Jackson Solutions - Solutions to jackson's Electrodynamics

Physics 742: Jackson, Classical Electrodynamics Russell Bloomer1 University of Virginia Note: There is no guarantee that these are correct, and they should not be copied

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email: [email protected]

Contents 1 Problem Set 1 1.1 Problem 1 . . . . . . . 1.2 Problem 2: Jackson 1.5 1.3 Problem 3: Jackson 1.6 1.4 Problem 4: Jackson 1.10

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1 1 2 3 4

2 Problem Set 2 2.1 Problem 1: Jackson 2.2 Problem 2: Jackson 2.3 Problem 3: Jackson 2.4 Problem 4: Jackson

1.7 1.9 2.1 2.2

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7 7 8 9 11

3 Problem Set 3 3.1 Problem 1: Jackson 3.2 Problem 2: Jackson 3.3 Problem 3: Jackson 3.4 Problem 4: Jackson

2.8 2.10 2.13 2.23

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13 13 14 16 17

4 Problem Set 4 19 4.1 Problem 1: Jackson 3.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 4.2 Problem 2: Jackson 3.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 4.3 Problem 3: Jackson 3.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 5 Problem Set 5 23 5.1 Problem 1: Jackson 3.17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 5.2 Problem 2: Jackson 4.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 6 Problem Set 6 27 6.1 Problem 1: Jackson 4.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 6.2 Problem 2: Jackson 4.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 6.3 Problem 3: Distressed Simple Cubic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 7 Problem Set 7 31 7.1 Problem 1: Jackson 4.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 8 Problem Set 8 8.1 Problem 1: Jackson 5.2 8.2 Problem 2: Semi-Infinite 8.3 Problem 3: Jackson 5.6 8.4 Problem 4: Jackson 5.13

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33 33 34 35 35

9 Problem Set 9 39 9.1 Problem 1: Jackson 5.17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 9.2 Problem 2: Jackson 5.19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 9.3 Problem 3: Jackson 5.22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 10 Problem Set 10 45 10.1 Problem 1: Jackson 5.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 10.2 Problem 2: Jackson 5.27 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 10.3 Problem 3: Jackson 5.29 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 11 Problem Set 11 51 11.1 Problem 1: Jackson 6.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 11.2 Problem 2: Jackson 6.20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 11.3 Problem 3: Jackson 6.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 12 Problem Set 12 55 12.1 Problem 1: Jackson 6.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 12.2 Problem 2: Jackson 6.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 A Special Functions

61

ii

Chapter 1 Problem Set 1

1.1

Problem 1

Consider a vacuum diode consisting of two parallel plates of area A which are large compared to their separation d. Neglecting edge effects, all quantities will be assumed to depend only on x a coordinate perpendicular to the plates. Electrons are “boiled” off the cathode plate which is heated and held at potential Φ = 0. They are attracted to the anode plate which is held at a positive potential Vo . The cloud of electrons within the gap (called the space charge) quickly builds up to a point where it reduces field at the cathode to zero. From then on a steady current flows between the plates. Show that the current in the diode is given by I = KVo3/2 where K =

4o A 9d2

r

2e m

(1.1)

The charge density is ρ = −o ∇2 V → ρ = −o

∂2V ∂x2

(1.2)

because the charge density only depends on the position x. Next the velocity of the electron at any position in the potential. r 2eV 1 mv 2 = eV → v = (1.3) 2 m Because this is a steady state, the current is independent of the position x. dq = ρAdx →

dq dx = ρA → I = ρAv dt dt

Combining equations 3 and 41 r I = ρA

2eV I →ρ= m A

r

m 2eV

Now adding in equation 2 I ρ= A

r

m ∂2V I =− → 2eV ∂x2 o A

r

m 2eV

All that remains is to solve the differential. Start by defining W = ∂V . The differential becomes ∂x r r ∂W I 2 −1/2 ∂V I 2 −1/2 W =− V ⇒ W ∂W = − V ∂V ∂x o A me ∂x o A me 1 If

the ‘e’ on equation is lower case, it is from the homework set, but if it is upper case, it is from ‘Jackson’ 3rd ed.

1

(1.4)

Integrating once to find 2I 1 2 W =− 2 o A

r

2 1/2 V me

Solving for W 1 2 2I W =− 2 o A Now reintroduce W =

r

2 1/2 V →W = me

8m e

I o A

2 !1/4

V 1/4

∂V ∂x

∂V = ∂x

8m e

I o A

2 !1/4

8m e

V 1/4 ⇒ V −1/4 ∂V =

I o A

2 !1/4 ∂x

Integrating

V

−1/4

∂V =

8m e

I o A

2 !1/4

4 ∂x ⇒ V 3/4 = 3

8m e

I o A

2 !1/4 x+C

The first boundary condition is at V (0) = 0, which determines the constant to be C = 0. The other boundary condition is at V (d) = Vo . The equation becomes 4 3/4 Vo = 3

8m e

I o A

2 !1/4 d

Solving for the current r 4 2 2 4 2 4 e 8m I I e 4 I 4 3 Vo3 = d4 ⇒ = V ⇒ = Vo3/2 o 3 e o A o A 8md4 3 o A 8m 3d r 4o A 2e 3/2 I= Vo → I = KVo3/2 X 9d2 m

1.2

Problem 2: Jackson 1.5

The time-averaged potential of a neutral hydrogen atom is given by Φ=

q e−αr αr 1+ 4πo r 2

(1.5)

where q is the magnitude of the electric charge, and α−1 = ao /2, ao being the Bohr radius. Find the distribution of charge (both continuous and discrete) that will give this potential and interpret your result physically. The charge density is related to the potential by αr q e−αr ρ = −o ∇ Φ → ρ = −o ∇ 1+ 4πo r 2 −αr q e αr = − ∇2 1+ 4π r 2 2

2

The Laplace in spherical coordinates for a spherically symmetrical object is 1 ∂ ∂ ∇2 = 2 r2 r ∂r ∂r

2

Then the charge density becomes −αr q 1 ∂ e αr 2 ∂ ρ=− 1 + r 4π r2 ∂r ∂r r 2 −αr −αr q 1 ∂ e αe ∂ 2 =− + r 4π r2 ∂r ∂r r 2 2 1 ∂ −αr −αr 1 q αr2 −αr 2 −αr ∂ e + r e e =− − 4π r2 ∂r 2 ∂r r 2 3 2 q 1 α r 1 1 2 −αr −αr −αr 2 2 −αr 2 −αr ∂ −αr ∂ 2 ∂ =− − α re + e − αe + α r e − αr e + e r 4π r2 2 ∂r r ∂r ∂r r q α3 −αr α 1 ∂ 1 αr2 e−αr −αr 2 ∂ =− e − 2 e−αr + 2 + e r 4π 2 r r r2 r2 ∂r ∂r r With ∇2 r1 = −4πδ(r), the factor in front of the delta function is 1 at the origin and 0 everywhere else. The above equation reduces to −q α3 −αr qα3 −αr ρ= e − 4πδ(r) = qδ(r) − e (1.6) 4π 2 8π The first term is a positive point charge at the origin, which would be the proton. The second term is a negative spherically symmetry object that is decaying off rapidly after the Bohr radius, which is a good description of the electric cloud. X

1.3


A simple capacitor is a device formed by two insulated conductors adjacent to each other. If equal and opposite charges are placed on the conductors, there will be a certain difference potential between them. The ratio of magnitude of the charge on one conductor to magnitude of the potential difference is called the capacitance. Using Gauss’s law, calculate the capacitance of (a) two large, flat, conducting sheets of area A, separated by a small distance d The electric field outside the plate two plates is zero, because if pillbox containing both plates has no charge therefore no electric field. Forming a pillbox around one plate by Gauss’s law has |E| = σ/2o , where σ = Q/A. There are two plates adding constructively in between, so |E| = σ/o . The potential is then σ V = d o Qd σd →V = V = o Ao E=

The capacitance is then C=

Q Q Ao = Qd → C = X V d A o

(b) two concentric conduction spheres with radii a, b (b > a) Gauss’s law for a sphere is E=

Q 1 4πo r2

The potential is then Z a Q dr ˆ · dˆ E l=− 2 4π o b b r Q 1 a Q 1 1 = = − 4πo r b 4πo a b Z

a

V =−

Now the capacitance C=

Q = V

Q Q 4πo

1 a

−

3

1 b

= 4πo

ab X b−a

(c) two concentric conducting cylinders of length L, large compared to their radii a, b (b > a) Using Gauss’s law for cylindrical surface is E=

Q 1 2πo r

The potential becomes Z b dr Q ˆ · dˆ E l=− 2π o a r a b Q Q Q =− ln r a = (ln b − ln a) = ln (b/a) 2πo 2πo 2πo Z

b

V =−

The capacitance is Q = V

C=

Q 2πo

Q 2πo = X ln (b/a) ln (b/a)

(1.7)

(d) What is the inner diameter of the outer conductor in an air-filled coaxial cable whose center conductor is a cylindrical wire of diameter 1 mm and whose capacitance is 3 × 10−11 F/m? 3 × 10−12 F/m? Using equation 7, and rearranging it b = ae

2πo C

For 3 × 10−11 F/m b = 1 × e2π·8.85×10

−12

/3×10−11

= e1.854 = 6.38 mm X

For 3 × 10−12 F/m b = 1 × e2π·8.85×10

1.4

−12

/3×10−12

= e18.54 = 1.12 × 108 mm X


Prove the mean value theorem: For charge-free space the value of the electrostatic potential at any point is equal to the average of the potential over the surface of any sphere centered on that point. From Equation 1.36 Φ(x) =

1 4πo

ρ(~ x0 ) 3 0 1 d x + R 4π

Z V

I S

1 dΦ d −Φ 0 R dn0 dn

1 R

da0

(1.8)

For a charge-free space ρ = 0, then 1 4πo

Z V

ρ(~ x0 ) 3 0 1 d x = R 4πo

Z V

0 3 0 d x =0 R

Using Equation 1.33 for the second part of equation 8 1 1 dΦ = ∇Φ · n ˆ0 R dn0 R From equation 1.16 1 1 ∇Φ · n ˆ0 = − E · n ˆ0 R R The second part of equation 8 becomes 1 4π

I S

1 1 dΦ 0 da = − R dn0 4π

4

I S

1 E·n ˆ 0 da0 R

(1.9)

Because this is over the surface it is independent of R, then I 1 E·n ˆ 0 da0 − 4πR S By the divergence theorem −

1 4πR

I

E·n ˆ 0 da0 = S

1 4πo R

Z

ρ(~ x)d3 x0

V

Again in a charge-free space 1 4πo R

Z

ρ(~ x)d3 x0 = 0

(1.10)

V

1 d For the last part of equation 8, dn = − R12 , because the differential is before the surface. At the surface the 0 R radius is constant so the last part of equation 8 becomes I I 1 d 1 1 − Φ 0 Φda0 (1.11) da0 = 4π S dn R 4πR2 S So combining equations 9, 10, and 11 to find Φ(x) =

1 4πR2

5

I

Φda0 X S

6


2.1


Two long, cylindrical conductors of radii a1 and a2 are parallel and separated by a distance d, which is large compared with either radius. Show that the capacitance per unit length is given approximately by −1 d C ' πo ln (2.1) a where a is the geometrical mean of the two radii. Approximate what gauge wire (state diameter in millimeters) would be necessary to make a two-wire transmission line with capacitance of 1.2 × 10−11 F/m in the separation of the wires was 0.5 cm? 1.5 cm? 5.0 cm? The electric field for a wire is given by ~ = E

Q rˆ 2πo r

The potential between the two wire is Z Φ=

The capacitance is given by C =

Q . Φ

Z d−a1 dr dr + r r a a2 1 d − a2 d − a1 ln + ln a1 a2 (d − a2 ) (d − a1 ) ln a1 a2

~ · d~l = E

=

Q 2πo

=

Q 2πo

Q 2πo

Z

d−a2

h

Then C=

Q = Φ

Q Q 2πo

ln

(d−a2 )(d−a1 ) a1 a2

i

2πo

= ln

(d−a2 )(d−a1 ) a1 a2

Since the distance between the wires is much large than their radii, d − a1 ≈ d − a2 ≈ d. Then capacitance reduces to 2πo

C= ln ≈ ln

(d−a2 )(d−a1 ) a1 a2

2π o = 2 d

a1 a2

πo

ln

d a1 a2

√

√ Defining the average radius as a = a1 a2 . The capacitance is then πo πo = C= d d ln ln √a1 a2 a

7

For C = 1.2 × 10−11 F/m and d = 0.5cm the radius is 1.2 × 10−11 =

πo .005 a

ln

→ a = .49mm

The diameter is .98 mm. Now for d = 1.5cm 1.2 × 10−11 =

πo .015 a

ln

→ a = 1.5mm

The diameter is 3 mm. Lastly for d = 5.0cm 1.2 × 10−11 =

πo .050 a

ln

→ a = 4.9mm

The diameter is 9.8 mm. X

2.2


Calculate the attractive force between conductors in the parallel plate capacitor (Problem 1.6a) and parallel cylinder capacitor (Problem 1.7) for (a) fixed charges on each conductor For parallel plate capacitor, the work is 1 CV 2 2

W = From the last problem set, C =

Ao , d

and V =

Qd . Ao

W =

(2.2)

Therefore

1 Ao 2 d

Qd Ao

2 =

1 Q2 d 2 Ao

Force is given by F =− Because the charge is constant

∂Q ∂l

∂W ∂l

+

Q

∂W ∂l

(2.3) V

= 0, then the force is F =−

∂W ∂ 1 Q2 d 1 Q2 =− =− X ∂l ∂l 2 Ao 2 Ao

−1 For parallel cylinder capacitor, the work is given by equation 2, with C = πo ln ad and V = is then −1 2 1 d Q d 1 Q2 d W = πo ln ln = ln 2 a πo a 2 πo a Because the charge is constant, the force becomes from equation 3 ∂ 1 Q2 d 1 Q2 ln =− X F =− ∂l 2 πo a 2 πo d (b) fixed potential difference between conductors. For parallel capacitors, the charge in terms of potential is V =

Ao V Qd →Q= Ao d

The work becomes W =

1 Q2 d 1 o AV 2 →W = 2 Ao 2 d

8

Q πo

ln

d a

. The work

From equation 3, the force is F =

∂W ∂l

=− V

1 o AV 2 X 2 d2

For parallel cylinder capacitor, the charge in terms of potential is Q d πo V V = ln →Q= πo a ln ad The work becomes W =

1 Q2 ln 2 πo

d 1 πo V 2 →W = a 2 ln ad

From equation 3, the force is F =

2.3

∂W ∂l

= V

∂ 1 πo V 2 1 πo V 2 = − X ∂l 2 ln ad 2 d ln d 2 a


A point charge q is brought to a position a distance d away from infinite plane conductor held at zero potential. Using the method of images find: (a) the surface-charge density induced on the plane, and plot it. From the method of images q/4πo q 0 /4πo + |~ x−~ y| |~ x − y~0 |

Φ(~ x) =

(2.4)

Because it is an infinite plane conductor, by symmetry of problem ~ y = d~ and y~0 = −d~ when defining the capacitor on the xy plane. The charge is then also q 0 = −q. The potential becomes Φ(~ x) =

q/4πo q/4πo − ~ ~ |~ x − d| |~ x + d|

Defining the potential in cylindrical coordinates Φ(ρ, θ, z) =

q/4πo 2

(z − d) −

ρ2

1/2 −

q/4πo (z + d)2 + ρ2

1/2

The surface-charge density induced on the plane is when z = 0 ∂ σ = −o Φ z=0 ∂z " # −q −q ∂ 4π 4π = 1/2 − 1/2 ∂z (z − d)2 − ρ2 (z + d)2 + ρ2 z=0 " # −1 −1 2(z − d) 2 2(z + d) 2 q =− 3/2 − 3/2 4π (z − d)2 + ρ2 (z + d)2 + ρ2 z=0 " # q q 2d 1 =− =− 2 3/2 4π (d2 + ρ2 )3/2 2πd2 1 + dρ See figure 1. X (b) the force between the plane and the charge by using Coulomb’s law for the force between the charge and its image.

9

Figure 2.1: s(r) = −1/(1 + r2 )3/2 The force between the charge and the image is x1 − ~ x2 ~ = q 1 q2 ~ F 4πo |~ x1 − ~ x2 |3 The separation ~ x1 − ~ x2 = −2dˆ z . The charges are q1 = −q2 = q. The force becomes z q2 zˆ q 2 zˆ ~ = q(−q) −2dˆ F = = X 4πo | − 2d|3 4πo (2d)2 16πo d2 (c) the total force acting on the plane by integrating σ 2 /2o over the whole plane. The force is 2



1  q 1 σ2  da → dF = dF =  3/2  da 2o 2o 2πo 2 1 + dρ =

q2 8π 2 o d4 1 +

ρ 2 d

3 da

Integrating over the entire plane F =

q2 8π 2 o d4

2π

Z 0

q2 2 8π o d4

∞

Z 0

Z

1 1+

ρ 2 d

3 ρdρdθ

∞

1 2 3 ρdρ 1 + dρ ∞ q2 −d6 q2 = = X 4πo d4 4(ρ2 + d2 )2 0 16πo d2 = 2π

0

(d) the work necessary to remove the charge q from its position to infinity. From part b, the force is F =

q 2 zˆ 16πo z 2

So the work is the amount of force to move a particle a distance, then the work is ∞ Z ∞ Z ∞ q 2 dz q 2 −1 ~ · dˆ W = F z= = 16πo z 2 16πo z d d d =

q2 X 16πo d

(e) the potential energy between the charge q and its image [compare the answer to part d and discuss]

10

The work is given by X qi qj 1 1 2 4πo |xi − xj |

W =

i,j,i6=j

For this system, W =

1 1 2 4πo

−q 2 q2 q2 + =− | − 2d| |2d| 8πo d

The difference between part d and e is that in d the image charge is being moved, and in e the image charge is stationary. X (f ) find the answer to part d in electron volts for an electron originally on angstrom from the surface. In part d, q2 16πo d

W = Then for this case

e · 1.60 × 10−19 = 3.60eV X 16π (8.85 × 10−12 ) 10−10

2.4


Using the method of images, discuss the problem of a point charge q inside a hollow, grounded, conducting sphere of inner radius a. Find (a) the potential inside the sphere; The potential is Φ(~ x) =

1 4πo

q q0 + |~ x−~ y| |~ x−~ y0 |

Now the boundary condition Φ(x = a) = 0 =

q/4πo q 0 /4πo + a| − n ˆ + y/aˆ n0 | y0 | − n ˆ 0 + a/y 0 n ˆ|

This is identical to the charge on the outside of the sphere, because the magnitude is the concern. Therefore q 0 = −q ay and y 0 =

a2 . y

Then the potential is Φ(~ x) =

1 4πo

q q0 + |~ x−~ y| |~ x−~ y0 |

X

(b) the induced surface-charge density; The surface-charge density is given by σ = o

∂Φ ∂x x=a

because it is opposite the normal to the inside. So " # o q/4πo o q 0 /4πo ∂ σ= + ∂x (x2 + y 2 − 2xy cos γ)1/2 (x2 + y 02 − 2xy 0 cos γ)1/2 " # q(−1/2)(2x − 2y cos γ) q 0 (−1/2)(2x − 2y 0 cos γ) 1 = + 4π (x2 + y 2 − 2xy cos γ)3/2 (x2 + y 02 − 2xy 0 cos γ)3/2 =

q y 2 − a2 X 4πa (y 2 + a2 − 2ay cos γ)3/2

11

(c) the magnitude and direction of the force acting on q. The separation between the charge and its image y − y 0 , so the magnitude is ~| = |F =

q 2 ay 1 qq 0 1 = 4πo (y − y 0 )2 4πo (y − a2 )2 y 1 q 2 ay 4πo (a2 − y 2 )

The direction is radially towards the image charge, because it is attractive. X (d) Is there any change in the solution if the sphere is kept at a fixed potential V ? If the sphere has a total charge Q on its inner and outer surfaces? Because this is inside of a sphere, there is no change if there is a fixed potential or a charge on the sphere. This is due to spherical symmetry. The only change will be a constant,V , add to the potential 1 q0 q Φ(~ x) = + +V X 4πo |~ x−~ y| |~ x−~ y0 |

12


3.1


A two-dimensional potential is defined by two straight parallel line charges separated by a distance R with equal and opposite linear charge densities λ and −λ, (a) Show by direct construction that the surface of constant potential V is a circular (circle in the transverse dimensions) and find the coordinates of the axis of the cylinder and its radius in terms of R, λ, and V . The potential for a line charge is λ r0 =V ln 2πo r

Φ(~r) = Solving for r0 ,

0 2 r r

(3.1)

~ 2 . Now = e4πo V /λ . But from the geometry r02 = (~r − R) ~ 2 = r2 e4πo V /λ r02 = r2 e4πo V /λ → (~r − R)

From this R 1 + e2πo V /λ R = 1 − e2πo V /λ

rmin = rmax This yields a radius of radius =

R . 2 sinh(2πo V /λ)

Now the offset will be

rcenter = −

~ R e2πo V /λ

−1

The circle becomes with the axis of the cylinder in the zˆ direction ~r +

!2

~ R e2πo V /λ − 1

=

R 2 sinh (2πo V /λ)

2 X

(b) Use the results of part a to show that the capacitance per unit length C of two right-circular conductors, with radii a and b, separated by a distance d > a + b, is C=

cosh−1

2π o

d2 −a2 −b2 2ab

In this case define d = R + d1 + d2 , where d1 =

R R ; d2 = −V /λ eV1 /λ e 2

13

(3.2)

where V1 = 4πo Va and V2 = 4πo Vb . From the previous part the radii can be defined as a=

ReV1 /λ Re−V1 /λ ; b = eV1 /λ − 1 e−V1 /λ − 1

Then d2 − a2 − b2 =

R2 e(V1 −V2 )/λ + 1 (eV1 /λ − 1) (e−V2 /λ − 1)

Then 2

2

2

d −a −b = 2ab (V1 −V2 )/2λ

=

e

eV1 /λ − 1 e−V2 /λ − 1 ×

2R2 eV1 /λ e−V2 /λ −(V1 −V2 )/2λ

+e 2

= cosh

R2 e(V1 −V2 )/λ + 1

(eV1 /λ − 1) (e−V2 /λ − 1) V1 − V2 2λ

Then the potential per length becomes d2 − a2 − b2 = cosh 2ab

V1 − V2 2λ

⇒ 2 cosh−1

d2 − a2 − b2 2ab

=

V1 − V2 λ

The capacitance per unit length becomes C=

1 λ = −1 d2 −a2 −b2 V1 − V2 2 cosh 2ab

= 2 cosh

3.2

2πo

−1

d2 −a2 −b2 2ab

X


A large parallel plate capacitor is made up of two plane conducting sheets with separation D, one of which has a small hemispherical boss of radius a on its inner surface (D a). The conductor with the boss is kept at zero potential, and the other is at a potential such that far from the boss the electric field between the plates is Eo . (a) Calculate the surface-charge densities at an arbitrary on the plane and on the boss, sketch their behavior as a function of distance (or angle). This is similar to a conducting sphere in an electric field, which from class is a3 Φ(r, θ) = −Eo r − 2 cos θ r For 0 ≤ θ ≤ π/2, the potential can be written in Cartesian coordinates Φ(x, y, z) = −Eo z +

Eo a3 z (x2 + y 2 + z 2 )3/2

The plate is ∂Φ σ(x, y) = o Ez = −o ∂z z=0 " # a3 = o Eo 1 − (x2 + y 2 )3/2 For the case of the boss, ∂Φ σ(r, θ) = o Er = −o ∂r r=a = 3o Eo cos θ

14

Figure 3.1: σ/o Eo = 1 − 1/x3 (b) Show that the total charge on the boss has the magnitude 3πo Eo a2 . For the total charge on the boss is found from the integral over the area. 2π

Z

π/2

Z

3o Eo cos θa2 sin θdθdφ

Q= 0

0 2π

Z

1 o Eo a2 dφ = 3πo Eo a2 X 2

= 0

(c) If, instead of the other conducting sheet at a different potential, a point charge q is placed directly above the hemispherical boss at a distance d from its center , show that the charge induced on the boss is d2 − a2 q 0 = −q 1 − √ d d2 + a2

(3.3)

The potential is 1 Φ(~r) = 4πo

"

q ~ |~r − d|

−

qa/d ~ |~r − a2 /dd|

−

q ~ |~r + d|

+

qa/d ~ |~r + a2 /dd|

Now in terms of r and θ q 1 √ 4πo r2 + d2 − 2rd cos θ qa/d q −p −√ 2 2 2 2 2 2 r + d + 2rd cos θ r + (a /d) − 2r(a /d) cos θ qa/d +p r2 + (a2 /d)2 − 2r(a2 /d) cos θ

Φ(r, θ) =

Now the surface-charge density needs to be found ∂Φ ∂r r=a q d2 − a2 1 = 2 2 4π a2 (a + d − 2ad cos θ)3/2 1 − (a2 + d2 + 2ad cos θ)3/2

σ = −o

15

#

So the induced charge on the boss is q0 =

3.3

π/2

q d2 − a2 1 1 − 2πa2 sin θdθ 2 3/2 3/2 2 2 2 2 4π a (a + d − 2ad cos θ) (a + d + 2ad cos θ) 0 −qa2 d2 − a2 1 1 1 1 1 −√ −√ = + 2a ad d − a a+d a2 + d2 a2 + d2 " # 2 2 d −a X = −q 1 − √ d a2 + d2

Z


(a) Two halves of a long hollow cylinder of inner radius b are separated by a small lengthwise gap on each side, and are kept at different potentials V1 and V2 . Show that the potential inside is given by V1 + V2 V1 − V2 Φ(ρ, φ) = + tan−1 2 π

2bρ cos φ b2 − ρ2

(3.4)

The starting point of this is Equation 2.71. In this case, the origin is located inside the cylinder, so bn = 0 to prevent divergence at the origin. Equation 2.71 becomes Φ(ρ, φ) = a0 +

∞ X

an ρn sin (nφ + αn )

n=1

Here Φ(ρ, θ) = Φ(ρ, −θ), so αn = 0. For this case, a0 Z

3π/2

Φ(b, φ) = 2πa0 = πV1 + πV2 → a0 = −π/2

V1 + V2 2

Now for the remaining terms in the sum 3π/2

Z

cos nθ cos mθdθ = δmn π −π/2

Then for Φ Z

3π/2

V (θ) cos mθdθ = −π/2

X

an bn

Z

3π/2

cos nθ cos mθdθ = an bn δmn π −π/2

n

The coefficients become an = =

V1 bn π

Z

π/2

cos(nθ)dθ − −π/2

V2 bn π

Z

3π/2

cos(nθ)dθ π/2

n+3 2(V1 − V2 ) 2(V1 − V2 ) sin (nπ/2) = (−1) 2 bn π nπbm

Because of Φ’s symmetry only odd values for n are allowed. The potential becomes 2(V1 − V2 ) X ρ n in cos(nθ) π b n n ! X Zn 2(V1 − V2 ) = −i iIm where Z = i(ρ/b)eiθ π n

Φ(ρ, θ) = −i

n odd

16

P

n odd

Zn n

=

1 2

ln

1+Z 1 . 1−Z

From Im ln(A + iB) = tan−1 (B/A), then 1+Z 2ρb cos θ Im ln = tan−1 1−Z b2 − ρ2 ! X Zn 2(V1 − V2 ) 2(V1 − V2 ) 2ρb cos θ iIm = tan−1 −i π n π b2 − ρ2 n odd

Combining the two parts of the potential to find Φ(ρ, θ) =

2(V1 − V2 ) V1 + V2 + tan−1 2 π

2ρb cos θ b2 − ρ2

X

(b) Calculate the surface-charge density on each half of the cylinder The surface-charge density ∂Φ σ = −o ∂ρ ρ=b V1 − V2 b2 + ρ2 b cos θ 4 2 2 π b − 2b ρ + ρ4 + 4ρ2 b2 cos2 θ V1 − V2 X = −o bπ cos θ

= −2o

3.4


A hollow cube has conducting walls defined by six planes x = 0, y = 0, z = 0, and x = a, y = a, z = a. The walls z = 0 and z = a are held at a constant potential V . The other four sides are at zero potential. (a) Find the potential Φ(x, y, z) at any point inside the cube. From class Φ(x, y, z)top =

∞ X ∞ X

Anm sin

nπx a

m=1 n=1

where γ =

q

Anm =

nπ 2 a

+

mπ 2 . a

4 a2 sinh (γnm a)

Z

sin

mπy a

sinh (γmn z)

Following Equation 2.58 a

a

Z dx

0

dyV sin

nπx

0

a

sin

mπy a

=

16V , where m,n are both odd π 2 nm sinh(γnm a)

By symmetry the bottom has to be the same except for z → a − z. Therefore the bottom is Φ(x, y, z)bottom

odd X

Anm sin

nπx

n,m=1

a

sin

mπy a

sinh (γmn (a − z))

The total potential is the sum of both top and bottom potential Φtop + Φbottom =

odd X n,m=1

Anm sin

nπx a

sin

mπy a

[sinh (γmn z) + sinh (γmn (a − z))] X

(b) Evaluate the potential at the center of the cube numerically, accurate to three significant figures. How many terms in the series is it necessary to keep in order to attain this accuracy? Compare your numerical results with the average value of the potential on the walls. See Problem 2.28. 1 Jackson

p74-5

17

To have three digits of accuracy, the first four terms need to be kept, which is a a a √ 16V √ (1)(1)(2 sinh(π 2/2)) Φ , , = 2 2 2 2 π sinh( 2π) √ 16V √ + (1)(−1)(2 sinh(π 10/2)) 2 3π sinh( 10π) √ 16V √ + (−1)(1)(2 sinh(π 10/2)) 3π 2 sinh( 10π) √ 16V √ + (1)(1)(2 sinh(3π 2/2)) 2 9π sinh(3 2π) ≈ .3329V = .333V From Problem 2.28 the potential should be

2 V 6

= .33¯ 3V X

(c) Find the surface-charge density on the surface z = a. The surface-charge density in the cube is given by ∂Φ σ = o ∂z z=a =

=

=

16o V π2 16o V π2 16o V π2

odd X n,m=1 odd X n,m=1 odd X n,m=1

Anm sin

nπx a

sin

mπy a

[cosh (γmn a) − 1]

nπx mπy cosh (γ a) − 1 1 mn sin sin nm a a sinh (γmn a) nπx mπy γ a 1 mn sin sin tanh X nm a a 2

18


4.1


A thin, flat, conducting, circular disc of radius R is located is located in the x − y plane with its center at the origin, and is maintained at a fixed potential V . With the information that the change density on a disc at fixed potential is proportional to (R2 − ρ2 )−1/2 , where ρ is the distance out from the center of the disc, (a) show that for r > R the potential is Φ(r, θ, φ) =

2l ∞ 2V R X (−1)l R p2l (cos θ) π r 2l + 1 r l=0

The surface charge is given by λ σ(ρ) = p R 2 − ρ2 The volume charge is found by consider a shell of radius r Z 2π Z dq = σ(r)2πrdr = dφ 0

1

d(cos θ)r2 drf (r)δ(cos θ) −1

σ(r)2πrdr = 2πr2 f (r)dr σ(r) δ(cos θ). r

The potential at the origin is Z R r R 1 λ2πrdr λ √ V = Φ(0) = = sin−1 4πo 0 r R2 − r2 2o R 0 λπ = 4o

The volume charge density is ρ(r) =

The potential at any point becomes Φ(~ x) =

Z ∞ l δ(cos θ0 )r02 dr0 dΩ0 1 4o V X X 1 ∗ √ Ylm (θ, φ) l+1 r0l Ylm (θ0 , φ0 ) 4πo π r r0 R2 − r02 v l=0 m=−l

Solving the integral Z RZ 1 ∞ δ(cos θ0 )dr0 d(cos θ0 ) V X 2π √ P (cos θ) r0l+1 Pl (cos θ0 ) l 2 l+1 π r R2 − r02 0 −1 l=0 Z ∞ P (0) R r0l+1 dr0 2V X √ = Pl (cos θ) l+1 π r R2 − r 2 0 l=0

Φ(~ x) =

=

∞ (−1)n (2n + 1)!! R2n+1 n!2n 2V X P2n (cos θ) π n=0 2n n! (2n + 1)!!r2n+1

19

(4.1)

Then 2n+1 ∞ 2V X (−1)n R P2n (cos θ) X Φ(~ x) = π n=0 2n + 1 r (b) find the potential for r < R. To find the potential for inside the circular radius, use the boundary condition when r = R. The potential inside and outside has to be equal, therefore Al Rl =

2V (−1)n = Bl R−l−1 π 2n + 1

where l = 2n. Then the equation for inside becomes Φ=

∞ 2V X (−1)n r 2n P2n (cos θ) X π n=0 2n + 1 R

(c) What is the capacitance of the disc? The capacitance is given by C = Q/V . From part (a) V = Z

R

Q= 0

λπ . 4o

The charge is

p R λ √ 2πrdr = −2πλ R2 − r2 0 = 2πλR 2 2 R −r

The capacitance becomes C=

2πλR Q = λπ → C = 8o R X V 4 o

4.2


Two point charges q and −q are located on the z axis at z = +a and z = −a, respectively. (a) Find the electrostatic potential as an expansion in spherical harmonics and powers of r for both r > a and r < a. From Equation 3.38 1 1 − |~ x − ~a| |~ x + ~a| " # l ∞ l r< r< q 1 X = Pl (cos θ) − Pl (cos(π − θ)) 4πo r> r> r>

Φ(~ x) =

q 4πo

l=0

From equation Equation 3.57 Φ=

l ∞ q 1 X r< 4πo r> r>

r

l=0

4π (Yl0 (θ, φ) − Yl0 (π − θ, φ)) 2l + 1

From parity of the spherical harmonics l ∞ q 1 X r< Φ(~ x) = 4πo r> r> l=0

r

4π 1 + (−1)l+1 Yl0 (θ, φ) 2l + 1

where r< is the smaller of mod[x, a] and r> is the larger of mod[x, a]. X (b) Keeping the product qa ≡ p/2 constant, take the limit of a → 0 and find the potential for r 6= 0. This is by definition a dipole along the z axis and its potential.

20

So for a < r Φ=

r ∞ 4π q 1 X a l 1 + (−1)l+1 Yl0 (θ, φ) 4πo r r 2l + 1 l=0

In the case a → 0 only the first power of a are examined. The potential is then r q 1 a 4π Φ= (1 + 1) Y10 (θ, φ) 4πo r r 2+1 r 3 q a 4π = (2) cos θ 4πo r2 3 4π p cos θ = X 4πo r2 (c) Suppose now that the dipole of part b is surrounded by a grounded spherical shell of radius b concentric with the origin. By linear superposition find the potential everywhere inside the shell. The potential is then Φ = Φd + Φs . Here only the Al terms because we are inside a sphere. Then ∞

Φ=

X p cos θ + Al rl Pl (cos θ) 4πo r2 l=0

At the boundary condition Φ(b) = 0, then ∞

0=

X p cos θ + Al bl Pl (cos θ) 2 4πo b l=0

The only term needed is l = 1 because the dipole, so p cos θ 4πo b2 p A1 = − 4πo b3

A1 b1 cos θ = −

The potential then becomes Φ=

4.3

p cos θ 4πo

1 − f racrb3 r2

X


For the cylinder in Problem 3.9 the cylindrical surface is made of two equal half-cylinder, one at potential V and the other at potential −V , so that " V for − π/2 < φ < π/2 V (φ, z) = (4.2) −V for π/2 < φ < 3π/2 (a) Find the potential inside the cylinder From class Φ=

∞ X ∞ X

(Anν sin νφ + Bnν cos νφ) Iν

n ν=0

nπρ L

sin

nπz L

Fourier analyst can yield the coefficients. For Anν Z L Z 2π nπz 2 1 sin νφdφdz Anν = V (φ, z) sin πL Iν 0 0 L

21

But this has two intervals from −π/2 to π/2 and π/2 to 3π/2. In these intervals cosine is symmetric so Anν = 0. Now for Bnν Z L Z 2π nπz 2 1 V (φ, z) sin Bnν = sin νφdφdz πL Iν 0 0 L RL 2L over the two ranges, only the odd terms survive for ν. For the z component 0 sin (nπz/L) dz = (2l+1)π , because only odd terms survive from n, also. So Bnν becomes 4(−1)k 2L 2V Bnν = (2l + 1)π 2k + 1 πLI2k+1 =

16V (−1)k X (2k + 1)(2l + 1)π 2 I2k+1

(b) Assuming L b, consider the potential at z = L/2 as a function of ρ and φ and compare it with two-dimensional Problem 2.13. From Jackson I2k+1 and for small angles sin

(2l+1)π L

(2l + 1)πρ L

k+1 (2l + 1)πρ 1 ≈ Γ(2k + 2) 2L

≈ (−1)l . Then

Φ=

∞ X l,k=0

ρ 2k+1 16(−1)k+l V cos((2k + 1)φ) + 1)(2k + 1) b

π 2 (2l

Using the identity ∞

X (−1)l π = tan−1 (1) = (1)2l+1 4 2l + 1 l=0

The potential becomes Φ=

∞ 4V X (−1)k ρ 2k+1 cos((2k + 1)φ) π 2k + 1 b k=0

From problem 2.13 ∞ X (−1)k ρ 2k+1 1 2bρ cos φ cos((2k + 1)φ) = tan−1 2k + 1 b 2 b2 − ρ2

k=0

The potential becomes Φ=

2V 2bρ cos φ tan−1 X π b2 − ρ2

22


5.1


The Dirichlet Green function for the unbounded space between the planes as z = 0 and z = L allows discussion of a point charge or a distribution of charge between parallel conducting planes held at zero potential. (a) Using cylindrical coordinates show that one form of the Green function is ∞ ∞ nπz nπ nπ 4 X X im(φ−φ0 ) nπz 0 G(x, x ) = e sin sin LIm ρ< Km ρ> L n=1 m=−∞ L L L L 0

(5.1)

We know that ∇2 G(x, x0 ) =

−4πδ(ρ − ρ0 ) δ(φ − φ)δ(z − z 0 ) ρ0

(5.2)

Then expanding the δ-functions ∞ ∞ nπz δ(ρ − ρ0 ) 4 X X −imφ0 imφ nπz 0 sin e e sin L m=−∞ n=1 L L ρ0 ∞ ∞ nπz X X 0 nπz 0 δ(ρ − ρ0 ) imφ 4 e sin = − e−imφ sin 0 L L ρ L m=−∞ n=1

∇2 G(x, x0 ) = −

(5.3)

Now expanding the Green function ∇2 G(x, x0 ) = ∇2

∞ ∞ X X

eimφ sin

m=−∞ n=1

nπz L

Ψ

where Ψ is the eigenfunction. Expanding the Laplacian ∞ ∞ nπz X X 1 ∂ ∂ m2 n2 π 2 2 0 ∇ G(x, x ) = ρ − 2 − Ψeimφ sin 2 ρ ∂ρ ∂ρ ρ L L m=−∞ n=1 Then equating equation 3 and equation 4 1 ∂ ∂ m2 n2 π 2 1 ρ − 2 − Ψ ρ ∂ρ ∂ρ ρ L2 e−imφ0 sin Then define gmn =

Ψ 0 . 0 e−imφ sin nπz L

nπz 0 L

=−

(5.4)

4 δ(ρ − ρ0 ) L ρ0

From Equation 3.98 and 3.99 gmn = CIm (kρ< )Km (kρ> ) where k = nπ/L. Now

to find C ∂gmn −4 ∂gmn −4 1 4 − = ⇒Eq3.147 = −Ck →C= ∂ρ0 ρ+ ∂ρ0 ρ− ρL ρL kρ L Now combining the above constant with gmn and equation 3 to find ∞ ∞ nπz nπ nπ 4 X X imφ −imφ0 nπz 0 G(x, x0 ) = e e sin ρ< Km ρ> X Im sin L m=−∞ n=1 L L L L

23

(b) Show that an alternative form of the Green function is Z ∞ X

G(x, x0 ) = 2

m=−∞

∞

0

dk eim(φ−φ ) Jm (kρ)Jm (kρ0 )

0

sinh(kz< ) sinh[k(L − z> )] sinh(kL)

(5.5)

Expanding equation 2 in φ and ρ ∇2 G(x, x0 ) = −2

Z ∞ X m=−∞

=

Z ∞ X m=−∞

∞

∞

0

e−imφ eimφ Jm (kρ0 )Jm (kρ)δ(z − z 0 )kdk

0

h

i 0 −2ke−imφ Jm (kρ0 )δ(z − z 0 ) eimφ Jm (kρ)dk

(5.6)

0

Now expanding the Green function Z ∞ X

∇2 G(x, x0 ) = ∇2

m=−∞

∞

eimφ Jm (kρ)Ψdk

0

where Ψ is the eigenfunction. Expanding the Laplacian Z ∞ 2 ∞ X m2 1 ∂ ∂ ∂ − + ρ eimφ Jm (kρ)Ψdk ∇2 G(x, x0 ) = 2 2 ∂z ρ ρ ∂ρ ∂ρ 0 m=−∞ Z ∞ 2 ∞ X ∂ 2 = − k Ψeimφ Jm (kρ)dk 2 ∂z 0 m=−∞

(5.7)

Equating equation 6 and 7, to find

∂2 1 2 = −2kδ(z − z 0 ) − k Ψ −imφ0 ∂z 2 e Jm (kρ0 )

0

Then gm,k = Ψ/e−imφ Jm (kρ0 ), but the expansion of the δ-function in z is gm,k = C sinh(kz< ) sinh(k(L − z> )). Now finding C ∂gm,k ∂gm,k − = −2k ∂z 0 z+ ∂z 0 z− −kC [sinh(kz) cosh(k(L − z)) + cosh(kz) sinh(k(L − z))] = −2k → −kC sinh(kL) = −2k → C =

2 sinh(kL)

Now combining the above equation with equation 6 and 7 Z ∞ ∞ X 0 sinh(kz< ) sinh(k(L − z>)) X G(x, x0 ) = 2 eimφ e−imφ Jm (kρ)Jm (kρ0 ) sinh(kL) 0 m=−∞

5.2


A localized distribution of charge has a charge density ρ(r) =

1 2 −r r e sin2 θ 64π

(5.8)

(a) Make a multipole expansion of the potential due to this charge density and determine all the nonvanishing multipole moments. Write down the potential at large distance as a finite expansion in Legendre polynomials. For Equation 4.1 to be solved first Equations 4.4-4.6 have to be solved. First notice that the charge density is independent of φ, therefore m = 0. Now notice sin2 θ = 1 − cos2 θ, because this is power 2, then l ≤ 2. Now beginning with Y20 r r r 1 2 4π 2 5 3 5 3 Y20 = cos2 θ − → Y20 = − cos2 θ + 1 → − Y20 = sin2 θ − 4π 2 2 4π 2 3 5 3

24

Now the next term has no power of θ, so the next term would be from Y00 . Therefore r 2√ 2 4π 2 sin θ = 4πY00 − Y20 3 3 5 Now using Equation 4.3 Z 1 2√ 1 4 −r 1 √ 4π r e sin θdrdθdφ = 3 64π 4π 4π V r r Z 5 −2 4π 1 6 −r −30 1 r e sin3 θdrdθdφ = 2 4π V 3 5 64π 4π Back to Equation 4.1 Φ(x) =

1 4πo

6P2 (cos θ) 4π 1 P0 P0 4π 30 P2 (cos θ) 1 − − → Φ(x) = X 1 4π r 5 4π r3 4πo r r3

(b) Determine the potential explicitly at any point in space, and show that near the origin, correct to r2 inclusive,

Φ(r) '

1 4πo

1 r2 − P2 (cos θ) 4 120

(5.9)

We know from Equation 4.2 Z 1 X 1 ∗ 0 0 0l 0 3 0 Ylm (θ, φ) Φ(x) = Ylm (θ , φ )r ρ(x )d x o 2l + 1 rl+1 l,m

Looking ahead to the deserved solution and combined with the previous part only l = 0, 2 and m = 0 will be considers. Then " # r Z ∞ Z r2 ∞ 1 1 3 −r 2 4π 1 2√ −r 4πY00 r e dr − Y20 re dr Φ(x) = o 3 64π 3 5 5 0 64π 0 " # r r 5 r2 P2 (cos θ) 1 1 2√ P0 1 2 4π = 4π √ 6− 1 o 3 3 5 4π 5 64π 4π 64π 1 P0 r2 = − P2 (cos θ) o 16π 480π 1 r2 1 − P2 (cos θ) X = 4πo 4 120 (c) If there exists at the origin a nucleus with a quadrupole moment Q = 10−28 m2 , determine the magnitude of the interaction energy, assuming that the unit of charge in ρ(r) above is the electronic charge and the unit of length is the hydrogen Bohr radius ao = 4πo /me2 = 0.529 × 10−10 m. Express your answer as a frequency by dividing by Planck’s constant h. The quadrupole interaction energy is given in Equation 4.24 W =−

1 XX ∂Ej Qij (0) 6 i j ∂xi

(5.10)

But as it has been shown earlier that m = 0 so the only terms that remain are i = j, but the trace has to be 0, which is eQ 1 ∂Ex 1 ∂Ey ∂Ez W = + − 6 2 ∂x 2 ∂y ∂z eQ 1 3 ∂Ez eQ 1 ρ 3 ∂Ez = ∇·E− = − 6 2 2 ∂z 6 2 o 2 ∂z

25

At the origin ρ(0) =, so the first term drops out, so the energy is then W = transferred into Cartesian coordinates. So for the quadrupole term Φ(x) =

eQ ∂ 2 Φ . 4 ∂z 2

Now the potential has to be

1 1 r2 P2 (cos θ) → Φ(x) = 2z 2 − x2 − y 2 480πo 960πo

from Merzbacher Quantum Mechanics. So

∂2Φ ∂z 2

=

1 , 240πo

W =

then the quadrupole term becomes

e2 Q 960πo a3o

Then the frequency is e2 Q W = = 0.98 MHz X h 960πo ha3o

26


6.1


A point charge q is located in free space a distance d from the center of a dielectric sphere of radius a (a < d) and dielectric constant /o . (a) Find the potential at all points in space as an expansion in spherical harmonics. Inside the sphere ∇ · D = 0, because there is no charge inside Φ(r, θ) =

∞ X

Al rl Pl (cos θ)

r
l=0

For outside the sphere, there are two regions, a < r < d and r < d, which will dictate the expansion of 1/|x − x0 , ie what is r< and r> . Then the outside region will be Φ(r, θ) =

∞ X Bl r−(l+1) Pl (cos θ) + l=0

1 q 4πo |x − x0 |

Using the typical spherical expansion for 1/|x − x0 , the expression for all space is Φ(r, θ) = P∞ Al rL Pl (cos θ) l=0 P∞ q −(l+1) rl Pl (cos θ) B r + l l+1 l=0 4πo d P∞ q −(l+1) dl + 4πo rl+1 Pl (cos θ) l=0 Bl r

rd

This is the potential in terms of Lagrange Polynomials, so in Spherical Harmonics Φ(r, θ) = P∞ q 2l+1 Al rL Yl,0 (θ, φ) r d Bl r−(l+1) + 4π l+1 l=0 4π r o Now to find the coefficients. Using the normal boundary condition ∂Φ ∂Φ = o ∂r r=a− ∂r r=a+ lqal−1 lAl al−1 = −(l + 1)Bl a−(l+2) + o 4πo dl+1 (l + 1) o q Al = − Bl a−(2l+1) + l 4πo dl+1

27

(6.1)

(6.2)

(6.3)

The other boundary condition is for tangential component ∂Φ ∂Φ = ∂θ r=a− ∂θ r=a+ Al al = Bl a−(l+1) +

al q 4πo dl+1

Then the coefficients can easily be found from equation 3 and 4 2l + 1 q 1 Al = /o + (l + 1)/l l 4πo dl+1 qa2l+1 1 Bl = 1− /o + (l + 1)/l o 4πo dl+1

(6.4)

(6.5) (6.6)

Combining equations 5 and 6 into equation 1 yields the potential at every point in space. X (b) Calculate the rectangular components of the electric field near the center of the sphere. Expanding the potential inside the sphere Φ(r, θ) = A1 rP1 (cos θ) + A2 r2 P2 (cos θ) + . . . From Quantum Mechanics, rP1 (cos θ) = z and r2 P2 (cos θ) = z 2 − x2 + y 2 , then the potential becomes " # 5 z 2 − x2 − y 2 3z q + + ... Φ(x, y, z) = 4πo (/o + 2)d2 2(2/o + 3)d3 R So then Ei = − Φdxi . Then the component in Cartesian space q 5 x Ex = + . . . 4πo d2 2/o + 3 d q 5 y Ey = + . . . 4πo d2 2/o + 3 d q 5 3 z Ez = − + + . . . X 4πo d2 /o + 2 2/o + 3 d (c) Verify that, in the limit /o → ∞, your result is the same as that for the conducting sphere. Examining equation 2, the equation would blow-up if Al 6= 0, therefore Al = 0. This is the same for the conductor. Then from that and equation 4, Bl = −

q a2l+1 4πo dl+1

which is the same as a conduction sphere. X

6.2


Two concentric conducting spheres of inner and outer radii a and b , respectively, carry charges ±Q. The empty space between the spheres is half-filled by a hemispherical shell of dielectric (of dielectric constant /o ), as shown in the figure.1 (a) Find the electric field everywhere between the spheres. Let’s assume that E ∝ rr3 . Using Gauss’s Law for a sphere between the two spheres: I D · da = Q 2πr2 (o E(r) − E(r)) = Q Q r = E(r) 2π(o + ) r3 Now checking the conditions on the solution. The curl of the electric field is obviously 0. The divergent of the field does yield the charge density. By the uniqueness theorem, this is the correct solution. X 1 Jackson.

pg 174

28

(b) Calculate the surface-charge distribution of the inner sphere. From Equation 4.40, in particular n · ((D2 − D1 ) = σfree . The half without the dielectric is σfree = r · D(a) = ˆ r·

r Qo Qo = a2 2π(o + ) 2πa2 (o + )

Now for part with the dielectric is directly σfree =

Q X 2πa2 (o + )

(c) Calculate the polarization-charge density induced on the surface of the dielectric at r = a The electric polarization is given by P = ( − o )E. Because in the free space = o , the electric polarization for the n ·(P1 −P2 ). free space is P = 0. Therefore σpol = 0 for the free space half. Now, the polarization density is σpol = −ˆ For the dielectric part σpol = −ˆ r · P(a) = −ˆ r σpol = −

6.3

( − o )Q 2πa2 (o + )

ˆ r ( − o )Q a2 2π(o + ) X

Problem 3: Distressed Simple Cubic

Repeat the analysis of section 4.5 for a cubic crystal which is subjected to a stress such that the lattice separation is elongated along one edge of the cube (the x axis) and contracted along the other cube edges (the y and z axes) by the amounts ax = a(1 + δ) 1 ay,z = a 1 − δ 2

(6.7) (6.8)

where a is the unstressed cubic lattice spacing. Find the susceptibility for two cases, considering that the molecular polarizabilty γmol is a constant scalar (Hint: Calculate Enear at a particular dipole by considering only the 6 nearest dipoles in the cubic lattice): (a) An electric field applied parallel to the x axis The polarization is given on page 161 in Jackson as P = N hpmol i and also hpmol i = o γmol (E + Ei ) From the P previous page Equation 4.63, Ei = 3 + Enear . All that remains is to find Enear . From Equation 4.64: Enear =

3ˆ n(ˆ n · p) − p r3

29

(6.9)

Considering the six nearest neighbors, which occur from the shape(from class notes): 3ˆ n(ˆ n˙pi ) − pi a3i 2px 2px 4px Ex = 3 − 3 − 3 ax ay az 4py 2py 2py Ey = 3 − 3 − 3 ay ax az 4pz 2pz 2pz Ez = 3 − 3 − 3 az ax ay Ei =

Now using the values for the expansion of the nearest neighbors and the fact that δ < 1 and the binomial expansion. 2px 4px 4px 2px 4px 3 18px δ − = Ex = 3 − (1 − 3δ) − 1 + δ =− 3 (6.10) 3 3 a (1 + δ)3 a3 a3 2 a a3 1 − 21 δ a3 1 − 21 δ 4py 2py 2py 2py 3 2py 9py δ Ey = − − = 1 + δ − 3 (1 − 3δ) = 3 (6.11) 3 3 3 (1 + δ)3 3 1 1 3 3 a a 2 a a a 1 − 2δ a 1 − 2δ 4pz 3 2pz 2pz 2pz 2pz 9pz δ Ez = − (6.12) 3 − 3 3 = 3 1 + δ − 3 (1 − 3δ) = 3 3 1 1 3 3 a (1 + δ) a 2 a a a 1 − 2δ a 1 − 2δ So now Enear known. To simplify the problem, the fact that P = o χE can be used. Since Ey = Ez = 0, then Py = Pz = 0. So only Px is needed. 18 18 1 N γmol Px → 1 + N γmol 3 δo − Px = N γmol o Ex Px = N γmol o Ex + o − 3 δ + a 3o a 3 18 N γmol N γmol 1 + N γmol 3 δo − X χo Ex = N γmol o Ex → χ = a 3 1 − 13 N γmol + a183 N γmol δo (b) An electric field applied parallel to the y axis. For an electric field in the y direction. As above, P = o χE, then Ex = Ez = Px = Pz = 0. So only Enear for the y direction is needed. Then 9 9 1 N γmol Py → 1 − N γmol 3 δo − Py = N γmol o Ey Py = N γmol o Ey + o δ+ a3 3o a 3 9 N γmol N γmol 1 − N γmol 3 δo − χo Ey = N γmol o Ey → χ = X a 3 1 − 13 N γmol − a93 N γmol δo

30


7.1


Two long, cylindrical conducting surfaces of radii a and b are lowered vertically into a liquid dielectric. If the liquid rises an average height h between the electrodes when a potential difference V is established between them, show that the susceptibility of the liquid is b2 − a2 ρgh ln(b/a) χe = (7.1) o V 2 where ρ is the density of the liquid, g is the acceleration due to gravity, and the susceptibility of air is neglected. The first thing to find is the electric field between the cylinders. There is no charge between the cylinders, so ∇2 Φ = 0. From the symmetry of the problem, cylindrical coordinates are the proper choice, Further there is radial symmetry, so there is no polar coordinate. So the Laplacian is 1 ∂ ∂Φ r =0 (7.2) r ∂r ∂r The familiar solution is Φ(r) = A ln (r/B) Now the boundary condition that the difference of the two cylinders is V or more exact Φ(b) − Φ(a) = A ln(b/a) = V ⇒ A =

V ln(b/a)

So the electric field is E=−

∂Φ −V ˆ r ˆ r→E= ∂r ln(b/a) r

Now that the electric field is known, the electric displacement is simply D = E for inside the liquid, and for above the liquid, the electric displacement is D = o E. The work to raise the liquid is Z z Z Z Z l 1 1 b W = E · Dd3 x = 2πrdr E2 dz + o E2 dz 2 2 a −d z where d is the length in the liquid and l is the length above. The work is then Z z Z Z l 1 b πV 2 2πrdr E2 dz + E2 dz = ( (z + d) + o (l − z)) W = 2 a ln(b/a) −d z From Equation 4.102, the force needed to left the liquid is ( − o ) πV 2 ∂W Fz = = ∂z V ln(b/a)

31

Now finding the volume of liquid that has been lifted is b2 − a2 πh. The mass is then m = b2 − a2 ρπh. At equilibrium, the force of gravity is equal to the force of the electric field. 2 2 (/o − 1) πV 2 ( − o ) πV 2 b − a ρπh = b2 − a2 ρπh = ln(b/a) o ln(b/a) The susceptibility is defined as χe = /o − 1. Then b2 − a2 ρgh ln(b/a) b2 − a2 ρπh χe πV 2 = ⇒ χe = o V 2 o ln(b/a)

32

X


8.1


A long, right cylindrical, ideal solenoid of arbitrary cross section is created by stacking a large number of identical current-carrying loops one above the other, with N coils per unit length and each loop carrying a current I. (a) In the approximation that the solenoidal coil is an ideal current sheet and infinitely long, use Problem 5.1 to establish that any point outside the coil that H = 0, and that any point inside the coil the magnetic field is axial and equal to H = NI

(8.1)

This can be solved using two ampere loops. Placing a loop outside the cylinder, it yields B = 0. Placing a second loop through a side of the surface perpendicular to the bottom surface. So I B · dl = µo Ienc From the first loop that was outside the cylinder, the induced magnetic field outside is 0. This means that the only field is inside the cylinder. The integral becomes I B · dl = µo Ienc ⇒ BL = µo N IL Using right hand rule, Bz = µo N I. In this system M = 0, then H = B/µo , therefore Bz = µo N I ⇒ H = N I. X (b) For a realistic solenoid of circular cross section of radius a (N a 1), but still infinite in length, show that the “smoothed” magnetic field just outside the solenoid (averaged axially over several turns) is not zero, but is the same in magnitude and direction as that of a single wire on the axis carrying a current I, even if N a → ∞. Compare fields inside and out. In a real solenoid, it is not simple current loops stacked on each other. There is a little bind in the loop to connect the loops. This means there is a current in the vertical direction. So what is the current? This can be found by the fact the total current moves up 1/N each loop. In each loop there is I/N in the vertical direction, so over a unit length there is N loops, so the amount of vertical current is then N I/N = I. Therefore this current in the vertical direction I B · dl ⇒ 2πρB = µo I ⇒ Bθ =

µo I 2πρ

This is the same as a wire along the z-axis. As the number of loops increase per unit length, the more ideal it will become. The size of inside is many orders of magnitude large than the field outside. To be more exact, µo N I/(µo I/2πρ) → 2πρN . So the field inside is 2πρN times larger. The directions of the two are perpendicular to each other. X

33

8.2

Problem 2: Semi-Infinite Cylinder

Consider a semi-infinite solenoid, which we idealize as a cylindrical current sheet of radius a and azimuthal current K per unit length, with one end at the origin and the other end at -∞ along the z-axis. (a) Use the Biot-Savart Law to show that the magnetic induction on the axis is " # µo K z Bz = 1− 2 (a2 + z 2 )1/2

(8.2)

Let’s begin by thinking of a single current loop at the origin. The magnetic-flux density is I dl × x µo I Bz = 4π |x|3 Because this has to end the zˆ direction, x = a cos φˆ ρ + a sin φφˆ + z zˆ, |x|3 = a2 + z 2 a cos φφˆ + 0ˆ z . Then Z 2π 1 µo I Bz = −az cos φ + az sin φ + a2 dφ 4π (a2 + z 2 )3/2 0 =

3/2

, and dl = −a sin φˆ ρ+

a2 µo I zˆ 2 (a2 + z 2 )3/2

Now this has to be integrated from start to finish of the cylinder remembering that I → K Z 0 µo a2 K dz zˆ Bz = 2 2 (a + z 2 )3/2 −∞ π 0 µo a2 K z µo K = zˆ = zˆ cos θ 1/2 2 2 2 2 (a + z ) θ1 −∞ where θ is the angle that the origin makes with the edge at the end of the cylinder and cos θ = " # z µo K 1− Bz = 2 (a2 + z 2 )3/2

z

(a2 +z2 )3/2

, then

X

(b) Use a Taylor series expansion of the field near the axis together with ∇ · B = 0 and ∇ × B = 0 to show that the magnetic induction near the axis is a2 ρ µo K 4 (a2 + z 2 )3/2 " # µo K z 3 a2 zρ2 Bz ≈ 1− − 2 4 (a2 + z 2 )5/2 (a2 + z 2 )1/2

Bρ ≈

(8.3) (8.4)

The two conditions, ∇·B and ∇×B, lead to Laplace’s Equation ∇2 Φmag = 0 and B = −∇Φmag . From electrostatics, Φmag = R(ρ)Z(z)Q(θ). The radial solutions are the Bessel and Neumann functions. Then Φmag = (AJν (kρ) + BNν (kρ)) Ce−kz Deiνθ From symmetry, ν = 0 and as ρ → 0 has to be well define B = 0. Then Φmag = CJ0 (kρ)e−kz . Then the induced field is ∂ ∂ B = −∇Φmag = − ρˆ + zˆ Φmag ∂ρ ∂z ∂Jo (x) −kx = −Ck e ρˆ + CkJo (kρ)e−kz zˆ ∂x

34

From Arfken’s book,

∂Jo (x) ∂x

= −J1 (x)1 . Expanding the two terms2 , kρ k 3 ρ3 Bρ = CkJ1 (kρ)e−kz ≈ Cke−kz − 2 16 2 2 k ρ Bz = CkJ0 (kρ)e−kz ≈ Cke−kz 1 − 4

Now, Bz (0, z) = Cke−kz , this has to be the case, because it has to be continuous. It follows that (−1)n Ckn+1 e−kz . With this substitution, kρ ρ ∂Bz (0, z) Bρ ≈ Cke−kz ≈ 2 2 ∂z 2 2 k ρ ρ2 ∂ 2 Bz (0, z) Bz ≈ Cke−kz 1 − ≈ Bz (0, z) − 4 4 ∂z

∂ n Bz (0,z) ∂z n

=

Using Bz from part one finally # " µo K a2 ρ ρ ∂ µo K z ⇒ Bρ ≈ 1− 3/2 2 ∂z 2 4 (a2 + z 2 )3/2 (a2 + z 2 ) " # " " ## µo K z ρ2 ∂ 2 µ o K z Bz ≈ 1− − 1 − 2 4 ∂z 2 (a2 + z 2 )3/2 (a2 + z 2 )3/2 " # µo K z 3 a2 zρ2 Bz ≈ 1− − X 1/2 2 2 2 2 4 (a + z ) (a + z 2 )5/2 Bρ ≈

8.3


A cylindrical conductor of radius a has a hole of radius b bored parallel to, and centered a distance d from, the cylinder axis (d + b < a). The current density is uniform throughout the remaining metal of the cylinder and is parallel to the axis. Use Amp`ere’s law and principle of linear superposition to find the magnitude and the direction of the magnetic-flux density in the hole. From the hint to use superposition, we will first find the magnetic-flux density for the cylindrical conductors. For the one with radius a applying Amp`ere’s law in the form of Equation 5.24 Z I B · dl = µo J · nda S

C

B2πr2 = µo Jπr2 → B =

µo Jrˆ z × rˆ 2

Now for the hole, it has to be the same except that it has a different coordinate system, that is related by a linear transformation, directly B0 = µ2o J 0 r0 zˆ × rˆ0 , because this is a hole the current density is in the opposite direction, J 0 = −J. Next is relating rˆ r to r0 rˆ0 . Defining the center of the hole on the −y-axis of the cylinder, the relationship 0 0 is r rˆ = rˆ r + dˆ y . Now linear superposition can be used to find Bf inal = B + B0 = =

8.4

µo J zˆ × (rˆ r − rˆ r − dˆ y) 2

µo µo Jd Jdˆ z × (−ˆ y ) ⇒ Bf inal = x ˆ 2 2

X


A sphere of radius a carries a uniform surface-charge distribution σ. The sphere is rotated about a diameter with constant angular velocity ω. Find the vector potential and magnetic-flux density both inside and outside the sphere. 1 Arfken. 2 Arfken.

MMP 5th ed. pg.673 MMP 5th ed. pg.670

35

The surface current is J~ = σ~v = σ(~ ω × ~r0 ) = σaω sin θ0 (r0 − a)φˆ0 . The surface current is then related to l = 1 and m = 1 only, because of the sine term. The

36

vector potential is then Z µo X Pl1 (cos θ) rl < σωaδ(r0 − a) l+1 Pl1 (cos θ0 )r02 sin2 θ0 dr0 dθ0 dφ 4π l(l + 1) r> l 1 3 l Z π X Pl (cos θ) σωa r< µo Pl1 (cos θ0 )P11 (cos θ) sin θ0 dθ0 =− l+1 2 l(l + 1) r 0 > l

Aφ (r, θ) =

=− =

4 µo σωa3 r< 1 P1 (cos θ) 2 4 r> 3

µo σωa3 r< sin θ 2 3r>

Now for the magnetic field, B = ∇×A. This needs to be done for the interior (a > r) and exterior (a < r). The vector sin θ 1 1 potential interior is Aint = µo σωar sin θφˆ . The vector potential for exterior is Aext (r, θ) = µo σωa4 2 φˆ So 3 3 r for the interior ∂ µo σωa 1 ∂ µo σωar2 sin θ 1 sin2 θ − θˆ Bint (r, θ) = rˆ r sin θ ∂θ 3 r ∂r 3 h i 2 r − sin θθˆ Bint (r, θ) = µo σωa cos θˆ 3 Now for the exterior 1 ∂ r sin θ ∂θ

µo σωa4 sin2 θ 3r2

1 ∂ − θˆ r ∂r i 1 µo σωa4 h 2 cos θˆ r + sin θθˆ Bext (r, θ) = X 3 3 r

Bext (r, θ) = rˆ

37

µo σωa4 sin θ 1 3 r

38


9.1


A current distribution J(x) exists in a medium of unit relative permeability adjacent to a semi-infinite slab of material having relative permeability µr and filling the half space, z < 0. (a) Show that for z > 0 the magnetic induction ca be calculated by replacing the medium of permeability µr by an image current distribution, J∗ , with components, µr − 1

!

µr + 1

µr − 1

Jx (x, y, −z),

µr + 1

! Jy (x, y, −z), −

µr − 1 µr + 1

! Jz (x, y, −z)

(9.1)

The field z > 0 is µo 4π

B=

Z

(J + J∗ ) ×

(x − x0 ) 3 0 d x |x − x0 |3

(9.2)

The field z < 0 is B=

µo 4π

Z J×

(x − x0 ) 3 0 d x |x − x0 |3

(9.3)

At the boundary of z = 0, the normal component of the field must be continuous, so above Bz+ (−z) =

µo 4π

Z (Jx + J ∗ )(y − y 0 ) − (Jy + J ∗ )(x − x0 ) x+ y+ d3 x0 3/2 2 2 (x − x0 ) + (y − y 0 ) + z 02

below Bz− (z) =

µo 4π

Jx∗− (y − y 0 ) − Jy∗− (x − x0 ) 3 0 3/2 d x 2 2 0 0 02 (x − x ) + (y − y ) + z

Z

Then Bz+ (−z) = Bz− (z). Therefore Jx (−z) + Jx∗+ (z) = Jx∗− (z)

(9.4)

Jy∗+ (z)

(9.5)

Jy (−z) +

=

Jy∗− (z)

Now for the tangential components of the magnetic field must be continuous, and H = B/µ. So for above Z −z 0 (Jy + Jy∗+ ) − (y − y 0 )(Jz + Jz∗+ ) 3 0 1 Hx+ (−z) = d x 4π ((x − x0 )2 + (y − y 0 )2 + z 02 )3/2 Z (x − x0 )(Jz + Jz∗+ ) + z 0 (Jx + Jx∗+ ) 3 0 1 Hy+ (−z) = d x 4π ((x − x0 )2 + (y − y 0 )2 + z 02 )3/2 Below Hx− (z) =

1 4πµr

Hy− (z) =

1 4πµr

Z

−z 0 (Jy∗− ) − (y − y 0 )(Jz∗− )

((x − x0 )2 + (y − y 0 )2 + z 02 )3/2 Z (x − x0 )(Jz∗− ) + z 0 (Jx∗− ) ((x − x0 )2 + (y − y 0 )2 + z 02 )3/2

39

d3 x0 d3 x0

From H+ = H− µr Jz (−z) + µr Jz+ (z) = Jz− (z)

(9.6)

µr Jy (−z) − µr Jy+ (z) = Jy− (z)

(9.7)

µr Jx (−z) − µr Jx+ (z) = Jx− (z)

(9.8)

From eq 4 and 7, and eq 5 and 8 Jx+ (z) =

µr − 1 Jx (−z) µr + 1

Jx− (z) =

2µr Jx (z) µr + 1

(9.9)

Jy+ (z) =

µr − 1 Jy (−z) µr + 1

Jy− (z) =

2µr Jy (z) µr + 1

(9.10)

Now for the z components. ∇ · J∗− = 0 =

∂Jy∗− ∂Jx∗− ∂Jz∗− 2µr + + = ∂x ∂y ∂z µr + 1

∂Jy ∂Jx + ∂x ∂y

+ µr

∂ ∂ 2µr ∂Jz (Jz + Jz+ ) = − + µr (Jz + Jz+ ) ∂z µr + 1 ∂z ∂z

Then solving and integrating −

µr − 1 µr + 1

Z

Z ∂Jz =

∂Jz+ ⇒ −

µr − 1 Jz (−z) = Jz+ (z) µr + 1

(9.11)

From equations 9, 10, 11 for J∗+ matches equation 1. X (b) Show that for z < 0 the magnetic induction appears to be due to a current distribution [2µr /(µr + 1)]J in a medium of unit relative permeability. Combining equation 6 with equation 11 to find Jz∗− (z) =

2µr Jz (z) µr + 1

(9.12)

Adding equation 12 with the J− parts of equations 9, 10 to find that J∗− = [2µr /(µr + 1)]J X

9.2


A magnetically “hard” material is in the shape of a right circular cylinder of length L and radius a. The cylinder has a permanent magnetization Mo , uniform throughout its volume and parallel to its axis. (a) Determine the magnetic field H and magnetic induction B at all points on the axis of cylinder, both inside and outside. Because there is no free current, this is a magnetostatic potential problem. This problem then can be treated as an electrostatic problem. Because the magnetization is uniform, ρM = −∇ · M = 0. The surface ‘charge’ density is then σM = n ˆ · M = ±Mo at the caps. The potential is then Z 1 σM ΦM = d3 x0 4π |x − x0 | This is evaluated at the caps only. Then Zρ

q a Mo 2πρdρ q = ρ2 + (z − L/2)2 2 0 ρ2 + (z − L/2)2 0 q q Mo = a2 + (z − L/2)2 − (z − L/2)2 2

Φ1 =

Mo 4π

This can easily be show for the bottom with symmetry that q q Mo Φ2 = − a2 + (z + L/2)2 − (z + L/2)2 2

40

The potential over all space is Φ = Φ1 + Φ2 . Φ =

v u Mo  u t  a2 + 2

z−

L

!2

v u u − ta2 +

L z+

2

2

!2

 L L  + z +  − z − 2 2

The magnetic field has two cases, one inside Hz =

∂Φ ∂z 



Mo  z + L/2 z − L/2  Hz = −q − 2 q 2 2 2 2 2 a + (z + L/2) a + (z − L/2)

And outside  Hz =



z + L/2 Mo  z − L/2  −q q  2 a2 + (z + L/2)2 a2 + (z − L/2)2

Now the induced field is Bz = µo (Hz + M ). In this case, M = Mo inside and M = 0 outside. Then the induced field both inside and outside is 



z − L/2 µo M o  z + L/2  −q Bz =  q 2 2 2 2 2 a + (z + L/2) a + (z − L/2)

X

(b) Plot the ratios B/µo Mo and H/Mo on the axis as functions of z for L/a = 5. See last page

9.3


Show that in general a long, straight bar of uniform cross-sectional area A with uniform lengthwise magnetization M , when placed with its flat end against as infinitely permeable flat surface, adheres with a force given approximately by F '

µo AM 2 2

(9.13)

Relate your discussion to the electrostatic considerations in Section 1.11. So consider this as two long straight bars with uniform cross-sectional areas A with uniform magnetization, M . One rod runs from z = 0 to z = −L and the other from z = 0 to z = L. These two dipoles are “head” to “tail”, so the components of the B-field cancel everywhere except at z = 0. Now, let’s introduce a small separation zo . The potential energy on the real rod from the image rod is Z Z W (zo ) = − M · Bi d3 x = −M Bz,i d3 x V

Because the image rod is not moved and the rod is much longer than it is any other direction Z Z zo +L W (zo ) = −M Bz,i d3 x ≈ −M A Bz,i (z)dz V

zo

The force from the change in the vertical separation dW (zo ) = M A(Bz,i (L) − Bz,i (0)) Fz = − dzo zo =0 From the previous problem, if the shape is circular, which is not that bad of an approximation, the induced field is ! µo M z+L z p Bz,i (z) = −√ 2 a2 + z 2 a2 + (z + L)2

41

Figure 9.1: H/Mo =

1 2

√

z/L+1/2 1/25+(z/L+1/2)2

−√

z/L−1/2 1/25+(z/L−1/2)2

-0 |z/L| > 1/2 or1 |z/L| ≤ 1/2

The force becomes Fz =

Now the condition that a L,

µo M 2 A 2

Fz = −

L L 2L √ −√ −√ +0 a2 + 4L2 a2 + L2 a2 + L2

µo M 2 A 2

X

Now comparing the problem with 1.11. From the previous problem inside the rod, σM = M on the end above the plain, and σM = −M below the plain. So µo M inside Bz = 0 outside Because of the constant induced field, this is similar to a constant electric field. In fact, the magnetic potential is similar to the electric potential of plane capacitors with σE /o → µo σM . To create a capacitor, lets separate the real rod, which is the one above the plane, and the image rod. Now on page 43, F/A = σ 2 /2o for a capacitor. Then µo 2 2 F/A = σM µ2o /2µo → F/A = µo σM /2. Finally the value of σM , F/A = µo M 2 /2 → F = AM 2 2

42

Figure 9.2: B/µo Mo =

1 2

√

z/L+1/2 1/25+(z/L+1/2)2

43

−√

z/L−1/2 1/25+(z/L−1/2)2

44


10.1


A long, hollow, right circular cylinder of inner (outer) radius a (b), and of relative permeability µr , is placed in a region of initially uniform magnetic-flux density Bo at right angles to the field. Find the flux density at all points in space, and sketch the logarithm of the ratio of the magnitudes of B on the cylinder axis to Bo as a function of log10 µr for a2 /b2 = 0.5, 0.1. Neglect end effects. Because there are no end effects, this problem can be reduced to a two dimensional problem. With the z-axis the length of the cylinder and the Bo in the x-direction. With those being set, this is much like the problem done in class. So ∇ · B = ∇ · (µH) = 0 and for inside and outside ∇ × H = 0. The solution is H = −∇Φm where ∞ ∞ P P Φ = ao + bo ln ρ + an ρn (cos nϕ + sin nϕ) + bn ρ−n (cos nϕ + sin nϕ). All that remains is to solve for the n=1

n=1

coefficients. By symmetry the sine terms can be dropped. For ρ > b, an = 0, and for ρ < a, bn = 0. So the potential is ( P∞ α ρn cos nϕ n Pn=1 ∞ βn ρn cos nϕ + γn ρ−n cos nϕ Φm (ρ, ϕ) = Pn=1 ∞ −n cos nϕ n=1 δn ρ

ρ b

H has to be continuous at the boundaries. At ρ = b, the differential with respect to the radial is ∞ ∞ X X Bo cos ϕ + −n βn bn−1 − γn b−(n+1) cos nϕ nb−(n+1) δn cos nϕ = µr µo n=1 n=1

From this n = 1, so Bo + δ1 b−2 = −µr β1 + µr γ1 b−2 µo

(10.1)

The tangential continuation at ρ = b −

Bo + δ1 b−2 = β1 + γ1 b−2 µo

(10.2)

Likewise at ρ = a α1 = µr β1 − µr γ1 a−2 −2

α1 = β1 + γ1 a

(10.3) (10.4)

Reducing the number of coefficients in equations 1 and 2 (µr − 1) −2 (µr + 1) Bo δ1 b − 2µr 2µr µo (1 − µr ) Bo 2 (µr + 1) γ1 = δ1 + b 2µr 2µr µo

β1 =

45

(10.5) (10.6)

Doing the same with equations 3 and 4 (µr + 1) α1 2µr (µr − 1) γ1 = α1 a2 2µr

β1 =

(10.7) (10.8)

Now combining equations 5 and 7 (µr − 1) −2 Bo δ1 b − (µr + 1) µo

(10.9)

(µr + 1) b2 Bo δ1 a−2 − 2 (µr − 1) a µo

(10.10)

α1 = Now for equations 6 and 8 α1 = Equating 9 and 10

µ2r − 1 b2 a2 Bo 2 δ1 = 1 − 2 b b (µr + 1)2 b2 − (µr − 1)2 a2 µo

(10.11)

With equation 9 and 11 α1 =

Bo −4µr b2 2 2 2 2 (µr + 1) b − (µr − 1) a µo

(10.12)

Placing equation 12 into equation 7 and also equation 8 β1 =

−2 (µr + 1) b2 Bo 2 2 2 2 (µr + 1) b − (µr − 1) a µo

(10.13)

γ1 =

−2 (µr − 1) b2 Bo 2 a (µr + 1)2 b2 − (µr − 1)2 a2 µo

(10.14)

Now the field is ( H=

−α ρ+α ˆ 1 cos ϕˆ 1 sin ϕϕ γ1 γ1 − β1 − ρ2 cos ϕˆ ρ + β1 + ρ2 sin ϕϕ ˆ δ1 δ1 Bo Bo ρ − µo + ρ2 sin ϕϕ ˆ + ρ2 cos ϕˆ µo

ρ b

See last page for graph X

10.2


A circuit consists of a long thin conducting shell of radius a and a parallel return wire of radius b on axis inside. Of the current is assumed distributed uniformly throughout the cross section of the wire, calculate the self-inductance per unit length. What is the self-inductance if the inner wire is a thin tube? First let’s apply Ampere’s law Z B · dl = µo Ienc So quickly, B is ( B=

µo I ρ 2π b2 µo I 1 2π ρ

0

46

ρa

(10.15)

The energy of the magnetic field per length of the system is given by Eq. 5.148 Z Z 1 1 H · Bd3 x = B 2 d3 x W = 2 2µo 2 Z b 3 Z a 2π µo I ρ dρ dρ = + 2µo 2π b4 ρ 0 b 2 a a π µo I 1 µo I 2 = + ln = 4 ln +1 µo 2π 4 b 16π b From Eq. 5.152, W = 12 LI 2 . The self-inductance per unit length is a a µo I 2 µo I 2 1 2 LI = 4 ln +1 ⇒ L= 4 ln +1 2 16π b 8π b

X

Now if the inner wire is a tube B = 0 for ρ < b, so there is only the middle term which yields a self-inductance per unit length of µo a L= ln X 2π b

10.3


The figure (See Jackson pg. 232) represents a transmission line consisting of two, parallel perfect conductors of arbitrary, but constant, cross section. Current flows down one conductor and returns via the other. Show that the product of the inductance per unit length L and the capacitance per unit length C is LC = µ

(10.16)

where µ and are the permeability and the permittivity of the medium surrounding the conductors. First using Ampere’s Law for outside the wires Z B · dl = µI ⇒ B = But inside the wire B = 0. The vector potential is Z Z Az = Bφ dρ =

µI 2πρ

µI µI dρ ⇒ Az = − ln 2πρ 2π

ρ ρo

where ρo = a, b. The work per unit length can be found from Eq. 5.149 Z Z 1 1 W = J · Ad3 x = (Ja Adaa + Jb Adab ) 2 2 Z a Z b d d 1 µI 2πI 2πI ln ρ dρ + ln ρ dρ = a a b b 2 2 2 2π b a 0 πa 0 πb 2 b2 πI 1 µI a πI d d = ln + ln 2 2π πa2 b πb2 a 2 µI 2 d = ln 4π ab Relating this to the self-inductance per unit length 1 2 µI 2 LI = ln 2 4π

d2 ab

⇒L=

µ ln 2π

d2 ab

From the second problem 1.7, the capacitance is C= ln

2π

47

d2 ab

(10.17)

Then µ ln 2π

LC =

d2 ab

ln

2π ⇒ LC = µ 2 d ab

X

Capacitance The electric field for a wire is given by Q rˆ 2πo r

~ = E The potential between the two wire is Z Φ=

The capacitance is given by C =

Q . Φ

=

Q 2πo

=

Q 2πo

Z

d−a2

h

Z d−a1 dr dr + r r a a2 1 d − a2 d − a1 ln + ln a1 a2 (d − a2 ) (d − a1 ) ln a1 a2

~ · d~l = E

Q 2πo

Then C=

Q = Φ

Q Q 2πo

ln

(d−a2 )(d−a1 ) a1 a2

i

2πo

= ln

(d−a2 )(d−a1 ) a1 a2

Since the distance between the wires is much large than their radii, d − a1 ≈ d − a2 ≈ d. Then capacitance reduces to C≈ ln

2π o

48

d2 a1 a2

Figure 10.1:

B Bo

=

µr 2 (µr +1)2 −(µr −1)2 a b2

49

50


11.1


A dielectric sphere of dielectric constant and radius a is located at the origin. There is a uniform applied electric field Eo in the x direction. The sphere rotates with an angular velocity ω about the a axis. Show that there is a magnetic field H = −∇ΦM , where ΦM =

3 5

− o + 2o

o Eo ω

a r>

5 · xy

(11.1)

where r> is the larger of r and a. The motion is nonrelativistic. First the polarization from Eq. 4.57 Px = 3

− o Eo x ˆ → Px = αEo x ˆ + 2o

So the volume charge density is −∇ · P = 0. The surface charge density is σ=n ˆ·M=n ˆ · ((r · P)ωˆ z ) = αo Eo ωa sin θ cos θ cos φ r 8π 1 = −αo Eo ωa [Y2,1 − Y2,−1 ] 15 2 The potential is σ 0 da0 |x − x0 | r l 8π 1 a2 X 4π r< = −αo Eo ωa Ylm (θ, φ)× l+1 15 2 4π 2l + 1 r> Z ∗ Ylm (θ0 , φ0 ) Y2,1 (θ0 , φ0 ) − Y2,−1 (θ0 , φ0 ) sin θ0 dθ0 dφ0 r 2 8π 1 a2 r< = −αo Eo ωa [Y2,1 (θ, φ) − Y2,−1 (θ, φ)] 3 15 2 5 r> " r # r 2 8π 1 a2 r< 15 −2 = −αo Eo ωa sin θ cos θ cos φ 3 15 2 5 r> 8π

ΦM =

1 4π

Z

r3 3 − o o Eo ωa3 < sin θ cos θ cos φ 5 5 + 2o r> 5 3 − o a ΦM = o Eo ωxz X 5 + 2o r>

=

51

(11.2)

11.2


An example of the preservation of causality and finite speed of propagation in spite of the use of the Coulomb gauge is afforded by a dipole source that is flashed on and off at t = 0. The effective charge and current densities are ρ(x, t) = δ(x)δ(y)δ 0 (z)δ(t)

(11.3)

Jz (x, t) = −δ(x)δ(y)δ(z)δ 0 (t)

(11.4)

where the prime means differentiation with respect to the argument. This dipole is of unit strength and it points in the negative Z direction. (a) Show that the instantaneous Coulomb potential (6.23) is Φ(x, t) = −

1 z δ(t) 3 4πo r

(11.5)

Using Eq. 6.23 1 Φ(x, t) = 4πo

Z

ρ |x − x0 |

3 d x =

1

Z

0 0 0 0 δ(x )δ(y )δ (z )δ(t)

4πo

d3 x |x − x0 |

1

= −

∂ δ(t)

4πo

1

∂z 0 |z − z 0 |

! = −

1 δ(t) 4πo

−

−z 0 − z |x − z ˆ0 |3

! = −

1

z δ(t)

4πo

r3

X

(b) Show that the transverse current Jt with a factor of 2/3 multiplying the delta function coming from treating the gradient of z/r3 according to (4.20). 2 3 3 Jt (x, t) = −δ (t) 3 δ(x) − + n(3 · n) 3 4πr3 4πr3 0

(11.6)

From Eq. 6.25 and 6.29 1 1 0 z ∂Φ = −δ(x)δ(y)δ(z)δ 0 (t)ˆ z+ ∇ δ (t) 3 ∇ 2 µo c ∂t 4π r 3ˆ r (ˆ z · r ˆ ) − zˆ 4π 1 δ 0 (t) δ(x)δ(y)δ(z)ˆ z− = −δ(x)δ(y)δ(z)δ 0 (t)ˆ z+ 4π 3 r3 2 1 3 = −δ 0 (t) δ(x)ˆ z− zˆ + rˆ(ˆ z · rˆ) X 3 4πr3 4πr3

Jt = J − Jl = J −

(c) Show that the electric and magnetic fields are causal and that the electric field components are (with Ey the same as Ex except with cos φ replaced by sin φ), 3 3 1 c −δ 00 (r − ct) + δ 0 (r − ct) − 2 δ(r − ct) sin θ cos θ cos φ 4πo r r r 0 δ (r − ct) δ(r − ct) 1 c sin2 θδ 00 (r − ct) + (3 cos2 θ − 1) · − Ez (x, t) = 4πo r r r2

Ex (x, t) =

From section 6.3, A is causal, ∴ B is causal. Now for E. From equation 6.48 Z µo 1 A= [J] d3 x 4π R ret Z ∂A µo 1 ∂ R 1 0 ∂ 1 =− δ0 t − δ(x0 )ˆ z+ ∇ d3 x0 ∂t 4π R ∂t c 4π ∂z 0 r0 Z µo 1 00 R µo dk 1 ∂ 2 ik(t−R/c) 1 0 ∂ 1 3 0 =− δ t− − e ∇ d x 4π r c 4π 2π R ∂t2 4π ∂z 0 r0 Now ∇02

1 eik(t−R/c) = (−ik/c)2 eik(t−R/c) − eik(t−R/c) 4πδ(x − x0 ) R R 1 ∂ 2 eik(t−R/c) − eikt 4πδ(x − x0 ) = 2 2 c ∂t R

52

(11.7) (11.8)

With that ∂A µo 1 00 µ o c2 =− δ (t − R/c)ˆ z− ∂t 4π r 4π

Z

dk 2π Z

eik(t−R/c) ∂ 1 ∇02 + eikt δ(x − x0 ) ∇0 0 0 d3 x0 4πR ∂z r dk eik(t−R/c) 0 ∂ 02 1 3 0 ∂ 1 ∇ ∇ d x + δ(t)∇ 2π 4πR ∂z 0 r0 ∂z r

µo µo 1 00 δ (t − R/c)ˆ z− 4π r 4πµo o Z µo 1 00 1 dk eik(t−R/c) 0 ∂ z 1 =− δ (t − R/c)ˆ z+ ∇ δ(t)∇ 3 δ(x0 )d3 x0 − 4π r 4πo 2π R ∂z 0 4πo r =−

Now integrating the second term by parts ∂ eik(t−R/c) 1 z ∇ δ(x0 )d3 x0 − δ(t)∇ 3 ∂z R 4πo r

µo 1 00 1 ∂A =− δ (t − R/c)ˆ z+ ∂t 4π r 4πo

Z

dk 2π

µo 1 00 1 δ (t − R/c)ˆ z+ 4π r 4πo

Z

dk ∂ eik(t−r/c) 1 z ∇ − δ(t)∇ 3 2π ∂z r 4πo r

=−

Again working with part of the second term ike−ikr/c e−ikr/c ∂ ik 1 ∂ e−ikr/c −ikr/c =∇ − − r = −∇ + e cos θ ∇ ∂z r cr r2 ∂z cr r2 2 2 2ik k ik 1 sin θ ˆ −ikr/c −ikr/c = + − cos θe r ˆ + e + θ r3 cr2 c2 r cr r2 r Now integrating over k-space, which is just a Fourrier transform µo 1 00 1 1 0 1 00 1 z 1 ∂A 2 cos θˆ r + sin θ θˆ + 2 δ (t − R/c) cos θˆ =− δ (t − R/c)ˆ z+ δ(t − R/c) − δ (t − R/c) r − δ(t)∇ 3 ∂t 4π r 4πo r3 cr 2 c r 4πo r µo c3 1 00 1 00 c 1 1 0 1 z =− 2 cos θˆ r + sin θ θˆ + δ (ct − R) cos rˆ − δ (ct − R) cos θˆ r − sin θ θˆ + δ(ct − R) − 2 δ (ct − R) δ(t)∇ 3 3 4π r 4πo r r r 4πo r 1 c 1 0 1 z 1 00 =− δ (ct − R) − 2 δ(ct − R) 2 cos θˆ r + sin θ θˆ − δ (ct − R) sin θ θˆ − δ(t)∇ 3 4πo r r r 4πo r

Now this can be used to find the electric field ∂A ∂t 1 z 1 c 1 0 1 1 z = δ(t)∇ 3 − δ (ct − R) − 2 δ(ct − R) 2 cos θˆ r + sin θθˆ − δ 00 (ct − R) sin θθˆ − δ(t)∇ 3 4πo r 4πo r r r 4πo r 1 0 1 1 c δ (ct − R) − 2 δ(ct − R) =− 2 cos θˆ r + sin θθˆ − δ 00 (ct − R) sin θθˆ 4πo r r r

E = −∇Φ −

Now the electric field in Cartesian coordinates 1 c 1 0 1 Ex = − δ (ct − R) − 2 δ(ct − R) (2 cos θ sin θ cos φ + sin θ cos θ cos φ) − δ 00 (ct − R) sin θ cos θ cos φ 4πo r r r 1 c 3 0 3 00 = −δ (r − ct) + δ (r − ct) − 2 δ(r − ct) sin θ cos θ cos φ 4πo r r r 1 c 1 0 1 Ey = − δ (ct − R) − 2 δ(ct − R) (2 cos θ sin θ sin φ + sin θ cos θ sin φ) − δ 00 (ct − R) sin θ cos θ sin φ 4πo r r r 1 c 3 3 = −δ 00 (r − ct) + δ 0 (r − ct) − 2 δ(r − ct) sin θ cos θ sin φ 4πo r r r 1 c 1 0 1 2 2 00 2 Ez = − δ (ct − R) − 2 δ(ct − R) 2 cos θ − sin θ + δ (ct − R) sin θ 4πo r r r 0 δ (r − ct) δ(r − ct) 1 c = sin2 θδ 00 (r − ct) + (3 cos2 θ − 1) · − 4πo r r r2 So the electric field is causal because the only time dependence is in the δ-functions. X

53

11.3


A transverse plane wave is incident normally in vacuum on a perfectly absorbing flat screen. (a) From the law of conservation of linear momentum, show that the pressure (called radiation pressure) exerted on the screen is equal to the field energy per unit volume in the wave. The total electromagnetic momentum in an infinitesimal volume for a small area, ∆A over a small time, ∆t, ∆P = g∆Ac∆t The force due to the momentum, F = ∆P/∆t = g∆Ac. The pressure is then perpendicular to the force per area. g∆Ac S = gc = = energy per unit volume or u X ∆A c

p=

(b) In the neighborhood of the earth the flux of electromagnetic energy from the sun is approximately 1.4 W/m2 . If an interplanetary “sailplane” had a sail of mass 1 g/m2 of area and negligible other weight, what would be its maximum acceleration in meters per second squared due to the solar radiation pressure? How does this compare with the acceleration due to the solar “wind” (corpuscular radiation)? The acceleration do the radiation pressure arad =

S/c 1.4 × 103 /(3 × 108 ) = ⇒ arad ≈ 4.67 × 10−3 m/s2 m/A 10−6

There are 105 protons/m2 s with an average velocity v = 4 × 105 m/s awind =

So arad awind

mv/tA 1.67 × 10−27 × 4 × 105 × 105 −14 2 = = 6.68 × 10 m/s m/A 10−6

X

54


12.1


A uniformly magnetized and conducting sphere of radius R and total magnetic moment m = 4πM R3 /3 rotates about its magnetization axis with angular speed ω. In the steady state no current flows in the conductor. The motion is nonrelativistic; the sphere has no excess charge on it. (a) By considering Ohm’s law in the moving conductor, show that the motion induces an electric field and a uniform volume charge density in the conductor, ρ = −mω/πc2 R3 . Here there are no free currents, so there is scalar potential that can describe the magnetic field. H = B/µo − M → ∇ · H = −∇ · M, then ∇2 ΦM = ∇ · M. Here M = M η (R − r) zˆ1 , so ∂ ∂r M η(R − r) = −M δ(r − R) ∂z ∂z z = −M δ(r − R) = −M δ(r − R) cos θ r

∇·M=

Then 2

∇ ΦM = −M δ(r − R) cos θ ⇒ Φ =

1 4π

Z

M δ(r − R) cos θ 0 3 0 d r |r − r0 |

This integral can be solved using a multipole expansion as done with electrostatics for inside a sphere. 1 ΦM =

∞ X X

Z M δ(r 0 − R) cos θ 0

4π

r0

4π l=0 m 2l + 1 1

=

∞ X

l X

4π

s

1 =

∞ X

l X

R2

4π M

4π l=0 m=−l 2l + 1 4π

4π l=0 m=−l 2l + 1

3 s

4π 3

Y

l+1

Rl+1

Z

∗

0 0 3 0 l lm(θ , ϕ )d r r Ylm (θ, ϕ)

l ∗ Y10 Ylm dΩr Ylm

1 1 l M δl,1 δm,0 r Ylm = M r cos θ = Mz 3 3

The magnetic field inside the sphere is H = −∇ · Φ = 13 M zˆ. The magnetic induction is 1 1 2 B = µo (H + M) = µo M zˆ + M zˆ ⇒ B = µo M zˆ 3 3 3 From Ohm’s Law E = −v × B = − (ω × r) × B =

2 3

E = −

2 3

µo M ω (xˆ x + y y) ˆ = −

µo M ω z ˆ × (ˆ z × r) =

2 3

µo M ω (ˆ z · rˆ z − r)

1 µo mω (xˆ x + y y) ˆ 2 R3

The charge density is mω ρ 4 4 Mω = − µo M ω = − =− o 3 3 o c2 πo πc2 R3 ρ mω mω ∇·E= =− ⇒ ρ=− X o πo πc2 R3 ππc2 R3

∇·E=

1 Here

η is a step function

55

(b) Because the sphere is electrically neutral, there is no monopole electric field outside. Use symmetry arguments to show that only a quadrupole field exists outside and that the quadrupole moment tensor has nonvanishing components, Q33 = −4mωR2 /3c2 , Q11 = Q22 = −Q33 /2. The surface of the sphere has to have positive charge some place on it, just as it has to have negative charge on it. These charges cancel out because there is no monopole term. Now considering a flip of the radial vector. This obviously does not change the sign of the angular momentum. This will change the sign of ∇ and J, so the magnetic field does not change signs. From this the magnetic induction does not change sign. Because ρ is independent of the radial vector, it does not change sign. This means that under a radial flip, the problem is the same, so the odd terms of l must vanish. Since the dipole is odd in l, it is zero, therefore the first nonvanishing term is the quadrupole. Because of the azimuthal symmetry about z axis, the Qxx term can not be distinguished from the Qyy term. From this, Qxx = Qyy . Because the quadrupole term is a traceless matrix, Qxx + Qyy + Qzz = 0 ⇒ Qxx = Qyy = −Qzz /2. For the off diagonal terms, under x → −x, y, z → y, z are the Qxy and Qxz terms. From arguments before, which are r → −r terms are 0, Qxy = Qxz = 0. Now for y → −y x, z → x, z, which are Qyx (which from before is 0) and Qyz , are 0 from the fact that the is equivalent to r → −r. I show that Qzz = −4mωR2 /3c2 in the next part. X (c) By considering the radial electric fields inside and outside the sphere, show that the necessary surface-charge density σ(θ) is σ(θ) =

5 1 4mω 1 − P (cos θ) 2 4πR2 3c2 2

(12.1)

For inside the sphere from part a 2 µo M ω 2 2 E = − µo M ω (xˆ x + y yˆ)) = −∇Φ1 ⇒ Φ1 = x +y +C 3 3

In terms of spherical harmonics, µo M ω 2 µo M ω 2 2 x + y 2 + C ⇒ Φ1 = r sin θ + C 3 3 ! r µo M ω 2 2 4π = r 1− Y20 + C 3 3 5 ! r 2 4π 2 = µo M ωr 1 − Y20 + C 9 5

Φ1 =

For outside the sphere, there is no charge and no dipole terms, so the first no vanishing term is the quadrupole term. The potential outside is Φ2 = A2

R r

3 Y20

The potential must be continuous at the boundaries, so 2 Φ1 (R) = Φ2 (R) ⇒ µo M ωR2 9

Then C = − 92 µo M ωR2 and A2 = − 29 ( D=

q

4π µo M ωR2 . 5

r 1−

4π Y20 5

! + C = A2 Y20

The displacement field is

Mω ω r sin2 θˆ r − 23 M r sin θ cos θθˆ c2 c2 2 M ω R5 ω R5 − 3 c2 r4 P2 (cos θ) rˆ − 23 M sin θ cos θθˆ c2 r 4

− 23

56

rR

From ∇ · D = ρ with ρ evaluated at r = R, the surface charge density becomes 2 Mω 2 Mω RP2 (cos θ) + R sin2 θ 3 c2 3 c2 2 2 Mω 2 Mω 2 =− RP (cos θ) + R − P (cos θ) 2 2 3 c2 3 c2 3 3 4 Mω 5 =− R 1 − P2 (cos θ) 9 c2 2 1 4mω 5 = 1 − P2 (cos θ) 4πR2 3c2 2

σ(θ, ϕ) = −

For Qzz Z Qzz = 2

2 3 ρr P2 (cos θ)d r

1 4mω 5 mω 2 3 η(R − r) + 1 − P2 (cos θ) δ(r − R) r d r πc2 R3 4πR2 3c2 2 ! R 5 4mω Z 4 1 − P2 (cos θ) δ(r − R) r drd(cos θ) = −η(R − r) + c2 R 3 3 2 Z = 2 −

=

10 mωR2 Z −1 3

c2

Qzz = −

1

2 (P2 (cos θ)) d(cos θ)

4mωR2 3c2

X

(d) The rotating sphere serves as a unipolar induction device if a stationary circuit is attached by a slip ring to the pole and a sliding contact to the equator. Show that the line integral of the electric field from the equator contact to the pole contact (by any path) is E = µo mω/4πR The EMF is given by r=R,θ=π/2 Z

r=R,θ=0 Z

E=

E · dl =

∇Φ · dl = Φ(R, π/2) − Φ(R, 0) r=R,θ=0

r=R,θ=π/2

2 2 µo M ωR2 µo M ωR2 − µo M ωR2 + µo M ωR2 = 3 9 9 3 µo mω E= X 4πR

=

12.2


An ideal circular parallel plate capacitor of radius a and plate separation d a is connected to a current source by axial leads. The current in the wire is I(t) = Io cos ωt. (a) Calculate the electric and magnetic fields between the plates to second order in powers of the frequency (or wave number), neglecting the effects of fringing fields. As in the static case Eθ = Eρ = 0 and Bρ = Bz = 0. From Gauss’s law in statics Ez =

σ Q Io = = o πa2 o −iωπa2 o

From Ampere’s law 2πρBθ = µo o (−iω)πρ2 Ez ⇒ Bθ =

µo Io ρ 2πa2

From Faraday’s law, the next order correction is ∇×E=

∂Ez ˆ ∂B θ=− = iωBθ θˆ ∂ρ ∂t

57

Then Z Io iωµo Io ρ2 Ez (ρ) = Ez (0) − iω Bθ dρ ⇒ Ez d = − 2 −iωπa o 4πa2 2 2 1ω ρ Io 1− Ez (ρ) = −iωπa2 o 4 c2 Now the correction to the B-field. ∂E 1 ∂(ρBθ ) = µo o = −iµo o ωEz zˆ ρ ∂ρ ∂t

∇×B= So Bθ (ρ) = −

Bθ (ρ) =

iµo o ω ρ

µo Io ρ 2πa2

Z Ez ρdρ = − 1 ω 2 ρ2 8 c2

1−

ρ2 1 ω 2 ρ4 − 2 16 c2

iµo o ω Io ρ −iωπa2 o

!

!

Now for the time dependence, because the time dependence is of the form e−iωt = cos ωt − i sin ωt, the real parts are

Bθ (ρ)

2 2 Io 1 − 14 ωc2ρ sin ωt ωπo a2 2 2 o Io ρ = µ2πa 1 − 18 ωc2ρ cos ωt 2

Ez (ρ) =

X

(b) Calculate the volume integrals of we and wm that enter the definition of the reactance X, (6.140), to second order in ω. Show that in terms of the input current Ii , defined by Ii = −iωQ, where Q is the total charge on one plate, these energies are Z

The energy in the fields is we =

  Z µo |Ii |2 d ω 2 a2 |Ii |2 d 3 1 +  , wm d x = 4πo ω 2 a2 4π 8 12c2 1

3 we d x =

o |E|2 4

and wm =

|B|2 . µo 4

(12.2)

From part a

o |E|2 o |Ii |2 = 4 4 (ωπo a2 )2 2 |B| µo |Ii |2 ρ2 1 ω 2 ρ2 = ≈ 1+ µo 4 16π 2 a4 8 c2

we = wm

The integral for we over all of space in between the plates Z Z o |Ii |2 2 |Ii |2 d a 1 |Ii |2 d ρdρdθdz = ρdρ = 2 2 4 2 4πo ω a 0 4πo ω 2 a2 4 (ωπo a ) For wm Z

1 ω 2 ρ2 8 c2

!

a4 1 ω 2 a6 + 4 48 c2

!

µo |Ii |2 ρ2 16π 2 a4

µo |Ii |2 d 4π 2a4

1+

ρdρdθdz =

=

µo |Ii |2 d 4π 8

µo |Ii |2 d 4π 2a4 1+

Z

ω 2 a2 12c2

3

ρ +

1 ω 2 ρ5 8 c2

! dρ

! X

2 (c) Show that the equivalent series circuit √ has C ' πo a /d, L ' µo d/8π, and that an estimate for the resonant frequency of the system is ωres ' 2 2c/a. Compare with the first root of Jo (x).

From section 6.9, the capacitance X=

1 4ω 1 |Ii |2 d d πo a2 = = ⇒C= 2 2 2 2 ωC |Ii | 4πo ω a ωπo a d

For the inductance X = ωL =

4ω µo |Ii |2 d ωµo d µo d = ⇒L= |Ii |2 4π 8 8π 8π

58

Resonance occurs when ωL = 1/ωC, ωL =

1 1 ⇒ω= √ = r ωC LC

1 µo d πo a2 8π

r =

8 = µo o a2

r

d

√ 8c2 = 2 2c/a a2

If the next term in the B-field is kept the inductance becomes µo d ω 2 a2 L= 1+ 8π 12c2 The resonance is ω2 =

8c2 a2

r −1 24 c 1 ω 2 a2 8c2 1 ω 2 a2 8c2 2 2 ω ⇒ ω = 1+ ≈ 1 − = − 12 c2 a2 12 c2 a2 3 5 a

So the further the electric field and the induced magnetic field are expand in power of ω, the closer the resonant frequency will be to Jo (x) = 0, where x = ω. This seems logical because the first resonant frequency should occur when half a wavelength is inside the cavity. X

59

60

Appendix A Special Functions

Hi

61

Index hydrogen, 2 mean value theorem, 4

62

Jackson Solutions - Solutions to jackson's Electrodynamics

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