r [R2 (vr sin ✓/c)2 ]3/2 [R2 (vr sin ✓/c)2 ] But ˆ r ⇥ ˆ = ✓ˆ = (cos ✓ cos x ˆ + cos ✓ sin y ˆ sin ✓ ˆ z), and the x and y components integrate to zero, so: q1 q2 v(1 v 2 /c2 ) ˆ z p(t) = (4⇡c)2 ✏0 q1 q2 v(1 v 2 /c2 ) ˆ z = 8⇡c2 ✏0

Z Z

sin2 ✓ r [r2 + (vt)2

2rvt cos ✓

3/2

(vr sin ✓/c)2 ]

r sin3 ✓ [r2 + (vt)2

2rvt cos ✓

3/2

(vr sin ✓/c)2 ]

r2 sin ✓ dr d✓ d

dr d✓.


227


I’ll do the r integral first. According to the CRC Tables,  Z 1 1 x 2(bx + 2a) 2 b 2a p p p dx = = 2 2 )3/2 2 2 4ac b c a (a + bx + cx (4ac b ) a + bx + cx 0 0 p p p 2 2 (2 ac b) 2 p p = p b 2 ac = p = p 2 ac + b 2 c(4ac b ) c (2 ac b) (2 ac + b) c In this case x = r, a = (vt)2 , b =

2vt cos ✓, and c = 1

2  q (v/c)2 sin2 ✓ 2vt 1 (v/c)2 sin2 ✓ q 1 (v/c)2 sin2 ✓ + cos ✓ q = ⇥ vt 1 (v/c)2 sin2 ✓ 1 (v/c)2 sin2 ✓ So

=

q 1

But Z ⇡ 0

q1 q2 v(1 v 2 /c2 ) ˆ z p(t) = 2 8⇡c ✏0 vt(1 8 Z q1 q2 ˆ z < ⇡ = sin ✓ d✓ + 8⇡c2 ✏0 t : 0

R⇡

2vt cos ✓

c v

Z

0

q vt 1

⇡

1 q (v/c)2 sin2 ✓ 1 v 2 /c2 )

2

41 + q 1

(v/c)2 sin2 ✓ cos ✓ (v/c)2 sin2 ✓

2 3 1 4 cos ✓ 5 sin3 ✓ d✓ 1+ q 2 2 2 0 sin ✓ 1 (v/c) sin ✓ 9 = cos ✓ sin ✓ q d✓ . ; (c/v)2 sin2 ✓ Z

.

(v/c)2 sin2 ✓. So the r integral is

1 ⇤ = vt sin2 ✓(1 cos2 ✓

1 v 2 /c2 )

1

cos ✓ 3

5.

⇡

sin ✓ d✓ = 2. In the second integral let u ⌘ cos ✓, so du = sin ✓ d✓: Z 1 cos ✓ sin ✓ u q p d✓ = du = 0 (the integrand is odd, and the interval is even). 2 (c/v) 1 + u2 1 (c/v)2 sin2 ✓ µ0 q1 q2 Conclusion: p(t) = ˆ z (plus a term constant in time). 4⇡t (d) ✓ ◆ 1 q1 q2 1 q1 q2 (1 v 2 /c2 ) q1 q2 v2 q1 q2 µ0 q1 q2 F12 + F21 = ˆ z ˆ z = 1 1 + ˆ z= ˆ z= ˆ z. 4⇡✏0 v 2 t2 4⇡✏0 v 2 t2 4⇡✏0 v 2 t2 c2 4⇡✏0 c2 t2 4⇡t2 dp µ0 q1 q2 = ˆ z = F12 + F21 . qed dt 4⇡t2 0

Since q1 is at rest, and q2 is moving at constant velocity, there must be another force (Fmech ) acting on them, to balance F12 + F21 ; what we have found is that Fmech = dpem /dt, which means that the impulse imparted to the system by the external force ends up as momentum in the fields. [For further discussion of this problem see J. J. G. Scanio, Am. J. Phys. 43, 258 (1975).]


228


Problem 10.33 Maxwell’s equations with magnetic charge read: @B , @t @E (ii) r · B = µ0 ⇢m , (iv) r ⇥ B = µ0 Je + µ0 ✏0 , @t (i) r · E =

1 ✏ 0 ⇢e ,

(iii) r ⇥ E =

µ0 Jm

where ⇢e is the electric charge density, ⇢m is the magnetic charge density, Je is the electric current density, and Jm is the magnetic current density. If there are only electric charges and currents, then the usual potential formulation applies (Section 10.1.1), with E = rVe @Ae /@t, B = r ⇥ Ae . If there are only magnetic charges and currents, (i) r · E = 0,

(iii) r ⇥ E =

@B , @t

µ0 Jm

(ii) r · B = µ0 ⇢m , (iv) r ⇥ B = µ0 ✏0

@E . @t

This time equation (i) says that E can be expressed as the curl of a vector potential: E = r ⇥ Am , and plugging this into (iv) yields r ⇥ (B + µ0 ✏0 @Am /@t) = 0, which tells us that the quantity in parentheses can be represented as the gradient of a scalar: B = rVm µ0 ✏0 @Am /@t. [The signs of Vm and Am are arbitrary, but I think this choice yields the most symmetrical formulation.] Putting these into (ii) and (iii) yields ✓ ◆ ✓ ◆ @ @Vm @Vm 22 Vm = µ0 ⇢m µ0 ✏0 r · Am + , 22 Am = µ0 Jm + r r · Am + . @t @t @t The “Lorenz gauge” for the magnetic potentials, r · Am = @Vm /@t, kills the terms in parentheses. In general, if there are both electric and magnetic charges and currents present, the total fields are the sums (by the superposition principle): E=

@Ae @t

rVe

B=

r ⇥ Am ,

rVm

µ0 ✏0

@Am + r ⇥ Ae . @t

If we choose to work in the Lorenz gauge: r · Ae =

µ0 ✏0

@Ve , @t

r · Am =

@Vm , @t

Maxwell’s equations become (in terms of the potentials) 22 V e =

1 ⇢e , ✏0

22 V m =

µ0 ⇢m ,

22 A e =

µ0 Je ,

22 A m =

µ0 4⇡

Z

⇢m (r0 , tr )

µ0 Am (r, t) = 4⇡

Z

Jm (r0 , tr )

µ0 Jm ,

and the retarded solutions are Ve (r, t) =

1 4⇡✏0

µ0 Ae (r, t) = 4⇡

Z

Z

⇢e (r0 , tr )

r

Je (r0 , tr )

r

d⌧ 0 ,

0

d⌧ ,

Vm (r, t) =

r

r

d⌧ 0 ,

d⌧ 0 .


229

CHAPTER 10. POTENTIALS AND FIELDS Problem 10.34 The retarded potentials are V (r, t) =

1 4⇡✏0

Z

⇢(r0 , tr )

r

d⌧ 0 ,

A(r, t) =

µ0 4⇡

Z

J(r0 , tr )

r

d⌧ 0 .

We want a source localized at the origin, so we expand in powers of r0 , keeping terms up to first order:

r2 r

t

= (r r0 ) · (r r0 ) = r2 2r · r0 + (r0 )2 , ✓ ◆ r · r0 ⇡r 1 , r2 ✓ ◆ 1 1 r · r0 ⇡ 1+ 2 , r r r r ⇡ t r + r · r0 = t + r · r0 , 0 c c rc rc

where t0 ⌘ t

r/c is the retarded time for a source at the origin. Thus (Taylor expanding about t0 ) ✓ ◆ r ⇡ ⇢(r0 , t ) + r · r0 ⇢(r ⇢(r0 , tr ) = ⇢ r0 , t ˙ 0 , t0 ), 0 c rc ✓ ◆ ⇢(r0 , tr ) 1 r · r0 r · r0 1 r · r0 r · r0 = 1+ 2 ⇢(r0 , t0 ) + ⇢(r ˙ 0 , t0 ) ⇡ ⇢(r0 , t0 ) + 3 ⇢(r0 , t0 ) + 2 ⇢(r ˙ 0 , t0 ), r r r rc r r r c

and hence

 Z Z Z 1 1 r r 0 0 0 0 0 V (r, t) = ⇢(r , t0 ) d⌧ + 3 · r ⇢(r , t0 ) d⌧ + 2 · r0 ⇢(r ˙ 0 , t0 ) d⌧ 0 . 4⇡✏0 r r r c

The first integral is the total charge of the dipole, which is zero; the second integral is the dipole moment, and the third is the time derivative of the dipole moment: V (r, t) =

i 1 ˆ r h r ˙ · p(t ) + p(t ) . 0 0 4⇡✏0 r2 c

By the same reasoning, the vector potential is  Z Z Z µ0 1 1 1 ˙ 0 , t0 ) d⌧ 0 . A(r, t) = J(r0 , t0 ) d⌧ 0 + 3 (r · r0 )J(r0 , t0 ) d⌧ 0 + 2 (r · r0 )J(r 4⇡ r r r c Z ˙ From Eq. 5.31 J(r0 , t) d⌧ 0 = p(t), and J = ⇢v0 = ⇢(dr0 /dt) is already first order in r0 , so the second and third integrals are second order, and we drop them.

A(r, t) =

 ˙ 0) µ0 p(t . 4⇡ r

In calculating the fields, remember that the dipole moments themselves depend on r, through their argument (t0 = t r/c). For a function of t0 alone, ✓ ◆ df 1 1 ˙ rf (t0 ) = rt0 = f˙ rr = fˆ r, dt0 c c c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

230


h i ˆ r d r . For a structure of the form f (t0 ) + f˙(t0 ) c dt c ✓ ◆ h i r ˆ r ˙ r ˆ r¨ f˙ r ¨ r f (t0 ) + f˙(t0 ) = f+ f + ˆ r= f. c c c c c c2

so you simply replace r by

Using the product rule for r(A · B): ⇢ ✓ ◆ ✓ ◆h ⇣ h ⌘ ˆ 1 ˆ r r i⌘ h r i ˆ r ˆ r r i ⇣h r i r ˙ ˙ ˙ ˙ ⇥ r ⇥ p + + · r p + rV = p + p + p ⇥ r ⇥ p + p + p · r 2 2 2 4⇡✏0 r c c r r c c r2 ⇢  ⇣ ⌘ ⇣ ⌘ ⇣h i ⇣ h i⌘⌘ 1 ˆ r r ˆ r r 1 r r ¨ +0+ 2 · ¨ + 3 p + p˙ = ⇥ ⇥p p 3ˆ r ˆ r · p + p˙ 2 2 2 4⇡✏0 r c r c r c c ⇢ ¨ ) [p + (r/c)p] ˙ ˙ 1 ˆ r(ˆ r·p 3ˆ r(ˆ r · [p + (r/c)p]) = + . 4⇡✏0 rc2 r3 The others are easier:  µ0 1 ˙ r⇥A = (r ⇥ p) 4⇡ r ✓ ◆ ¨ @A µ0 p = . @t 4⇡ r

✓ ◆  ✓ ◆ ✓ ◆ 1 µ0 1 ˆ r ˆ r ¨ + p˙ ⇥ 2 p˙ ⇥ r = ⇥p = r 4⇡ r c r

µ0 4⇡

⇢

ˆ r ⇥ [(p˙ + (r/c)¨ p] 2 r

The fields are therefore E(r, t) =

rV

@A = @t

B(r, t) = r ⇥ A =

µ0 4⇡

µ0 4⇡

⇢

⇢

¨ p

¨) ˙ ˆ r(ˆ r·p [p + (r/c)p] + c2 r

˙ 3ˆ r(ˆ r · [p + (r/c)p]) 3 r

,

ˆ r ⇥ [p˙ + (r/c)¨ p] 2 r

(with all dipole moments evaluated at the retarded time t0 = t

r/c).


,

231

CHAPTER 11. RADIATION

Chapter 11

Radiation Problem 11.1 µ0 p0 ! 1 ˆ so From Eq. 11.17, A = sin[!(t r/c)](cos ✓ ˆ r sin ✓ ✓), 4⇡ r ⇢   µ0 p0 ! 1 @ 1 @ 1 21 r·A = r sin[!(t r/c)] cos ✓ + sin2 ✓ sin[!(t r/c)] 2 4⇡ r @r r r sin ✓ @✓ r ⇢ ⌘ µ0 p0 ! 1 ⇣ !r 2 sin ✓ cos ✓ = sin[!(t r/c)] cos[!(t r/c)] cos ✓ sin[!(t r/c)] 4⇡ r2 c r2 sin ✓ ⇢ ✓ ◆ p0 ! 1 ! = µ0 ✏0 sin[!(t r/c)] + cos[!(t r/c)] cos ✓ . 2 4⇡✏0 r rc Meanwhile, from Eq. 11.12, ⇢ @V p0 cos ✓ !2 ! = cos[!(t r/c)] sin[!(t r/c)] @t 4⇡✏0 r c r ⇢ p0 ! 1 ! sin[!(t r/c)] + = cos[!(t r/c)] cos ✓. So r · A = 4⇡✏0 r2 rc

Problem 11.2

! p0 · ˆ r sin[!(t 4⇡✏0 c r

Eq. 11.14: V (r, t) =

r/c)]. Eq. 11.17: A(r, t) =

Now p0 ⇥ ˆ r = p0 sin ✓ ˆ and ˆ r ⇥ (p0 ⇥ ˆ r) = p0 sin ✓(ˆ r ⇥ ˆ) = Eq. 11.18: E(r, t) = Eq. 11.21: hSi =

µ0 ! ˆ r ⇥ (p0 ⇥ ˆ r) cos[!(t 4⇡ r 2

µ0 ✏0

@V . qed @t

µ0 ! p0 sin[!(t 4⇡ r

r/c)].

ˆ so p0 sin ✓ ✓,

r/c)]. Eq. 11.19: B(r, t) =

µ0 ! 2 (p0 ⇥ ˆ r) cos[!(t 4⇡c r

r/c)].

µ0 ! 4 (p0 ⇥ ˆ r)2 ˆ r. 32⇡ 2 c r2

Problem 11.3 P = I 2 R = q02 ! 2 sin2 (!t)R (Eq. 11.15) ) hP i = 12 q02 ! 2 R. Equate this to Eq. 11.22:

1 2 2 µ0 q02 d2 ! 4 µ0 d2 ! 2 2⇡c q0 ! R = ) R= ; or, since ! = , 2 12⇡c 6⇡c ✓ ◆2 ✓ ◆2 ✓ ◆2 µ0 d2 4⇡ 2 c2 2 d 2 d d 7 8 2 R= = ⇡µ c = ⇡(4⇡ ⇥ 10 )(3 ⇥ 10 ) = 80⇡ ⌦ = 789.6(d/ )2 ⌦. 0 2 6⇡c 3 3


232


For the wires in an ordinary radio, with d = 5 ⇥ 10 2 m and (say) which is negligible compared to the Ohmic resistance.

= 103 m, R = 790(5 ⇥ 10

) = 2 ⇥ 10

5 2

6

⌦,

Problem 11.4 By the superposition principle, we can add the✓potentials of the ◆ two dipoles. Let’s first express V (Eq. 11.14) p0 ! z sin[!(t r/c)]. That’s for an oscillating dipole in Cartesian coordinates: V (x, y, z, t) = 4⇡✏0 c x2 + y 2 + z 2 along the z axis. For one along x or y, we just change z to x or y. In the present case, p = p0 [cos(!t) x ˆ + cos(!t sin[!(t

r/c)] ! sin[!(t

p0 ! 4⇡✏0 c

V = = A =

⇢

⇡/2) y ˆ], so the one along y is delayed by a phase angle ⇡/2: ⇡/2] =

r/c)

cos[!(t

x sin[!(t x2 + y 2 + z 2

r/c)]

p0 ! sin ✓ {cos sin[!(t 4⇡✏0 c r

r/c)]

µ0 p0 ! {sin[!(t 4⇡r

cos[!(t

r/c)] x ˆ

r/c)] (just let !t ! !t

y cos[!(t x2 + y 2 + z 2

sin cos[!(t

r/c)]} .

⇡/2). Thus

r/c)] Similarly,

r/c)] y ˆ} .

We could get the fields by di↵erentiating these potentials, but I prefer to work with Eqs. 11.18 and 11.19, ˆ and cos ✓ = z/r, Eq. 11.18 can be written using superposition. Since ˆ z = cos ✓ ˆ r sin ✓ ✓, ⇣ µ0 p0 ! 2 z ⌘ E= cos[!(t r/c)] ˆ z ˆ r . In the case of the rotating dipole, therefore, 4⇡r r E= B =

S=

µ0 p0 ! 2 n cos[!(t 4⇡r 1 (ˆ r ⇥ E). c

⇣ r/c)] x ˆ

x ⌘ ˆ r + sin[!(t r

1 1 1 ⇥ 2 (E ⇥ B) = [E ⇥ (ˆ r ⇥ E)] = E ˆ r µ0 µ0 c µ0 c

E2 =

✓

where a ⌘ x ˆ

µ0 p0 ! 2 4⇡r

◆2

a2 cos2 [!(t

(x/r)ˆ r and b ⌘ y ˆ

⇣ r/c)] y ˆ

y ⌘o ˆ r , r

⇤ E2 (E · ˆ r)E = ˆ r (notice that E · ˆ r = 0). Now µ0 c

r/c)] + b2 sin2 [!(t

r/c)] + 2(a · b) sin[!(t

r/c)] cos[!(t

r/c)] ,

(y/r)ˆ r. Noting that x ˆ · r = x and y ˆ · r = y, we have


233

CHAPTER 11. RADIATION a2 = 1 +

x2 r2

E =

✓

2

= =

= =

2

x2 =1 r2

x2 2 ; b =1 r2

y2 ; a·b= r2

yx rr

x y xy + 2 = rr r

xy . r2

◆2 ⇢✓ ◆ ✓ ◆ µ0 p0 ! 2 x2 y2 2 1 cos [!(t r/c)] + 1 sin2 [!(t r/c)] 4⇡r r2 r2 o xy 2 2 sin[!(t r/c)] cos[!(t r/c)] r ✓ ◆2 ⇢ µ0 p0 ! 2 1 1 x2 cos2 [!(t r/c)] + 2xy sin[!(t r/c)] cos[!(t r/c)] + y 2 sin2 [!(t 4⇡r r2 ✓ ◆2 ⇢ µ0 p0 ! 2 1 2 1 (x cos[!(t r/c)] + y sin[!(t r/c)]) 4⇡r r2 But x = r sin ✓ cos and y = r sin ✓ sin . ✓ ◆2 n o µ0 p0 ! 2 2 1 sin2 ✓ (cos cos[!(t r/c)] + sin sin[!(t r/c)]) 4⇡r ✓ ◆2 n o µ0 p0 ! 2 2 1 (sin ✓ cos[!(t r/c) ]) . 4⇡r

S= hSi = P = =

r/c)]

✓ ◆2 n o µ0 p0 ! 2 2 1 (sin ✓ cos[!(t r/c) ]) ˆ r. c 4⇡r ✓ ◆2  µ0 p0 ! 2 1 1 sin2 ✓ ˆ r. c 4⇡r 2 ✓ ◆2 Z ✓ ◆ Z µ0 p0 ! 2 1 1 2 hSi · da = 1 sin ✓ r2 sin ✓ d✓ d c 4⇡ r2 2 Z ⇡ ✓ ◆ Z µ0 p20 ! 4 1 ⇡ 3 µ0 p20 ! 4 1 4 µ0 p20 ! 4 2⇡ sin ✓ d✓ sin ✓ d✓ = 2 · = . 2 16⇡ c 2 0 8⇡c 2 3 6⇡c 0

This is twice the power radiated by either oscillating dipole alone (Eq. 11.22). In general, S =

1 (E ⇥ B) = µ0

1 1 [(E1 + E2 ) ⇥ (B1 + B2 )] = [(E1 ⇥ B1 ) + (E2 ⇥ B2 ) + (E1 ⇥ B2 ) + (E2 ⇥ B1 )] = S1 + S2 + cross terms. µ0 µ0 In this particular case the fields of 1 and 2 are 90 out of phase, so the cross terms go to zero in the time averaging, and the total power radiated is just the sum of the two individual powers. Problem 11.5 Go back to Eq. 11.33: A=

µ0 m0 4⇡

✓

sin ✓ r

◆⇢

1 cos[!(t r

r/c)]

! sin[!(t c


r/c)]

ˆ.

234


Since V = 0 here,

✓ ◆⇢ µ0 m0 sin ✓ 1 ! E= ( !) sin[!(t r/c)] ! cos[!(t 4⇡ r r c ✓ ◆⇢ µ0 m0 ! sin ✓ 1 ! = sin[!(t r/c)] + cos[!(t r/c)] ˆ. 4⇡ r r c @A = @t

r/c)]

ˆ

1 @ 1 @ B = r⇥A= (A sin ✓) ˆ r (rA ) ✓ˆ r sin ✓ @✓ r @r ⇢  µ0 m0 1 2 sin ✓ cos ✓ 1 ! = cos[!(t r/c)] sin[!(t r/c)] ˆ r 4⇡ r sin ✓ r r c  sin ✓ 1 ! ! ⇣ !⌘ cos[!(t r/c)] + sin[!(t r/c)] cos[!(t r/c)] ✓ˆ r r2 rc c c  µ0 m0 n 2 cos ✓ 1 ! cos[!(t r/c)] sin[!(t r/c)] ˆ r 4⇡ r2  r c = ⇣ ! ⌘2 o sin ✓ 1 ! cos[!(t r/c)] + sin[!(t r/c)] + cos[!(t r/c)] ✓ˆ . 2 r r rc c These are precisely the fields we studied in Prob. 9.35, with A !

the solution to that problem) is

µ0 m0 ! 2 . The Poynting vector (quoting 4⇡c

✓ ◆ ✓ ◆ µ0 m20 ! 3 sin ✓ n 2 cos ✓ c2 c S= 1 sin u cos u + cos2 u sin2 u ✓ˆ 2 r2 16⇡ 2 c2✓ r2 r ! !r ◆ o 2 c2 ! c sin ✓ + 2 3 sin u cos u + cos2 u + 2 sin2 u cos2 u ˆ r , r ! r c !r where u ⌘

!(t

r/c). The intensity is hSi =

µ0 m20 ! 4 sin2 ✓ ˆ r, the same as Eq. 11.39. 32⇡ 2 c3 r2

Problem 11.6 I 2 R = I02 R cos2 (!t) ) hP i =

1 2 µ0 m20 ! 4 µ0 ⇡ 2 b4 I02 ! 4 µ0 ⇡b4 ! 4 2⇡c I0 R = = , so R = ; or, since ! = , 3 3 2 12⇡c 12⇡c 6c3

µ0 ⇡b4 16⇡ 4 c4 8 R= = ⇡ 5 µ0 c 3 4 6c 3

✓ ◆4 b

=

8 5 (⇡ )(4⇡ ⇥ 10 3

7

)(3 ⇥ 108 )(b/ )4 = 3.08 ⇥ 105 (b/ )4 ⌦.

Because b ⌧ , and R goes like the fourth power of this small number, R is typically much smaller than the electric radiative resistance (Prob. 11.3). For the dimensions we used in Prob. 11.3 (b = 5 cm and = 103 m), R = 3 ⇥ 105 (5 ⇥ 10 5 )4 = 2 ⇥ 10 12 ⌦, which is a millionth of the comparable electrical radiative resistance. Problem 11.7 0 0 With ↵ = 90 , Eq. 7.68 ) E0 = cB, B0 = E/c, qm = cqe ) m0 ⌘ qm d = cqe d = cp0 . So ⇢ ✓ ◆ ✓ ◆ µ0 ( m0 /c)! 2 sin ✓ µ0 m0 ! 2 sin ✓ E0 = c cos[!(t r/c)] ˆ = cos[!(t r/c)] ˆ. 4⇡c r 4⇡c r ⇢ ✓ ◆ ✓ ◆ 1 µ0 ( m0 /c)! 2 sin ✓ µ0 m0 ! 2 sin ✓ ˆ B0 = cos[!(t r/c)] ✓ˆ = cos[!(t r/c)] ✓. c 4⇡ r 4⇡c2 r


235


These are identical to the fields of an Ampére dipole (Eqs. 11.36 and 11.37), which is consistent with our general experience that the two models generate identical fields except right at the dipole (not relevant here, since we’re in the radiation zone). Problem 11.8 (a) The power radiated (Eq. 11.60) is so

dWr µ0 2 ¨ = Q0 = p¨ . In this case p = Qd, so p¨ = Qd dt 6⇡c

dWr µ0 (Q0 d)2 2t/RC = e , and the total energy radiated is dt 6⇡c (RC)4  Z µ0 (Q0 d)2 1 2t/RC µ0 (Q0 d)2 RC 2t/RC Wr = e dt = e 6⇡c (RC)4 0 6⇡c (RC)4 2

1

✓

1 RC

◆2

e

t/RC

d,

µ0 (Q0 d)2 RC µ0 (Q0 d)2 = . 4 6⇡c (RC) 2 12⇡c (RC)3

=

0

The fraction of the original energy that is radiated is therefore f=

Wr µ0 (Q0 d)2 2C µ0 d2 = = . 2 3 W0 12⇡c (RC) Q0 6⇡c R3 C 2

˙ ¨ picks up a delta function. But any real circuit has some [Technically, Q(t) is discontinuous at t = 0, and Q ˙ (self-)inductance, which smoothes out the sudden change in Q.] (b) With the parameters given, f=

4⇡ ⇥ 10 7 10 8 6⇡(3 ⇥ 108 ) (106 )(10

24 )

About a millionth of the total energy is radiated away, so Problem 11.9 ¨ (t) = p(t) = p0 [cos(!t) x ˆ + sin(!t) y ˆ] ) p

= 2.22 ⇥ 10

6

.

yes, this is negligible.

! 2 p0 [cos(!t) x ˆ + sin(!t) y ˆ] )

µ0 p20 ! 4 sin2 ✓ ˆ r. (This appears to disagree 16⇡ 2 c r2 ¨ (t0 ); with the answer to Prob. 11.4. The reason is that in Eq. 11.59 the polar axis is along the direction of p as the dipole rotates, so do the axes. Thus the angle ✓ here is not the same as in Prob. 11.4.) Meanwhile, 2

[¨ p(t)] = ! 4 p20 [cos2 (!t) + sin2 (!t)] = p20 ! 4 . So Eq. 11.59 says S =

µ0 p20 ! 4 . (This does agree with Prob. 11.4, because we have now integrated over all angles, 6⇡c and the orientation of the polar axis irrelevant.) Eq. 11.60 says P =

Problem 11.10 At t = 0 the dipole moment of the ring is Z Z p0 = r dl = ( 0 sin )(b sin y ˆ + b cos x ˆ)b d = = b (⇡ y ˆ + 0x ˆ) = ⇡b 2

2

0

y ˆ.

Therefore (Eq. 11.60) P =

µ0 4 2 ! (⇡b 6⇡c

2 0)

=

6c

2

✓ Z y ˆ

2⇡

sin2 d + x ˆ

0

¨ = sin(!t) x ˆ], so p

As it rotates (counterclockwise, say) p(t) = p0 [cos(!t) y ˆ ⇡µ0 ! 4 b4 20

0b

.


Z

0

2⇡

sin cos d

◆

! 2 p, and hence (¨ p)2 = ! 4 p20 .

236


Problem 11.11 Here V = 0 (since the ring is neutral), andI the current depends only on t (not on position), so the retarded µ0 I(t r /c) 0 vector potential (Eq. 11.52) is A(r, t) = dl . But in this case it does not suffice to replace r 4⇡ r by r in the denominator—that would ✓ ◆ lead to Eq. 11.54, and hence to A = 0 (since p = 0). Instead, use 1 ⇠1 b 0 0 0ˆ 0 0 0 Eq. 11.30: r = r 1 + r sin ✓ cos . Meanwhile, dl = b d = b( sin xˆ + cos yˆ) d , and I(t

r

/c) ⇠ = I(t

r/c + (b/c) sin ✓ cos

0

) = I(t0 + (b/c) sin ✓ cos

0

˙ 0 ) b sin ✓ cos )⇠ = I(t0 ) + I(t c

0

(carrying all terms to first order in b). As always, t0 = t r/c. (From now on I’ll suppress the argument: I, ˙ etc. are all to be evaluated at t0 .) Then I, ✓ ◆✓ ◆ I µ0 1 b b A(r, t) = 1 + sin ✓ cos 0 I + I˙ sin ✓ cos 0 b( sin 0 x ˆ + cos 0 y ˆ) d 0 4⇡ r r c  Z µ0 b 2⇡ b b ⇠ I + I˙ sin ✓ cos 0 + I sin ✓ cos 0 ( sin 0 x ˆ + cos 0 y ˆ) d 0 . = 4⇡r 0 c r Z 2⇡ Z 2⇡ Z 2⇡ Z 2⇡ 0 0 0 0 0 0 0 But sin d = cos d = sin cos d = 0, while cos2 0 d 0 = ⇡. 0 0 0 0  ⇣ µ0 b b b µ0 b2 r ˙⌘ = (⇡ y ˆ) I˙ sin ✓ + I sin ✓ = sin ✓ I + I y ˆ. 4⇡r c r 4r2 c

In general (i.e. for points not on the x z plane) y ˆ ! ˆ; moreover, in the radiation zone we are not interested h i sin ✓ µ0 b2 ˙ ˆ. in terms that go like 1/r2 , so A(r, t) = I(t r/c) 4c r E(r, t) = B(r, t) = = S= P = =

i sin ✓ µ0 b2 h ¨ ˆ. I(t r/c) 4c r 1 @ 1 @ r⇥A= (A sin ✓) ˆ r (rA ) ✓ˆ r sin ✓ @✓ r @r " # ✓ ◆ µ0 b2 I˙ 1 1¨ 1 µ0 b2 ¨sin ✓ ˆ ˆ 2 sin ✓ cos ✓ ˆ r I sin ✓ ✓ = I ✓. 4c r sin ✓ r r c 4c2 r ✓ ◆2 1 1 µ0 b2 ¨sin ✓ ⇣ ˆ ˆ⌘ µ0 ⇣ 2 ¨⌘2 sin2 ✓ (E ⇥ B) = I ⇥✓ = b I ˆ r. µ0 µ0 c 4c r 16c3 r2 ✓ ◆ Z Z µ0 ⇣ 2 ¨⌘2 sin2 ✓ 2 µ0 ⇣ 2 ¨⌘2 4 µ0 ⇡ ⇣ 2 ¨⌘2 S · da = b I r sin ✓ d✓ d = b I (2⇡) = b I 16c3 r2 16c3 3 6c3 @A = @t

µ0 m ¨2 . 6⇡c3

¨ 2 .) (Note that m = I⇡b2 , so m ¨ = I⇡b

Problem 11.12 p = ey y ˆ, y = 12 gt2 , so p =

µ0 ge y ˆ. Therefore (Eq. 11.60) :P = (ge)2 . Now, the time 6⇡c p it takes to fall a distance h is given by h = 12 gt2 ) t = 2h/g, so the energy radiated in falling a distance h 1 2 2 get

¨= y ˆ; p


237


µ0 (ge)2 p 2h/g. Meanwhile, the potential energy lost is Upot = mgh. So the fraction is 6⇡c s s r µ0 g 2 e2 2h 1 µ0 e2 2g (4⇡ ⇥ 10 7 )(1.6 ⇥ 10 19 )2 (2)(9.8) = = . = = 2.76 ⇥ 10 6⇡c g mgh 6⇡mc h 6⇡(9.11 ⇥ 10 31 )(3 ⇥ 108 ) (0.01)

is Urad = P t = f=

Urad Upot

22

.

Evidently almost all the energy goes into kinetic form (as indeed I assumed in saying y = 12 gt2 ). Problem 11.13

dWr µ0 q 2 a2 1 qQ k qQ = . Here F = = ma, so a = 2 , where k ⌘ . dt 6⇡c 4⇡✏0 x2 x 4⇡✏0 m The energy radiated (twice that radiated on the way out) is Z 2 Z µ0 q 2 k µ0 q 2 2 1 1 1 Wr = dt = 2k dx, 4 6⇡c x4 6⇡c x0 x v The power radiated (Eq. 11.70) is

where x0 is the distance of closest approach. Conservation of energy (ignoring radiative losses, as suggested) says 1 1 1 qQ qQ 1 2k 2k mv 2 = mv 2 + ) v 2 = v02 2 = v02 , and x0 = 2 2 0 2 4⇡✏0 x 4⇡✏0 m x x v0 (note that q is at x0 when v = 0). So Z Z 1 µ0 q 2 2 1 1 µ0 q 2 2k 2 1 p p p Wr = 2k dx = 2 4 4 6⇡c 6⇡c 2k x (1/x x v 2k/x x0 x0 0) 0

(1/x)

dx =

µ0 q 2 2k 2 16 p . 6⇡c 2k 15x5/2 0

(I used Mathematica to do the integral.) Simplifying, Wr =

2µ0 q 2 v05 2µ0 q 2 v05 4⇡✏0 m 8qmv05 Wr 8qmv05 2 16q ⇣ v0 ⌘3 = = ) f= = = . 2 3 3 45⇡ck 45⇡c qQ 45c Q W0 45c Q mv0 45Q c

Problem 11.14 s 1 q2 v2 1 q2 F = = ma = m ) v = . At the beginning (r0 = 0.5 ˚ A), 2 4⇡✏0 r r 4⇡✏0 mr  v (1.6 ⇥ 10 19 )2 = 12 c 4⇡(8.85 ⇥ 10 )(9.11 ⇥ 10 31 )(5 ⇥ 10

1/2 11 )

1 = 0.0075, 3 ⇥ 108

and when the radius is one hundredth of this v/c is only 10 times greater (0.075), so for most of the trip the velocity is safely nonrelativistic. ✓ ◆2 ✓ ◆2 µ0 q 2 v 2 µ0 q 2 1 q2 From the Larmor formula, P = = (since a = v 2 /r), and P = dU/dt, 6⇡c r 6⇡c 4⇡✏0 mr2 where U is the (total) energy of the electron: ✓ ◆ 1 1 q2 1 1 q2 1 q2 1 q2 2 U = Ukin + Upot = mv = = . 2 4⇡✏0 r 2 4⇡✏0 r 4⇡✏0 r 8⇡✏0 r So

dU = dt

1 q 2 dr q2 = P = 8⇡✏0 r2 dt 6⇡✏0 c3

✓

1 q2 4⇡✏0 mr2

◆2

, and hence

dr = dt

1 3c


✓

q2 2⇡✏0 mc

◆2

1 , or r2

238


✓

◆2 ✓ ◆2 Z 0 ✓ ◆2 2⇡✏0 mc 2⇡✏0 mc 2⇡✏0 mc 2 2 dt = 3c r dr ) t = 3c r dr = c r03 q2 q2 q2 r0  2 2⇡(8.85 ⇥ 10 12 )(9.11 ⇥ 10 31 )(3 ⇥ 108 ) 8 = (3 ⇥ 10 ) (5 ⇥ 10 11 )3 = 1.3 ⇥ 10 (1.6 ⇥ 10 19 )2

11

s. (Not very long!)

Problem 11.15

 d sin2 ✓ According to Eq. 11.74, the maximum occurs at = 0. Thus d✓ (1 cos ✓)5 2 sin ✓ cos ✓ 5 sin2 ✓( sin ✓) = 0 ) 2 cos ✓(1 cos ✓) = 5 sin2 ✓ = 5 (1 cos2 ✓); (1 cos ✓)5 (1 cos ✓)6 2 cos ✓ 2 cos2 ✓ = 5 5 cos2 ✓, or 3 cos2 ✓ + 2 cos ✓ 5 = 0. So p ⌘ 2 ± 4 + 60 2 1 ⇣ p cos ✓ = = ± 1 + 15 2 1 . We want the plus sign, since ✓m ! 90 (cos ✓m = 0) when 6 3 ! p 1 + 15 2 1 1 ! 0 (Fig. 11.11): ✓max = cos . 3 For v ⇡ c, p 1 + 15 3

2

⇡ 1; write 1

!

=1

✏ (where ✏ ⌧ 1), and expand to first order in ✏:

hp

i 1 hp i 1 ⇠ 1 + 15(1 2✏) 1 = (1 + ✏) 3(1 ✏) 3  ✓ h p i 1 ⇥p ⇤ 1 1 = (1 + ✏) 16 30✏ 1 = (1 + ✏) 4 1 (15✏/8) 1 = (1 + ✏) 4 1 3 3 3 ✓ ◆ 1 15 5 ⇠ 5 1 = (1 + ✏) 3 ✏ = (1 + ✏)(1 ✏) = 1 + ✏ ✏=1 ✏. 3 4 4 4 4 1

=

1 + 15(1

✏)2

2 2 Evidently ✓max ⇡ 0, so cos ✓max ⇠ = 1 14 ✏ ) ✓max = 12 ✏, or ✓max ⇠ = 1 12 ✓max =  2 (dP/d⌦|✓m )ur sin ✓max Let f ⌘ = . Now sin2 ✓max ⇠ = ✏/2, and (dP/d⌦|✓m )rest (1 cos ✓max )5 ur

(1

cos ✓max ) ⇠ =1 1

=p

1

2

=p 1

(1

✏)(1

1 (1

✏)2

1 4 ✏)

⇠ =p 1

✏/2 (1 ✏ = So f = = (5✏/4)5 1 1 1 = p ) ✏ = 2 . Therefore 2 2✏ (1 2✏)

⇠ =1

✓ ◆5 4 1 f= (2 5 2

1 4 ✏)

5 4 ✏.

1 ) = 4

2 4

✓ ◆5 8 5

8

= 2.62

8

p p ✏/2 = (1 ✓ ◆5 4 1 . 5 2✏4

◆ 15 ✏ 16

)/2.

But

.

Problem 11.16

rˆ ⇥ (u ⇥ a) 2 dP q2 Equation 11.72 says = . d⌦ 16⇡ 2 ✏0 ( rˆ · u)5

u = c rˆ

v = c rˆ

vˆ z ) rˆ · u = c

v( rˆ · ˆ z) = c

Let

⌘ v/c. ⇣ v cos ✓ = c 1

⌘ v cos ✓ = c(1 c

cos ✓);


1

239

CHAPTER 11. RADIATION a · u = ac(ˆ x · rˆ )

av(ˆ x·ˆ z) = ac sin ✓ cos ;

u2 = u · u = c2

2cv( rˆ · ˆ z) + v 2 = c2 + v 2

2cv cos ✓.

rˆ ⇥ (u ⇥ a) = ( rˆ · a)u ( rˆ · u)a; ˆ | r ⇥ (u ⇥ a)|2 = ( rˆ · a)2 u2 2(u · a)( rˆ · a)( rˆ · u) + ( rˆ · u)2 a2 = (c2 + v 2 2cv cos ✓)(a sin ✓ cos )2 2(ac sin ✓ cos )(a sin ✓ cos )(c v cos ✓) + a2 c2 (1 ⇥ ⇤ = a2 c2 (1 cos ✓)2 + (sin2 ✓ cos2 )(c2 + v 2 2cv cos ✓ 2c2 + 2cv cos ✓ ⇥ ⇤ 2 = a2 c2 (1 cos ✓)2 (1 )(sin ✓ cos )2 . ⇥ ⇤ 2 cos ✓)2 (1 ) sin2 ✓ cos2 dP µ0 q 2 a2 (1 = . d⌦ 16⇡ 2 c (1 cos ✓)5

cos ✓)2

The total power radiated (in all directions) is: Z Z Z ⇥ (1 cos ✓)2 dP dP µ0 q 2 a2 d⌦ = sin ✓ d✓ d = 2 d⌦ d⌦ 16⇡ c (1 Z 2⇡ Z 2⇡ But d = 2⇡ and cos2 d = ⇡. 0 0 ⇤ Z ⇡⇥ 2 2(1 cos ✓)2 (1 ) sin2 ✓ µ0 q 2 a2 = ⇡ sin ✓ d✓. 16⇡ 2 c (1 cos ✓)5 0

P =

Z

Let w ⌘ (1

cos ✓). Then (1 (1

2w2

2 2

)

(

2

w)/ = cos ✓;

1 + 2w

w2 ) = =

dw =

sin ✓ d✓ ) sin ✓ d✓ =

1

Z (1+ ) µ0 q 2 a2 1 1 ⇥ (1 3 5 16⇡c (1 ) w Z 1 2 2 Int = (1 ) dw 2(1 w5  ✓ ◆ 1 2 2 = (1 ) 2(1 4w4 1+ 1 1+ 1 1+ 1

1 (1 + )2 1 = (1 + )3 1 = (1 + )4 =

1 (1

)2 1 )3

(1 1 (1

)4

1 ⇥ 2 2w 2 1 ⇥ (1 2

2

= = =

(1 (1 (1

⇥

)

2 2

⇤ w)2 /

(1

2

+ (1

When ✓ = 0, w = (1

dw.

P =

1 w2 1 w3 1 w4

sin2 ✓ =

)

2

2

2(1

sin ✓ d✓ d .

, and the numerator becomes

2

)w + w2 (1

)w + (1 +

2

⇤ )w2 ;

2

⇤ )

); when ✓ = ⇡, w = (1 + ).

⇤ )w + (1 + 2 )w2 dw. Z Z 1 1 2 2 ) dw + (1 + ) dw 4 w w3 ✓ ◆ ✓ ◆ 1 1 2 2 ) + (1 + ) 3w3 2w2 )

2 2

⇤

) sin2 ✓ cos2 cos ✓)5 2

2(1

2 2

2(1

(1

The integral is

2

1+

. 1

2 + 2 ) (1 + 2 + 2 ) 4 = . 2 2 2 )2 (1 + ) (1 ) (1 3 3 +3 2 ) (1 + 3 + 3 2 + 3 ) 2 (3 + 2 ) = . 2 )3 (1 + )3 (1 )3 (1 4 + 6 2 4 3 + 4 ) (1 + 4 + 6 2 + 4 3 + 4 ) 8 (1 + 2 ) = . 4 4 2 )4 (1 + ) (1 ) (1


240

✓ ◆ 8 (1 + 2 ) 1 2 (3 + 2 ) 2 2(1 ) + (1 + 2 4 2 )3 (1 ) 3 (1  3 2 2 8 = (1 + 2 ) (3 + 2 ) + (1 + 2 ) = . 2 2 2 )2 (1 ) 3 3 (1

Int = (1

P =

)

2 2

✓


1 4

µ0 q 2 a2 1 8 16⇡c 3 3 (1

◆

3 2 )2

=

µ0 q 2 a2 6⇡c

4

,

where

=p 1

1

2

2

)

✓

1 2

◆

4 (1

2 )2

.

Is this consistent with the Liénard formula (Eq. 11.73)? Here v ⇥ a = va(ˆ z⇥x ˆ) = va y ˆ, so ✓ ◆ ⇣v ⌘2 2 2 6 2 v 1 µ q a 0 2 2 a2 ⇥ a = a2 1 = (1 )a = 2 a2 , so the Liénard formula says P = .X 2 2 c c 6⇡c Problem 11.17

µ0 q 2 ˙ a. 6⇡c For circular motion, r(t) = R [cos(!t) x ˆ + sin(!t) y ˆ] , v(t) = r˙ = R! [ sin(!t) x ˆ + cos(!t) y ˆ] ; (a) To counteract the radiation reaction (Eq. 11.80), you must exert a force Fe =

a(t) = v˙ =

R! 2 [cos(!t) x ˆ + sin(!t) y ˆ] =

Pe = Fe · v =

µ0 q 2 2 2 ! v . 6⇡c

! 2 r; a˙ =

! 2 v. So Fe =

µ0 q 2 2 2 ! v , and the two expressions agree. 6⇡c

µ0 q 2 a2 , and a2 = ! 4 r2 = ! 4 R2 = ! 2 v 2 , so 6⇡c

(b) For simple harmonic motion, r(t) = A cos(!t) ˆ z; v = r˙ = ! 2 r; a˙ =

! 2 r˙ =

µ0 q 2 2 ! v. 6⇡c

This is the power you must supply.

Meanwhile, the power radiated is (Eq. 11.70) Prad = Prad =

! 2 r˙ =

! 2 v. So Fe =

A! sin(!t) ˆ z; a = v˙ =

A! 2 cos(!t) ˆ z=

µ0 q 2 2 µ0 q 2 2 2 ! v; Pe = ! v . But this time a2 = ! 4 r2 = ! 4 A2 cos2 (!t), 6⇡c 6⇡c

whereas ! 2 v 2 = ! 4 A2 sin2 (!t), so Prad =

µ0 q 2 4 2 µ0 q 2 4 2 2 ! A cos2 (!t) 6= Pe = ! A sin (!t); 6⇡c 6⇡c

the power you deliver is not equal to the power radiated. However, since the time averages of sin2 (!t) and cos2 (!t) are equal (to wit: 1/2), over a full cycle the energy radiated is the same as the energy input. (In the mean time energy is evidently being stored temporarily in the nearby fields.) (c) In free fall, v(t) = 12 gt2 y ˆ; v = gt y ˆ; a = g y ˆ; a˙ = 0. So Fe = 0; the radiation reaction is zero, and µ0 q 2 2 g . Evidently energy is being continuously extracted from 6⇡c the nearby fields. This paradox persists even in the exact solution (where we do not assume v ⌧ c, as in the Larmor formula and the Abraham-Lorentz formula)—see Prob. 11.34. hence Pe = 0. But there is radiation: Prad =

Problem 11.18 (a) From Eq. 11.80, Frad =

µ0 q 2 ... a˙ = m⌧ x (Eq. 11.82). The equation of motion is 6⇡c

F = m¨ x = Fspring + Frad =

... kx + m⌧ x, or x ¨ + !02 x

p ... ⌧ x = 0, with !0 = k/m.


241


... Since the damping is small, it oscillates at the natural frequency !0 , and hence x = 2 2 or x ¨ + x˙ + !0 x = 0, with = !0 ⌧. The solution with x(0) = 0 is (for ⌧ !0 ) x(t) = Ae

t/2

and x(t) =

v0 e !0

sin(!0 t); v(t) = t/2

2

Ae

t/2

sin(!0 t) + !0 Ae

t/2

cos(!0 t), so v(0) = A!0 = v0 ) A =

a=

!02 x =

!02

v0 e !0

Averaging over a full cycle (holding e Z

0

1

hP i dt =

v0 , !0

sin(!0 t). µ0 q 2 a2 . In this case (still assuming 6⇡c

According to the Larmor formula (Eq. 11.70), the power radiated is P = ⌧ !0 )

2 !02 x, ˙ so x ¨ +!02 ⌧ x+! ˙ 0 x = 0,

µ0 q 2 !02 v02 12⇡c



1

e

t

t

t/2

sin(!0 t),

constant) hP i = 1 0

=

P =

µ0 q 2 2 (!0 v0 ) e 6⇡c

µ0 q 2 !02 v02 e 6⇡c

t1

2

t

sin2 (!0 t).

, and the total energy radiated is

µ0 q 2 !02 v02 µ0 q 2 !02 v02 µ0 q 2 v02 = = 2 12⇡c 12⇡c!0 ⌧ 12⇡c

✓

6⇡mc µ0 q 2

◆

=

1 mv 2 . X 2 0

(b) This “equivalent” single oscillator has twice the charge, and twice the mass, so ⌧ (and hence ) is R doubled. Since hP i dt goes like q 2 / , it also doubles. The power radiated is indeed four times as great, but the oscillations die away faster, and the total energy radiated is just twice as much as for one oscillator. Problem 11.19 Z Z Z F dv da F dv da 1 (a) a = ⌧ a˙ + ) =⌧ + ) dt = ⌧ dt + F dt. m dt dt m dt dt m 2✏ [v(t0 + ✏) v(t0 ✏)] = ⌧ [a(t0 + ✏) a(t0 ✏)] + Fave , where Fave is the average force during the interm val. But v is continuous, so as long as F is not a delta function, we are left (in the limit ✏ ! 0) with [a(t0 + ✏) a(t0 ✏)] = 0. Thus a, too, is continuous. Z Z qed da da 1 da 1 t (b) (i) a = ⌧ a˙ = ⌧ ) = dt ) = dt ) ln a = + constant ) a(t) = Aet/⌧ , where A dt a ⌧ a ⌧ ⌧ is a constant. F da F da 1 t F (ii) a = ⌧ a˙ + )⌧ =a ) = dt ) ln(a F/m) = + constant ) a = Bet/⌧ ) m dt m a F/m ⌧ ⌧ m F a(t) = + Bet/⌧ , where B is some other constant. m (iii) Same as (i): a(t) = Cet/⌧ , where C is a third constant. (c) At t = 0, A = F/m + B; at t = T , F/m + BeT /⌧ = CeT /⌧ ) C = (F/m)e 8 [(F/m) + B] et/⌧ , t  0; > > > > > > h i < (F/m) + Bet/⌧ , 0  t  T; a(t) = > > > > h i > > : (F/m)e T /⌧ + B et/⌧ , t T.

To eliminate the runaway in region (iii), we’d need B = (F/m)e (i), we’d need B = (F/m). Obviously, we cannot do both at once.

T /⌧

T /⌧

+ B.

So

; to avoid preacceleration in region


242


(d) If we choose to eliminate the runaway, then

h

(i) v = (F/m) 1

e

T /⌧

iZ

8 h > (F/m) 1 > > > > > < h a(t) = (F/m) 1 > > > > > > : 0,

h et/⌧ dt = (F ⌧ /m) 1

e

T /⌧

e(t

i

et/⌧ , t  0;

T )/⌧

i

,

0  t  T; t

e

T /⌧

i

T.

et/⌧ + D, where D is a constant determined by the

condition v( 1) =h0 ) D = 0. i (ii) v = (F/m) t ⌧ e(t T )/⌧ + E, where E is a constant determined by the continuity of v at t = 0: h i h i (F ⌧ /m) 1 e T /⌧ = (F/m) ⌧ e T /⌧ + E ) E = (F ⌧ /m). (iii) v is a constant determined by the continuity of v at t = T : v = (F/m)[T + ⌧ ⌧ ] = (F/m)T.

(e)

8 h i T /⌧ > (F ⌧ /m) 1 e et/⌧ , t  0; > > > > > < h i v(t) = (F/m) t + ⌧ ⌧ e(t T )/⌧ , 0  t  T ; > > > > > > : (F/m)T, t T.

Problem 11.20

 ✓ ◆ µ0 (q/2)2 µ0 q 2 1 1 µ0 q 2 int end a, ˙ so Frad = Frad + 2Frad = a˙ +2 = a. ˙ X 6⇡c 6⇡c 2 4 6⇡c (b) Following the suggested method: ✓ ◆ 1 1 µ0 q 2 a˙ F (q) = F int (q) + 2 F (q) ) F (q) = F int (q) ) F (q) = 2F int (q) = . X 4 2 6⇡c Z L ⇢Z y1 µ0 (c) Frad = a˙ 2 dy2 2 dy1 . (Running the y2 12⇡c 0 0 integral up to y1 insures that y1 y2 , so we don’t count the same pair twice. Alternatively, run both integrals from 0 to L—intentionally double-counting—and divide the result by 2.) end (a) From Eq.11.80, Frad =


243


Frad

µ0 a˙ = (4 12⇡c

2

)

Z

L

y1 dy1 =

0

µ0 a˙ (4 12⇡c

2

)

L2 µ0 µ0 q 2 = ( L)2 a˙ = a. ˙ X 2 6⇡c 6⇡c

Problem 11.21 (a) The total torque is twice the torque on +q; we might as well calculate it at time t = 0. First we need the electric field at +q, due to q when it was at the retarded point P (Eq. 10.72). From the figure, rˆ = cos ✓ xˆ sin ✓ yˆ, r = 2R cos ✓. The velocity of q (at P ) was v = !R (sin 2✓ xˆ + cos 2✓ yˆ), and its acceleration was a = ! 2 R (cos 2✓ x ˆ sin 2✓ y ˆ) . Quantities we will need in Eq. 10.72 are:

r

u = c rˆ v = (c cos ✓ + !R sin 2✓) x ˆ (c sin ✓ !R cos 2✓) y ˆ, · u = 2R cos ✓(c + !R sin ✓), r · a = 2(!R cos ✓)2 .

y

P

r -q

E=

q R

q

x

n q 2R cos ✓ [c2 3 3 4⇡✏0 (2R cos ✓) (c + !R sin ✓)

(!R)2 + 2(!R cos ✓)2 ][(c cos ✓ + !R sin 2✓) x ˆ o 2R cos ✓(c + !R sin ✓)! 2 R (cos 2✓ x ˆ sin 2✓ y ˆ)

(c sin ✓

!R cos 2✓) y ˆ]

The total torque (about the origin) is N = 2(R x ˆ) ⇥ (qE) =

= =

n 2q 2 R ˆ z 2 3 4⇡✏0 (2R cos ✓) (c + !R sin ✓)

⇥ 2 ⇤ c (!R)2 + 2(!R cos ✓)2 (c sin ✓ o + 2(!R)2 cos ✓(c + !R sin ✓) sin 2✓

⇥ 3 q2 ˆ z c sin ✓ + c2 !R(2 cos2 ✓ 2 3 4⇡✏0 2R cos ✓(c + !R sin ✓) ⇥ q2 1 sin ✓ + (2 cos2 ✓ 1) + 2 3 4⇡✏0 2R cos ✓(1 + sin ✓)

!R cos 2✓)

1) + c(!R)2 (2 cos2 ✓ + 1) sin ✓ + (!R)3

2

(2 cos2 ✓ + 1) sin ✓ +

3

⇤

ˆ z,

⇤

where ⌘ !R/c. [Since E and r both lie in the xy plane, B = (1/c) rˆ ⇥ E is along the z direction, v ⇥ B is radial, and hence the magnetic contribution to the torque is zero.] The angle ✓ is determined by the retarded time condition, r = ctr (note that tr is negative, here), and 2✓ is the angle through which the dipole rotates in time tr , so 2R cos ✓ = ctr = c(2✓/!), or ✓ = cos ✓. We c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

244


can use this to eliminate the trig functions: N=

q2 4⇡✏0

⇣ p 2R✓2 1 +

2

✓2

Meanwhile, expanding in powers of ✓:

⌘3

h

(

4

+ 2✓2 ) +

2

p

✓2 (

2

2

i 1) ˆ z.

+ 2✓2

1 5 61 7 = ✓ sec ✓ = ✓ + ✓3 + ✓5 + ✓ + .... 2 24 720 This can be “solved” (for ✓ as a function of ) by reverting the series: 1 2

✓= Then ✓2 =

h

(

✓

4 3

4

1 1 = 2 1+ ✓2

2

2

4

1

2

2

+

+ 2✓ ) + 2

To leading order in , then,

p

N=

3

+

13 24

◆ p + ... , + ... , ⇣

✓2 =

2

1+

✓2 ( 2

2

541 720

5

1 p

+ 2✓

2

2

q2 1 8 4 ˆ z= 4⇡✏0 2R 2 3

(b) The radiation reaction force on +q is (Eq. 11.80) F = net torque (counting both ends) is N=

2R

µ0 q 2 3 ! Rˆ z= 6⇡c

7

2

+ ....

✓

2 3

1

⌘3 = 1

✓2 i 8 1) = 3

4

✓

2

+

3

2

1

4 5

4

◆ + ... ,

+ ...,

4 5

2

◆ + ... .

q2 4 3 ˆ z. 4⇡✏0 3R µ0 q 2 ˙ In this case a˙ = a. 6⇡c

!2 v =

!3 R y ˆ, so the

q2 4 3 ˆ z. 4⇡✏0 3R

Adding this to the interaction torque from (a), the total is N= ¨ = 2q¨r = 2q( ! 2 )r = (c) p

q2 8 3 ˆ z= 4⇡✏0 3R

µ0 p2 ! 3 ˆ z. 6⇡c

! 2 p, so Eq. 11.60 says the power radiated is P =

associated with the torque in (b) is N ! =

µ0 p2 ! 3 !, so they are in agreement. X 6⇡c

µ0 4 2 ! p . The power 6⇡c

Problem 11.22 p (a) This is an oscillating electric dipole, with amplitude p0 = qd and frequency ! = k/m. The (averaged) ✓ ◆ µ0 p20 ! 4 sin2 ✓ Poynting vector is given by Eq. 11.21: hSi = ˆ r, so the power per unit area of floor is 32⇡ 2 c r2 ✓ ◆ µ0 p20 ! 4 sin2 ✓ cos ✓ R h If = hSi · ˆ z= . But sin ✓ = , cos ✓ = , and r2 = R2 + h2 . 32⇡ 2 c r2 r r ✓ ◆ µ0 q 2 d2 ! 4 R2 h = . 2 2 32⇡ c (R + h2 )5/2 c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

245


 dIf d R2 2R 5 R2 =0) =0) 2R = 0 ) 2 2 5/2 2 2 5/2 2 dR dR (R + h ) 2 (R + h2 )7/2 (R + h ) p 5 2 3 (R2 + h2 ) R = 0 ) h2 = R2 ) R = 2/3h, for maximum intensity. 2 2 (b) ✓ ◆ Z 1 Z Z µ0 (qd)2 ! 4 R3 P = If (R) da = If (R) 2⇡R dR = 2⇡ h dR. 32⇡ 2 c (R2 + h2 )5/2 0 Z 1 Z R3 1 1 x 1 (2) (1/2) 2 dR = dx = = . 2 2 5/2 2 5/2 2 0 (x + h ) 2h (5/2) 3h (R + h ) 0 ✓ ◆ µ0 q 2 d2 ! 4 2 µ0 q 2 d2 ! 4 = 2⇡ h = , 32⇡ 2 c 3h 24⇡c

Let x ⌘ R2 :

which should be (and is) half the total radiated power (Eq. 11.22)—the rest hits the ceiling, of course. (c) The amplitude is x0 (t), so U = 12 kx20 is the energy, at time t, and dU/dt = 2P is the power radiated: 1 d 2 µ0 ! 4 2 2 d µ0 ! 4 q 2 2 k (x0 ) = q x0 ) (x20 ) = (x0 ) = x20 ) x20 = d2 e t or x0 (t) = de t/2 . 2 dt 12⇡c dt 6⇡kc 2 12⇡kc 2 12⇡cm2 ⌧= = m = .  µ0 q 2 k 2 µ0 q 2 k ✓ ◆ Problem 11.23 µ0 m20 ! 4 sin2 ✓ (a) From Eq. 11.39, hSi = ˆ r. Here sin ✓ = 32⇡ 2 c3 r2 p R/r, r = R2 + h2 , and the total radiated is ✓ power ◆ (Eq. 11.40) µ0 m20 ! 4 12P R2 P = . So the intensity is I(R) = = 12⇡c3 32⇡ (R2 + h2 )2

3P R2 . 8⇡ (R2 + h2 )2 (b) The intensity directly below the antenna (R = 0) would (ideally) have been zero. The engineer should have measured it at the position of maximum intensity:  dI 3P 2R 2R2 3P 2R = 2R = R2 + h2 2R2 = 0 ) R = h. dR 8⇡ (R2 + h2 )2 (R2 + h2 )3 8⇡ (R2 + h2 )3 At this location the intensity is I(h) = (c) Imax =

3P h2 3P = . 2 2 8⇡ (2h ) 32⇡h2

3(35 ⇥ 103 ) = 0.026 W/m2 = 2.6 µW/cm2 . 32⇡(200)2

Yes, KRUD is in compliance.


246


Problem 11.24

(a)

✓ ◆ p0 ! cos ✓± V± = ⌥ sin[!(t r± /c)]. Vtot = V+ + V . 4⇡✏0 c r± ✓ ◆ p p d 2 2 ⇠ ⇠ r± = r + (d/2) ⌥ 2r(d/2) cos ✓ = r 1 ⌥ (d/r) cos ✓ = r 1 ⌥ cos ✓ . 2r ✓ ◆ 1 ⇠ 1 d 1± cos ✓ . = r± r 2r ✓ ◆ ✓ ◆ r cos ✓ ⌥ (d/2) d 1 d d d cos ✓± = = r cos ✓ ⌥ 1± cos ✓ = cos ✓ ± cos2 ✓ ⌥ r± 2r r 2r 2r 2r d d = cos ✓ ⌥ (1 cos2 ✓) = cos ✓ ⌥ sin2 ✓. 2r 2r ⇢  ✓ ◆ ✓ ◆ r d !d sin[!(t r± /c)] = sin ! t 1⌥ cos ✓ = sin !t0 ± cos ✓ , where t0 ⌘ t r/c. c 2r 2c ✓ ◆ ✓ ◆ !d !d !d = sin(!t0 ) cos cos ✓ ± cos(!t0 ) sin cos ✓ ⇠ cos ✓ cos(!t0 ). = sin(!t0 ) ± 2c 2c 2c ⇢✓ ◆✓ ◆ p0 ! d d !d V± = ⌥ 1± cos ✓ cos ✓ ⌥ sin2 ✓ sin(!t0 ) ± cos ✓ cos(!t0 ) 4⇡✏0 cr 2r 2r 2c ⇢✓ ◆ p0 ! d d !d =⌥ cos ✓ ⌥ sin2 ✓ ± cos2 ✓ sin(!t0 ) ± cos ✓ cos(!t0 ) 4⇡✏0 cr 2r 2r 2c  p0 ! !d d =⌥ cos ✓ sin(!t0 ) ± cos2 ✓ cos(!t0 ) ± cos2 ✓ sin2 ✓ sin(!t0 . 4⇡✏0 cr 2c 2r  p0 ! !d d Vtot = cos2 ✓ cos(!t0 ) + cos2 ✓ sin2 ✓ sin(!t0 ) 4⇡✏0 cr c r i p0 ! 2 d h 2 c 2 2 = cos ✓ cos(!t ) + cos ✓ sin ✓ sin(!t 0 0 . 4⇡✏0 c2 r !r

In the radiation zone (r Meanwhile

!/c) the second term is negligible, so V =

p0 ! 2 d cos2 ✓ cos[!(t 4⇡✏0 c2 r

r/c)].

µ0 p0 ! 4⇡r± µ0 p0 ! =⌥ 4⇡r µ0 p0 ! =⌥ 4⇡r

A± = ⌥

Atot

sin[!(t r± /c)] ˆ z ⇢✓ ◆ d !d 1± cos ✓ sin(!t0 ) ± cos ✓ cos(!t0 ) ˆ z 2r 2c  !d d sin(!t0 ) ± cos ✓ cos(!t0 ) ± cos ✓ sin(!t0 ) ˆ z. 2c 2r  µ0 p0 ! !d d = A+ + A= cos ✓ cos(!t0 ) + cos ✓ sin(!t0 ) ˆ z 4⇡r c r h i µ0 p0 ! 2 d c = cos ✓ cos(!t0 ) + sin(!t0 ) ˆ z. 4⇡cr !r c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

247

CHAPTER 11. RADIATION µ0 p0 ! 2 d cos ✓ cos[!(t r/c)] ˆ z. 4⇡cr µ0 p0 ! 2 d (b) To simplify the notation, let ↵ ⌘ . Then 4⇡

In the radiation zone, A =

cos2 ✓ cos[!(t r/c)]; r ⇢ @V 1 @V ˆ 1 ! 2 rV = ˆ r+ ✓ = ↵ cos ✓ cos[!(t r/c)] + sin[!(t r/c)] ˆ r 2 @r r @✓ r rc 2 2 cos ✓ sin ✓ ˆ = ↵ ! cos ✓ sin[!(t r/c)] ˆ +↵ cos[!(t r/c)] ✓ r (in the radiation zone). r2 c r ⇣ ⌘ ⇣ ⌘ ↵ cos ✓ @A ↵! cos ✓ A = cos[!(t r/c)] cos ✓ ˆ r sin ✓ ✓ˆ . = sin[!(t r/c)] cos ✓ ˆ r sin ✓ ✓ˆ . c r @t c r V =↵

E=

rV

@A = @t

↵! sin[!(t cr

⇣ r/c)] cos2 ✓ ˆ r

⌘ cos2 ✓ ˆ r + sin ✓ cos ✓ ✓ˆ

↵! ˆ sin ✓ cos ✓ sin[!(t r/c)] ✓. cr  1 @ @Ar ˆ B = r⇥A= (rA✓ ) r @r @✓ ⇢  ↵ @ @ cos2 ✓ ˆ = (cos ✓ cos[!(t r/c)]( sin ✓)) cos[!(t r/c)] cr @r @✓ r ↵ ! ↵! = ( sin ✓ cos ✓) sin[!(t r/c)] ˆ (in the radiation zone) = sin ✓ cos ✓ sin[!(t cr c c2 r =

Notice that B =

r/c)] ˆ.

1 (ˆ r ⇥ E) and E · ˆ r = 0. c

⇤ 1 1 1 ⇥ 2 E2 (E ⇥ B) = E ⇥ (ˆ r ⇥ E) = E ˆ r (E · ˆ r)E = ˆ r µ0 µ0 c µ0 c µ0 c o2 ⌘2 1 n ↵! 1 ⇣ ↵! = sin ✓ cos ✓ sin[!(t r/c)] ˆ r. I= sin ✓ cos ✓ . µ0 c rc 2µ0 c rc Z Z Z ⇡ ⇣ ⌘ 1 ↵! 2 1 ⇣ ↵! ⌘2 P = hSi · da = sin2 ✓ cos2 ✓ sin ✓ d✓ d = 2⇡ (1 µ0 c c 2µ0 c c 0 cos3 ✓ ⇡ cos5 ✓ ⇡ 2 2 4 The integral is : + = = . 3 0 5 0 3 5 15 1 ! 2 µ20 4 µ0 = (p0 d)2 ! 4 2⇡ = (p0 d)2 ! 6 . 2µ0 c c2 16⇡ 2 15 60⇡c3 S=

cos2 ✓) cos2 ✓ sin ✓ d✓.

Notice that it goes like ! 6 , whereas dipole radiation goes like ! 4 . Problem 11.25 (a) m(t) = M cos ˆ z + M sin [cos(!t) x ˆ + sin(!t) y ˆ]. As in Prob. 11.4, the power radiated will be twice that of an oscillating magnetic dipole with dipole moment of amplitude m0 = M sin . Therefore (quoting Eq. 11.40): P =

µ0 M 2 ! 4 sin2 6⇡c3

. (Alternatively, you can get this from the answer to Prob. 11.11.)


248


(b) From Eq. 5.88, with r ! R, m ! M , and ✓ = ⇡/2: B = M=

4⇡R3 4⇡(6.4 ⇥ 106 )3 (5 ⇥ 10 B = µ0 4⇡ ⇥ 10 7

µ0 M , so 4⇡ R3 5

)

= 1.3 ⇥ 1023 A m2 .

✓ ◆4 )(1.3 ⇥ 1023 )2 sin2 (11 ) 2⇡ = 4 ⇥ 10 5 W (not much). 6⇡(3 ⇥ 108 )3 24 ⇥ 60 ⇥ 60 µ0 (4⇡R3 B/µ0 )2 ! 4 sin2 8⇡ 2 = ! 2 R3 B sin . Using the average value (1/2) for sin2 , (d) P = 3 3 6⇡c 3µ0 c "✓ #2 ◆2 8⇡ 2⇡ 1 4 3 8 P = (10 ) (10 ) = 2 ⇥ 1036 W (a lot). 3(4⇡ ⇥ 10 7 )(3 ⇥ 108 )3 10 3 2 (c) P =

(4⇡ ⇥ 10

7

Problem 11.26 (a) Write p(t) = q(t)d ˆ z, with q(t) = kt2 , where kd = (1/2)¨ p0 . As in Eq. 11.5,  1 k(t r + /c)2 k(t r /c)2 V (r, t) = r+ r 4⇡✏0 " # 2 2 t (2t/c) r + + r + /c2 t2 (2t/c) r + r 2+ /c2 k = r+ r 4⇡✏0  ✓ ◆ k 1 1 1 + 2 (r + r ) . = t2 r+ r 4⇡✏0 c From Eqs. 11.8 and 11.9,

r

±

✓ ◆ d =r 1⌥ cos ✓ , 2r

1

r

±

=

1 r

✓

1±

◆ d cos ✓ , 2r

so  2✓ ◆ k t d V (r, t) = cos ✓ 4⇡✏0 r r

r c2

✓

d cos ✓ r

◆

k = d cos ✓ 4⇡✏0 c2

"✓

ct r

◆2

#

µ0 p¨0 1 = cos ✓ 8⇡

"✓

ct r

◆2

#

1 .

As in Eq. 11.15, I(t) = (dq/dt) ˆ z = 2kt ˆ z, so (following Eqs. 11.16, and 11.17), µ0 A(r, t) = ˆ z 4⇡

Z

d/2 d/2

2k(t

r

r

/c)

✓

µ0 (t r/c) µ0 p¨0 dz = 2k dˆ z= 4⇡ r 4⇡c

( "✓ ◆  2 @A µ0 p¨0 (ct)2 1 ct E = rV = cos ✓ 2 3 ˆ r sin ✓ @t 8⇡ r r r " ✓ ◆2 ✓ ◆2 µ0 p¨0 ct 1 ct 1 = cos ✓ ˆ r + sin ✓ ✓ˆ sin ✓ ✓ˆ (cos ✓ ˆ r 4⇡r r 2 r 2 ("✓ ◆ # "✓ ◆ # ) 2 2 µ0 p¨0 ct 1 ct = 1 cos ✓ ˆ r+ + 1 sin ✓ ✓ˆ . 4⇡r r 2 r

# ) 1 ✓ˆ

#

ct r

◆

1 ˆ z.

µ0 p¨0 ⇣ c ⌘ ˆ z 4⇡c r

ˆ sin ✓ ✓)


249

CHAPTER 11. RADIATION ⇢✓ ◆ µ0 p¨0 ct ˆ r⇥A= r⇥ 1 (cos ✓ ˆ r sin ✓ ✓) 4⇡c r ⇢ ✓ ✓ ◆ ◆ ✓✓ ◆ ◆ µ0 p¨0 @ ct @ ct = r 1 sin ✓ 1 cos ✓ 4⇡cr @r r @✓ r ⇢ ✓ ◆ µ0 p¨0 ct µ0 p¨0 t = sin ✓ + 1 sin ✓ ˆ = sin ✓ ˆ. 4⇡cr r 4⇡r2

B =

ˆ

(b) The Poynting vector is 1 µ0 p¨20 t S= (E ⇥ B) = sin ✓ µ0 16⇡ 2 r3

("✓

µ0 p¨20 t S · da = 32⇡ 2 µ0 p¨20 t P (r, t) = 32⇡ 2 r

"✓

ct r

◆2

"✓

ct r

ct r

#

#

◆2

◆2

+ 1 2⇡

ˆ +1 1 cos ✓( ✓) 2 #

+1

Z

⇡

0

"✓

ct r

◆2

#

)

+ 1 sin ✓ ˆ r .

sin2 ✓ 2 (r sin ✓ d✓ d ). r3

µ0 p¨20 t sin ✓ d✓ = 12⇡r 3

"✓

ct r

◆2

#

+1 .

 ✓ ◆ ✓ ◆" ✓ ◆2 # µ0 p¨20 ⇣ r ⌘ c2 2 r r2 µ0 p¨20 ct0 ct0 ct0 (c) P (r, t0 + r/c) = t0 + t + 2t0 + 2 + 1 = 1+ 2+2 + . 12⇡r c r2 0 c c 12⇡c r r r Prad (t0 ) = lim P (r, t0 + r/c) = r!1

µ0 p¨20 , 6⇡c

in agreement with Eq. 11.60. Problem 11.27 The momentum flux density is (minus) the Maxwell stress tensor (Section 8.2.3), ✓ ◆ ✓ ◆ 1 1 1 2 2 Tij = ✏0 Ei Ej + Bi Bj , ij E ij B 2 µ0 2 (Eq. 8.17) so the momentum radiated per unit (retarded) time is I dp 1 $ = T · dA, dtr (@tr /@t) where (Eq. 10.78) @tr = @t

rc r ·u =

✓

1

rˆ

·v c

◆

1

.

(This factor is 1, when v = 0; it can be ignored in deriving the Larmor formula, but it does contribute to the momentum radiated.) The integration is over a large spherical surface centered on the charge: dA =

r

2

sin ✓ d✓ d

rˆ .

As in the case of the power radiated, only the radiation fields contribute (Eq. 11.66): Erad =

r [ r ⇥ (u ⇥ a)], q 4⇡✏0 ( r · u)3

Brad =

1 rˆ ⇥ Erad . c


250


Thus  $ T · dA = ✏0 E(E · dA)

 1 2 1 E dA + B(B · dA) 2 µ0

1 2 B dA = 2

 1 1 2 ✏0 E 2 + B dA = 2 µ0

✏0 E 2 dA

(note that B · dA = 0, and, for radiation fields, E · dA = 0, (1/µ0 )B 2 = (1/µ0 c2 )( rˆ ⇥ E) · ( rˆ ⇥ E) = ✏0 rˆ · [E ⇥ ( rˆ ⇥ E)] = ✏0 rˆ · [ rˆ E 2 E( rˆ · E)] = ✏0 E 2 ). So ◆ I ✓ I r ·u 2 r [ r ⇥ (u ⇥ a)]2 dA. dp q2 = ✏0 E dA = 2 rc dtr 16⇡ ✏0 c ( r · u)5 We expand the integrand to first order in

r

⌘ v/c: u = c rˆ

1 1 ˆ = r 5 c5 [1 + 5( r · )], ( r · u)5 ⇥ (u ⇥ a) = u( r · a) a( r · u) = r c rˆ ⇥ = r c rˆ ( rˆ · a) a + a( rˆ · )

[ r ⇥ (u ⇥ a)]2 =

r

( r · u)5

r

h c ( rˆ · a)2

= =

r r r

( rˆ · a) ⇤ ( rˆ · a) .

a r c(1

,

rˆ

r

· u = c r (1

rˆ

· ),

· )

2( rˆ · a)2 + a2 + 2( rˆ · a)2 ( rˆ · ) 2( rˆ · )( rˆ · a)2 2a2 ( rˆ · ) i ⇥ ⇤ + 2(a · )( rˆ · a) = r 2 c2 a2 ( rˆ · a)2 2a2 ( rˆ · ) + 2( rˆ · a)(a · ) . 2 2

[ r ⇥ (u ⇥ a)]2 = =

v = c rˆ

1 2 c3

1

2 c3

1

2 c3

⇥ 2 a ⇥ 2 a

r

1

2 c3

( rˆ · a)2

⇥ [1 + 5( rˆ · )] a2

( rˆ · a)2

⇤ 2a2 ( rˆ · ) + 2( rˆ · a)(a · )

2a2 ( rˆ · ) + 2( rˆ · a)(a · ) + 5( rˆ · )a2

( rˆ · a)2 + 3a2 ( rˆ · ) + 2( rˆ · a)(a · )

⇥ ( rˆ ⇥ a)2 + rˆ · [3a2 + 2(a · )a]

5( rˆ · )( rˆ · a)2

5( rˆ · )( rˆ · a)2

⇤ 5( rˆ · )( rˆ · a)2 .

⇤

⇤

ˆ+ To integrate the first term we set the polar axis along a, so ( rˆ ⇥ a)2 = a2 sin2 ✓, while rˆ = sin ✓ cos x ˆ + cos ✓ z ˆ. The integral kills the x ˆ and y ˆ components, leaving sin ✓ sin y I Z ⇡ 1 2 2 ˆ ˆ ( r ⇥ a) dA = 2⇡a z sin3 ✓ cos ✓ d✓ = 0. 2

r

0

[Note that if v = 0 then dp/dtr = 0—a particle instantaneously at rest radiates no momentum. That’s why we had to carry the expansion to first order in , whereas in deriving the Larmor formula we could a↵ord to set v = 0.] The second term is of the form ( rˆ · g), for a constant vector g. Setting the polar axis along g, so rˆ · g = g cos ✓, the x and y components again vanish, leaving I Z ⇡ 1 4⇡ ˆ ( rˆ · g) dA = 2⇡g z cos2 ✓ sin ✓ d✓ = g. 2 r 3 0 The last term involves ( rˆ · )( rˆ · a)2 ; this time we orient the polar axis along a and let v lie in the xz plane: ˆ+ zz ˆ, so ( rˆ · ) = x sin ✓ cos + z cos ✓. Then = xx I Z ⇥ ⇤ 1 ˆ + sin ✓ sin y ˆ + cos ✓ z ˆ sin ✓ d✓ d ( rˆ · )( rˆ · a)2 dA = a2 ( x sin ✓ cos + z cos ✓) cos2 ✓ sin ✓ cos x

r

2


251

CHAPTER 11. RADIATION = a2

✓

ˆ xx

Z

sin3 ✓ cos2 ✓ cos2 d✓ d +

ˆ zz

Putting all this together, dp q2 1 = dtr 16⇡ 2 ✏0 c4

⇢

Z

cos4 ✓ sin ✓ d✓ d

⇤ 4⇡ ⇥ 2a(a · ) + 3a2 3

The angular momentum radiated is

I

dL = dtr

5

◆

=

4⇡ 2 a ( 15

x

ˆ+3 x

⇤ 4⇡ ⇥ 2 a + 2(a · )a 15

=

z

ˆ) = z

4⇡ 2 a + 2(a · )a . 15

µ0 q 2 2 a v. 6⇡c3

⇣ ⌘ 1 r ⇥$ T · dA. (@tr /@t)

Because of the “extra” r in the integrand, it seems at first glance that the radiation fields alone will ⇣ produce⌘a $ result that grows without limit (as r ! 1); however, the coefficient of this term is precisely zero: r ⇥ T · ✏0 E 2 ( r ⇥ dA) = 0 (for radiation fields). The finite contribution comes from the cross terms, ◆ in which ✓ r q $ one field (in T ) is a radiation field and the other a Coulomb field Ecoul ⌘ (c2 v 2 )u : 4⇡✏0 ( r · u)3 dA =

⇣ (r) (c) (c) (r) Tij = ✏0 Ei Ej + Ei Ej

This time

(r) ij E

h $ T · dA = ✏0 E(r) (E(c) · dA)

so

⌘ 1 ⇣ (r) (c) (c) (r) · E(c) + Bi Bj + Bi Bj µ0 ⇣

dL = dtr

q2 c 16⇡ 2 ✏0

I

r

n

r

(r)

⌘ · B(c) .

⌘ i E(r) · E(c) dA

$ ( r ⇥ T ) · dA = ✏0

Thus

ij B

⇣

r

⌘ 1 ⇣ (r) B · B(c) dA, µ0 ⌘⇣ ⌘ ⇥ E(r) E(c) · dA .

o ⇥ [ r ⇥ (u ⇥ a)] u · dA, where

1

⌘p

. (v/c)2 ⇥ ⇤ Now, { r ⇥ [ r ⇥ (u ⇥ a)]} = r [ r ·(u⇥a)] (u⇥a) r 2 , and (u⇥a) = c( rˆ )⇥a = c ( rˆ ⇥ a) ( ⇥ a) , ⇥ ⇤ rˆ · (u ⇥ a) = c rˆ · ( ⇥ a), so { r ⇥ [ r ⇥ (u ⇥ a)]} = r 2 c ( rˆ ⇥ a) ( ⇥ a) + rˆ rˆ · ( ⇥ a) . Meanwhile u · dA = (u · rˆ ) r 2 d⌦, where d⌦ ⌘ sin ✓ d✓ d . Expanding to first order in , I r ( r 2 c) ( rˆ ⇥ a) ( ⇥ a) + rˆ ⇥ rˆ · ( ⇥ a)⇤ r ( r · u) d⌦ dL µ0 q 2 c3 = 2 dtr 16⇡ ( r · u)5 Z ⇥ ⇤ ⇥ ⇤ µ0 q 2 = 1 + 4( rˆ · ) ( rˆ ⇥ a) ( ⇥ a) + rˆ rˆ · ( ⇥ a) d⌦ 2 16⇡ Z n ⇤o µ0 q 2 ˆ ⇥ a) ( ⇥ a) + rˆ [ rˆ · ( ⇥ a)] + 4( rˆ · )( rˆ ⇥ a) d⌦ r = ( 16⇡ 2 2

( r · u)5

The first integral is Z rˆ sin ✓ d✓ d = a⇥

a⇥

Z

the second is ( ⇥ a)

1

ˆ + sin ✓ sin y ˆ + cos ✓ z ˆ) sin ✓ d✓ d = 0, (sin ✓ cos x Z

sin ✓ d✓ d =

4⇡( ⇥ a),


252 for the third we set the polar axis along g ⌘ g

Z

rˆ

CHAPTER 11. RADIATION ⇥ a and get

ˆ cos ✓ sin ✓ d✓ d = 2⇡g z

Z

cos2 ✓ sin ✓ d✓ =

and for the last we put the polar axis along a and let 4a

Z

= 4a =

( Z

x

sin ✓ cos +

(

z

=

ˆ xx

+

ˆ zz

4⇡ 4⇡ g= ( ⇥ a), 3 3

lie in the xz plane:

cos ✓) sin ✓( ˆ) sin ✓ d✓ d

sin ✓ cos + z cos ✓)(sin x ˆ cos y ˆ) sin2 ✓ d✓ d Z 16⇡ 16⇡ ˆ cos2 sin3 ✓ d✓ d = ˆ= 4a x y a xy ( ⇥ a). 3 3 x

Putting this all together, we conclude dL µ0 q 2 = dtr 16⇡ 2



4⇡( ⇥ a) +

4⇡ 16⇡ µ0 q 2 ( ⇥ a) + ( ⇥ a) = (v ⇥ a). 3 3 6⇡c

Problem 11.28

Z µ0 K(tr ) (a) A(x, t) = r da 4⇡ Z µ0ˆ z K(tr ) p = 2⇡r dr 4⇡ r2 + x2 p Z µ0ˆ z K(t r2 + x2 /c) p = r dr. 2 r2 + x2 p The maximum r is given by t r2 + x2 /c = 0; p rmax = c2 t2 x2 (since K(t) = 0 for t < 0). (i)

µ0 K0ˆ z A(x, t) = 2 E(x, t) =

Z

0

@A = @t

B(x, t) = r ⇥ A =

rm

p

r µ0 K0ˆ zp 2 r + x2 dr = 2 r2 + x2

rm 0

=

µ0 K0 c ˆ z, for ct > x, and 0, for ct < x. 2

µ0 K0ˆ z ⇣p 2 rm 2

x2

⌘ µ K (ct 0 0 x = 2

x)

@Az µ0 K0 y ˆ= y ˆ, for ct > x, and 0, for ct < x. @x 2 c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

ˆ z.

253

CHAPTER 11. RADIATION (ii)

p  Z rm Z Z r2 + x2 /c µ0 ↵ ˆ z rm t µ0 ↵ˆ z r 1 rm p p A(x, t) = r dr = t dr r dr 2 2 c 0 r2 + x2 r2 + x2 0 0  µ0 ↵ ˆ z 1 2 2 µ0 ↵ ˆ z 2 µ0 ↵(x ct)2 = t(ct x) (c t x2 ) = (x 2ctx + c2 t2 ) = ˆ z. 2 2c 4c 4c @A µ0 ↵(x = @t 2

E(x, t) =

B(x, t) = r ⇥ A =

t

ct)

@Az y ˆ= @x

ˆ z, for ct > x, and 0, for ct < x. µ0 ↵ (x 2c

ct) y ˆ, for ct > x, and 0, for ct < x.

 ⌘ 1 ⇣p 2 1 1 1 1 r 2 p (b) Let u ⌘ r +x x , so du = 2r dr = p dr, and 2 2 2 c c 2 r +x c r + x2 p Z r2 + x2 x µ0 c ˆ z 1 ⇣ =t u, and as r : 0 ! 1, u : 0 ! 1. Then A(x, t) = K t c c 2 0 Z @A µ0 c ˆ z 1 E(x, t) = = @t 2 0 Z 1 ⇣ µ0 c @ = ˆ z K t 2 @u 0 µ0 c = K(t x/c) ˆ z, 2

⌘ @ ⇣ x K t u du. But @t c ⌘ x µ0 c h ⇣ u du = ˆ z K t c 2

[if K( 1) = 0].

x c

⌘ u du. qed

⌘ ⇣ ⌘ @ ⇣ x @ x K t u = K t u . @t c @u c ⌘i 1 x µ0 c u = [K(t x/c) K( 1)] ˆ z c 2 0

Note that (i) and (ii) are consistent with this result. Meanwhile

B(x, t) = = = S=

Z 1 ⇣ ⌘ ⇣ ⌘ 1 @ ⇣ ⌘ @Az µ0 c @ x @ x x y ˆ= y ˆ K t u du. But K t u = K t u . @x c @x c @x c c @u c 0 Z 1 ⇣ ⌘ h ⇣ ⌘i 1 µ0 @ x µ0 x µ0 y ˆ K t u du = y ˆ K t u = [K(t x/c) K( 1)] y ˆ 2 @u c 2 c 2 0 0 µ0 K(t x/c) y ˆ, [if K( 1) = 0]. 2 1 1 ⇣ µ0 c ⌘ ⇣ µ0 ⌘ µ0 c 2 (E ⇥ B) = K(t x/c) [ ˆ z⇥y ˆ] = [K(t x/c)] x ˆ. µ0 µ0 2 2 4

This is the power per unit area that reaches x at time t; it left the surface at time (t x/c). Moreover, an µ0 c 2 equal amount of energy is radiated downward, so the total power leaving the surface at time t is [K(t)] . 2 Problem 11.29 With ↵ = 90 , Eq. 7.68 gives E0 = cB, B0 =

1 0 E, qm = c

cqe . Use this to “translate” Eqs. 10.72, 10.73,


254


and 11.70:

E = 0

B0 = = P =

✓

◆ 1 ˆ r ⇥ E = rˆ ⇥ ( cB0 ) = c( rˆ ⇥ B0 ). c c r ⇥(c2 v2 )u + r ⇥ (u ⇥ a)⇤ 1 1 qe E= c c 4⇡✏0 ( r · u)3 0 r ⇥(c2 v2 )u + r ⇥ (u ⇥ a)⇤ = µ0 qm0 r ⇥(c2 1 ( qm /c) 3 c 4⇡✏0 ( r · u) 4⇡ ( r · u)3 ✓ ◆ 2 µ0 a2 2 µ0 a2 1 0 µ0 a2 0 2 qe = qm = (q ) . 6⇡c 6⇡c c 6⇡c3 m

v 2 )u +

r

⇤ ⇥ (u ⇥ a) .

Or, dropping the primes,

B(r, t) = E(r, t) = =

P

Problem 11.30 Z Z (a) Wext = F dx = F

T

r ⇥(c2 µ0 qm 4⇡ ( r · u)3

v 2 )u +

c( rˆ ⇥ B).

Wext

=

2

F m

"Z

T

v(t) dt. From Prob. 11.19, v(t) =

t dt + ⌧

0



1 2 T + ⌧T 2

Z

T

dt

⌧e

T /⌧

0

⌧ 2e

⇤ ⇥ (u ⇥ a) .

2 2 µ0 qm a . 6⇡c3

0

F2 = m

r

Z

T t/⌧

e

0

T /⌧

⇣

eT /⌧

#

dt =

F h t+⌧ m

 F 2 t2 + ⌧t m 2

✓ ⌘ F2 1 2 1 = T + ⌧T m 2

(b) From Prob. 11.19, the final velocity is vf = (F/m)T , so Wkin = (c) Wrad =

R

P dt. According to the Larmor formula, P =

a(t) =

8 ⇥ < (F/m) 1 :

⇥ (F/m) 1

e e(t

T /⌧

⇤

⇤

T )/⌧

i

. So

T

⌧e

T /⌧

⌧ et/⌧ 0

⌧ 2 + ⌧ 2e

T /⌧

◆

.

1 1 F2 F 2T 2 mvf2 = m 2 T 2 = . 2 2 m 2m

µ0 q 2 a2 , and (again from Prob. 11.19) 6⇡c

et/⌧ ,

T )/⌧

⌧ e(t

,

(t  0); (0  t  T ).


255


Wrad = = = = =

( ) Z Th ⌘2 Z 0 i2 µ0 q 2 F 2 ⇣ T /⌧ 2t/⌧ (t T )/⌧ 1 e e dt + 1 e dt 6⇡c m2 1 0 ( ) Z T Z T Z T ⌘2 ⇣ ⌧ ⌘0 F2 ⇣ 1 e T /⌧ e2t/⌧ + dt 2e T /⌧ et/⌧ dt + e 2T /⌧ e2t/⌧ dt ⌧ m 2 0 0 0 1  ⌘2 ⇣ ⌘T ⇣⌧ ⌘T ⌧F2 ⌧ ⇣ 1 e T /⌧ + T 2e T /⌧ ⌧ et/⌧ + e 2T /⌧ e2t/⌧ m 2 2 0 0 ⌘ ⇣ ⌘ ⇣ ⌘i 2 h ⇣ ⌧F ⌧ ⌧ 1 2e T /⌧ + e 2T /⌧ + T 2⌧ e T /⌧ eT /⌧ 1 + e 2T /⌧ e2T /⌧ 1 m 2 2 2 h ⌧F ⌧ ⌧ ⌧ ⌧ 2T /⌧ i ⌧F2 ⇣ ⌧ e T /⌧ + e 2T /⌧ + T 2⌧ + 2⌧ e T /⌧ + e = T ⌧ + ⌧e m 2 2 2 2 m

T /⌧

⌘

.

Energy conservation requires that the work done by the external force equal the final kinetic energy plus the energy radiated:

Wkin + Wrad

F 2T 2 ⌧F2 ⇣ = + T 2m m

Problem 11.31 k (a) a = ⌧ a˙ + (t) ) m

Z

✏

⌧ + ⌧e

a(t) dt = v(✏)

T /⌧

⌘

F2 = m

v( ✏) = ⌧

✏

If the velocity is continuous, so v(✏) = v( ✏), then a(✏)

Z

✏ ✏

✓

1 2 T + ⌧T 2

da k dt + dt m

a( ✏) =

Z

◆

= Wext . X

(t) dt = ⌧ [a(✏)

a( ✏)] +

⌧ +⌧ e 2

✏

k , so the general solution is a(t) = m⌧

⇢

T /⌧

✏

k . m

k . m⌧

When t < 0, a = ⌧ a˙ ) a(t) = Aet/⌧ ; when t > 0, a = ⌧ a˙ ) a(t) = Bet/⌧ ; )B=A

2

a=B

A=

k m⌧

Aet/⌧ , (t < 0); [A (k/m⌧ )] et/⌧ , (t > 0).

To eliminate the runaway we’d need A = k/m⌧ ; to eliminate preacceleration we’d need A = 0. Obviously, ⇢ (k/m⌧ )et/⌧ , (t < 0); you can’t do both. If you choose to eliminate the runaway, then a(t) = 0, (t > 0). Z t k k ⇣ t/⌧ ⌘ t k et/⌧ dt = ⌧e = et/⌧ (for t < 0); m⌧ m⌧ m 1 1 1 ⇢ Z t k (k/m)et/⌧ , (t < 0); for t > 0, v(t) = v(0) + a(t) dt = v(0) = . So v(t) = (k/m), (t > 0). m 0 ⇢ Z t k 0, (t < 0); For an uncharged particle we would have a(t) = (t), v(t) = a(t) dt = (k/m), (t > 0). m 1 v(t) =

Z

t

a(t) dt =

The graphs:


256


(b) Wext =

Z

F dx =

Z

F v dt = k

Z

k2 . m

(t)v(t) dt = kv(0) =

✓ ◆2 1 1 k k2 mvf2 = m = . 2 2 m 2m ✓ ◆2 Z Z Z µ0 q 2 k 2 = Prad dt = [a(t)] dt = ⌧ m 6⇡c m⌧

Wkin = Wext

Clearly, Wext = Wkin + Wrad . X

0

e2t/⌧ dt =

1

Problem 11.32 U0 Our task is to solve the equation a = ⌧ a˙ + [ (x) + (x m (1) x continuous at x = 0 and x = L; (2) v continuous at x = 0 and x = L; (3) a = ±U0 /m⌧ v (plus at x = 0, minus at x = L).

k 2 ⇣ ⌧ 2t/⌧ ⌘ e m⌧ 2

0

= 1

k2 ⌧ k2 = . m⌧ 2 2m

L)], subject to the boundary conditions

The third of these follows from integrating the equation of motion: Z Z Z dv da U0 dt = ⌧ dt + [ (x) + (x L)] dt, dt dt m Z U0 dt v = ⌧ a+ [ (x) + (x L)] dx = 0, m dx Z U0 1 U0 a== [ (x) + (x L)] dx = ± . m⌧ v m⌧ v

In each of the three regions the force is zero (it acts only at x = 0 and x = L), and the general solution is a(t) = Aet/⌧ ;

v(t) = A⌧ et/⌧ + B;

x(t) = A⌧ 2 et/⌧ + Bt + C.

(I’ll put subscripts on the constants A, B, and C, to distinguish the three regions.) Region iii (x > L): To avoid the runaway we pick A3 = 0; then a(t) = 0, v(t) = B3 , x(t) = B3 t + C3 . Let the final velocity be vf (= B3 ), set the clock so that t = 0 when the particle is at x = 0, and let T be the time it takes to traverse the barrier, so x(T ) = L = vf T + C3 , and hence C3 = L vf T . Then a(t) = 0;

v(t) = vf ,

x(t) = L + vf (t

T ),

(t < T ).

Region ii (0 < x < L): a = A2 et/⌧ , v = A2 ⌧ et/⌧ + B2 , x = A2 ⌧ 2 et/⌧ + B2 t + C2 . c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

257


U0 U0 ) A2 = e T /⌧ . m⌧ vf m⌧ vf U0 U0 (2) ) vf = A2 ⌧ eT /⌧ + B2 = + B2 ) B2 = vf . mvf mvf U0 ⌧ U0 T U0 (1) ) L = A2 ⌧ 2 eT /⌧ + B2 T + C2 = + vf T + C2 = vf T + (⌧ mvf mvf mvf U0 C2 = L vf T + (T ⌧ ). mvf (3) ) 0

A2 eT /⌧ =

U0 (t T )/⌧ e ; m⌧ vf h i U0 v(t) = vf + e(t T )/⌧ 1 ; mvf U0 h (t x(t) = L + vf (t T ) + ⌧e mvf

T ) + C2 )

a(t) =

(0 < t < T ). T )/⌧

t+T

i

⌧ ;

[Note: if the barrier is sufficiently wide (or high) the particle may turn around before reaching L, but we’re interested here in the régime where it does tunnel through.] In particular, for t = 0 (when x = 0): i i U0 h U0 h 0 = L vf T + ⌧ e T /⌧ + T ⌧ ) L = vf T ⌧ e T /⌧ + T ⌧ . qed mvf mvf Region i (x < 0): a = A1 et/⌧ , v = A1 ⌧ et/⌧ + B1 , x = A1 ⌧ 2 et/⌧ + B1 t + C1 . Let vi be the incident velocity (at t ! 1); then B1 = vi . Condition (3) says U0 e m⌧ vf

A1 =

T /⌧

U0 , m⌧ v0

where v0 is the speed of the particle as it passes x = 0. From the solution in region (ii) it follows that ⌘ U0 ⇣ T /⌧ v0 = vf + e 1 . But we can also express it in terms of the solution in region (i): v0 = A1 ⌧ + vi . mvf Therefore ⌘ ⌘ U0 ⇣ T /⌧ U0 ⇣ T /⌧ U0 U0 vi = vf + e 1 A1 ⌧ = vf + e 1 + e T /⌧ mvf mvf mv0 mvf ( ) ✓ ◆ U0 U0 U0 vf U0 vf ⇥ ⇤ = vf + = vf 1 = vf 1 mvf mv0 mvf v0 mvf vf + (U0 /mvf ) e T /⌧ 1 ( ) U0 1 ⇥ ⇤ . qed = vf 1 mvf 1 + (U0 /mvf2 ) e T /⌧ 1 If 12 mvf2 = 12 U0 , then L = vf T

h vf ⌧ e

vi = vf

T /⌧

+T

 vf 1

i h ⌧ = vf T 1

1+e

T /⌧

1

⌧e

T /⌧

⇣ = vf 1

i ⇣ T + ⌧ = ⌧ vf 1 1 + eT /⌧


⌘

e

= vf eT /⌧ .

T /⌧

⌘

;

258


Putting these together, L =1 ⌧ vf

e

T /⌧

)e

T /⌧

=1

L ) eT /⌧ = ⌧ vf 1

1 ) vi = (L/⌧ vf ) 1

vf . qed (L/vf ⌧ )

✓ ◆2 1 2 vf 4 KEi vi 16 2 mvi In particular, for L = vf ⌧ /4, vi = = vf , so = 1 = = ) 2 1 1/4 3 KEf v 9 f 2 mvf 16 16 1 8 KEi = KEf = U0 = U0 . 9 9 2 9 Problem 11.33 r ⇥(c2 v2 )u + ( r · a)u ( r · u)a⇤. Here u = c rˆ v, r = (q/2) (a) From Eq. 10.72, E1 = 4⇡✏0 ( r · u)3 r · v = c r lv. We want only the x lx ˆ + dy ˆ, v = v x ˆ, a = a x ˆ, so r · v = lv, r · a = la, r · u = c r component. Noting that ux = (c/ r )l v = (cl v r )/ r , we have:  r q 1 2 2 3 r (cl v r )(c v + la) a(c r lv) 8⇡✏0 (c r lv) ⇥ ⇤ q 1 = (cl v r )(c2 v 2 ) + cl2 a v r la ac r 2 + alv r . But r 8⇡✏0 (c r lv)3 ⇥ ⇤ q 1 = (cl v r )(c2 v 2 ) acd2 . 3 8⇡✏0 (c r lv) 2 ⇥ ⇤ q 1 = (cl v r )(c2 v 2 ) acd2 x ˆ. (This generalizes Eq. 11.90.) 3 8⇡✏0 (c r lv)

E1x =

Fself

Now x(t) x(tr ) = l = vT + 12 aT 2 + 16 aT ˙ 3 + · · · , where T = t retarded time tr . (cT )2 =

c2 T 2 (1

r

2

2

= l 2 + d2 .

tr , and v, a, and a˙ are all evaluated at the

1 1 3 2 1 1 = l2 + d2 = d2 + (vT + aT 2 + aT ˙ ) = d2 + v 2 T 2 + vaT 3 + v aT ˙ 4 + a2 T 4 ; 2 6 3 4

v 2 /c2 ) = c2 T 2 /

2

= d2 + vaT 3 +

✓

◆ 1 1 v a˙ + a2 T 4 . Solve for T as a power series in d: 3 4

✓ ◆ 4 3 3 d c2 2 d2 d v a˙ a2 2 2 2 2 2 T = 1 + Ad + Bd + · · · ) 2 2 1 + 2Ad + 2Bd + A d = d +va 3 (1+3Ad)+ + d4 . c c c 3 4 c4 Comparing like powers of d: A =

3 1 3va va 3 ; 2B + A2 = 2 c c3

3

A+

✓

v a˙ a2 + 3 4

◆

4

c4

.

✓ ◆ 3 6 2 3va 3 1 1 2 2 6 v a˙ 4 a2 4 v a˙ 4 a 1 v2 3 v 2 a2 6 2B = va v a + + = + + 2 c3 2 c3 4 c6 3 c4 4c4 3 c4 4c4 c2 2 c6  ✓ ◆  ✓ ◆ 4 2 2 2 2 2 4 2 2 2 v a˙ a v v v v a˙ a v = 4 + 1 +6 2 )B= 4 + 1+4 2 . 2 2 c 3 4 c c c 2c 3 4 c ⇢  ✓ ◆ 4 2 2 d va 3 v a˙ a v2 T = 1+ d + + 1 + 4 d2 + ( ) d4 + · · · (generalizing Eq. 11.93). c 2 c3 2c4 3 4 c2 c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


259

1 1 3 l = vT + aT 2 + aT ˙ + ··· 6 ⇢2  ✓ ◆  4 2 2 3 v d va 3 v a˙ a v2 1 2 d2 1 3 2 = 1+ d + + 1 + 4 d + a 1 + va 3 d + a˙ 3 d3 3 4 2 2 c 2 c 2c 3 4 c 2 c c 6 c ✓ ◆ ⇢  ✓ ◆ ⇣v ⌘ 4 2 2 4 2 2 2 2 3 a v v v v a˙ a v 1 1 3 = d+ 1 + 2 d2 + + 1+4 2 + a 2 va 3 + a˙ 3 2 2 4 c 2c c c 2c c 3 4 c 2 c c 6 c ✓ 4◆  ✓ ◆ ✓ ◆ ⇣v ⌘ 3 2 4 2 2 2 a a˙ v v a 1 v v = d+ d2 + 3 1+ 2 2 + + 2 +1 d3 c 2c2 2c 3 c c2 4 c c2 ✓ 4◆  ⇣v ⌘ 5 a˙ 5 v 2 a2 3 a 2 = d+ d + + d + ( ) d4 + · · · c 2c2 2c3 3 4 c2

d3

⇢  ✓ ◆ 4 va 3 v a˙ 1 v2 2 2 = cT = d 1 + d + + a + d2 + ( ) d4 + · · · 2 c3 2c4 3 4 c2  ✓ ◆  5 5 va 4 2 v a˙ 1 v2 v2 av 4 2 v a˙ 5 v 2 a2 3 2 2 3 lv = c d + d + + a + d d d + d + ··· 2 3 2 2 3 2c 2c 3 4 c c 2c 2c 3 4 c2 ✓ ◆  ✓ ◆ 5 v2 v a˙ 1 v2 v a˙ 5 v 2 2 a2 3 2 2 =c d 1 + + a + d + ··· 2 3 2 c 2c 3 4 c 3 4 c2 5 2 c a 3 = d+ d + ( ) d4 + · · · 8c3

r cr

cl

vr = = = =

(c r

lv)

3

=

✓ ◆  ✓ ◆ 5 a 4 2 a˙ 5 v 2 a2 v 2 a 4 2 v 5 v a˙ 1 v2 3 2 2 v d+ d + 2 + d v d d + a + 2 d3 2c 2c 3 4 c2 2 c3 2c4 3 4 c ✓ ◆  ✓ ◆ 5 a 4 v2 a˙ 5 v 2 a2 v 2 a˙ v 2 a2 1 v 2 2 1 d + 2 + + 2 d3 + ( ) d4 + · · · 2c c2 2c 3 4 c2 c2 3 c2 4 c ✓ 2◆  ✓ ◆ 5 a a˙ v 2 a2 5 1 v 2 d2 + 2 + d3 + ( ) d4 + · · · 2c 2c 3 2 c2 4 4 c2 ✓ 2◆ ✓ ◆ 3 a a˙ v 2 a2 d2 + 2 + d3 + ( ) d4 + · · · 2c 2c 3 c2  ✓ ◆ 3 ⇣ ⌘ ✓ ◆ 6 2 6 2 3 cd a 2 a 2 1+ d = 1 3 4 d + ··· 8c4 cd 8c

✓ ◆ ⇢✓ 2 ◆ ✓ ◆ 6 2 3 q 2 ⇣ ⌘3 a a a˙ v 2 a2 c2 1 3 4 d2 d2 + 2 + d3 2 acd2 x ˆ 2 8⇡✏0 cd 8c 2c 2c 3 c ✓ ◆ ✓ ◆ 3 q2 3 6 a2 2 ac a˙ v 2 a2 = 1 d + + d x ˆ 8⇡✏0 c3 d 8 c4 2 2 3 c2  ✓ ◆ 3 q2 1 a˙ v 2 a2 = ac + + d + ( ) d2 + · · · x ˆ 8⇡✏0 c3 d 2 3 c2  ✓ ◆ 4 q2 a˙ v 2 a2 3 a = + + + ( )d + ··· x ˆ (generalizing Eq. 11.95). 4⇡✏0 4c2 d 4c3 3 c2

Fself =

Switching to t: v(tr ) = v(t) + v(t)(t ˙ t) + · · · = v(t) a(t)T = v(t) r doesn’t matter—to this order—whether we evaluate at t or at tr .) c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

a d/c. (When multiplied by d, it

260




 ✓ ◆ 2 v(tr ) [v(t)2 2va d/c] v(t)2 2av 3 d 1 =1 = 1 1+ , so c c2 c2 c3 " # ✓ ◆2 1/2 ✓ ◆ v(tr ) va 3 a˙ = 1 = (t) 1 d ; a(tr ) = a(t) T a˙ = a(t) d. c c3 c Evaluating everything now at time t: " # ✓ ◆ 4 3va 3 d/c3 (a a˙ d/c) q2 a˙ v 2 a2 3 1 2 Fself = + 3 + + ( )d + ··· x ˆ 4⇡✏0 4c2 d 4c 3 c2  ✓ ◆ ✓ ◆ 3 3 4 q2 a a˙ va2 2 a˙ v 2 a2 = + +3 3 + 3 + + ( )d + ··· x ˆ 4⇡✏0 4c2 d 4c2 c c 4c 3 c2  ✓ ◆ 3 4 q2 a a˙ va2 2 v 2 a2 = + a ˙ + + 3 + + ( )d + ··· x ˆ 4⇡✏0 4c2 d 4c3 3 c2 c2  ✓ ◆ 3 4 q2 a va2 2 = + a ˙ + 3 + ( )d + ··· x ˆ (generalizing Eq. 11.96). 4⇡✏0 4c2 d 3c3 c2 The first term mass; the radiation reaction itself is the second term: ✓ is the electromagnetic ◆ 2 2 2 µ q va 0 int 4 Frad = a˙ + 3 2 (generalizing Eq. 11.99), so the generalization of Eq. 11.100 is 12⇡c c µ0 q 2 6⇡c

Frad =

(b) Frad = A

4

Z

✓

a˙ +

t2

3

2 2

a v

c2

Frad v dt =

t1

◆

✓

a˙ + 3

va2 c2

, where A ⌘

µ0 q 2 . P = Aa2 6⇡c

Z

or

t2

P dt,

t1

Z

t2

t1

(except for boundary terms—see Sect. 11.2.2). Z t2 Z t2 4 Rewrite the first term: av ˙ dt = ( t1

4

t1

4

v)

4

✓

av ˙ +3

da dt = dt !

4

va

6

2

◆

.

(Eq. 11.75). What we must show is that

v 2 a2 c2 t2 t1

2

Z

◆ t2

dt =

d ( dt

4

v) = 4

Problem 11.34

3d

dt

v+

4

a;

t2

a2

6

dt

t1

d ( dt

4

v)a dt. ✓ ◆ 1 2va va 3 = 2 . So 2 2 2 3/2 c c v /c ) t1

d d 1 1 p = = dt dt 2 (1 1 v 2 /c2 ✓ ◆ ✓ ◆ 3 2 2 d 4 va v v v2 6 ( v) = 4 3 v 2 + 4 a = 6 a 1 + 4 = a 1 + 3 . dt c c2 c2 c2 ✓ ◆ Z t2 Z t2 t2 v2 4 6 2 av ˙ dt = 4 va a 1 + 3 2 dt, and hence c t1 ◆ t1 ✓ ◆ Zt1t2 ✓ Z t2  t2 3 2 a2 v 2 v2 a2 v 2 4 4 6 2 av ˙ + dt = va + a 1 + 3 +3 6 2 2 2 c c c t1 t1 t1 Now

Z

dt =

4

va

t2 t1

Z

t2

6 2

a dt.

qed

t1

p µ0 q 2 a2 6 c2 t (Eq. 11.75). w = b2 + c2 t2 (Eq. 10.52); v = w˙ = p ; 6⇡c b2 + c2 t2 2 2 2 2 2 2 c c t(c t) c b c a = v˙ = p = 2 b2 + c2 t2 c2 t2 = 2 ; 2 2 2 3/2 2 2 3/2 2 2 2 (b + c t ) (b + c t ) (b + c2 t2 )3/2 b +c t (a) P =


261

CHAPTER 11. RADIATION 2

=

1 = v 2 /c2 1

1

1 [c2 t2 /(b2

+

c2 t2 )]

=

b2

b2 + c2 t2 1 = 2 b2 + c2 t2 . So + c2 t2 c2 t2 b

µ0 q b c (b + c t ) q2 c = . Yes, it radiates (in fact, at a constant rate). 6⇡c (b2 + c2 t2 )3 b6 6⇡✏0 b2 ✓ ◆ ✓ ◆ µ0 q 2 4 3 2 a2 v 3 b2 c2 (2c2 t) 3b2 c4 t 3 2 a2 v (b) Frad = a˙ + ; a ˙ = = ; a ˙ + = 6⇡c c2 2 (b2 + c2 t2 )5/2 c2 (b2 + c2 t2 )5/2 3b2 c4 t 3 (b2 + c2 t2 ) b4 c4 c2 t p + = 0. Frad = 0. No, the radiation reaction is zero. c2 b2 (b2 + c2 t2 )3 b2 + c2 t2 (b2 + c2 t2 )5/2 P =

2

4 4

2

2 2 3

Problem 11.35 The fields of a nonstatic ideal dipole (Problem 10.34) contain terms that go like 1/r, 1/r2 , and 1/r3 , for fixed t0 ⌘ t r/c; radiation comes from the first of these: Er =

µ0 [¨ p 4⇡r

¨ )], ˆ r(ˆ r·p

Br =

µ0 ¨) (ˆ r⇥p 4⇡rc

where the dipole moments are evaluated at time t0 . The 1/r2 term in the Poynting vector is 1 µ0 µ0 ¨ )] ⇥ (ˆ ¨) = ¨ ) (ˆ ¨ )[ˆ ¨ )]} (Er ⇥ Br ) = [¨ p ˆ r(ˆ r·p r⇥p {¨ p ⇥ (ˆ r⇥p r·p r ⇥ (ˆ r⇥p 2 2 µ0 16⇡ cr 16⇡ 2 cr2 ⇥ 2 ⇤ µ0 µ0 ¨ (ˆ ¨ ) (ˆ ¨ )[ˆ ¨) p ¨] = ¨ )2 ˆ = ˆ r p¨2 p r·p r·p r(ˆ r·p p¨ (ˆ r·p r. 16⇡ 2 cr2 16⇡ 2 cr2 ⇥ ⇤ ¨ , p¨2 (ˆ ¨ )2 = p¨2 sin2 ✓, and the power radiated is Setting the z axis along p r·p Sr =

P =

I

Sr · da =

µ0 2 p¨ 16⇡ 2 c

I

1 µ0 2 sin2 ✓r2 sin ✓ d✓ d = p¨ 2 r 8⇡c

Z

⇡

0

sin3 ✓ d✓ =

µ0 2 p¨ . 6⇡c

¨ = ! 2 p0 cos !t, p¨2 = ! 4 p20 cos2 !t, and its time average For the sinusoidal case, p = p0 cos !t, we have p is (1/2)p20 ! 4 , so µ0 2 4 hP i = p¨ ! , 12⇡c 0 in agreement with Eq. 11.22. And in the case of quadratic time dependence our result agrees with the answer to Problem 11.26.


262

CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY

Chapter 12

Electrodynamics and Relativity Problem 12.1 ¯ and v the velocity of S¯ with respect to S. ¯ its velocity in S, Let u be the velocity of a particle in S, u ¯ + v. For a free particle, u is constant (that’s Newton’s first Galileo’s velocity addition rule says that u = u law in S. ¯ and hence S¯ is inertial. ¯ = u v is also constant, so Newton’s first law holds in S, (a) If v is constant, then u ¯ ¯ is also constant, so v = u u ¯ is constant. (b) If S is inertial, then u Problem 12.2 ¯ i + v. (a) mA uA + mB uB = mC uC + mD uD ; ui = u mA (¯ uA + v) + mB (¯ uB + v) = mC (¯ uC + v) + mD (¯ uD + v), ¯ A + mB u ¯ B + (mA + mB )v = mC u ¯ C + mD u ¯ D + (mC + mD )v. mA u Assuming mass is conserved, (mA + mB ) = (mC + mD ), it follows that ¯ ¯ A + mB u ¯ B = mC u ¯ C + mD u ¯ D , so momentum is conserved in S. mA u

(b) 12 mA u2A + 12 mB u2B = 12 mC u2C + 12 mD u2D ) 1 u2A + 2¯ uA · v + v 2 ) + 21 mB (¯ u2B + 2¯ uB · v + v 2 ) = 12 mC (¯ u2C + 2¯ uC · v + v 2 ) + 12 mD (¯ u2D 2 mA (¯ 1 ¯ A + mB u ¯ B ) + 12 v 2 (mA + mB ) ¯2A + 12 mB u ¯2B + v · (mA u 2 mA u 1 1 2 2 ¯ C + mD u ¯ D ) + 12 v 2 (mC + mD ). = 2 mC u ¯C + 2 mD u ¯D + v · (mC u

+ 2¯ uD · v + v 2 )

But the middle terms are equal by conservation of momentum, and the last terms are equal by conservation of mass, so 12 mA u ¯2A + 12 mB u ¯2B = 12 mC u ¯2C + 12 mD u ¯2D . qed Problem 12.3 vAB +vBC vAB vBC (a) vG = vAB + vBC ; vE = 1+v ; ) vGvGvE = 2 ⇡ vG 1 c2 AB vBC /c 8 In mi/h, c = (186, 000 mi/s) ⇥ (3600 sec/hr) = 6.7 ⇥ 10 mi/hr.

)

(b)

vG vE vG

1 2c

=

(5)(60) (6.7⇥108 )2

+ 34 c / 1 +

1 2

·

3 4

= 6.7 ⇥ 10 =

5 4c

(c) To simplify notation, let 2

=

2 1

+2 1 (1 + 2

+2 1 2+ 2 2 1 2 + 1 2)

2

/

16

. ) 6.7 ⇥ 10

11 8

= vAC /c, 2 2

=

1+2 (1 + 2

= 1

% error (pretty small!)

10 c (still less than c) 11 = vAB /c, + 2+

1 2 1

14

vAB vBC . c2

2 1 2 1

2 2 2 2)

2

= vBC /c. Then Eq. 12.3 says: (1 + 12 (1 + 2

2 2 1 2

2 1

+

2 2) 2 2 1 2)

=1

(1

=

1+ 2

1+

2 1 )(1

(1 +

1

2

2 1 2)

, or: 2 2)

=1


,

263

CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY where

⌘ (1

2 1 )(1

2 2 )/(1

+

2 1 2)

is clearly a positive number. So

2

Problem 12.4

< 1, and hence |vAC | < c. qed.

(a) Velocity of bullet relative to ground is 12 c + 13 c = 56 c = 10 12 c. 9 Velocity of getaway car is 34 c = 12 c. Since vb > vg , bullet does reach target . (b) Velocity of bullet relative to ground is Velocity of getaway car is 34 c =

21 28 c.

1 1 2 c+ 3 c 1+ 12 · 13

=

5 6c 7 6

= 57 c =

20 28 c.

Since vg > vb , bullet does not reach target .

Problem 12.5 (a) Light from 90th clock took

90⇥109 m 3⇥108 m

= 300 sec = 5 min to reach me, so the time I see on the clock is

11:55 am . (b) I observe 12 noon . Problem 12.6 ( light signal leaves a at time t0a ; arrives at earth at time ta = t0a + dca light signal leaves b at time t0b ; arrives at earth at time tb = t0b + dcb h ( v t0 cos ✓) = t0 1 c c (Here da is the distance from a to earth, and db is the distance from b to earth.) )

t = tb

s = v t0 sin ✓ =

⇥ v (1 du = d✓

✓max

v c

ta = t0b

t0a +

v sin ✓ t (1 (1t

(db

da )

v cos ✓) c

=

t0 +

) u=

(1

i v cos ✓ c

v sin ✓ is the the apparent velocity. v c cos ✓)

⇤ cos ✓)(cos ✓) sin ✓( vc sin ✓) = 0 ) (1 (1 vc cos ✓)2

v v cos ✓) cos ✓ = sin2 ✓ c c v v ) cos ✓ = (sin2 ✓ + cos2 ✓) = c c p 2 2 v 1 v /c = cos 1 (v/c) At this maximal angle, u = 1 v2 /c2 = p v 2 2 1 v /c

As v ! c, u ! 1 , because the denominator ! 0 — even though v < c.

Problem 12.7 The student has not taken into account time dilation of the muon’s “internal clock”. In the laboratory, the muon lasts ⌧ = p ⌧ 2 2 , where ⌧ is the “proper” lifetime, 2 ⇥ 10 6 sec. Thus 1 v /c

v=

d

p t/ 1 ⇣ ⌧ ⌘2 v2 = 1 d

v2 ; c2

v2 1 = 2; c2 1 + (⌧ c/d)

r

v2 , where d = 800 meters. c2 ⇣⇣ ⌧ ⌘2 1⌘ 1 v2 + 2 = 1; v 2 = . 2 d c (⌧ /d) + (1/c)2

d = 2 2 ⌧ v /c

1

⌧c (2 ⇥ 10 6 )(3 ⇥ 108 ) 6 3 = = = ; d 800 8 4

v2 1 16 = = ; c2 1 + 9/16 25


v=

4 c. 5

264


Problem 12.8 (a) Rocket clock runs slow; so earth clock reads t = p

1 1 v 2 /c2

· 1 hr. Here

=p

=p

1 1 v 2 /c2

1 1 9/25

= 54 .

) According to earth clocks signal was sent 1 hr, 15 min after take-o↵. (b) By earth observer, rocket is now a distance 35 c 54 (1 hr) = 34 c hr (three-quarters of a light hour) away. Light signal will therefore take 34 hr to return to earth. Since it left 1 hr and 15 min after departure, light signal reaches earth 2 hrs after takeo↵ (c) Earth clocks run slow: trocket = Problem 12.9 Lc = 2Lv ;

Lc c

=

Lv v

; so

2 c

=

1 v

· (2 hrs) =

q = 1

1 2 2

5 4

· (2 hrs) = 2.5 hrs

=

q

3 1 4 ; c2

v2 c2

=1

=

v2 c2

=

3 v2 16 . c2

=1

3 16

=

13 16 ;

v=

p

13 c 4

Problem 12.10 ¯ horizontal projection Say length of mast (at rest) is ¯l. To an observer on the boat, height of mast is ¯l sin ✓, ¯ To observer on dock, the former is una↵ected, but the latter is Lorentz contracted to 1 ¯l cos ✓. ¯ is ¯l cos ✓. Therefore: ¯l sin ✓¯ tan ✓¯ ¯ tan ✓ = 1 ¯ = tan ✓, or tan ✓ = p ¯ l cos ✓ 1 v 2 /c2

Problem 12.11 p Naively, circumference/diameter = 1 (2⇡R)/(2R) = ⇡/ = ⇡ 1 (!R/c)2 — but this is nonsense. Point is: an accelerating object cannot remain rigid, in relativity. To decide what actually happens here, you need a specific model for the internal forces holding the disc together. Problem 12.12 ¯ (iv) ) t = t + vx c2 . Put this into (i), and solve for x: x ¯= x

x

Similarly, (i) ) x =

v x ¯

t¯ = t

⇣ t¯

+

⇣ vx ⌘ = x 1 c2

v2 ⌘ c2

v t¯ = x

1

v t¯ =

2

x

v t¯; x = (¯ x + v t¯) X

+ vt. Put this into (iv) and solve for t: v⇣x ¯

c2

⌘ ⇣ + vt = t 1

v2 ⌘ c2

v t x ¯ c2

v x ¯; t = c2

⇣

v ⌘ t¯ + 2 x ¯ X c

Problem 12.13 Let brother’s accident occur at origin, time zero, in both frames. In system S (Sophie’s), the coordinates v of Sophie’s cry are x = 5 ⇥ 105 m, t = 0. In system S¯ (scientist’s), t¯ = (t vx/c2 . Since c2 x) = ¯ this is negative, Sophie’s cry occurred before the accident, in S. = p 1 = p16913 144 = 13 5 . So 2 1 (12/13)

t¯ =

13 5

12 13 c

(5 ⇥ 10 )/c = 5

2

12 ⇥ 10 /3 ⇥ 10 = 10 5

8

3

. 4 ⇥ 10

3

seconds earlier

Problem 12.14 ¯ meanwhile, it moves a distance d¯ (a) In S it moves a distance dy in time dt. In S, y = dy in time dt¯ = v (dt c2 dx). )

d¯ y dy = = ¯ dt (dt cv2 dx)

(dy/dt) ; or u ¯y = 1 cv2 dx dt

uy 1

vux c2

; u ¯z =

uz 1

vux c2


265


(b) S = dock frame; S 0 = boat frame; we need reverse transformations (v ! v): u ¯y / 1 + vcu¯2x uy 1 u ¯y ¯ u ¯ so tan ✓ = = . In this case u ¯x = c cos ✓; ¯y = c sin ✓, vu ¯x = ux (¯ ux + v) (¯ ux + v)/ 1 + c2 ✓ ◆ 1 sin ✓¯ 1 c sin ✓¯ tan ✓ = = ¯ ( c cos ✓+v) cos ✓¯ v/c ¯

sin ✓ [Contrast tan ✓ = cos in Prob. 12.10. The point is that velocities are sensitive not only to the transfor✓¯ mation of distances, but also of times. That’s why there is no universal rule for translating angles — you have to know whether it’s an angle made by a velocity vector or a position vector.] That’s how the velocity vector of an individual photon transforms. But the beam as a whole is a snapshot of many di↵erent photons at one instant of time, and it transforms the same way the mast does.

Problem 12.15

3 1 c 1c c 5 2 c. Outlaws relative to police: 4 3 2 1 = 45 = c. 7 5 1 4·2 8 5 3 1 c c c 1 28 Bullet relative to outlaws: 7 5 4 3 = 13 = c . [Velocity of A relative to B is minus the velocity of 13 1 7·4 28 B relative to A, so all entries below the diagonal are trivial. Note that in every case vbullet < voutlaws , so no matter how you look at it, the bad guys get away.]

Bullet relative to ground:

speed of !

Ground relative to # Police Outlaws Bullet

Ground

Police

Outlaws

0

1 2c

3 4c 2 5c

1 2 3 4c 5 7c

0

2 5c 1 3c

Bullet

Do they escape?

5 7c 1 3c 1 13 c

0

0

1 13 c

Yes Yes Yes Yes

Problem 12.16 (a) Moving clock runs slow, by a factor 5 3

= p

1 1 (4/5)2

=

5 3.

Since 18 years elapsed on the moving clock,

⇥ 18 = 30 years elapsed on the stationary clock. 51 years old

(b) By earth clock, it took 15 years to get there, at 45 c, so d = 45 c ⇥ 15 years = 12c years (12 light years) (c) t = 15 yrs, x = 12c yrs ¯ and rode along on S, ¯ so she’s still at the origin. If you (d) t¯ = 9 yrs, x ¯ = 0. [She got on at the origin in S, doubt these values, use the Lorentz Transformations, with x and t in (c).] ⇢ ˜ = (x + vt) (e) Lorentz Transformations: x [note that v is negative, since S˜ us going to the left] t˜ = (t + cv2 x) )x ˜ = 53 (12c yrs + 45 c · 15 yrs) = 53 · 24c yrs = 40c years. t˜ = 5 (15 yrs + 4 c2 · 12c yrs) = 5 15 + 48 yrs = (25 + 16)yrs = 41 years. 3

5c

3

5

(f) Set her clock ahead 32 years, from 9 to 41 (t¯ ! t˜). Return trip takes 9 years (moving time), so her clock will now read 50 years at her arrival. Note that this is 53 · 30 years—precisely what she would calculate if the stay-at-home had been the traveler, for 30 years of his own time. c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

266


(g) (i) t¯ = 9 yrs, x = 0. What is t? t =

+

v c2 x

t¯

=

3 5

· 9 yrs =

= 5.4 years, and he started at age 21, so he’s

27 5

26.4 years old (Younger than traveler (!) because to the traveller it’s the stay-at-home who’s moving.) (ii) t˜ = 41 yrs, x = 0. What is t? t =

t˜

=

3 5

45.6 years old.

· 41 yrs, or

yrs, or 24.6 yrs, and he started at 21, so he’s

123 5

(h) It will take another 5.4 years of earth time for the return, so when she gets back, she will say her twin’s age is 45.6 + 5.4 = 51 years—which is what we found in (a). But note that to make it work from traveler’s point of view you must take into account the jump in perceived age of the stay-at-home when she changes ˜ coordinates from S¯ to S.) Problem 12.17

a ¯0¯b0 + a ¯1¯b1 + a ¯2¯b2 + a ¯3¯b3 = =

2

=

2 0 0

= Problem 12.18 0 1 0 ct¯ 1 Bx C B ¯ C B (a) B @ y¯ A = @ 0 z¯ 0 0

0 1 0 0

0 1 0 0

B 0 (b) ⇤ = B @ 0

0 0 1 0

2

(a0

(a0 b0

a1 )(b0 a1 b0 +

a0 b1

a b (1

2

b1 ) +

)+

2

2 1 1

a b

a b (1

2 1 1

2

(a1

a0 )(b1

a0 ) + a2 b2 + a3 b3

a1 b1 + a1 b0 + a0 b1

a b ) + a2 b2 + a3 b3

2 0 0

) + a2 b2 + a3 b3

a0 b0 + a1 b1 + a2 b2 + a3 b3 . qed [Note:

2

(1

2

) = 1.]

10 1 0 ct BxC 0C C B C (using the notation of Eq. 12.24, for best comparison) 0A @ y A 1 z

0 0

1 0 0C C 0A 1

0

¯ B 0 (c) Multiply the matrices: ⇤ = B @ ¯¯ 0

0 1 0 0

¯¯ 0 ¯ 0

10 0 B 0C CB 0A @ 0 1 0

0 0

0 0 1 0

1 0 0C C= 0A 1

0

¯

B B @ ¯ ¯ 0

¯ ¯ ¯ 0

¯¯ 0 ¯ 0

1 0 0C C 0A 1

Yes, the order does matter. In the other order “bars” and “no-bars” would be switched, and this would yield a di↵erent matrix. Problem 12.19 2 sinh ✓ (a) Since tanh ✓ = cosh sinh2 ✓ = 1, we have: ✓ , and cosh ✓ 1

=p

1

0

v 2 /c2

=p 1

cosh ✓ sinh ✓ B sinh ✓ cosh ✓ ) ⇤=B @ 0 0 0 0

0 0 1 0

1 2

tanh ✓

cosh ✓

=p

cosh2 ✓

1 0 0 cos C 0C @ sin Compare: R = 0A 0 1

sinh2 ✓ sin cos 0

= cosh ✓ ;

= cosh ✓ tanh ✓ = sinh ✓.

1 0 0A 1


267


u v u v u ¯ tanh tanh ✓ c c = ) tanh ¯ = , where tanh = u/c, tanh ✓ = v/c; uv ) u v 1 c2 c 1 tanh tanh ✓ 1 c c tanh ¯ = u ¯/c. But a “trig” formula for hyperbolic functions (CRC Handbook, 18th Ed., p. 204) says:

(b) u ¯ =

tanh tanh ✓ = tanh( 1 tanh tanh ✓ Problem 12.20 (a) (i) I = c2 t2 + x2 + y 2 + z 2 =

) tanh ¯ = tanh(

✓).

(5

15)2 + (10

$ "#

4

1

100 + 25 + 25 =

50

5

)v=

5ˆ x 5ˆ y = 10/c

c x ˆ 2

c y ˆ 2

!

❂ A!

2

(3

0)2 =

✓

S¯ travels in the direction from B toward A, making the trip in time 10/c.

B

6

(b) (i) I =

3)2 + (0

t¯ = 0, so I would have to be positive, which it isn’t.)

(ii) No. (In such a system y (iii) Yes. ✻ 8

2 1)2 + (5

5)2 + (8

✓), or: ¯ =

"# 5

4 6 2)2 + 0 + 0 =

$

2

Note that vc2 = less than c.

✲x 8 10 4+9= 5

1 4

+

1 4

= 12 , so v =

p1 c, 2

safely

(ii) Yes. By Lorentz Transformation: (ct¯) = (ct) ( x) . We want t¯ = 0, so (ct) = ( x); or ✒ v (ct) (3 1)ct 2 2 = = = ✻ . So v = c in the +x direction. c ( x) (5 2) 3 3 ■ (iii) No. (In such a system x = y = z = 0 so I would be negative, which it isn’t.) Problem 12.21 Using Eq. 12.18 (iv): t¯ = ( t world line of player 1 # ✒ ✮ ✛ ■

10 ft

✲

v c2

x) = 0 )

t=

v c2

x, or v =

t 2 xc

=

tB xB

tA 2 c xA

world line of player 2

✲x world line of the ball

✠ ct ✻ ✒

✐ ✶ A


B ✲x

B $

8

"#

6 4 2 2

!

❂ A!

"# 5

268 6

4

5

$

8

✲x CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY 10

Problem 12.22 (a) ✒ ct ✻ ■


Truth is, you never do communicate with the other person right now —you communicate with the person he/she will be when the message gets there; and the response comes back to an older and wiser you.

#

✛ ■


✒ ✮

10 ft

✲

✲x world line of the ball

(b) No way It is true that a moving observer might say she arrived at B before she left A, but for the round trip everyone must agree that she arrives back after she set out.

✠ ct ✻ ✒

✐ ✶ A

B ✲x


Problem 12.24 (a) 1

u2 c2

⌘ 2 = u2 ; u2 1 +

⌘2 c2

1 = ⌘2 ; u = p ⌘. 1 + ⌘ 2 /c2 c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

269

CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY (b) p

=p

1 1 u2 /c2

=p

1 1 tanh2 ✓

cosh ✓ cosh2 ✓ sinh2 ✓

Problem 12.25 (a) ux = uy = u cos 45 = (b) p

=p

1 1 u2 /c2

(c) ⌘0 = c =

(f) p

1 1 u ¯2 /c2

1 4/5

=

p p 5 5 4

r

=

=

p

= cosh ✓ c tanh ✓ = c sinh ✓.

1 u 1 u2 /c2

2 c. 5

5. ) ⌘ = p

u 1 u2 /c2

) ⌘x = ⌘y =

p

2

p

(d) Eq. 12.45)

(e) ⌘¯x = (⌘x

1

p1 p2 c 2 5

)⌘= p

= cosh ✓

5 c. 8 > ¯x = > > ¯y = :u

ux v 1 ucx2v 1

⇣

=

uy 1

ux v c2

p2 5

⌘

c 1

=

p2 5

2 5

q

1

c

= 0. p2

2 5 1

5

c

2 5

r p 2 2/5 p = c= c. 3/5 3

q q p p p 2 1 25 ( 2 c ⌘¯y = ⌘y = 2 c. 5 5 c) = 0. p ⇢ p p ⌘ ¯ = ¯x = p 0X x 1 p3 u =p = 3; ) ⌘ ¯ = 3u ¯) 1 2/3 ⌘¯y = 3 u ¯y = 2 c X ⌘0 ) =

Problem 12.26 ⌘ µ ⌘µ = (⌘ 0 )2 + ⌘ 2 =

1 (1 u2 /c2 ) (

c2 + u2 ) =

Problem 12.27 Use the result of Problem 12.24(a): u = p

c2 (1 (1

u2 /c2 ) u2 /c2 )

1

=

⌘. Here

c2 .

Timelike.

⌘ 4 1 3 = , so p = , and hence c 3 5 1 + 16/9

1+ 3 8 8 u = (4 ⇥ 10 ) = 2.4 ⇥ 10 m/s. Innocent. 5 Problem 12.28 p p R R (a) From Prob. 11.34 we have = 1b b2 + c2 t2 . ) ⌧ = 1 dt = b pb2dt = cb ln(ct + b2 + c2 t2 ) + k; at +c2 t2  p b 1 b b t = 0 we want ⌧ = 0: 0 = c ln b + k, so k = c ln b; ⌧ = ln (ct + b2 + c2 t2 ) c b p p (b) x2 b2 + x = bec⌧ /b ; x2 b2 = bec⌧ /b x; x2 b2 = b2 e2c⌧ /b 2xbec⌧ /b + x2 ; 2xbec⌧ /b = b2 (1 + e2c⌧ /b ); p c⌧ /b c⌧ /b x = b e +e = b cosh(c⌧ /b) . Also from Prob. 11.34: v = c2 t/ b2 + c2 t2 . 2 p q ⇣ c⌧ ⌘ p 2 c 2 cosh2 (c⌧ /b) 2 = c cosh (c⌧ /b) 1 = c sinh(c⌧ /b) = c tanh v = xc x2 b2 = b cosh(c⌧ b b . /b) cosh(c⌧ /b) cosh(c⌧ /b) b ⇣ ⌘ c⌧ c⌧ (c) ⌘ µ = (c, v, 0, 0). = xb = cosh c⌧b , so ⌘ µ = cosh c⌧b c, c tanh c⌧b , 0, 0 = c cosh , sinh , 0, 0 . b b Problem 12.29 (a) mA uA + mB uB = mC uC + mD uD ;

⌘ 2 /c2

u ¯i + v . 1 + (¯ ui v/c2 ) u ¯A + v u ¯B + v u ¯C + v u ¯D + v mA + mB = mC + mD . 2 2 2 1 + (¯ uA v/c ) 1 + (¯ uB v/c ) 1 + (¯ uC v/c ) 1 + (¯ uD v/c2 ) This time, because the denominators are all di↵erent, we cannot conclude that mA u ¯A + mB u ¯B = mC u ¯ C + mD u ¯D . As an explicit counterexample, suppose all the masses are equal, and uA = uB = v, uC = uD = 0. This is a symmetric “completely inelastic” collision in S, and momentum is clearly conserved (0=0). But the Einstein ui =


270


velocity addition rule gives u ¯A = 0, u ¯B = momentum is not conserved:

2v/(1 + v 2 /c2 ), u ¯C = u ¯D = ✓ ◆ 2v m 6= 2mv. 1 + v 2 /c2

v, so in S¯ the (incorrectly defined)

(b) mA ⌘A + mB ⌘B = mC ⌘C + mD ⌘D ; ⌘i = (¯ ⌘i + ⌘¯i0 ). (The inverse Lorentz transformation.) 0 0 0 0 mA (¯ ⌘A + ⌘¯A ) + mB (¯ ⌘B + ⌘¯B ) = mC (¯ ⌘C + ⌘¯C ) + mD (¯ ⌘D + ⌘¯D ). The gamma’s cancel: 0 0 0 0 mA ⌘¯A + mB ⌘¯B + (mA ⌘¯A + mB ⌘¯B ) = mC ⌘¯C + mD ⌘¯D + (mC ⌘¯C + mD ⌘¯D ). 0 0 ¯ ¯ ¯ ¯ ¯ But mi ⌘i = pi = Ei /c, so if energy is conserved in S (EA + EB = EC + ED ), then so too is the momentum (correctly defined): mA ⌘¯A + mB ⌘¯B = mC ⌘¯C + mD ⌘¯D . qed Problem 12.30 2 1 mc2 mc2 = nmc2 ) = n + 1 = p1 2 2 ) 1 uc2 = (n+1) 2 1 u /c p 2 n(n + 2) 1 n2 +2n+1 1 ) uc2 = 1 (n+1) = n(n+2) c 2 = (n+1)2 (n+1)2 ; u = n+1 Problem 12.31 ET = E1 + E2 + · · · ; pT = p1 + p2 + · · · ; p¯T = (pT v = c pT /ET = c (p1 + p2 + · · · )/(E1 + E2 + · · · ) 2

2

ET /c) = 0 )

= v/c = pT c/ET

Problem 12.32 Eµ =

(m2⇡ + m2µ ) 2 c = mµ c2 ) 2m⇡

v2 =1 c2

1 2

=1

=

(m2⇡ + m2µ ) 1 =p ; 2m⇡ mµ 1 v 2 /c2

1

v2 1 = 2; 2 c

4m2⇡ m2µ m4⇡ + 2m2⇡ m2µ + m4µ 4m2⇡ m2µ (m2⇡ m2µ )2 = = ;v = 2 2 2 2 2 2 (m⇡ + mµ ) (m⇡ + mµ ) (m2⇡ + m2µ )2

(m2⇡ m2µ ) (m2⇡ + m2µ )

!

Problem 12.33 p Initial momentum: E 2 p2 c2 = m2 c4 ) p2 c2 = (2mc2 )2 m2 c4 = 3m2 c4 ) p = 3 mc. Initial energy: 2mc2 + mc2 = 3mc2 . p Each is conserved, so final energy is 3mc2 , final momentum is 3 mc. E2

p2 c2 = (3mc2 )2

p ( 3 mc)2 c2 = 6m2 c4 = M 2 c4 .

) M=

p

6 m ⇡ 2.5m

(In this process some kinetic energy was converted into rest energy, so M > 2m.) p pc2 3 mc c2 c v= = = p = v. E 3mc2 3 Problem 12.34 9 5 2 4 2 First calculate pion’s energy: E 2 = p2 c2 + m2 c4 = 16 m2 c4 + m2 c4 = 25 16 m c ) E = 4 mc . 5 2 Conservation of energy: 4 mc = EA + EB 2EA = 2mc2 EB 3 3 2 Conservation of momentum: 4 mc = pA + pB = EcA ) mc = E E A B c 4 1 2 2 ) EA = mc ; EB = mc . 4


c.

✒✻

✒✻

−1¯t = 2 = ¯ t c −c2 1 ¯= c¯t = −c3t 0 ¯= c¯t = ct −1 c¯t = −2 ✲x c¯t = −3 c¯t =

9.2RELATIVITY CHAPTER 12. ELECTRODYNAMICS AND

271

✲x

✛ ✲❄ 8.7 Problem 12.35 9.2 Classically, E = 12 mv 2 . In a colliding beam experiment, the relative velocity (classically) is twice the velocity of either one, so the relative energy is 4E. ✛ ✲❄ 8.7 v ✻ ✻ =⇒ Let S¯ be the system in which 1 is at rest. Its ¯ E E E ⃝ 1 ✲ ✛⃝ ⃝ 1 ✛ ⃝ 2 2 speed v, relative to S, is just the speed of 1 ✲S

✻

v ✲ S¯ ✻ =⇒ ¯ E ⃝ 1 ✛ ⃝ 2

E E ⃝ 1 ✲ ✛⃝ 2

in S.

¯ E ¯ the momentum of 2 in S. p¯0 = (p0 p1 ) ✲ ) SEc = p , where✲p S is c E E 2 ¯= E = M c , so = M c2 ; p = MV = M c. ) E M c c = (E + M c2 c + h i 2 2 1 1 E E 2 2 ¯ = 11 2 )1 = 12 ) 2 = 1 = . ) E = E + 1 M c2 2 2 M c2 M c2

¯= E

E2 M c2

+

E2 M c2

2E 2 ¯✻ M c2 ; x E = M c2

)

2 M c✒ .

¯ = (2)(900) For E = 30 GeV and M c = ✒ 1 GeV, we have E 1 A x Problem 12.36 ✻ ✒ ✲ ct One photon is impossible, B because✒in the “center of momentum” frame (Prob. 12.31) A we’d be left with a photon at rest, whereas photons have to travel at speed ✲c.ct 2

2

EA

✒ 1 = 1800 =◦ 60E. ✲ 1 = 1799 GeV 60 θ m m EB⑦ EA (before) (after) ✒◦ 60 ✲ θ m m EB⑦ (before) (after)

B p 8 < Cons. of energy: ⇢ p0 c2 + m2 c4 + mc2 = EA + EB horizontal: p0 = EcA cos 60 + EcB cos ✓ ) EB cos ✓ =pp0 c : Cons. of mom.: EB 3 vertical: 0 = EcA sin 60 c sin ✓ ) EB sin ✓ = 2 EA

1 2 3 2 p0 cEA + EA + EA 4 4 q 2 p0 cEAc + EA = p20 c2 + m2 c4 + mc2

1 2 EA

square and add:

2 EB (cos2 ✓ + sin2 ✓) = p0 c2 2 ) EB = p0 c2

= p0 c2 +

2

EA

⃝2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be q reproduced, in any form or by any means, without permission in writing from the publisher. 2 4 2 m c + 2 p20 c2 + m2 c4 (mc2 EA ) + m2 c4 2EA mc2 + EA . Or:

p0 cEA = 2m2 c4 + 2mc2 q ) EA (mc2 + p20 c2 + m2 c4

q ⃝2005 q c Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is 2 4 currently as2 c they p20protected c2 + m2under c4 all 2Ecopyright p20 c2laws +m 2EA mcexist. ; No portion of this material may be A reproduced, in any form or by any means, without permission in writing from the publisher.

p p0 c/2) = m2 c4 + mc2 p0 c2 + m2 c4 ; p p (mc2 + p20 c2 + m2 c4 ) (mc2 p2 c2 + m2 c4 p0 c/2) 2 p p 0 EA = mc · (mc2 + p20 c2 + m2 c4 p0 c/2) (mc2 p20 c2 + m2 c4 p0 c/2) p p 2 4 p20 c2 m2 c4 12 p0 mc3 p20 c p20 c2 + m2 c4 ) mc2 (mc + 2p0 + p20 + m2 c2 ) 2 (m c = mc = . 2 2 (mc + 34 p0 ) (m2 c4 p0 mc3 + p04c p0 c2 m2 c4 )


9.2 ✛

v CHAPTER ✻ 12. ELECTRODYNAMICS ✻ =⇒ AND RELATIVITY ¯ E E E ⃝ 1 ✲ ✛⃝ ⃝ 1 ✛ ⃝ 2 2

272 Problem 12.37 (

du dt

dp d mu p = =m p +u dt dt 1 u2 /c2 1 u2 /c2 ⇢ m u(u·a) =p a+ 2 . qed 2 2 (c u2 ) 1 u /c

F=

✲❄

8.7

✓

1 2

◆

✲S

2u· du dt u2 /c2 )3/2

1 c2

(1

✲ S¯

)

Problem 12.38 At constant force you go in “hyperbolic” motion. Photon A, which left the origin at t < 0, catches up with you, but photon B, which passes the origin at t > 0, never does.

x✻ A

✒

✒ ✲ ct

B

" !# Problem 12.39 d⌘ d⌘ dt d c 1 0 0 0 (a) p p ↵ = = = 2 2 d⌧ dt d⌧ dt 1 u /c 1 u2 /c2 ✓ ◆ 1 c 1 1 u·a c2 2u·a =p = . 2 (1 u2 /c2 )3/2 c (1 u2 /c2 )2 1 u2 /c2 d⌘ dt d⌘ 1 d ↵= = =p 2 2 d⌧ d⌧ dt dt 1 u /c  1 u(u·a) = a+ 2 . (1 u2 /c2 ) (c u2 )

(b)

p 1

u

u2 /c2

!

=p 1

1

u2 /c2

(

p

a 1

u2 /c2

+ u( t)

(1

1 c2 2u·a u2 /c2 )3/2

c ⃝2005 Pearson Education, Inc., Upper Saddl protected under all copyright laws as they cur reproduced, in any form or by any means, wit

 2 1 (u·a)2 1 u2 1 ↵ ↵ ↵µ ↵ = (↵ ) + ·↵ = + a(1 ) + u(u·a) c2 (1 u2 /c2 )4 (1 u2 /c2 )4 c2 c2 ( ) ✓ ◆ ✓ ◆ 2 1 u2 u2 1 2 1 2 2 2 2 2 = (u·a) + a 1 + 2 1 (u·a) + 4 u (u·a) (1 u2 /c2 )4 c2 c2 c c2 c ⇢ ✓ ◆ 2 1 u2 (u·a)2 u2 u2 2 = a 1 + + 1 + 2 2 2 (1 u2 /c2 )4 c2 c2 c2} | {z c µ

0 2

(1

=

(c) ⌘ µ ⌘µ = (d) K µ =

(1

c2 , so

d⇢µ d⌧

=

 1 (u·a)2 2 a + . u2 /c2 )2 (c2 u2 )

d µ d⌧ (⌘ ⌘µ )

d µ d⌧ (m⌘ )

u2 c2 )

= ↵µ ⌘µ + ⌘ µ ↵µ = 2↵µ ⌘µ = 0, so ↵µ ⌘µ = 0.

= m↵µ .

)

K µ ⌘µ = m↵µ ⌘µ = 0.


273

CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY Problem 12.40 Kµ K µ = (K 0 )2 + K·K. From Eq. 12.69, K·K = 1 d 1 dE K = = p c d⌧ c 1 u2 /c2 dt 0

p 1

But Eq. 12.74: u·F = uF cos ✓ = p

mc2 u2 /c2 m

1

uF cos ✓ K = p . c 1 u2 /c2 0

u2 /c2

!

F2 (1 u2 /c2 ) .

=p 1

 (u·a) +

F2 ) Kµ K = (1 u2 /c2 ) µ

mc u2 /c2

From Eq. 12.70: 

1 ( 1/c2 ) (u·a) m 2u·a = 2 (1 u2 /c2 )3/2 c (1 u2 /c2 )2

u2 (u·a) m(u·a) = , c2 (1 u2 /c2 ) (1 u2 /c2 )3/2

so:

 u2 F 2 cos2 ✓ 1 (u2 /c2 ) cos2 ✓ = F 2 . qed c2 (1 u2 /c2 ) (1 u2 /c2 )

Problem 12.41

 u(u·a) u(u·a) qp F= p a+ 2 = q(E + u⇥B) ) a + = 1 u2 /c2 (E + u⇥B). c u2 (c2 u2 ) m 1 u2 /c2 ⇥ ⇤ u2 (u·a) u·a qp = = Dot in u: (u·a) + 2 1 u2 /c2 u·E + u·(u⇥B) ; 2 2 2 2 | {z } c (1 u /c ) (1 u /c ) m =0 r r u(u·a) q u2 u(u·E) q u2 1 ) 2 = 1 . So a = 1 E + u⇥B u(u·E) . qed (c u2 ) m c2 c2 m c2 c2 m

Problem 12.42 One way to see it is to look back at the general formula for E (Eq. 10.36). For a uniform infinite plane of charge, moving at constant velocity in the plane, J˙ = 0 and ⇢˙ = 0, while ⇢ (or rather, ) is independent of t (so retardation does nothing). Therefore the field is exactly the same as it would be for a plane at rest (except that itself is altered by Lorentz contraction). A more elegant argument exploits the fact that E is a vector (whereas B is a pseudovector). This means that any given component changes sign if the configuration is reflected in a plane perpendicular to that direction. But in Fig. 12.35(b), if we reflect in the x y plane the configuration is unaltered, so the z component of E would have to stay the same. Therefore it must in fact be zero. (By contrast, if you reflect in a plane perpendicular to the y direction the charges trade places, so it is perfectly appropriate that the y component of E should reverse its sign.) Problem 12.43 (a) Field is 0 /✏0 , and it points perpendicular to the positive plate, so: E0 =

(b) From Eq. 12.109, Ex = Ex0 =

0

✏0 p

(cos 45 x ˆ + sin 45 y ˆ) = p 0

2 ✏0

; Ey = E y0 =

(c) From Prob. 12.10: tan ✓ = , so ✓ = tan

1

p

0

2 E0

0

2 ✏0

( x ˆ+y ˆ).

. So E = p

.

(d) Let n ˆ be a unit vector perpendicular p to the plates in S — evidently n ˆ = sin ✓ x ˆ + cos ✓ y ˆ; |E| = p20✏ 1 + 2. 0


0

2 ✏0

( x ˆ+ y ˆ).

y

B

A

274

CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY y✻ 3

ˆ ❨n θ

So the angle

between n ˆ and E is:

E·ˆ n = cos |E| But

= tan ✓ =

✲x

sin ✓ cos ✓

=

p

1 =p 1+ 1 cos2 ✓ cos ✓

=

2

q

cos ✓ (sin ✓ + cos ✓) = p 1+

1 cos2 ✓

1)

1 2 cos ✓

=

2

2

2 (tan ✓ + ) = p 1+

+ 1 ) cos ✓ = p

Evidently the field is not perpendicular to the plates in S.

1 1+

2

2

cos ✓

. So cos

=

✓

2 1+

2

◆

.

y 12.44 Problem y ✻v ˆ s x x ˆ + y✻ ˆ 0 0y ✲ (a) E = = . 2 2 qB 2⇡✏0 s 2⇡✏0 (x0 +qBy0 ) d d x0 y0 (x0 x ˆ + y0 y ˆ ✲x ¯ = ¯ ¯y = ✲ ¯z = Ez = 0, E ✛, E (b)qE = E = Eyx , E . ¯= x x 2 2 2 2 2 2 A q 2⇡✏0 (x0 + y0 v) A 2⇡✏0 (x0 + y0 ) 2⇡✏0 (x0 + y0 ) Using the inverse Lorentz transformations (Eq. 12.19), x0 = (x + vt), y0 = y,

(x + vt) x ˆ+ yy ˆ 1 y (x ˆ+yy ˆ ✻ + vt) x = . 2 2 2 2 2 2⇡✏0 [ (x + vt)3 + y ] 2⇡✏0 [(x + vt) + y / 2] n ˆ ❨ θ 2 2 2 Now S = (x+vt) x ˆ +y y ˆ, and y = S sin ✓, so [(x+vt) +y / ] = [(x+vt)2 +y 2 (1✲ vx2 /c2 ] = S 2 (v/c)2 S 2 sin2 ✓ = 2 2 2 S [1 (v/c) sin ✓], so p ˆ 1 (v/c)2 S ¯ E= . 2 2 2 2⇡✏0 1 v sin ✓/c S y¯ ✻ This is reminiscent of Eq. 10.75. Yes, the field does point away fromY¯the present location of the wire. ❑ y✻ ¯ Problem 12.45 ✯X Y❑ y ✻v 1 qA qB 1 qA ✲ (a) Fields of A at B: E = 4⇡✏ ˆ; B = 0. So force on qB is F = y ˆ. ✯ X 2 y 2 0 d φ qB✲ 4⇡✏0 d ❄ ¯ ✯ dx ✲x vt y qA φ ✻ ✲x qA qB ❑ ¯ ¯ (b) (i) From Eq. 12.67: F = y ˆ . (Note: here the particle is at rest in S.) qB 4⇡✏0 d2 d 2 2 ✛ ✲x qA ¯ ¯ = 1 qA (1 v /c ) 1 y (ii) From Eq. 12.93, with ✓ = 90 : E ˆ= y ˆ v qA 2 4⇡✏0 (1 v /c2 )3/2 d2 4⇡✏0 d2 (this also follows from Eq. 12.109). qA qB ¯ there is no magnetic force anyway, and F ¯ 6= 0, but since vB = 0 in S, ¯= B y ˆ (as before). 4⇡✏0 d2 ¯ = E

✻v ✲ d

✲x


c ⃝2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is ¯this material may be protected under all copyright laws as they currently exist. No portion of y ✻ the publisher. reproduced, in any form or by any means, without permission in writing from

Y❑

y✻

Y¯❑

¯ ✯X ✯X φ

q

✛ v

275

CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY Problem 12.46 System A: Use Eqs. 12.93 and 12.112, with ✓ = 90 , R = d y ˆ, and ˆ = ˆ z: q y ˆ; 4⇡✏0 d2

E=

q v ˆ z; 4⇡✏0 c2 d2

B= 2

2

where 2

=p 1

1 v 2 /c2

.

2 [Note that (E 2 B 2 c2 ) = 4⇡✏q0 d2 1 vc2 = 4⇡✏q0 d2 is invariant, since it doesn’t depend on v (see Prob. 12.47b for the general proof). We’ll use this as a check.]

F = q E + ( vˆ x)⇥B = System B : The speed of

B

1

=q 1

⇥ Check : (E 2 F = qE =

q is

4v 2 /c2

(1+v 2 /c2 )2

)E= B 2 c2 ) =

vB =

q2 y ˆ 4⇡✏0 d2

v2 (ˆ x⇥ˆ z) = c2

v+v 2v = 1 + v 2 /c2 (1 + v 2 /c2 )

(1 + v 2 /c2 ) (1 + v 2 /c2 ) =q = = 2 4 (1 v 2 /c2 ) 1 2 vc2 + vc4

q 1 4⇡✏0 d2

2

2 4 q 4⇡✏0 d2

q2 v2 1 + y ˆ. 4⇡✏0 d2 c2

1+

1+

v2 y ˆ; c2 2v 2 c2

+

B=

v4 c4

4v 2 c2

2

1+

v2 ; vB c2

B

= 2v

2

.

q 2v 2 ˆ z. 4⇡✏0 c2 d2 =

2 4 1 q 4 4⇡✏0 d2

=

2 q 4⇡✏0 d2

X

⇤

q2 2 v2 1+ 2 y ˆ (+q at rest ) no magnetic force). [Check : Eq. 12.67 ) FA = 1 FB . X] 2 4⇡✏0 d c

q 1 q2 1 y ˆ ; B = 0. F = qE = y ˆ. 4⇡✏0 d2 4⇡✏0 d2 [The relative velocity of B and C is 2v/(1 + v 2 /c2 ), and corresponding is 2 (1 + v 2 /c2 ). So Eq. 12.67 1 ) FC = 2 (1+v X] 2 /c2 ) FB . Summary: ✓ ◆ ✓ ◆ ✓ ◆ q q q 2 2 2 y ˆ 1 + v /c y ˆ y ˆ 2 2 4⇡✏ 4⇡✏0 d2 ✓ 4⇡✏0 d ◆ ✓ 0d ◆ q v q 2v 2 ˆ z ˆ z 0 2 2 2 4⇡✏ d c 4⇡✏ d c2 0 0 ✓ ◆ ✓ ◆ ✓ ◆ q2 q2 q2 2 2 2 2 2 1 + v /c y ˆ 1 + v /c y ˆ y ˆ 4⇡✏0 d2 4⇡✏0 d2 4⇡✏0 d2 System C :

vC = 0.

E=

Problem 12.47 (a) From Eq. 12.109: ¯x B ¯x + E ¯y B ¯y + E ¯z B ¯z = Ex Bx + ¯ B ¯ =E E·

2

(Ey

v {Ey By + 2 Ey Ez vBy Bz c ✓ ⇣ ⇣ v2 ⌘ 2 = Ex Bx + Ey By 1 + E B 1 z z c2 = Ex Bx +

2

v Ez ) + (Ez + vBy )(Bz c2 v2 v Ez Bz + Ez Bz Ey Ez + vBy Bz c2 c2 ◆ v2 ⌘ = Ex Bx + Ey By + Ez Bz = E·B. c2 vBz )(By +


v Ey ) c2 v2 Ey By } c2 qed

276 (b)

¯2 E

⇥ ¯ 2 = Ex2 + c2 B

2

= Ex2 + c2 =

2

(Ey

Ex2

2

c

vBz )2 +

c2 Bz2 + c2 2

Bx2

2

(Ez + vBy )2

⇤

⇥ c2 Bx2 +

By +

2

2Ey vBz + v 2 Bz2 + Ez2 + 2Ez vBy + v 2 By2

Ey2

v2 2 E c4 z


+

2

✓

Ey2

= (Ex2 + Ey2 + Ez2 )

⇣

v Bz Ey c2

1

c2

v2 2 E c4 y

⇣ v2 ⌘ 2 + E 1 z c2

v Ez c2

c2 By2

2

+

c2 2

2

⇤ v Ey 2 c

Bz

v By Ez c2

c2 Bx2

v2 ⌘ c2

c2 (Bx2 + By2 + Bz2 ) = E 2

2

c

⇣

(By2 )

1

v2 ⌘ c2

2

c

Bz2

⇣

v2 ⌘ c2

1

◆

B 2 c2 . qed

(c) No. For if B = 0 in one system, then (E 2 c2 B 2 ) is positive. Since it is invariant, it must be positive in any system. ) E 6= 0 in all systems. Problem 12.48 (a) Making the appropriate modifications in Eq. 9.48 (and picking = 0 for convenience), E(x, y, z, t) = E0 cos(kx

!t) y ˆ,

B(x, y, z, t) =

E0 cos(kx c

!t) ˆ z,

where k ⌘

! . c

(b) Using Eq. 12.109 to transform the fields: ¯x = E ¯z = 0, E ¯x = B ¯y = 0, B where

↵⌘

⇣

1

¯y = (Ey E ¯z = (Bz B

v⌘ = c

s

h vBz ) = E0 cos(kx

 v 1 Ey ) = E 0 cos(kx c2 c

!t) !t)

v cos(kx c v cos(kx c2

i !t) = ↵E0 cos(kx !t) = ↵

1 v/c . 1 + v/c

Now the inverse Lorentz transformations (Eq. 12.19) ) x = (¯ x + v t¯) and t =

!t),

v ⌘ t¯ + 2 x ¯ , so c

⇣ h⇣ i v ⌘i !v ⌘ ¯x ! ¯ = k¯ ! t¯ + 2 x ¯ = k x ¯ (! kv) t ¯ t¯, c c2 ⇣ !v ⌘ where, recalling that k = !/c): k¯ ⌘ k = k(1 v/c) = ↵k and ! ¯ ⌘ !(1 v/c) = ↵!. c2 ¯ ¯x ! ¯x ! ¯ x, y¯, z¯, t¯) = E ¯ x, y¯, z¯, t¯) = E0 cos(k¯ ¯0 cos(k¯ E(¯ ¯ t¯) y ˆ, B(¯ ¯ t¯) ˆ z, c s Conclusion: 1 v/c ¯0 = ↵E0 , k¯ = ↵k, ! where E ¯ = ↵!, and ↵ ⌘ . 1 + v/c s 1 v/c ¯ = 2⇡ = 2⇡ = . The velocity of the wave (c) ! ¯=! . This is the Doppler shift for light. 1 + v/c ↵k ↵ k¯ ! ¯ ¯ ! in S¯ is v¯ = = = c. Yup, this is exactly what I expected (the velocity of a light wave is the same 2⇡ 2⇡ in any inertial system). kx

!t =

h

⇣

E0 cos(kx c

!t),

k(¯ x + v t¯)


277


¯2 I¯ E 1 v/c = 02 = ↵2 = . I E0 1 + v/c

(d) Since the intensity goes like E 2 , the ratio is

Dear Al, The amplitude, frequency, and intensity of the light will all decrease to zero as you run faster and faster. It’ll get so faint you won’t be able to see it, and so red-shifted even your night-vision goggles won’t help. But it’ll still be going 3 ⇥ 108 m/s relative to you. Sorry about that. Sincerely, David Problem 12.49 t¯02 = ⇤0 ⇤2 t t¯03 = ⇤0 ⇤3 t t¯23 = ⇤2 ⇤3 t t¯31 = ⇤3 ⇤1 t t¯12 = ⇤1 ⇤2 t

= ⇤00 ⇤22 t02 + ⇤01 ⇤22 t12 = ⇤00 ⇤33 t03 + ⇤01 ⇤33 t13 = ⇤22 ⇤33 t23 = t23 . = ⇤33 ⇤10 t30 + ⇤33 ⇤11 t31 = ⇤10 ⇤22 t02 + ⇤11 ⇤22 t12

= t02 + ( = t03 + ( =( =(

Problem 12.50 Suppose t⌫µ = ±tµ⌫ (+ for symmetric, t¯ = ⇤µ ⇤⌫ tµ⌫ t¯  = ⇤µ ⇤⌫ tµ⌫ = ⇤⌫ ⇤µ t⌫µ

)t12 = (t02 )t13 = (t03

t12 ). t13 ) = (t03 + t31 ).

)t30 + t31 = (t31 )t02 + t12 = (t12

t03 ). t02 ).

for antisymmetric).

[Because µ and ⌫ are both summed from 0 ! 3,

it doesn’t matter which we call µ and and which call ⌫.]

=

⇤µ ⇤µ (±tµ⌫ ) ¯

= ±t .

[Using symmetry of tµ⌫ , and writing the ⇤’s in the other order.]

qed

Problem 12.51 F µ⌫ Fµ⌫ = F 00 F 00

F 01 F 01

F 02 F 02

F 03 F 03

F 10 F 10

F 20 F 20

F 30 F 30

+ F 11 F 11 + F 12 F 12 + F 13 F 13 + F 21 F 21 + F 22 F 22 + F 23 F 23 + F 31 F 31 + F 32 F 32 + F 33 F 33 =

(Ex /c)2

= 2B 2

(Ey /c)2

⇣ 2E 2 /c2 = 2 B 2

which, apart from the constant factor Gµ⌫ Gµ⌫ = 2(E 2 /c2 F µ⌫ Gµ⌫ = = =

(Ez /c)2

(Ex /c)2

(Ey /c)2

(Ez /c)2 + Bz2 + By2 + Bz2 + Bx2 + By2 + Bx2

E2 ⌘ , c2 2/c2 , is the invariant we found in Prob. 12.47(b).

B 2 ) (the same invariant).

2(F 01 G01 + F 02 G02 + F 03 G03 ) + 2(F 12 G12 + F 13 G13 + F 23 G23 ) ✓ ◆ 1 1 1 2 Ex Bx + Ey By + Ez Bz + 2[Bz ( Ez /c) + ( By )(Ey /c) + Bx ( Ex /c)] c c c 2 (E · B) c

2 (E · B) = c

4 (E · B), c


278


which, apart from the factor 4/c, is the invariant we found of Prob. 12.47(a). [These are, incidentally, the only fundamental invariants you can construct from E and B.] Problem 12.52 E=

1 2 4⇡✏0 x

x ˆ=

B=

µ0 2 v ˆ 4⇡ x y

=

9

µ0 c2 = ˆ 2⇡ x x

µ0 v ˆ 2⇡ x y

;

F µ⌫

0

0 µ0 B B c = 2⇡x @ 0 0

c 0 0 v

1 0 0 0 vC C 0 0A 0 0

Gµ⌫

0

0 µ0 B B0 = 2⇡x @ v 0

0 0 0 0

1 v 0 0 0C C 0 cA c 0

Problem 12.53 @⌫ F µ⌫ = µ0 J µ . Di↵erentiate: @µ @⌫ F µ⌫ = µ0 @µ J µ . But @µ @⌫ = @⌫ @µ (the combination is symmetric) while F ⌫µ = F µ⌫ (anti symmetric). ) @µ @⌫ F µ⌫ = 0. [Why? Well, these indices are both summed from 0 ! 3, so it doesn’t matter which we call µ, which ⌫: @µ @⌫ F µ⌫ = @⌫ @µ F ⌫µ = @µ @⌫ ( F µ⌫ ) = @µ @⌫ F µ⌫ . But if a quantity is equal to minus itself, it must be zero.] Conclusion: @µ J µ = 0. qed Problem 12.54 We know that @⌫ Gµ⌫ = 0 is equivalent to the two homogeneous Maxwell equations, r·B = 0 and r⇥E = @B @t . All we have to show, then, is that @ Fµ⌫ + @µ F⌫ + @⌫ F µ = 0 is also equivalent to them. Now this equation stands for 64 separate equations (µ = 0 ! 3, ⌫ = 0 ! 3, = 0 ! 3, and 4 ⇥ 4 ⇥ 4 = 64). But many of them are redundant, or trivial. Suppose two indices are the same (say, µ = ⌫). Then @ Fµµ + @µ Fµ = @µ F µ = 0. But Fµµ = 0 and Fµ = F µ , so this is trivial: 0 = 0. To get anything significant, then, µ, ⌫, must all be di↵erent. They could beall spatial (µ, ⌫, = 1, 2, 3 = x, y, z — or some permutation thereof), or one temporal and two spatial (µ = 0, ⌫, = 1, 2 or 2, 3, or 1, 3 — or some permutation). Let’s examine these two cases separately. All spatial : say, µ = 1, ⌫ = 2,

= 3 (other permutations yield the same equation, or minus it.)

@3 F12 + @1 F23 + @2 F31 = 0 ) One temporal : say, µ = 0, ⌫ = 1,

@ @ @ (Bz ) + (Bx ) + (By ) = 0 ) r·B = 0. @z @x @y

= 2 (other permutations of these indices yield same result, or minus it).

@2 F01 + @0 F12 + @1 F31 = 0 ) @E

@ @y

Ex @ @ Ey = (Bz ) + + = 0. c @(ct) @x c

@Ex y z or: @B = 0, which is the z-component of @t + @y @x y component; for ⌫ = 2, = 3 we get the x component.)

Conclusion: @ Fµ⌫ + @µ F⌫ + @⌫ F @⌫ Gµ⌫ = 0. qed

µ

@B @t

= r⇥E. (If µ = 0, ⌫ = 1,

= 0 is equivalent to r·B = 0 and

@B @t

=

= 2, we get the

r⇥E, and hence to

Problem 12.55

q u·E. Now from Eq. 12.70 we know that c 1 K 0 = 1c dW d⌧ , where W is the energy of the particle. Since d⌧ = dt, we have: K 0 = q⌘⌫ F 0⌫

q(⌘1 F 01 + ⌘2 F 02 + ⌘3 F 03 ) = q(⌘·E)/c =

1 dW q dW = (u·E) ) = q(u·E) c dt c dt This says the power delivered to the particle is force (qE) times velocity (u) — which is as it should be.


279

CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY Problem 12.56 @ @0 = = @x ¯0

1 @ 1 = c @ t¯ c @t From Eq. 12.19, we have: = @ t¯ 1 @ @ So @ 0 = +v or c @t @x

@ @t @ @x @ @y @ @z + + + @t @ t¯ @x @ t¯ @y @ t¯ @z @ t¯ ,

@x @ t¯

= v,

@y @ t¯

=

(since ct = x0 =

@z @ t¯

= 0.

x0 ): @ 0 =

@ @x0

@1 =

@ @ @t @ @x @ @y @ @z v @ @ = + + + = 2 + = 1 @x ¯ @t @x @x @ x ¯ @y @ x ¯ @z @ x ¯ c @t @x

@2 =

@ @ @t @ @x @ @y @ @z @ = + + + = = @2 . @ y¯ @t @ y¯ @x @ y¯ @y @ y¯ @z @ y¯ @y

@3 =

@ @ @t @ @x @ @y @ @z @ = + + + = = @3 . @ z¯ @t @ z¯ @x @ z¯ @y @ z¯ @z @ z¯ @z

v @ = c @x1 @ @x1

⇥ 0 (@ )

v @ = c @x0

⇤ (@ 1 ) .

⇥ 1 (@ )

⇤ (@ 0 ) .

Conclusion: @ µ transforms in the same way as aµ (Eq. 12.27)—and hence is a contravariant 4-vector. qed Problem 12.57 According to Prob. 12.54, of Prob. 12.56):

@Gµ⌫ @x⌫

= 0 is equivalent to Eq. 12.130. Using Eq. 12.133, we find (in the notation

@Fµ⌫ @F⌫ @F µ + + = @ Fµ⌫ + @µ F⌫ + @⌫ F µ @x @xµ @x⌫ = @ (@µ A⌫ @⌫ Aµ ) + @µ (@⌫ A @ A⌫ ) + @⌫ (@ Aµ @µ A ) = (@ @µ A⌫ @µ @ A⌫ ) + (@µ @⌫ A @⌫ @µ A ) + (@⌫ @ Aµ @ @⌫ Aµ ) = 0. qed [Note that @ @µA⌫ =

@ 2 A⌫ @x @x⌫

=

@ 2 A⌫ @x⌫ @x

Problem 12.58 From Eqs. 12.40 and 12.42, ⌘ µ = cr + v · r = ( r c r · v).

= @⌫ @ A⌫ , by equality of cross-derivatives.] (c, v), while

r

µ

= (ct

ctr , r

w(tr )) = ( r ,

q ⌘0 q c 1 qc 1 = = = V ⌫ 4⇡✏0 c (⌘ r ⌫ ) 4⇡✏0 c ( r c r · v) 4⇡✏0 c ( r c r · v) c (Eq. 10.46),

q ⌘ q v 1 qv = = =A ⌫ 4⇡✏0 c (⌘ r ⌫ ) 4⇡✏0 c ( r c r · v) 4⇡✏0 c ( r c r · v)

(Eq. 10.47), so (Eq. 12.132)

q ⌘µ = Aµ . X ⌫ 4⇡✏0 c (⌘ r ⌫ )


r

), so ⌘ ⌫ r

⌫

=

✛ v qA

qA

280

✲x ¯


Problem 12.59

y¯

Step 1 : rotate from xy to XY , using Eq. 1.29:

y✻

Y❑

X = cos x + sin y Y = sin x + cos y

✻

Y¯❑

¯ ✯X ✯X φ ❄

✯

¯ Y¯ , using Step 2 : Lorentz-transform from XY to X Eq. 12.18:

✲x ¯

vt φ ❑

✲x

¯ = (X vt) = [cos x + sin y X ct] ¯ Y = Y = sin x + cos y Z¯ = Z = z ⇥ ⇤ ct¯ = (ct X) = ct (cos x + sin y)

¯ Y¯ to x Step 3 : Rotate from X ¯y¯, using Eq. 1.29 with negative : ¯ x ¯ = cos X

sin Y¯ =

cos [cos x + sin y

sin [ sin x + cos y]1 cos (ct) ct) + cos ( sin c + cos y) ct]

= ( cos + sin )x + ( 1) sin cos y ¯ ¯ y¯ = sin X + cos Y = sin (cos x + sin y 2

=(

2

1) sin cos x + ( sin2

+ cos2 )y

sin (ct)

sin 40

(

1) sin cos 0

( sin2

Problem 12.60 In center-of-momentum system, threshold occurs when incident energy is just sufficient to cover the rest energy of the resulting particles, with none “wasted” as kinetic energy. Thus, in lab system, we want the outgoing K and ⌃ to have the same velocity, at threshold: ✲ π

p ✲

0Inc., 1Upper 0Saddle River, NJ. All rights reserved. This material is c ⃝2005 Pearson Education, cos of this material may sin be ct¯laws as they currently exist. No portion protected under all copyright reproduced, in any form or any means, writing Bby B without x ¯C cos permission ( cos2 in+ sin2 from ) ( the publisher. 1) sin cos C B In matrix form: B @ y¯ A = @ z¯

π

+ cos2 0

1 0 0C C ) 0A 1

before (CM)

✛

0 1 ct K Σ BxC B C @yA z π

✲

✛

p

K Σ

after (CM)

before (CM) after (CM)

✲ K Σ

p

After

Before

Initial momentum: p⇡ ; Initial energy of ⇡: E p2 c2 = m2 c4 ) E⇡2 = m2⇡ c4 + p2⇡ c2 . p Total initial energy: mp c2 = m2⇡ c4 + p2⇡ c2 . These are also the final energy and momentum: E 2 (mK + m⌃ )2 c4 . 2

⇣

mp c2 +

p

m2⇡ c4 + p2⇡ c2

⌘2

p2 c2 =

p2⇡ c2 = (mK + m⌃ )2 c4

p m2p c4 + 2mp c2 m2⇡ c2 + p2⇡ c + m2⇡ c4 + p2⇡ c2 p2⇡ c2 = (mK + m⌃ )2 c4 c4 2mp p 2 2 m⇡ c + p2⇡ = (mK + m⌃ )2 m2p m2⇡ c ✲ ✲ π p Σ c 2012 PearsonK Education, Inc., Upper Saddle River, NJ. All rights reserved.

This material is protected under all copyright laws as they currently exist. No portion of this material may be After Before reproduced, in any form or by any means, without permission in writing from the publisher.

✲ ✲ π p CHAPTER 12. ELECTRODYNAMICS AND K Σ RELATIVITY After

Before

(m2⇡ c2 + p2⇡ )

4m2p = (mK + m⌃ )4 c2

4m2p 2 p = (mK + m⌃ )4 c2 ⇡ p⇡ =

281

c q (mK + m⌃ )4 2mp

2(m2p + m2⇡ )(mK + m⌃ )2 + m4p + m4⇡ + 2m2p m2⇡

2(m2p + m2⇡ )(mK + m⌃ )2 + (m2p

2(m2p + m2⇡ )(mK + m⌃ )2 + (m2p

m2⇡ )2

m2⇡ )2

q (mK c2 + m⌃ c2 )4 2 (mp c2 )2 + (m⇡ c2 )2 (mK c2 + m⌃ c2 )2 + (mp c2 )2 (m⇡ c2 )2 (2mp q 2 1 = 2c(900) (1700)4 2 (900)2 + (150)2 (1700)2 + (900)2 (150)2 p 1 1 (8.35 ⇥ 1012 ) (4.81 ⇥ 1012 ) + (0.62 ⇥ 1012 ) = 1800c (2.04 ⇥ 106 ) = 1133 MeV/c = 1800c =

1

2

c2 )c

Problem 12.61

p ✲

In CM :

rµ ❘ p ✒ φ

p ✛

y ✻ ✲x

(p = magnitude of 3-momentum in CM, φ = CM scattering angle)

p✠ ■ sµ After

Before

5

Outgoing 4-momentua: rµ = ✲

In Lab:

E E µ c , p cos , p sin , 0 ; s = c , p cos , p sin , 0 . µ Upper Saddle River, NJ. All rights reserved. This material is c ⃝2005 Pearson Education, r ¯Inc., protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form ✒ or by any means, without permission in writing from the publisher.

θ ❘

Before

Lorentz transformation: r¯x = (rx Now E = mc2 ; p = ) r¯x =

p cos +

cos ✓ =

¯ r·¯ s = q⇥ r¯s¯

s¯µ

r0 ); r¯y = ry ; s¯x = (sx

mv (v here is to the left; E 2

pc E E c

Problem: calculate θ, in terms of p, φ.

s0 ); s¯y = sy .

p2 c2 = m2 c4 , so

= p(1 + cos ); r¯y = p sin ; s¯x = p(1

pc E.

=

cos ); s¯y =

p sin .

p (1

cos2 ) p2 sin2 ⇤⇥ ⇤ 2 p2 (1 + cos )2 + p2 sin2 2 p2 (1 cos )2 + p2 sin2 2 2

2 ( 2 1) sin2 φ) = q⇥ / sin ⇤⇥ 2 (1 (τ2 τ / sin )2 φ+ sin2 √ 1+ (1 + cos θ ( 2 1) = rh 1 ih 2 2 1+cos 2 1 cos + 1 sin sin

cos )2 + sin2 i=q 2 +1

2

⇤

cot

( 2 2

2

1)

+ 1 tan2

2

+1


d/2

y ✻

l +q!"#$ ✕

282

cos ✓ = q =q =q

sin ✓ =

⌧ sin

Or, since (

=q


1 + cot2

2

+ ! cot2

1 + tan2

2

2

+ ! tan2

! csc2

2

+ ! cot

sec2

2 2

+ ! tan

2

2

2

1 2 ! sin

1+

! 12 (1

+ cos ) 1 +

sin 2 !

+1

.

⌧2 =

2

1) =

2

cos2

4 ! 2 (1 2

⇣

1

+ !) = 1

2

⌘

=q (

=

2

4

! 12 (1 sin

4 !2

1)2

2 v2 c2 ,

+ 2

4 !

+ sin2

, so tan ✓ =

tan ✓ =

! sin 1+

= q⇥

2

! cos2

2 !

=q 1+ (

Before

2

=q

cos )

✲

In Lab: (where ! ⌘ 2 1)

!

cos

2

2

1 + ! sin2

sin ⇤⇥ + 1 + cos

1

2 1) sin

2

⌧2 sin2

2 !

2

+1

, where ⌧ 2 =

cos

4 4 + . 2 ! !

.

2c2 2 v sin

⇤

2

√

1+

(τ

θ

in /s

φ)

τ / sin φ

1

Problem 12.62 dp dp dt dt p 1 2 2 ; p = p mu2 2 . d⌧ = K (a constant) ) dt d⌧ = K. But d⌧ = 1 u /c 1 u /c ⇣ ⌘ p d p u K 2 /c2 . Multiply by dt = 1 : ) dt = 1 u m dx u 2 2 1 u /c

p ✓ ◆ ✓ ◆ dt d u d u K 1 u2 /c2 u p p = = . Let w = p . dx dt dx m u 1 u2 /c2 1 u2 /c2 1 u2 /c2 dw K 1 = ; dx mw

) w2 =

Let

2K m x+

w

dw 1 d 2 k = w = ; dx 2 dx m

d(w2 ) 2K 2K = ) d(w2 ) = (dx). dx m m

constant. But at t = 0, x = 0 and u = 0 (so w = 0), and hence the constant is 0.

w2 =

2K x= m 1

u2 =

2Kx/m c2 = ; mc2 1 + 2Kx 1 + 2Kx mc2

u2 ; u2 /c2

u2 =

2Kx m

2Kx 2 u ; mc2

dx c =q dt 1+

mc2 2Kx

2Kx 2Kx = . 2 mc m Z r mc2 ct = 1+ dx 2Kx

u2 1 + ;

R px+a2 p p = a2 ; ct = dx. Let x = y 2 ; dx = 2y dy; x = y. x Z p 2 Z h p i p p y + a2 ct = 2y dy = 2 y 2 + a2 dy = y y 2 + a2 + a2 ln(y + y 2 + a2 ) + constant. y At t = 0, x = 0 ) y = 0. ) 0 = a2 ln a+ constant, so constant = a2 ln a. r⇣ ⌘ ⇣ ⌘ r⇣ ⌘ ✓ ◆ p p 2 y y y y 2 2 2 ) ct = y y 2 + a2 = a ln y/a + (y/a)2 + 1 = a + 1 + ln + +1 a a a a mc2 2K

Let: z =

y a

=

q p p p q 2K 2Kt 2 + ln(z + x mc2 = 2Kx = z. Then = z 1 + z 1 + z 2 ). 2 mc mc

c ⃝2005 Pearson Education, Inc., Upper Saddle River, NJ protected under all copyright laws as they currently exist reproduced, any formThis or by any means, without permi c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rightsinreserved. material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

283

CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY Problem 12.63 hp i F (a) x(t) = ↵c 1 + (↵t)2 1 , where ↵ = mc . The force of +q on q will be the mirror image of the force of q on +q (in the x-axis), so the net force is in the x direction (the net magnetic force is zero). So all we need is the x-component of E. The field at +q due to q is: (Eq. 10.72) r ⇥u(c2 v2 ) + u( r ·a) q E= 4⇡✏0 ( r ·u)3 u = cr

v ) ux = c rl

Ex =

=

q 4⇡✏0 (c r

v=

r vl)3

1

r

⇥ 1

r

⇥ q 1 (cl 3 4⇡✏0 (c r vl)

(cl

(cl

v r );

r

v r )(c2

v r )(c2

v2 )

·u = c r v2 ) =

1

r

d/2

y ✻

l +q!"#$

−d/2

⇤ a( r ·u) .

·v = (c r

lv);

r

✲x

−q

·a = la. So: ⇤

(cl v r )la a(c r lv) |r {z } 1 2 2 2 r ca(l r ) = cad / r

⇤ cad2 .

The force on +q is qEx , and there is an equal force on

q, so the net force on the dipole is:

c ⃝2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is 2 protected under all2q copyright laws No portion of⇤this material may beto determine ⇥ currently exist. 1 as they It remains reproduced, F in = any form or by any means, (clwithout v permission )(c2 v 2 )in writing cad2 from x ˆ the publisher. 3 v, and a, and plug these 4⇡✏0 (c lv)

r

r

✕r

d

r

, l, in.

p dx c1 1 c↵t c↵tr p = 2↵2 t = p ; v = v(tr ) = , where T = 1 + (↵tr )2 . 2 2 dt ↵ 2 1 + (↵t) T 1 (↵t) ✓ ◆ 2 dv c↵ 1 2↵ tr c↵ c↵ a(tr ) = = + c↵tr = 3 1 + (↵tr )2 (↵tr )2 = 3 dtr T 2 T3 T T p ⇥p ⇤ Now calculate tr : c2 (t tr )2 = r 2 = l2 + d2 ; l = x(t) x(tr ) = ↵c 1 + (↵t)2 1 + (↵tr )2 , so p p ⇥ ⇤ t2 2ttr + t2r = ↵12 1 + (↵t)2 + 1 + (↵tr )2 2 1 + (↵t)2 1 + (↵tr )2 + (d/c)2 p p 2 (F) 1 + (↵t)2 1 + (↵tr )2 = 1 + ↵2 ttr + 12 ↵d . Square both sides: c v(t) =

1 + (↵t)2 + (↵tr )2 + ↵4 t2 t2r = 1 + ↵4 t2 t2r + t2 + t2r

2ttr

ttr

⇣ ↵d ⌘2 c

⇣ d ⌘2 c

⇣ ↵d ⌘2 ⇣ ↵d ⌘2 1 ⇣ ↵d ⌘4 + 2↵2 ttr + + ↵2 ttr 4 c c c

↵ 2 ⇣ d ⌘4 =0 4 c

At this point we could solve for tr (in terms of t), but since v and a are already expressed in terms of tr , it is simpler to solve for t (in terms of tr ), and express everything in terms of tr :   ⇣ ↵d ⌘2 ⇣ d ⌘2 ↵ 2 ⇣ d ⌘4 t2 ttr 2 + + t2r = 0 =) c c 4 c s  (  ) ⇣ ↵d ⌘2 ⇣ ↵d ⌘2 ⇣ ↵d ⌘4 ⇣ d ⌘2 ⇣ d ⌘4 1 t= tr 2 + ± t2r 4 + 4 + 4t2r + 4 + ↵2 2 c c c c c s   ⇣ ↵d ⌘ ⇥ ⇤⇣ d ⌘2 1 ⇣ ↵d i 2 = tr 1 + ) ± 1 + (↵tr )2 1+ 2 c c 2c c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

284


Which sign? For small ↵ we want t ⇡ tr + d/c, so we need the + sign: r  ⇣ ↵d ⌘2 1 ⇣ ↵d ⌘2 d t = tr 1 + + T D, where D = 1 + 2 c c 2c So

r

p 2 = ct2r ↵d + dT D. Now go back to Eq. (F) and solve for 1 + (↵t)2 : c ⇢  ⇣ p 1 1 ⇣ ↵d ⌘2 1 ⇣ ↵d ⌘2 d 2 2 1 + (↵t) = 1+ + ↵ tr t r 1 + + TD T 2 c 2 c c ⇢   ⇣ ⌘ 2 ⇤ 1 ⇥ 1 ↵d 2 ↵ tr d 1 ⇣ ↵d ⌘2 ↵ 2 tr d 2 = 1 + (↵tr ) 1 + + TD = 1 + T+ D T | {z } 2 c c 2 c c T2 ⇢ i ⇣d ⌘ p c hp c 1 ⇣ ↵d ⌘2 ↵ 2 tr d l= 1 + (↵t2 ) 1 + (↵tr )2 = 1+ T+ D T = ↵d T + tr D ↵ ↵ 2 c c 2c = c(t

tr ) )

r

Putting all this in, the numerator in square brackets in F becomes: [

n ⇣d ⌘ c↵t h ct ⇣ ↵d ⌘2 ioh c2 ↵2 t2r i c↵ r r ] = c↵d T + tr D + dT D c2 c d2 2 2c T 2 c T T3 hd i c2 ⇥ 2 2 ⇤ d(atr )2 c ↵d = c↵d T + tr D tr D 1 + (↵tr )2 (↵tr )2 2c 2cT T2 T3 i i 2 2h 2 2h c ↵d 1 2 1 c ↵d c2 ↵d2 2 2 2 = T (↵t ) 1 = 1 + (↵t ) (↵t ) 2 = r r r T3 2 2 2T 3 2T 3

)F= (c r

q2 c2 ↵d2 ˆ. ⇥ ⇤3 x 4⇡✏0 (c r lv)T

It remains to compute the denominator:

⇢  ⇣ ⌘ ⇣d ⌘ c↵t ctr ↵d 2 r c + dT D ↵d T + tr D T 2 c 2c T h1 ⇥ ⇤ 1 2 2 cd(↵tr )2 i = ↵2 tr d2 + cdT D ↵ tr d D T = cdD T 2 (↵tr )2 = dcD | {z } 2 2 T

lv)T =

1+(↵tr )2

)F=

q 2 c2 d2 ↵ q2 ↵ x ˆ = 4⇡✏0 c3 d3 D3 4⇡✏0 cd⇥1 + ↵d 2c

x ˆ 2 ⇤3/2

⇣

↵=

(↵tr )2

F ⌘ mc

Energy must come from the “reservoir” of energy stored in the electromagnetic fields. h ⇣ ↵d ⌘2 i3/2 ⇣ µ q2 ⌘ 1 q2 ↵ q2 0 (b) F = mc↵ = ) 1+ = = . ⇥ ⇤ 2 2 3/2 2 4⇡✏0 cd 1 + ↵d 2c 8⇡✏0 mc d 8⇡md 2c

"

(force on one end only) r 2c ⇣ µ0 q 2 ⌘2/3 )↵= 1, d 8⇡md

so

2mc2 F = d

r ⇣ µ q 2 ⌘2/3 0 8⇡md

1.


285


Problem 12.64 (a) Aµ = (V /c, Ax , Ay , Az ) is a 4-vector (like xµ = (ct, x, y, z)), so (using Eq. 12.19): V = (V¯ + v A¯x ). But V¯ = 0, and µ0 (m⇥¯ r)x A¯x = 4⇡ r¯3 Now (m⇥¯ r)x = my z¯ mz y¯ = my z mz y. So V = v

µ0 (my z mz y) 4⇡ r¯3

Now x ¯ = (x vt) = Rx , y¯ = y = Ry , z¯ = z = Rz , where R is the vector (in S) from the (instantaneous) location of the dipole to the point of observation. Thus r¯2 =

2

Rx2 + Ry2 + Rz2 =

2

(Rx2 + Ry2 + Rz2 ) + (1

2

)(Ry2 + Rz2 ) =

2

R2

v2 2 2 R sin ✓ c2

(where ✓ is the angle between R and the x-axis, so that Ry2 + Rz2 = R2 sin2 ✓). )V =

µ0 v (my Rz mz Ry ) ; v·(m⇥R) = v(m⇥R)x = v(my Rz 4⇡ 3 R3 1 v22 sin2 ✓ 3/2 c 2

V =

or, using µ0 =

1 ✏0 c2

µ0 v·(m⇥R) 1 vc2 , 4⇡ R3 1 v22 sin2 ✓ 3/2 c

2 b R·(v⇥m) 1 vc2 1 and v·(m⇥R) = R·(v⇥m): V = 4⇡✏0 c2 R2 1 v22 sin2 ✓ 3/2 c

(b) In the nonrelativistic limit (v 2 ⌧ c2 ): V =

b b 1 R·(v⇥m) 1 R·p = , 2 2 4⇡✏0 c R 4⇡✏0 R2

which is the potential of an electric dipole. Problem 12.65 (a) B = µ20 Kˆ y (Eq. 5.58); N = m⇥B (Eq. 6.1), so N = µ0 N= mKˆ x = µ20 ( vl2 )( v)ˆ x = µ20 v 2 l2 x ˆ. 2

with p =

µ0 z⇥ˆ y). 2 mK(ˆ


v⇥m , c2

mz Ry ),

so

286


(b)

Charge density in the front side: 0 ( = 0 ); Charge density on the back side: ¯ = ¯ 0 , where v¯ = 1

so ¯ = q

4v 2 /c2 (1+v 2 /c2 )2

1

=q

(1 + v 2 /c2 ) 2

1 + 2 vc2 +

v4 c4

2

4 vc2

1 + v 2 /c2 =q 2 1 2 vc2 +

v4 c4

=

(1 + v 2 /c2 ) = (1 v 2 /c2 )

2v 1+v 2 /c2 , 2

⇣

1+

v2 ⌘ c2

Length of front and back sides in this frame: l/ . So net charge on back side is: l q+ = ¯ =

2

Net charge on front side is:

⇣

q =

v2 ⌘ c2

1+ l 0

l

l

=

=

⇣ v2 ⌘ = 1+ 2 l c 1 2

l

So dipole moment (note: charges on sides are equal): ⇣ l v2 ⌘ l (q ) y ˆ = 1+ 2 l 2 c 2

l p = (q+ ) y ˆ 2

1 2

l

l l2 ⇣ v2 y ˆ= 1+ 2 2 2 c

1+

v2 ⌘ y ˆ= c2

l2 v 2 y ˆ. c2

l2 v 2 1 µ0 (ˆ y⇥ˆ z) = l2 v 2 x ˆ. c2 2✏0 2 So apart from the relativistic factor of the torque is the same in both systems — but in S it is the torque exerted by a magnetic field on a magnetic dipole, whereas in S¯ it is the torque exerted by an electric field on an electric dipole. Problem 12.66 Choose axes so that E points in the z direction and B in the yz plane: E = (0, 0, E); B = (0, B cos , B sin ). Go to a frame moving at speed v in the x direction: E=

z, 2✏0 ˆ 0

where

=

0,

¯ = 0, E

so N = p⇥E =

¯ 0, (B cos + v E), B sin vB sin , (E + vB cos ) ; B c2

(I used Eq. 12.109.) Parallel provided

vB sin (E + vB cos ) = , v (B cos + c2 E) B sin

v E (E + vB cos ) = EB cos + vB 2 cos2 c2 ⇣ v v2 ⌘ v EB cos = 0 = vB 2 + 2 E 2 + EB cos 1 + 2 ; 2 2 c c 1 + v /c B 2 + E 2 /c2 vB 2 sin2

Now E⇥B =

= B cos +

x ˆ y ˆ 0 0 0 B cos

ˆ z E B sin

=

EB cos . So

+

E⇥B v = 2 . 1 + v 2 /c2 B + E 2 /c2

.

or v 2 v2 E + 2 EB cos c2 c

qed

No, there can be no frame in which E ? B, for (E·B) is invariant, and since it is not zero in S it can’t be ¯ zero in S.


v☛ v☛

✕v ✕v

287

CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY Problem 12.67 ❖ ✻ ✗✻ ✗ ❑ ❑ ❖ ✕ ✕ ✎☛ ❯❲ ❄ ❯✎☛❲ ❄ ✲✲ −q−q ✕✗ ✻ ✕❖❑✗ ✻ ❖❑

✛✛

+q+q

Just before: Field lines emanate from present position of particle.

✲✲ x x

☛ ☛ ❯ ✎ ✎❄ ❲❄ ❲ ❯ ❖ ✻✻ ▼ ▼❖ ✗ ✗ ✍ ◆✍❲ ◆❄ ❲✎ ❄ ✌ ✎✌

!!

E E t

c ✶ct✶ −−

++

E

✲✲ x x

E !!

❖▼✻❖▼ ◆ ✍◆✗ ✍✻ ✗ ✌ ✌ ❲ ❲ ✎ ❄ ✎ ❄

Just after : Field lines outside sphere of radius ct emanate from position particle would have reached, had it kept going on its original “flight plan”. Inside the sphere E = 0. On the surface the lines connect up (since they cannot simply terminate in empty space), as suggested in the figure. This produces a dense cluster of tangentially-directed field lines, which expand with the spherical shell. This is a pictorial way of understanding the generation of electromagnetic radiation.

Problem 12.68 ¯ So Equation 12.67 assumes the particle is (instantaneously) at rest in S. Here the particle is at rest in S. 1¯ ¯ ¯ ¯ F? = F? , Fk = Fk . Using F = q E, then, ¯x , Fx = F¯x = q E

Fy =

1¯ 1 ¯ Fy = q E y,

Fz =

1¯ 1 ¯ Fz = q E z.

Invoking Eq. 12.109: Fx = qEx , But

v⇥B=

1

q (Ey

vBz x ˆ + vBy ˆ z,

Problem 12.69 z ✻

z¯ ✻ =⇒v

✻ E ✰ ✰B

Fy =

✲y

✰ x ¯

✲ y¯

so

vBz ) = q(Ey

vBz ) Fz =

F = q(E + v ⇥ B).

1

q (Ez + vBy ) = q(Ez + vBy ).

qed

1

c ⃝2005 ⃝2005 Pearson Education, Inc., Upper NJ.zNJ. All rights reserved. This material is is c Rewrite Pearson Education, Inc., Upper River, reserved. This material Eq. 12.109 with xSaddle !Saddle y, River, y! z, !All x:rights protected under all copyright lawslaws as they currently exist. No No portion of this material maymay be be protected under all copyright as they currently exist. portion of this material reproduced, in any form or by means, without permission in writing from the the publisher. reproduced, form or any by any without permission from publisher. ¯iny any ¯z = means, ¯xin=writing E = Ey E (Ez vBx ) E (Ex + vBz )

¯y = By B

¯z = B

⇣

Bz +

⌘ v E x c2

¯x = B

⇣

Bx

This gives the fields in system S¯ moving in the y direction at speed v.

x

¯y = 0, E ¯z = (E0 vB0 ), E ¯x = 0. Now E = (0, 0, E0 ); B = (B0 , 0, 0), so E ¯ = 0, we must pick v so that E0 vB0 = 0; i.e. v = E0 /B0 If we want E (The condition Ez¯0 /B0 < c guarantees that there is no problem getting to such a system.) ✻ c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is Rlaws protected under all copyright ✶ as they currently exist. No portion of this material may be ✒ reproduced, in any form or by any means, without permission in writing from the publisher. ωt

q ✰ x ¯

⌘ v E z c2

✲ y¯

z

z¯ ✻ =⇒v

✻ ✻ E

✲ ✲ y y¯ ✰ AND RELATIVITY CHAPTER 12. ELECTRODYNAMICS ¯ ✰ ✰B x x

288 ¯y = 0, B ¯z = 0, B ¯x = (B0 With this, B

v c2 E0 )

= B0 1

v2 c2

= B0

1

2

¯ = 1 B0 x = 1 B0 ; B ˆ.

¯ Since the particle started out at rest at the origin The trajectory in S: ¯ According to Eq. 12.71 in S, it started out with velocity vˆ y in S. it will move in a circle of radius R, given by p = qBR, or mv = q

⇣1

B0

⌘

z¯

✻ R✶

✒

m 2v R) R= . qB0

ωt

✲ y¯

q ✰ x ¯

v . R The trajectory in S: The Lorentz transformations Eqs. 12.18 and 12.19, for the case of relative motion in the y-direction, read: The actual trajectory is given by x ¯ = 0 ; y¯ =

R sin ! t¯; z¯ = R(1

cos ! t¯); where ! =

x ¯=x y¯ = (y z¯ = z ⇣ t¯ = t

x=x ¯ vt) y = (¯ y + v t¯) z = z¯ ⇣ ⌘ v ⌘ ¯ + v y¯ y t = t c2 c2 So the trajectory in S is given by: ⇢ h ⇣ ¯ ¯ x = 0; y = ( R sin ! t + v t) = R sin ! t ⇣ v2 ⌘ y 1+ 2 2 | {z c }

2 y(1

v2 c2

z = R(1

2 + vc2

)=

=

2

vt

2y

h cos2 ! t¯) = R 1

cos !

⇣

t

R

v c2 y

v ⌘i y . c2

h ⇣ v ⌘i v ⌘i y ; z = R R cos ! t . c2 c2 We can get rid of the trigonometric terms by the usual trick: ⇥ ⇤ v (y vt) = R sin ⇥ ! (t v c2⇤ y) ) 2 (y vt)2 + (z z R = R cos ! (t c2 y) So: x = 0; y = vt

h ⇣ sin ! t

h ⇣ R sin ! t

⇣ v ⌘ v ⌘i y + v t y , or c2 c2 v ⌘i) h ⇣ y c2 (y vt) = R sin ! t

⌘i

;

R)2 = R2 .

Absent the 2 , this would be the cycloid we found back in Ch. 5 (Eq. 5.9). The 2 makes it, as it were, an elliptical cycloid — same picture as Fig. 5.7, but with the horizontal axis stretched out. Problem 12.70 (a) D = ✏0 E + P suggests E ! ✏10 D but it’s a little cleaner if we divide by µ0 while we’re at it, so that H = µ10 B M suggests B ! µ0 H c ⃝2005 Pearson Education, Inc., Upper Saddle River, NJ. Al protected under all copyright laws as they currently exist. N 8 9 in any form or by any means, without permissio cDx cDy cDz reproduced, > > E ! µ01✏0 D = c2 D, B ! H. Then: > 0 > < = cDx 0 Hz Hy µ⌫ D = cDy Hz 0 Hx > > > > : ; cDz Hy Hx 0 c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

289

CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY Then (following the derivation in Sect. 12.3.4):

@ @ 1 @ @Dµ⌫ D0⌫ = cr·D = c⇢f = Jf0 ; D1⌫ = ( cDx ) + (r⇥H)x = (Jf )x ; so = Jfµ , ⌫ ⌫ @x @x c @t @x⌫ where Jfµ = (c⇢f , Jf ). Meanwhile, the homogeneous Maxwell equations r·B = 0, E = and hence (b)

@Gµ⌫ = 0. @x⌫

@B @t

are unchanged,

8 9 0 Hx Hy Hz > > > > < = Hx 0 cDz cDy = Hy cDz 0 cDx > > > > : ; Hz cDy cDx 0

H µ⌫

(c) If the material is at rest, ⌘⌫ = ( c, 0, 0, 0), and the sum over ⌫ collapses to a single term: Dµ0 ⌘0 = c2 ✏F µ0 ⌘0 ) Dµ0 = c2 ✏F µ0 ) H µ0 ⌘0 =

1 µ0 1 G ⌘0 ) H µ0 = Gµ0 ) µ µ

cD = H=

c2 ✏

E ) D = ✏E (Eq. 4.32), X c

1 1 B ) H = B (Eq. 6.31). X µ µ

(d) In general, ⌘⌫ = ( c, u), so, for µ = 0: D0⌫ ⌘⌫ = D01 ⌘1 + D02 ⌘2 + D03 ⌘3 = cDx ( ux ) + cDy ( uy ) + cDz ( uz ) = c(D · u), Ex Ey Ez ( ux ) + ( uy ) + ( uz ) = (E · u), so c c c c ⇣ ⌘ 0⌫ 2 0⌫ 2 D ⌘⌫ = c ✏F ⌘⌫ ) c(D · u) = c ✏ (E · u) ) D · u = ✏(E · u). c

F 0⌫ ⌘⌫ = F 01 ⌘1 + F 02 ⌘2 + F 03 ⌘3 =

[1]

H 0⌫ ⌘⌫ = H 01 ⌘1 + H 02 ⌘2 + H 03 ⌘3 = Hx ( ux ) + Hy ( uy ) + Hz ( uz ) = (H · u),

G0⌫ ⌘⌫ = G01 ⌘1 + G02 ⌘2 + G03 ⌘3 = Bx ( ux ) + By ( uy ) + Bz ( uz ) = (B · u), so H 0⌫ ⌘⌫ =

1 0⌫ 1 1 G ⌘⌫ ) (H · u) = (B · u) ) H · u = (B · u). µ µ µ

[2]

Similarly, for µ = 1: D1⌫ ⌘⌫ = D10 ⌘0 + D12 ⌘2 + D13 ⌘3 = ( cDx )( ⇥ 2 ⇤ = c D + (u ⇥ H) x ,

c) + Hz ( uy ) + ( Hy )( uz ) = (c2 Dx + uy Hz

Ex ( c) + Bz ( uy ) + ( By )( uz ) = (Ex + uy Bz c D1⌫ ⌘⌫ = c2 ✏F 1⌫ ⌘⌫ )

F 1⌫ ⌘⌫ = F 10 ⌘0 + F 12 ⌘2 + F 13 ⌘3 = =

[E + (u ⇥ B)]x ,

so

⇥ 2 ⇤ 1 c D + (u ⇥ H) x = c2 ✏ [E + (u ⇥ B)]x ) D + 2 (u ⇥ H) = ✏ [E + (u ⇥ B)] . c

H 1⌫ ⌘⌫ = H 10 ⌘0 + H 12 ⌘2 + H 13 ⌘3 = ( Hx )( c) + ( cDz )( uy ) + (cDy )( uz ) = c(Hx uy Dz + uz Dy ) = c [H (u ⇥ D)]x , c 2012 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

u z Hy )

uz By )

[3]

290


◆ ✓ ◆ Ez Ey G ⌘⌫ = G ⌘0 + G ⌘2 + G ⌘3 = ( Bx )( c) + ( uy ) + ( uz ) c c ⇥ 2 ⇤ 1 = (c2 Bx uy Ez + uz Ey ) = c B (u ⇥ E) x , so H 1⌫ ⌘⌫ = G1⌫ ⌘⌫ ) c c µ  ⇤ 1 ⇥ 2 1 1 c [H (u ⇥ D)]x = c B (u ⇥ E) x ) H (u ⇥ D) = B (u ⇥ E) . µc µ c2 1⌫

10

12

✓

13

Use Eq. [4] as an expression for H, plug this into Eq. [3], and solve for D: ⇢  1 1 1 D + 2 u ⇥ (u ⇥ D) + B (u ⇥ E) = ✏ [E + (u ⇥ B)] ; c µ c2 D+

1 ⇥ (u · D)u c2

1 1 (u ⇥ B) + 4 [u ⇥ (u ⇥ E)] . 2 µc µc

⇤ u2 D = ✏ [E + (u ⇥ B)]

Using Eq. [1] to rewrite (u · D): ✓ ◆ ⇤ u2 ✏ 1 1 ⇥ D 1 = (E · u)u + ✏[E + (u ⇥ B)] (u ⇥ B) + 4 (E · u)u u2 E 2 2 2 c c µc µc ⇢   2 u 1 1 1 =✏ 1 E 1 (E · u)u + (u ⇥ B) 1 . ✏µc4 c2 ✏µc2 ✏µc2 Let

⌘p 1

1 u2 /c2

,

1 v⌘p . ✏µ D=

2

✏

⇢✓

Then

u2 v 2 c4

1

◆

✓ E+ 1

v2 c2

◆ (u ⇥ B)

1 (E · u)u c2

.

Now use Eq. [3] as an expression for D, plug this into Eq. [4], and solve for H: ⇢  1 1 1 H u⇥ (u ⇥ H) + ✏[E + (u ⇥ B)] = B (u ⇥ E) ; 2 c µ c2  ⇤ 1 ⇥ 1 1 2 H + 2 (u · H)u u H = B (u ⇥ E) + ✏(u ⇥ E) + ✏[u ⇥ (u ⇥ B)]. c µ c2

Using Eq. [2] to rewrite (u · H): ✓ ◆  ⇥ u2 1 1 1 H 1 = (B · u)u + B (u ⇥ E) + ✏(u ⇥ E) + ✏ (B · u)u 2 2 2 c µc µ c ⇢  ⇥ ⇤ 1 1 = 1 µ✏u2 B + ✏µ [(u ⇥ E) + (B · u)u] . µ c2 H=

2

µ

⇢✓

1

u2 v2

◆

B+

✓

1 v2

1 c2

◆

u2 B

⇤

[(u ⇥ E) + (B · u)u] .


[4]

291


Problem 12.71 We know that (proper) power transforms as the zeroth component of a 4-vector K 0 = 2

1 dW c d⌧

. The Larmor

1 2q 2 4⇡✏0 3 c3 a

formula says that for v = 0, = (Eq. 11.70). Can we think of a 4-vector whose zeroth component reduces to this when the velocity is zero? Well, a2 smells like (↵⌫ ↵⌫ ), but how do we get a 4-vector in here? How about ⌘ µ , whose zeroth component is just c, when v = 0? Try, then: 1 2 q2 ⌫ Kµ = (↵ ↵⌫ )⌘µ 4⇡✏0 3 c5 dW d⌧

This has the right transformation properties, but we must check that it does reduce to the Larmor formula when v = 0: dW 1 dW 1 1 µ0 q 2 ⌫ dW µ0 q 2 ⌫ 0 0 = = cK 0 = c (↵ ↵ )⌘ , but ⌘ = c , so = (↵ ↵⌫ ) . [Incidentally, this tells ⌫ dt d⌧ 6⇡c3 dt 6⇡c us that the power itself (as opposed to proper power) is a scalar. If this had been obvious from the start, we could simply have looked for a Lorentz scalar that generalizes the Larmor formula.] In Prob. 12.39(b) we calculated (↵⌫ ↵⌫ ) in terms of ordinary velocity and acceleration: h i (v·a)2 i 1 6 2 2 2 = (v·a) a + (c2 v 2 ) c2 h ⇣ ⌘ i n v2 1 1⇥ 2 2 = 6 a2 1 + 2 (v·a)2 = 6 a2 v a 2 c c c2

↵⌫ ↵⌫ =

4

h

a2 +

(v·a)2

Now v·a = va cos ✓, where ✓ is the angle between v and a, so:

⇤o .

(v·a)2 = v 2 a2 (1 cos2 ✓) = v 2 a2 sin2 ✓ = |v⇥a|2 . ⇣ v⇥a 2 ⌘ ↵⌫ ↵⌫ = 6 a2 . c dW µ0 q 2 6 ⇣ 2 v⇥a 2 ⌘ = a , which is Liénard’s formula (Eq. 11.73). dt 6⇡c c v 2 a2

Problem 12.72 (a) It’s inconsistent with the constraint ⌘µ K µ = 0 (Prob. 12.39(d)). µ

⌫

µ µ (b) We want to find a 4-vector bµ with the property that d↵ about bµ =  d↵ d⌧ +b ⌘µ = 0. How d⌧ ⌘⌫ ⌘ ? Then d↵⌫ d↵µ d↵⌫ d↵µ d↵⌫ µ µ µ 2 2 c , so this becomes d⌧ ⌘µ c  d⌧ ⌘⌫ , which is zero, d⌧ + b ⌘µ = d⌧ ⌘µ +  d⌧ ⌘⌫ (⌘ ⌘µ ). But ⌘ ⌘µ = ⌘ 2⇣ µ ⌫ µ0 q d↵ 1 d↵ µ if we pick  = 1/c2 . This suggests Krad = + 2 ⌘⌫ ⌘ µ . Note that ⌘ µ = (c, v) , so the spatial 6⇡c d⌧ c d⌧ components of bµ vanish in the nonrelativistic limit v ⌧ c, and hence this still reduces to the Abraham-Lorentz ⌫ d d↵⌫ ⌫ d⌘⌫ formula. [Incidentally, ↵⌫ ⌘⌫ = 0 ) d⌧ (↵⌫ ⌘⌫ ) = 0 ) d↵ ↵⌫ ↵⌫ , and hence bµ can d⌧ ⌘⌫ + ↵ d⌧ = 0, so d⌧ ⌘⌫ = 1 ⌫ µ just as well be written c2 (↵ ↵⌫ )⌘ .]

Problem 12.73 Define the electric current 4-vector as before: Jeµ = (c⇢e , Je ), and the magnetic current analogously: µ Jm = (c⇢m , Jm ). The fundamental laws are then @⌫ F µ⌫ = µ0 Jeµ ,

@⌫ Gµ⌫ =

µ0 µ J , c m

⇣ qm µ⌫ ⌘ K µ = qe F µ⌫ + G ⌘⌫ . c


292


The first of these reproduces r·E = (1/✏0 )⇢e and r⇥B = µ0 Je + µ0 ✏0 @E/@t, just as before; the second yields r·B = (µ0 /c)(c⇢m ) = µ0 ⇢m and (1/c)[@B/@t + r⇥E] = (µ0 /c)Jm , or r⇥E = µ0 Jm @B/@t (generalizing Sec. 12.3.4). These are Maxwell’s equations with magnetic charge (Eq. 7.44). The third says " ✓ ◆ ✓ ◆# qe qm c uy Ez uz Ey 1 p K = p [E + (u ⇥ B)]x + ( Bx ) + p +p , c c c 1 u2 /c2 1 u2 /c2 1 u2 /c2 1 u2 /c2 ⇢  1 1 K = p qe [E + (u ⇥ B)] + qm B (u ⇥ E) , or 2 2 c2 1 u /c  1 F = qe [E + (u ⇥ B] + qm B (u ⇥ E) , c2 which is the generalized Lorentz force law (Eq. 7.69).


293

CONCORDANCE 4th ed 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.12 1.13 1.14 1.15 1.16 1.17 1.18 1.19 1.20 1.21 1.22 1.23 1.24 1.25 1.26 1.27 1.28 1.29 1.30 1.31 1.32 1.33 1.34 1.35 1.36 1.37 1.38 1.39 1.40 1.41 1.42 1.43 1.44 1.45

Same? X X X X X X X X X X X X X X X X X X new X X X X X X X X X X X X X X X X X X X X X X X X X X

3rd ed 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.12 1.13 1.14 1.15 1.16 1.17 1.18 – 1.19 1.20 1.21 1.22 1.23 1.24 1.25 1.26 1.27 1.28 1.29 1.30 1.31 1.32 1.33 1.34 1.35 1.36 1.37 1.38 1.39 1.40 1.41 1.42 1.43 1.44

4th ed 1.46 1.47 1.48 1.49 1.50 1.51 1.52 1.53 1.54 1.55 1.56 1.57 1.58 1.59 1.60 1.61 1.62 1.63 1.64 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12 2.13 2.14 2.15 2.16 2.17 2.18 2.19 2.20 2.21 2.22 2.23 2.24 2.25 2.26

Same? X X X X X X X X X X X X X X X X X X new X mod X X X X X X X X X X X X mod X X X X X X X X X X X

3rd ed 1.45 1.46 1.47 1.48 1.49 1.50 1.51 1.52 1.53 1.54 1.55 1.56 1.57 1.58 1.59 1.60 1.61 1.62 – 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12 2.13 2.14 2.15 2.16 2.17 2.18 2.19 2.20 2.21 2.22 2.23 2.24 2.25 2.26

4th ed 2.27 2.28 2.29 2.30 2.31 2.32 2.33 2.34 2.35 2.36 2.37 2.38 2.39 2.40 2.41 2.42 2.43 2.44 2.45 2.46 2.47 2.48 2.49 2.50 2.51 2.52 2.53 2.54 2.55 2.56 2.57 2.58 2.59 2.60 2.61 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10

Same? X X X X X new new X X X new X X new X X X X X new X X X X new X X mod X X X new new new new X X X new X X X X X X


3rd ed 2.27 2.28 2.29 2.30 2.31 – – 2.32 2.33 2.34 – 2.35 2.36 – 2.37 2.38 2.39 2.40 2.41 – 2.43 2.44 2.45 2.46 – 2.47 2.48 2.49 2.50 2.51 2.52 – – – – 3.1 3.2 3.3 – 3.4 3.5 3.6 3.7 3.8 3.9

294 4th ed 3.11 3.12 3.13 3.14 3.15 3.16 3.17 3.18 3.19 3.20 3.21 3.22 3.23 3.24 3.25 3.26 3.27 3.28 3.29 3.30 3.31 3.32 3.33 3.34 3.35 3.36 3.37 3.38 3.39 3.40 3.41 3.42 3.43 3.44 3.45 3.46 3.47 3.48 3.49 3.50 3.51 3.52 3.53 3.54 3.55

Same? X X X X X mod X X X X X X X X X X X new X X X X X X new X new new X X new new X X X X X X new X X mod X X X

3rd ed 3.10 3.11 3.12 3.13 3.14 3.15 3.16 3.17 3.18 3.19 3.20 3.21 3.22 3.23 3.24 3.25 3.26 – 3.27 3.28 3.29 3.30 3.31 3.32 – 3.33 – – 3.35 3.36 – – 3.37 3.38 3.39 3.40 3.41 3.42 – 3.43 3.44 3.45 3.46 3.47 3.48

CONCORDANCE 4th ed 3.56 3.57 3.58 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11 4.12 4.13 4.14 4.15 4.16 4.17 4.18 4.19 4.20 4.21 4.22 4.23 4.24 4.25 4.26 4.27 4.28 4.29 4.30 4.31 4.32 4.33 4.34 4.35 4.36 4.37 4.38 4.39 4.40 4.41 4.42

Same? X new new X X X X X X X X X X X X X X X mod X X X X X X X X X X X X X X new new X new X X X X X X X X

3rd ed 3.49 – – 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11 4.12 4.13 4.14 4.15 4.16 4.17 4.18 4.19 4.20 4.21 4.22 4.23 4.24 4.25 4.26 4.27 4.28 4.29 4.30 – – 4.31 – 4.32 4.33 4.34 4.35 4.36 4.37 4.38 4.39

4th ed 4.43 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10 5.11 5.12 5.13 5.14 5.15 5.16 5.17 5.18 5.19 5.20 5.21 5.22 5.23 5.24 5.25 5.26 5.27 5.28 5.29 5.30 5.31 5.32 5.33 5.34 5.35 5.36 5.37 5.38 5.39 5.40 5.41 5.42 5.43 5.44

Same? X X X X X X X X X X X X new X X X X X X X X X X X X X X X X X X X X X X X X mod X new X X X X X

3rd ed 4.40 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10 5.11 – 5.12 5.13 5.14 5.15 5.16 5.17 5.18 5.19 5.20 5.21 5.22 5.23 5.24 5.25 5.26 5.27 5.28 5.29 5.30 5.31 5.32 5.33 5.34 5.37 5.35, 5.36 5.60 – 5.38 5.39 5.40 5.41 5.42


295

CONCORDANCE 4th ed 5.45 5.46 5.47 5.48 5.49 5.50 5.51 5.52 5.53 5.54 5.55 5.56 5.57 5.58 5.59 5.60 5.61 5.62 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9 6.10 6.11 6.12 6.13 6.14 6.15 6.16 6.17 6.18 6.19 6.20 6.21 6.22 6.23 6.24 6.25

Same? X X X mod X X new X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X new X

3rd ed 5.43 5.44 5.46 5.47 5.48 5.49 – 5.50 5.51 5.52 5.53 5.54 5.55 5.56 5.57 5.58 5.59 5.61 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9 6.10 6.11 6.12 6.13 6.14 6.15 6.16 6.17 6.18 6.19 6.20 6.21 6.22 6.25 – 6.23

4th ed 6.26 6.27 6.28 6.29 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10 7.11 7.12 7.13 7.14 7.15 7.16 7.17 7.18 7.19 7.20 7.21 7.22 7.23 7.24 7.25 7.26 7.27 7.28 7.29 7.30 7.31 7.32 7.33 7.34 7.35 7.36 7.37 7.38 7.39

Same? X X X X X X X X X X X X X X X X X X X X X X X new new X X X X X X X X X X X new X X X X X X

3rd ed 6.24 6.26 6.27 6.28 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10 7.11 7.12 7.13 7.14 7.15 7.16 7.17 7.18 7.19 – – 7.20 7.21 7.22 7.23 7.24 7.25 7.26 7.27 7.28 7.29 7.30 – 7.31 7.32 7.33 7.34 7.35 7.36

4th ed 7.40 7.41 7.42 7.43 7.44 7.45 7.46 7.47 7.48 7.49 7.50 7.51 7.52 7.53 7.54 7.55 7.56 7.57 7.58 7.59 7.60 7.61 7.62 7.63 7.64 8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 8.9 8.10 8.11 8.12 8.13 8.14 8.15 8.16 8.17 8.18

Same? X X X X X X X X mod mod X new X X new X X X X new X X X X X X X X X new mod mod X new X new new X new new X X new


3rd ed 7.37 7.39 7.41 7.57 7.42 7.43 7.44 7.45 7.46 7.47 7.48 – 7.49 7.50 – 7.51 7.52 7.53 7.54 – 7.55 7.56 7.58 7.59 7.60 8.1 8.2 8.3 8.4 – 8.6 8.5 8.7 – 8.8 – – 8.9 – – 8.10 8.11 –

296 4th ed 8.19 8.20 8.21 8.22 8.23 8.24 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 9.10 9.11 9.12 9.13 9.14 9.15 9.16 9.17 9.18 9.19 9.20 9.21 9.22 9.23 9.24 9.25 9.26 9.27 9.28 9.29 9.30 9.31 9.32 9.33 9.34 9.35 9.36 9.37 9.38 9.39 9.40

Same? X new mod X X new X X X X X X X X X X new X X X X X X X X X X X X X X new X X X X X X X new X X X X X X

3rd ed 8.12 – 8.13 8.14 8.15 – 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 9.10 – 9.11 9.12 9.13 9.14 9.15 9.16 9.17 9.18 9.19 9.20 9.21 9.22 9.23 9.24 – 9.26 9.27 9.28 9.29 9.30 9.31 9.32 – 9.33 9.34 9.35 9.36 9.37 9.38

CONCORDANCE 4th ed 10.1 10.2 10.3 10.4 10.5 10.6 10.7 10.8 10.9 10.10 10.11 10.12 10.13 10.14 10.15 10.16 10.17 10.18 10.19 10.20 10.21 10.22 10.23 10.24 10.25 10.26 10.27 10.28 10.29 10.30 10.31 10.32 10.33 10.34 11.1 11.2 11.3 11.4 11.5 11.6 11.7 11.8 11.9 11.10 11.11 11.12

Same? X X mod X X X new new new X X X X X X X X X X X mod X X X X new X X new new X X new new X X X X X X X new X X X X

3rd ed 10.1 10.2 10.3, 10.5 10.4 10.6 10.7 – – – 10.8 10.9 10.10 10.11 10.12 10.13 10.14 10.15 10.16 10.17 10.18 12.43a 10.19 10.20 10.21 10.22 – 10.23 10.24 – – 10.25 10.26 – – 11.1 11.2 11.3 11.4 11.5 11.6 11.7 – 11.8 11.9 11.12 11.10

4th ed 11.13 11.14 11.15 11.16 11.17 11.18 11.19 11.20 11.21 11.22 11.23 11.24 11.25 11.26 11.27 11.28 11.29 11.30 11.31 11.32 11.33 11.34 11.35 12.1 12.2 12.3 12.4 12.5 12.6 12.7 12.8 12.9 12.10 12.11 12.12 12.13 12.14 12.15 12.16 12.17 12.18 12.19 12.20 12.21 12.22 12.23

Same? new X X X X new X mod new X X X X new new X X X X X X X new X X X X X X X X X X X X X mod X X X X X X X X X

3rd ed – 11.14 11.15 11.16 11.17 – 11.19 11.20 – 11.21 11.22 11.11 11.23 – – 11.24 11.26 11.27 11.28 11.29 11.30 11.31 – 12.1 12.2 12.3 12.4 12.5 12.6 12.7 12.8 12.9 12.10 12.11 12.12 12.13 12.14 12.15 12.16 12.17 12.18 11.19 12.20 12.21 12.22 12.23


CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY 4th ed 12.24 12.25 12.26 12.27 12.28 12.29 12.30 12.31 12.32 12.33 12.34 12.35 12.36 12.37 12.38 12.39 12.40 12.41 12.42 12.43 12.44 12.45 12.46 12.47 12.48

Same? X X mod new X X X X X X X X X X X X X X X X new X X X X

3rd ed 12.24 12.25 12.26 – 12.27 12.28 12.29 12.30 12.31 12.32 12.33 12.34 12.35 12.36 12.37 12.38 12.39 12.40 12.41 12.42 – 12.44 12.45 12.46 12.47

4th ed 12.49 12.50 12.51 12.52 12.53 12.54 12.55 12.56 12.57 12.58 12.59 12.60 12.61 12.62 12.63 12.64 12.65 12.66 12.67 12.68 12.69 12.70 12.71 12.72 12.73

Same? X X X X X X X X X new X X X X X X X X X X X X X X X

3rd ed 12.48 12.49 12.50 12.51 12.52 12.53 12.54 12.55 12.56 – 12.57 12.58 12.59 12.60 12.61 12.62 12.63 12.64 12.65 12.66 12.67 12.68 12.69 12.70 12.71


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Instructor's Solution Manual Introduction to Electrodynamics Fourth Edition

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