CHAPTER 4
4.1. The value of E at P ( P (ρ = 2, φ = 40◦ , z = 3) is given as E = 100aρ 200aφ + 300az V/m. Determine the incremental work required to move a 20 µC charge a distance of 6 µm:
−
a) in the directio direction n of aρ : The incremental incremental work is given by dW = 6 − case, dL = dρ aρ = 6 10 aρ . Thus
× dW = −(20 × 10−
6
C)(100 C)(100 V/m)(6
6
× 10−
m) =
b) in the directio direction n of aφ : In this case dL = 2 dφ aφ = 6
−q E · dL, where in this
−12 × 10−
9
J=
−12nJ
6
× 10− aφ, and so dW = −(20 × 10− )(−200)(6 × 10− ) = 2.4 × 10− J = 24 nJ 6
6
c) in the directio direction n of az : Here, dL = dz az = 6 dW =
6
6
az , and so
6
−(20 × 10− )(300)(6 × 10− ) = −3.6 × 10−
d) in the directio direction n of E: Here, dL = 6 aE =
× 10−
8
× 10−
6
J=
−36 nJ
aE , where
100aρ 200aφ + 300az = 0.267 aρ [1002 + 2002 + 3002 ]1/2
−
8
− 0.535 aφ + 0.0.802 az
Thus dW = =
6
6
[0.267 aρ − 0.535 aφ + 0. 0.802 az ](6 × 10− ) −(20 × 10− )[100aρ − 200aφ + 300az ] · [0. 44..9 nJ −44 6
e) In the directio direction n of G = 2 ax
− 3 ay + 4 az : In this case, dL = 6 × 10− aG, where 2ax − 3ay + 4az = 0.371 ax − 0.557 ay + 0. 0.743 az aG = [22 + 32 + 42 ]1/2
So now 6
6
[0.371 ax − 0.557 ay + 0. 0.743 az ](6 × 10− ) −(20 × 10− )[100aρ − 200aφ + 300az ] · [0. = −(20 × 10− )[37. )[37.1(aρ · ax ) − 55 55..7(aρ · ay ) − 74 74..2(aφ · ax ) + 111. 111.4(aφ · ay ) + 222. 222.9](6 × 10− ) where, at P , P , (aρ · ax ) = (aφ · ay ) = cos(40◦ ) = 0.766, (aρ · ay ) = sin(40◦ ) = 0.643, and (aφ · ax ) = − sin(40◦ ) = −0.643. Substituting these results in dW = −(20 × 10− )[28. )[28.4 − 35 35..8 + 47. 47.7 + 85. 85.3 + 222. 222.9](6 × 10− ) = −41 41..8 nJ dW =
6
6
6
6
42
4.2. An electric electric field is given given as E = 10 10eey (sin2z (sin2z ax + x sin2z sin2z ay + 2x 2x cos2z cos2z az ) V/m. a) Find E at P (5 P (5,, 0, π/12): π/12): Substituting this point into the given field produces
−
EP =
[sin((π/6) π/6) ax + 5 sin( sin(π/6) π/6) ay + 10 cos( cos(π/6) π/6) az ] = − −10 [sin
√
5 ax + 25 ay + 50 3 az
b) How How much much work work is done done in mov moving ing a charge charge of 2 nC an increme increment ntal al distance distance of 1 mm from P in the direction of ax ? This will be 9
3
11
dW x =
−qE · dL ax = −2 × 10− (−5)(10− ) = 10−
dW y =
−qE · dL ay = −2 × 10− (−25)(10− ) = 50−
c) of ay ?
9
3
d) of az ? dW z =
9
−qE · dL az = −2 × 10− (−50
11
J = 10 pJ
J = 50 pJ
√
√
3)(10−3 ) = 100 3 pJ
e) of (ax + ay + az )? dW xyz xyz =
−qE · dL
ax + ay + az )
√3
√
10 + 50 + 100 3 = = 135 pJ 3
√
4.3. If E = 120 aρ V/m, find the incremental amount of work done in moving a 50 µm charge a distance of 2 mm from: a) P (1 P (1,, 2, 3) toward Q(2, (2, 1, 4): The vector vector along along this directi direction on will be Q from which aP Q = [ax ay + az ]/ 3. We now write
√
−
dW = =
−qE · dL = −(50 ×
(ax 10−6 ) 120aρ
·
−√ay + az 3
(2
(1, −1, 1) − P = (1, 3
× 10− )
1 )(120) [(aρ · ax ) − (aρ · ay )] √ (2 × 10− ) −(50 × 10− )(120) 3 6
3
At P , P , φ = tan−1 (2/ (2/1) = 63. 63.4◦ . Thu Thus (aρ ax ) = cos(63. cos(63.4) = 0. 0.447 and (aρ ay ) = sin(63. sin(63.4) = 0. 0.894. Substituting these, we obtain dW = 3.1 µJ.
·
·
b) Q(2, (2, 1, 4) toward P (1 P (1,, 2, 3): A little thought is in order here: Note that the field has only a radial component and does not depend on φ or z . No Note te also also that P and Q are at the same radius ( 5) from the z axis, but have different φ and z coordinates. We could just as well position the two points at the same z location and the problem would not change. If this were so, then moving along a straight line between P and Q would thus involve moving along a chord of a circle whose radius is 5. Halfway along this line is a point of symmet symmetry ry in the field (make (make a sketc sketch h to see this). this). This means means that when starting starting from either point, the initial force will be the same. Thus the answer is dW = 3.1 µJ as in part a. This is also found found by going going through through the same procedure procedure as in part a, but with the direction (roles of P of P and Q) reversed.
√
√
43
4.4. It is found that the energy expended in carrying a charge of 4 µC from the origin to (x,0,0) along the x axis axis is directly directly proportional proportional to the square square of the path length. length. If E If E x = 7 V/m at (1,0,0), determine determine E x on the x axis as a function of x. The work done is in general given by
−
x
W =
q
E x dx = Ax2
0
where A is a constant. constant. Therefore Therefore E x must be of the form E x = E 0 x. At x = 1, E x = 7, so E 0 = 7. There Therefo fore re E x = 7x V/m. Note that that with the positivepositive-x x-directed field, the expended energy in moving the charge from 0 to x would be negative.
P P
4.5. Compute the value value of A G dL for G = 2y ax with A(1, (1, 1, 2) and P (2 P (2,, 1, 2) using the path: a) straight-line straight-line segments segments A(1, (1, 1, 2) to B (1, (1, 1, 2) to P (2 P (2,, 1, 2): In general we would have
·
−
−
P
P
G dL =
·
A
2y dx
A
The change in x occurs when moving between B and P , P , during which y = 1. Thus
P
P
G dL =
·
A
2
2y dx =
B
2(1)dx 2(1)dx = 2
1
b) straight-line straight-line segments segments A(1, (1, 1, 2) to C (2, (2, 1, 2) to P (2 P (2,, 1, 2): In this case case the change change in x occurs when moving from A to C , during which y = 1. Thus
−
−
P
C
G dL =
·
A
−
2
2y dx =
A
2( 1)dx 1)dx =
−
1
−2
4.6. Determine the work done in carrying a 2-µ 2-µC charge from (2,1,-1) to (8,2,-1) in the field E = y ax + xay along a) the parabola parabola x = 2y 2 : As a loo look k ahea ahead, d, we can show show (by (by takin takingg its curl) curl) that E is conserv conservativ ative. e. We theref therefore ore expect expect the same answe answerr for all three paths. paths. The general general expression for the work is
· − − × √ − − − × − − − × − − − − − −
B
W =
q
8
E dL =
q
A
8
−×
3y): We find y = 7/3 8
10−6
2
2
2 3/2 x 3
2 10−6
1
b) the hyperbol hyperbolaa x = 8/(7
=
x/2. x/2. Substi Substituti tuting ng these and the charge, charge, we
2y2 dy =
x/2 x/2 dx +
2
1
2
2
W 2 =
x dy
2
In the present case, x = 2y2 , and so y = get W 1 = 2 10−6
2
y dx +
10−6
7 (8 3
7 3
8 dx + 3x 8 8 2) ln 3 2 44
2 + y3 3 2 8
2 1
8/3x, and the work is 2
1
8
7
3y
8 ln(7 3
dy
2
3y )
1
=
28 µJ
=
−28 µJ
4.6c. the straight straight line x = 6y
x/6 + 2/ 2 /3, and the work is − 4: Here, y = x/6
8
W 3 =
−2 ×
10−6
2
x 2 + 6 3
2
dx +
(6y (6y
1
− 4) dy
=
−28 µJ
4.7. Let G = 3xy 3 ax + 2z ay . Given Given an initia initiall point point P (2 P (2,, 1, 1) and a final point Q(4, (4, 3, 1), find G dL using the path:
·
a) straight straight line: y = x
− 1, z = 1: We obtain:
4
G dL =
·
3
2
3xy dx +
2
4
2z dy =
1
3
3x(x
2
2
− 1)
dx +
2(1) dy = 90
1
b) parabola: 6y 6y = x2 + 2, z = 1: We obtain:
4
G dL =
·
3
2
3xy dx +
2
4
2z dy =
1
2
1 x(x2 + 2)2 dx + 12
3
2(1) dy = 82
1
4.8. Given Given E = xax + yay , find the work involved in moving a unit positive charge on a circular arc, the circle centered at the origin, from x = a to x = y = a/ 2.
−
√
In moving along the arc, we start at φ = 0 and move to φ = π/4. π/4. The setup is W =
− − q
−
π/4
E dL =
·
π/ 4
E adφ aφ =
·
0
( x ax aφ + y ay aφ )a dφ
−
0
π/ 4
=
−
−
− sin φ
π/ 4
2a2 sin φ cos φ dφ =
0
· ·
a2 sin(2φ sin(2φ) dφ =
0
cos φ
2
−a / 2
where q = 1, x = a cos φ, and y = a sin φ. Note that the field is conservative, so we would get the same result by integrating along a two-segment path over x and y as shown: W =
− ·
−
√
a/
E dL =
a
√
a/
2
( x) dx +
−
0
2
y dy =
2
−a /2
4.9. A uniform surface charge density of 20 nC/ nC/m2 is present on the spherical surface r = 0.6 cm in free space. a) Find Find the absolute absolute potential potential at P ( P (r = 1 cm, θ = 25◦ , φ = 50◦ ): Since the charge charge density density is uniform uniform and is spheric spherically ally-sy -symme mmetric tric,, the angular angular coordina coordinates tes do not matter. matter. The 2 potential function for r > 0.6 cm will be that of a point charge of Q of Q = 4πa ρs , or V ( V (r) =
4π (0. (0.6
2 2
9
× 10− ) (20 × 10− ) = 0.081 V 4π 0 r
r
At r = 1 cm, this becomes becomes V ( V (r = 1 cm cm)) = 8. 8.14 V 45
with with r in meters
◦ ◦ ◦ ◦ b) Find V AB AB given points A(r = 2 cm, θ = 30 , φ = 60 ) and B (r = 3 cm, θ = 45 , φ = 90 ): Again, Aga in, the angles angles do not matter because because of the spheric spherical al symmetr symmetry y. We use the part part a result to obtain V AB AB = V A
−
1 V B = 0.081 0.02
−
1 = 1.36 V 0.03
4.10. Express Express the potential field of an infinite line charge a) with zero zero reference reference at ρ = ρ0 : We write in general: V (ρ) =
−
ρL dρ + C 1 = 2π 0 ρ
Therefore C 1 =
− 2ρπL
ln(ρ ln(ρ) + C 1 = 0 at ρ = ρ0 0
ρL ln(ρ ln(ρ0 ) 2π 0
and finally ρL V (ρ) = [ln(ρ [ln(ρ0 ) 2π0
−
ρL ln(ρ ln(ρ)] = ln 2π0
ρ0 ρ
b) with V = V 0 at ρ = ρ0 : Using the reasoning of part a, we have V (ρ0 ) = V 0 =
ρL ln(ρ ln(ρ0 ) + C 2 2π0
and finally ρL V (ρ) = ln 2π 0
⇒
C 2 = V 0 +
ρ0 ρ
ρL ln(ρ ln(ρ0 ) 2π 0
+ V 0
c) Can the zero reference reference be placed at infinity? infinity? Why? Why? Answer: Answer: No, because we would would have have a potential that is proportional to the undefined ln( /ρ). /ρ).
∞
4.11. Let a uniform surface surface charge charge density density of 5 nC/ nC/m2 be present at the z = 0 plane, a uniform line charge charge densit density y of 8 nC/ nC/m be located at x = 0, z = 4, and a point charge of 2 µC be present at P (2 P (2,, 0, 0). If V If V = 0 at M (0 M (0,, 0, 5), find V at N (1 N (1,, 2, 3): We need to find a potential function for the combined charges which is zero at M . M . That for the point charge we know to be V p (r) =
Q 4π 0 r
Potential functions for the sheet and line charges can be found by taking indefinite integrals of the electric fields for those distributions. For the line charge, we have V l (ρ) =
−
ρl dρ + C 1 = 2π0 ρ
ρl − 2π
ln(ρ ln(ρ) + C 1 0
For the sheet charge, we have V s (z ) =
−
ρs dz + C 2 = 20 46
− 2ρs z + C
2
0
The total potential function function will be the sum of the three. Combining Combining the integration integration constants, constants, we obtain: Q ρl ρs V = ln(ρ ln(ρ) z + C 4π 0 r 2π 0 20
−
−
The terms in this expression are not referenced to a common origin, since the charges are at different positions. The parameters r , ρ, and z are scalar scalar distances distances from the charges, and will be treated as such here. To evaluate the constant, C , we first look at p oint oint M , M , where V T T = 0. 2 2 At M , M , r = 2 + 5 = 29, ρ = 1, and z = 5. We thus have
√
√
2 10−6 4π 0 29
9
9
× √ − 8 × 10− ln(1) − 5 × 10− 5 + C ⇒ C = −1.93 × 10 V 2π 2 √ √ √ At point N , N , r = 1 + 4 + 9 = 14, ρ = 2, and z = 3. The potential at N is thus 0=
2 10−6 V N N = 4π0 14
3
0
9
× √ − 8 × 10− 2π
ln(
0
√
0
9
5
10− × 2) − 2
(3)
0
− 1.93 × 10
3
= 1.98
× 10
3
V = 1.98kV
4.12. In spherical spherical coordinates, E = 2r/( r/ (r 2 + a2 )2 ar V/m V/m.. Find Find the potential potential at any any point, point, using using the reference a) V = 0 at infinity: We write in general V ( V (r) = With a zero reference at r
−
2r dr 1 + C = + C (r 2 + a2 )2 r 2 + a2
V (r) = 1/(r → ∞, C = 0 and therefore V (
2
+ a2 ).
b) V = 0 at r = 0: Using the general expression, we find V (0) V (0) = Therefore
1 + C = 0 a2
1 V ( V (r) = 2 r + a2
−
⇒
C =
− a1
2
1 r2 = 2 2 a2 a ( r + a2 )
−
c) V = 100V at r = a: Here, we find V ( V (a) = Therefore
1 + C = 100 2 a2
1 V ( V (r) = 2 r + a2
−
⇒
C = 100
− 21a
2
1 a2 r 2 + 100 = 2 2 + 100 2a2 2a (r + a2 )
−
4.13. Three identical identical point charges charges of 4 pC each are located at the corners of an equilateral equilateral triangle 0.5 mm on a side in free space. How much work must be done to move one charge to a point equidis equidistan tantt from the other two and on the line joining joining them? This This will be the magnitud magnitudee of the charge times the potential difference between the finishing and starting positions, or W =
(4
12 2
× 10− 2π0
)
1 2.5
− × 1 5
47
104 = 5.76
× 10−
10
J = 576pJ
4.14. Given Given the electric electric field E = (y + 1)ax + (x 1)ay + 2az , find the potential difference between the points a) (2,-2,-1 (2,-2,-1)) and (0,0,0): (0,0,0): We choose choose a path along along which which motion motion occurs occurs in one coordinat coordinatee directio direction n at a time. time. Startin Startingg at the origin, origin, first move move along along x from 0 to 2, where y = 0; then along y from 0 to 2, where x is 2; then along z from 0 to 1. The setup is
−
−
−
− − 2
V b
− V a =
(y + 1)
0
y =0
dx
−
−2
(x
0
− 1) x
b) (3,2,-1) and (-2,-3,4): (-2,-3,4): Following ollowing similar reasoning, reasoning, 3
V b
− V a =
(y + 1)
2
y =−3
−2
−
dx
(x
−3
=2
− 1) x
=3
dy
−
2 dz = 2
−
2 dz = 10
−1
0
dy
−1
4
4.15. Two uniform line charges, 8 nC/m each, are located at x = 1, z = 2, and at x = 1, y = 2 in free space. space. If the potential potential at the origin is 100 V, find V at P (4 P (4,, 1, 3): The net potential potential function for the two charges would in general be:
−
ρl − 2π
V = At the origin, R1 = R2 =
ln(R ln(R1 ) 0
ρl − 2π
ln(R ln(R2 ) + C 0
√5, and V = 100 V. Thus, with ρ
l
=8
9
× 10− ,
√ 10− ) × 100 = −2 ln( 5) + C ⇒ C = 331. 331.6 V 2π √ √ = |(4, (4, 1, 3) − (1, (1, 1, 2)| = 10 and R = |(4, (4, 1, 3) − (−1, 2, 3)| = 26. Therefore 9
(8
0
At P (4 P (4,, 1, 3), R1
2
9
(8
10− ) × V P P = − 2π 0
√
√
ln( 10) + ln( 26) + 331. 331.6 =
68..4 V −68
4.16. The potential at any point in space is given in cylindrical coordinates by by V = (k/ρ 2 )cos(bφ )cos(bφ)) V/m, where k and b are constants. a) Where Where is the the zero zero referen reference ce for poten potentia tial? l? This This will occur occur at ρ cos(bφ cos(bφ)) = 0, which gives φ = (2m (2m 1)π/ 1)π/22b, where m = 1, 2, 3...
−
→ ∞,
or when whenev ever er
b) Find the vector vector electric field intensity intensity at any point (ρ,φ,z (ρ,φ,z). ). We use ρ,φ,z) = E(ρ,φ,z)
1 ∂V k −∇V = − ∂V aρ − aφ = ∂ρ ρ ∂φ ρ
3
[2 cos( cos(bφ bφ)) aρ + b sin(bφ sin(bφ)) aφ ]
4.17. Uniform Uniform surface charge charge densities densities of 6 and 2 nC/ nC/m2 are present at ρ = 2 and 6 cm respectively, in free space. Assume V = 0 at ρ = 4 cm, and calculate V at: a) ρ = 5 cm: Sinc Sincee V = 0 at 4 cm, the potential at 5 cm will be the potential difference between points 5 and 4:
−
5
V 5 =
4
−
5
E dL =
·
4
aρsa dρ = 0 ρ 48
−
(.02)(6 10−9 ) ln 0
×
5 4
=
−3.026V
b) ρ = 7 cm: Here we integrate piecewise from ρ = 4 to ρ = 7:
−
6
V 7 =
4
aρsa dρ 0 ρ
−
7
6
(aρsa + bρsb ) dρ 0 ρ
With the given values, this becomes V 7 = =
−
−
(.02)(6 10−9 ) ln 0
×
−9.678V
6 4
(.02)(6
9
9
× 10− ) + (.( .06)(2 × 10− ) 0
ln
7 6
4.18. Find the potential potential at the origin produced by a line charge ρL = kx/( kx/(x2 + a2 ) extending along the x axis from x = a to + , where a > 0. Assume a zero reference at infinity.
∞
Think of the line charge as an array of point charges, each of charge dq = ρL dx, dx, and each having potential at the origin of dV of dV = ρL dx/(4 dx/(4π π 0 x). The total total p otent otential ial at the origin origin is then the sum of all these potentials, or V =
∞ ρ dx L 4π0 x
a
=
∞
a
−
k dx k π −1 x ∞ = k = tan 4π 0 (x2 + a2 ) 4π 0 a a a 4π 0 a 2
π k = 4 16 160 a
4.19.. The annula 4.19 annularr surface, surface, 1 cm < ρ < 3cm, z = 0, carries the nonuniform surface charge density ρs = 5ρ nC/ nC/m2 . Find V at P (0 P (0,, 0, 2 cm cm)) if V if V = 0 at infinity: We use the superposition integral form: ρs da V P P = 4π0 r r
| − |
where r = z az and r = ρaρ . We integrate integrate over over the surface surface of the annular annular region, region, with da = ρdρdφ. ρdρdφ. Substituting the given values, we find
2π
V P P =
0
.03
(5
9
× 10− )ρ
4π 0
.01
2
dρdφ
ρ2 + z 2
Substituting z = .02, and using tables, the integral evaluates as V P P =
(5
9
× 10− ) 20
ρ 2
2
2
ρ + (. (.02)
−
(.02)2 ln(ρ ln(ρ + 2
.03
2
2
ρ + (. (.02) )
= .081V
.01
4.20. A point charge Q is located at the origin. origin. Expres Expresss the potential potential in both rectang rectangular ular and cylindrical coordinates, and use the gradient operation in that coordinate system to find the electric field intensity. The result may be checked by conversion to spherical coordinates. The potential is expressed in spherical, rectangular, and cylindrical coordinates respectively tively as: Q Q Q V = = = 4π0 r 2 4π 0 (x2 + y 2 + z 2 )1/2 4π 0 (ρ2 + z 2 )1/2 Now, working with rectangular coordinates E=
−∇V = −
∂V ax ∂x
−
∂V ay ∂y 49
−
∂V Q x ax + y ay + z az az = ∂z 4π 0 (x2 + y2 + z 2 )3/2
4.20. (continu (continued) ed) Now, converting this field to spherical components, we find
·
·
Q r sin θ cos φ(ax ar ) + r sin θ sin φ(ay ar ) + r cos θ(az ar ) E r = E ar = 4π 0 r3
·
·
·
Q sin2 θ cos2 φ + sin2 θ sin2 φ + cos2 θ Q = = 2 4π 0 r 4π 0 r 2
Continuing: Q r sin θ cos φ(ax aθ ) + r sin θ sin φ(ay aθ ) + r cos θ(az aθ ) E θ = E aθ = 4π 0 r3
·
·
Q sin θ cos θ cos2 φ + sin θ cos θ sin2 φ = 4π 0 r2
− cos θ sin θ
·
=0
Finally
Q r sin θ cos φ(ax aφ ) + r sin θ sin φ(ay aφ ) + r cos θ(az aφ ) E φ = E aφ = 4π0 r3 Q sin θ cos φ( sin φ) + sin θ sin φ cos φ + 0 = = 0 check eck 4π0 r2
·
·
·
·
−
Now, in cylindrical we have in this case E=
−∇V = −
∂V aρ ∂ρ
∂V Q ρ aρ + z az az = ∂z 4π 0 (ρ2 + z 2 )3/2
−
Converting to spherical components, we find
Q r sin θ(aρ ar ) + r cos θ(az ar ) Q sin2 θ + cos2 θ Q E r = = = 3 2 4π0 r 4π 0 r 4π 0 r 2
·
·
Q r sin θ(aρ aθ ) + r cos θ(az aθ ) Q sin θ cos θ + cos θ( sin θ ) E θ = = =0 4π0 r3 4π 0 r2
·
·
−
Q r sin θ(aρ aφ ) + r cos θ(az aφ ) E φ = = 0 check heck 4π 0 r3
·
·
4.21. Let V = 2xy 2 z 3 +3ln(x +3ln(x2 + 2y2 + 3z 2 ) V in free space. Evaluate Evaluate each each of the following following quantities quantities at P (3 P (3,, 2, 1):
−
a) V : V : Substitute P directly to obtain: V = b) V . This will be just 15. 15.0 V. c) E: We have
| |
−∇ −
E
P
=
+
15..0 V −15
6x 12 12yy 3 V = 2y 2 z 3 + 2 + 4 xyz + a x x + 2y 2y2 + 3z 3z2 x2 + 2y 2y 2 + 3z 3z 2 P 18 18zz 6xy 2 z 2 + 2 = 7.1ax + 22. 22.8ay 71 71..1az V/m az x + 2y 2y 2 + 3z 3z 2 P
50
−
ay
4.21d) E P : taking the magnitude of the part c result, we find E P = 75 75..0 V/m.
| |
| |
e) aN : By definition, this will be
aN
f ) D: This is D
P
= 0 E
P
P
=
− |EE| = −0.095 ax − 0.304 ay + 0.0.948 az
= 62 62..8 ax + 202 ay
2
pC/m . − 629 az pC/
4.22. A certain potential potential field is given given in spherical coordinates by V = V 0 (r/a r/a)sin )sin θ. Find the total charge contained within the region r < a: We first find the electric field through E=
1 ∂V V =− −∇V = − ∂V ar − ∂r r ∂θ a
0
[sin θ ar + cos θ aθ ]
The requested charge is now the net outward flux of D = 0 E through the spherical shell of radius a (with outward normal ar ): Q=
2π
D dS =
·
S
0
π
π
2
0 E ar a sin θdθdφ =
·
0
−2πaV
0 0
sin2 θ dθ =
0
2
−π a V 0
0
C
The same result can be found (as expected) by taking the divergence of D and integrating over the spherical volume:
∇·D=− =
−
−
1 ∂ 0 V 0 r2 sin θ 2 r ∂r a 0 V 0 2sin2 θ + 1 ra sin θ
Now
− 2π
Q=
0
π
a
0
0
−
1 ∂ r sin θ ∂θ
0 V 0 cos θ sin θ a 0 V 0 2sin2 θ = = ρv ra sin θ
−
0 V 0 2 r sin θ drdθdφ = ra sin θ
2
−2π V 0
0
a
=
−
0 V 0 cos(2θ cos(2θ) 2sin θ + ra sin θ
a
r dr =
0
2
−π a V 0
C
0
4.23.. It is known 4.23 known that the potential potential is given given as V = 80 80ρ ρ.6 V. Assuming free space conditions, find: a) E: We find this through E=
48ρ ρ−. −∇V = − dV aρ = −48 dρ
4
V/m
b) the volume volume charge charge density density at ρ = .5 m: Using sing D = 0 E, we find the charge density through ρv
.5
=[
∇ · D].
5
=
1 ρ
d (ρDρ ) dρ
.5
= 28 28..80 ρ−1.4
−
.5
=
3
673pC/m −673pC/
c) the total charge lying lying within the closed surface surface ρ = .6, 0 < z < 1: The easiest way to do this calculation is to evaluate evaluate Dρ at ρ = .6 (noting that it is constant), and then multiply 51
by the cylinder area: Using part a, we have Dρ Q=
−2π(.6)(1)521 × 10−
12
C=
−1.96nC.
.6
=
48 (.6)−. −48 0
4
=
2
521pC/m . Thus −521pC/
4.24. The surface defined defined by the equation x3 + y 2 + z = 1000, where x, y , and z are positive, is an equipotential equipotential surface on which the poten p otential tial is 200 V. If E = 50 V/m at the point P (7 P (7,, 25 25,, 32) on the surface, find E there:
| |
First, the potential function will be of the form V ( V (x,y,z) x,y,z) = C 1 (x3 + y 2 + z ) + C 2 , where C 1 and C 2 are constants to be determined (C (C 2 is in fact irrelevant for our purposes). The electric field is now V = C 1 (3x (3x2 ax + 2y 2y ay + az ) E=
−∇ − And the magnitude of E is |E| = C 9x + 4y 4y + 1, which at the given point will be 155.27 27C C = 50 ⇒ C = 0.322 |E|P = C 9(7) + 4(25) + 1 = 155. 1
1
4
4
2
2
1
1
Now substitute C 1 and the given point into the expression for E to obtain EP =
(47.34 ax + 16. 16.10 ay + 0. 0.32 az ) −(47.
The other constant, C 2 , is needed to assure a potential of 200 V at the given point. 4.25. Within the cylinder cylinder ρ = 2, 0 < z < 1, the potential is given by V = 100 + 50ρ 50ρ + 150ρ 150ρ sin φ V. ◦ a) Find V , V , E, D, and ρv at P (1 P (1,, 60 , 0.5) in free space: First, substituting the given point, we find V P 279.9V. Then, P = 279. E=
1 ∂V 150 sinφ sin φ] aρ − [15 [1500 cos cos φ] aφ −∇V = − ∂V aρ − aφ = − [50 + 150 ∂ρ ρ ∂φ
Evaluate the above at P to find EP = Now D = 0 E, so DP = ρv =
∇·D =
1 ρ
At P , P , this is ρvP =
179..9aρ − 75 75..0aφ V/m −179 2
nC/m . Then −1.59aρ − .664aφ nC/
d 1 ∂D φ (ρDρ ) + = dρ ρ ∂φ
−
1 1 (50 + 150 150 sinφ sin φ) + 150 sin sin φ 0 = ρ ρ
− 50ρ
0
C
3
pC/m . −443 pC/
b) How much much charge charge lies within the cylinder? cylinder? We will integrate integrate ρv over the volume to obtain:
− 2π
1
Q=
0
0
2
0
50 500 ρdρdφdz = ρ
52
(50) (2) = −5.56nC −2π(50) 0
4.26. Let us assume that we have a very thin, square, imperfectly conducting plate 2m on a side, located in the plane z = 0 with one corner at the origin such that it lies entirely within the first quadrant. The potential at any point in the plate is given as V = e−x sin y .
−
a) An electron electron enters enters the plate at x = 0, y = π/3 π/3 with zero initial velocity; in what direction is its initial movemen movement? t? We first find the electric field associated with the given potential: potential:
−∇V = −e−x[sin y ax − cos y ay ]
E=
Since we have an electron, its motion is opposite that of the field, so the direction on entry is that of E at (0, (0, π/3), π/3), or 3/2 ax 1/2 ay .
√
−
−
b) Because Because of collisions collisions with the particles particles in the plate, the electron achieves achieves a relatively relatively low velocity and little acceleration (the work that the field does on it is converted largely into heat). heat). The electron electron therefore therefore moves moves approxim approximatel ately y along along a streamli streamline. ne. Where Where does it leave leave the plate and in what directio direction n is it mo movin vingg at the time? time? Conside Considering ring the result result of part a, we would would expect expect the exit to occur along the bottom bottom edge of the plate. plate. The equation of the streamline is found through E y dy = = E x dx
−
cos y sin y
⇒
x=
−
tan y dy + C = ln(cos y) + C
At the entry point (0, (0, π/3), π/3), we have 0 = ln[cos(π/ ln[cos(π/3)] 3)] + C , from which C = 0.69. Now, Now, along the bottom edge (y (y = 0), we find x = 0.69, and so the exit point is (0. (0 .69 69,, 0). From the field expression evaluated at the exit point, we find the direction on exit to be ay .
−
4.27.. Two 4.27 Two point charge charges, s, 1 nC at (0, (0, 0, 0.1) and a) Calculat Calculatee V at P (0 P (0..3, 0, 0.4): Use
(0, 0, −0.1), are in free space. −1nC at (0, q
q − 4π |R | 4π |R− | = (.3, 0, .3) and R− = (.3, 0, .5), so that |R | = 0.424 and |R− | = 0.583. Thus V P P =
0
where R+
+
0
+
10−9 1 V P P = 4π0 .424
−
1 = 5.78 V .583
b) Calculat Calculatee E at P : P : Use
| |
q (.3ax + .3az ) EP = 4π 0 (.424)3
−
q (.3ax + .5az ) 10−9 = [2. [2.42ax + 1. 1 .41az ] V/m 4π 0 (.583)3 4π 0
Taking the magnitude of the above, we find EP = 25 25..2 V/m.
| |
c) Now Now treat the two charge chargess as a dipole at the origin origin and find V at P : P : In spherica sphericall coor1 − ◦ 2 2 dinates, P is located at r = .3 + .4 = .5 and θ = sin (.3/.5) /.5) = 36. 36.9 . Assumin Assumingg a dipole in far-field, we have
√
qd cos θ 10−9 (.2) cos(36 cos(36..9◦ ) = = 5.76 V V P P = 4π 0 r 2 4π 0 (.5)2 53
4.28. 4.2 8. Use Use the elect electric ric field field inte intens nsit ity y of the dipol dipolee (Sec. (Sec. 4.7 4.7,, Eq. Eq. (36)) (36)) to find find the differ differen ence ce in potential between points at θa and θb , each point having the same r and φ coordinates. Under what conditions does the answer agree with Eq. (34), for the potential at θa ? We perform a line integral of Eq. (36) along an arc of constant r and φ:
−
θa
V ab ab =
θb
qd [2cos θ ar + sin θ aθ ] aθ r dθ = 4π 0 r 3
·
qd [cos θa 4π 0 r 2
=
−
θa
θb
qd sin θ dθ 4π 0 r 2
− cos θb ]
This result agrees with Eq. (34) if θ if θa (the ending point in the path) is 90 ◦ (the xy plane). Under this condition, condition, we note that if θ if θb > 90◦ , positive work is done when moving (against the field) to the xy plane; if θ if θb < 90◦ , negative work is done since we move with the field. 4.29. A dipole having having a moment moment p = 3ax 5ay + 10az nC m is located at Q(1, (1, 2, 4) in free space. Find V at P (2 P (2,, 3, 4): We use the general expression for the potential in the far field:
−
·
V = where r
p (r 4π 0 r
−
· − r ) | − r |
3
(1, 1, 8). So − r = P − Q = (1, (3ax − 5ay + 10az ) · (ax + ay + 8az ) × 10− V P = P
9
4π 0 [12 + 12 + 82 ]1.5
= 1.31 V
4.30. A dipole for which p = 10 100 az C m is located located at the origin origin.. What What is the the equat equation ion of the surface on which E z = 0 but E = 0?
·
First we find the z component: E z = E az =
·
10 5 [2cos θ (ar az ) + sin θ (aθ az )] = 2cos2 θ 3 3 4πr 2πr
·
This will be zero when 2cos2 θ
− sin
2cos2 θ
2
·
− sin
θ = 0. Using identities, we write
− sin
2
θ=
2
θ
1 [1 + 3 cos(2 cos(2θθ)] 2
The above becomes zero on the cone surfaces, θ = 54 54..7◦ and θ = 125. 125.3◦ . 4.31. A potential field in free space is expressed as V = 20 20//(xyz) xyz ) V. a) Find Find the total energy energy stored stored within the cube 1 < x, x, y, z < 2. We integrate integrate the energy energy density over the cube volume, where wE = (1/ (1/2) 2)0 E E, and where E=
·
1 1 1 V = 20 2 ax + 2 ay + az V/m x yz xy z xyz 2
−∇
The energy is now
2
W E = 200 2000
1
2
1
2
1
1 1 1 + + x4 y2 z 2 x2 y 4 z 2 x2 y2 z 4 54
dxdydz
4.31a. (continu (continued) ed) The integral evaluates as follows:
− − − − − − 2
W E = 200 2000
2
1
1
2
= 200 2000
1 3
2
1
7 24
1
2
= 200 2000
7 24
1
2
= 200 2000
7 48
1
= 200 2000 (3)
1 x3 y 2 z 2
1 + y2z2
1 yz 2
1 + z2
1 6
7 48
1 xy 4 z 2
1 2
1 + y4 z 2
1 y3z2
1 + z2
2
1 xy 2 z 4
1 4
1 2
dy dz
1
1 y2 z 4
1 2
1 yz 4
1 z4
dz
dy dz
2
dz
1
7 = 387pJ 96
b) What value value would be obtained obtained by assuming a uniform energy density equal equal to the value value at the center of the cube? At C (1. (1.5, 1.5, 1.5) the energy density is
1 wE = 200 2000 (3) = 2.07 (1. (1.5)4 (1. (1.5)2 (1. (1.5)2
10
× 10−
J/m3
This, multiplied multiplied by a cube volume of 1, produces an energy value value of 207 pJ. 4.32.. Using 4.32 Using Eq. (36), (36), a) find the energy stored stored in the dipole dipole field in the region region r > a: We start with E(r, θ) =
qd [2cos θ ar + sin θ aθ ] 4π 0 r 3
Then the energy will be W e =
vol
=
− − 2π
1 0 E E dv = 2
·
0
π
0
∞
a
(qd) qd )2 4cos2 θ + sin2 θ r 2 sin θdrdθdφ 2 6 32 32π π 0 r 3 cos cos2
qd ) −2π(qd) 32 32π π 2 0
2
1 ∞ 3r 3 a
π
3cos2 θ + 1 sin θ dθ =
0
θ+1
(qd) qd )2 48 48π π 2 0 a3
cos3 θ
cos θ
π 0
4
2
=
(qd) qd ) J 12 12π π0 a3
b) Why can can we not not let a approach approach zero as a limit? From the abov ab ovee result, a singularity singularity in the energy occurs as a 0. More importantly, a cannot be too small, or the original far-field assump assumption tion used used to deriv derivee Eq. (36) (a (a >> d) d) will not hold, and so the field expression will not be valid.
→
4.33. A copper sphere of radius 4 cm carries a uniformly-distributed total charge of 5 µC in free space. a) Use Gauss’ law to find D extern external al to the sphere sphere:: with a spheric spherical al Gaussi Gaussian an surface surface at radius r , D will be the total charge divided by the area of this sphere, and will be ar directed. Thus Q 5 10−6 D= a = ar C/m2 2 r 2 4πr 4πr
×
55
4.33b) Calculate Calculate the total energy stored in the electrostatic electrostatic field: field: Use
· ×
2π 1 W E = D E dv = 0 vol 2 − 1 (5 10 6 )2 = (4π (4π) 2 16 16π π 2 0
π
0
∞ 1 (5
6 2
× 10− ) 2
r 2 sin θdrdθdφ
4
2 16 16π π 0 r ∞ dr 25 10−12 1 = = 2.81 J 2 8π 0 .04 .04 r .04
×
c) Use W E = Q2 /(2C (2C ) to calculate the capacitance of the isolated sphere: We have (5 10−6 )2 Q2 C = = = 4.45 2W E 2(2. 2(2.81)
×
12
× 10−
F = 4.45pF
4.34. A sphere sphere of radius radius a contains volume charge of uniform density ρ0 C/m3 . Find the total stored energy by applying a) Eq. (43): (43): We first need the potential potential every everywhe where re inside inside the sphere. sphere. The electric electric field field inside and outside is readily found from Gauss’s law: ρ0 r E1 = ar r 30
≤
ρ0 a 3 a and E2 = ar r 30 r 2
≥a
The potential at position r inside the sphere is now the work done in moving a unit positive point charge from infinity to position r :
−
a
V ( V (r) =
−
r
E2 ar dr
·
∞
a
−
a
E1
·
ar dr =
ρ0 a 3 dr ∞ 3 0 r 2
−
r
a
ρ0 r ρ0 dr = 3a2 30 6 0
−r
2
Now, using this result in (43) leads to the energy associated with the charge in the sphere: 1 W e = 2
2π
π
0
a
0
0
ρ20 3a2 60
−r
2
πρ 0 r sin θdrdθdφ = 30 2
a
2 2
3a r
0
−r
4
4πa5 ρ20 dr = 15 150
b) Eq. (45): (45): Using Using the given given fields we find the energy densiti densities es we1 =
1 ρ2 r 2 0 E1 E1 = 0 r 2 18 180
·
≤a
and we2 =
1 ρ2 a 6 0 E2 E2 = 0 4 r 2 18 180 r
·
≥a
We now integrate these over their respective volumes to find the total energy:
2π
W e =
0
π
0
a
0
ρ20 r2 2 r sin θdrdθdφ + 18 180
56
2π
0
π
0
∞ ρ2 a 6 0
a
4πa5 ρ20 r sin θdrdθdφ = 18 180 r 4 15 150 2
4.35. Four 0. 0.8 nC point charges charges are located located in free free space at the corners corners of a square square 4 cm on a side. a) Find Find the total total potential potential energy energy stored: stored: This This will be given by 4
1 W E = qn V n 2 n=1
where V n in this case is the potential at the location of any one of the point charges that arises arises from the other other three. three. This will be (for charge charge 1)
√
q 1 1 1 V 1 = V 21 + + 21 + V 31 31 + V 41 41 = 4π 0 .04 .04 .04 2
Taking the summation produces a factor of 4, since the situation is the same at all four points. Consequently,
√
1 (.8 10−9 )2 1 W E = (4)q (4)q1 V 1 = 2+ = 7.79 2 2π 0 (.04) 2
×
× 10−
7
J = 0.779 µJ
b) A fifth 0. 0.8 nC charge charge is install installed ed at the center center of the square. square. Agai Again n find the total stored stored energy: This will be the energy found in part a plus the amount of work done in moving the fifth charge into into position from infinity. infinity. The latter is just the p otential otential at the square center arising from the original four charges, times the new charge value, or 4(. 4(.8 10−9 )2 ∆W E = = .813 µJ 4π 0 (.04 2/2)
×√
The total energy is now W E net = W E (part (part a) + ∆W ∆W E = .779 + .813 = 1. 1.59 µJ
57
