GRIFFITHS ELECTRODYNAMICS SOLUTIONS

KALMAN KNIZHNIK

Problem 2.1a) Twelve equal charges, q are situated at the corners of a regular 12-sided polygon(for instance, one on each numeral of a clock face.) What is the net force on a testcharge Q at the center?The force on Q must be 0 by symmetry. The x, y, and z components of the fields due to eachcharge cancel, so the force vanishes.b) Suppose one of the 12 q’s is removed (the one at ”6 o’clock”). What is the forceon Q? Explain your reasoning carefully.Now it is like 10 of the charges cancel, the only one that doesn’t cancel being the one directlyopposite the missing charge. Thus, the force is

F =1

4πε0

qQ

r22 (1)

pointing in the direction of the missing charge. Here, r is the distance from the center to eachcharge.c) Now 13 equal charges, q, are placed at the corners of a regular 13-sided polygon.What is the force on a test charge Q at the center.Again, by symmetry, the answer is still 0.d) If one of the 13 q’s is removed, what is the force on Q? Explain your reasoning.The same answer as in part b, equation 1, and for the same reason.

Problem 2.2a) Find the electric field (magnitude and direction) a distance z above the midpointbetween two equal charges, q, a distance d apart. Check that your result is consistentwith what you’d expect when z � d.The force would not have components in the y or x direction, if we put the charges in the x − zplane, there is no y-component of the field at the point in question, and the x-components cancel.The z-component of the field is

E =1

4πε0

2q

z2 + (d/2)2sin(θ)z =

1

4πε0

2qz

(z2 + (d/2)2)3/2z 2 (2)

where the factor of 2 comes from adding up the two charges. When z � d, the electric field becomes

E =1

4πε0

2q

z2z (3)

which is what you would expect, since the system acts like a point charge of charge 2q.b) Repeat part (a), only this time make the right hand charge −q instead of +q.Now the z-components cancel and the x-components don’t. Mathematically, this means replacingsin(θ) with cos(θ).

E =1

4πε0

2q

z2 + (d/2)2cos(θ)z =

1

4πε0

qd

(z2 + (d/2)2)3/2x 2 (4)

where the 2 has canceled out with the d/2 that comes from the cosine. If z � d, then from faraway it looks like a dipole, and the field reduces to that of a dipole:

E =1

4πε0

qd

z3x (5)

Problem 2.3Find the electric field a distance z above one end of a straight line segment of lengthL, which carries a uniform line charge λ. Check that your formula is consistent withwhat you would expect for the case z � L.The electric field is

E =1

4πε0

∫ L

0

dq

r2cos(θ)x +

1

4πε0

∫ L

0

dq

r2sin(θ)z (6)

replacing dq by λdx, r2 by x2 + z2, cos(θ) by x/√x2 + z2, and sin(θ) by z/

√x2 + z2, we get

E =1

4πε0λ

∫ L

0

xdx

(x2 + z2)3/2x +

1

4πε0λ

∫ L

0

zdx

(x2 + z2)3/2z (7)

The first integral is a u-substitution, while the second can be evaluated using your favorite computerprogram.

E =1

4πε0λ{(−1

z+

1√L2 + z2

)x +L

z√L2 + z2

z} 2 (8)

In the limit z � L, we obtain

E =1

4πε0

λL

z2z (9)

which is what you would expect from a point charge q = λL.

Problem 2.5Find the electric field a distance z above the center of a circular loop of radius r, whichcarries a uniform line charge λ.

E =

∫1

4πε0

dq

R2(10)

Now, dq = λdl, where dl = rdθ, while R =√r2 + z2 is the distance from the edge of the loop

to the point z. Furthermore, clearly the horizontal components will cancel out, leaving only a zcomponent of the electric field, meaning we need to include a factor of cos(θ) = z/

√r2 + z2. Thus,

E =

∫ L

0

λ

4πε0

dl

(r2 + z2)cos(θ)z =

∫ 2π

0

1

4πε0λ

zrdθ

(r2 + z2)3/2z (11)

meaning the answer is

E =λrz

2ε0(r2 + z2)3/2z 2 (12)

Problem 2.6Find the electric field a distance z above the center of a flat circular dis ok radius R,which carries a uniform surface charge σ. What does your formula give in the limitR→∞? Also, check the case z � R.The method is the same as in problem 2.5, except that now dq = σda = σrdrdθ. Thus, the integralis

E =σ

4πε0

∫ 2π

0

∫ R

0

rzdrdθ

(r2 + z2)3/2z =

zσ

2ε0

−1

(r2 + z2)1/2z

∣∣∣∣R0

=zσ

2ε0(1

z− 1√

R2 + z2)z 2 (13)

In the limit R→∞, we recover the field due to an infinite conducting plane, Ez = σ/2ε0, while inthe limit that z � R, by Taylor expanding the square root, we recover the field between two pointcharges: E = Q/4πε0z

2, where Q = πR2σ.

2

Problem 2.9Suppose the electric field in some region is found to be E = kr3r, in spherical coordi-nates (k is some constant).a) Find the charge density ρ.This part is simple.

ρ = ε0∇ ·E = ε01

r2∂(r2Er)

∂r= ε0k

1

r2∂(r5)

∂r= 5ε0kr

2 (14)

b) Find the total charge contained in a sphere of radius R, centered at the origin. (Doit in two different ways).Method 1:

Qenc =

∫VρdV =

∫ 2π

0dφ

∫ π

0dθ

∫ R

05ε0kr

2r2dr = 4πε0kR5 2 (15)

Qenc = ε0

∮E · ds = ε0kR

34πR2 = 4πε0kR5 2 (16)

Problem 2.10A charge q sits at the back corner of a cube. What is the flux of E through one side?We need to surround the charge symmetrically, since otherwise it is unclear what fraction of theflux is going through each face. Naively, you might think that the flux through one face is 1/6 ofthe total flux, but this is an assumption that we cannot justify. For instance, no flux goes throughthe three faces adjacent to the charge, since they are parallel to the electric field. However, ifmake a larger cube, one centered on the charge, we can determine the flux through each face bysymmetry. If we center the cube on the charge, then we are effectively adding 7 cubes of the samedimensions as the original cube, yielding 8 total cubes. Now, the charge is centered in the cube,with the original face that we were examining being 1/4 of one face of this cube. Since the cubehas six sides, the original face composes 1/24 of the cube. The flux is thus q/24ε0. 2

Problem 2.21Find the potential inside and outside a uniformly charged solid sphere whose radiusis R and whose total charge is q. Use infinity as your reference point. Compute thegradient of V in each region, and check that it yields the correct result.Inside the sphere, the electric field∫

E · da =1

ε0

∫ρdV ⇒ E4πr2 =

3q

4πε0R3

∫dV =

3q

4πε0R3

4πr3

3⇒ E =

qr

4πε0R3r (17)

whereas outside the electric field is just

E =q

4πε0r2r (18)

Thus,

Vout =

∫ ∞r

E · dl =1

4πε0

∫ ∞r

q

r′2r · dr′r =

1

4πε0

q

r2 (19)

Vin =

∫ ∞r

E · dl =

∫ R

r

qr

4πε0R3r · drr +

1

4πε0

∫ ∞R

q

r2r · drr =

q

8πε0R− qr2

8πε0R3+

1

4πε0

q

R(20)

this can be rewritten as

Vin =3q

8πε0R− qr2

8πε0R3=

q

8πε0R(3− r2

R2) 2 (21)

Taking the gradient of equations 19 and 21 trivially yields in equation 18 and 17, respectively.

3

Problem 2.22Find the potential a distance s from an infinitely long straight wire that carries auniform line charge λ. Compute the gradient of your potential, and check that ityields the correct field.Let’s calculate the electric field for this wire:∫

E · da = E2πsL =q

ε0=λL

ε0⇒ E =

λ

2πsε0s (22)

Now we integrate this over the path ds s to find the potential. We don’t integrate from s to ∞,since the integral will blow up. Instead we integrate from s to some point a.

V =

∫ a

sE · dl =

∫ a

s

λ

2πsε0ds =

λ

2πε0lna

s2 (23)

Taking the gradient of -V trivially reproduces equation 22.

Problem 2.31a) Three charges are situated at the corners of a square (side a), as shown in Fig. 2.41.How much work does it take to bring in another charge +q, from far away and placeit in the fourth corner?The potential at the corner is

V = V1 + V2 + V3 =1

4πε0

−qa

+1

4πε0

−qa

+1

4πε0

q

a√

2=

1

4πε0

q

a(−2 +

1√2

) (24)

So then the work W = qV is, since the potential is zero at infinity,

W =1

4πε0

q2

a(−2 +

1√2

) 2 (25)

b) How much work does it take to assemble the whole configuration of four charges?It takes no work to bring in the first charge. The second charge requires (supposing it is the negativecharge) work W = −qV = −q2/4πε0a, since, once again, the potential is zero at infinity. The thirdcharge requires

W =1

4πε0

q(−q)a

+1

4πε0

−q(−q)a√

2=

q2

4πε0a(

1√2− 1) (26)

and we have already calculated the work required for the fourth charge. Therefore, the work forthe total configuration is:

W = 0 +1

4πε0

−q2

a+

q2

4πε0a(

1√2− 1) +

1

4πε0

q2

a(−2 +

1√2

) =q2

2√

2πε0a− q2

πε0a2 (27)

Problem 2.35A metal sphere of radius R, carrying charge q, is surrounded by a thick concentricmetal shell (inner radius a, outer radius b, as in Fig. 2.48). The shell carries no netcharge.a) Find the surface charge density σ at R, at a, and at b.All the charge has gone to the outer radius of the metal sphere, since it is a conductor, so at R,the surface charge density is σR = q/A = q/4πR2. At a, there will be induced a charge -q, so thesurface density is σa = −q/4πa2. In order to keep the shell neutral, the remaining charge +q willgo to the outer edge of the shell, causing a surface charge density σb = q/4πb2. 2

b) Find the potential at the center, using infinity as the reference point.We’ll use V =

∫∞0 E · dl.

V =

∫ ∞0

E · dl =

∫ R

0(0)dr +

∫ a

R

1

4πε0

q

r2dr +

∫ b

a(0)dr +

∫ ∞b

1

4πε0

q

r2dr (28)

4

where the (0) corresponds to regions of E = 0. Thus,

V =q

4πε0(1

b+

1

R− 1

a) 2 (29)

c) Now the outer surface is touched to a grounding wire, which lowers its potential tozero (same as at infinity). How do your answers to (a) and (b) change?If the outer surface is grounded, the charge on the outer surface will go to ground, since it will gofrom a region of high potential to region of low potential. Thus, σb = 0. As a result, the potentialat the center will have no contribution from the outer part of the shell.

V =

∫ ∞0

E · dl =

∫ R

0(0)dr +

∫ a

R

1

4πε0

q

r2dr +

∫ b

a(0)dr +

∫ ∞b

(0)dr =q

4πε0(

1

R− 1

a) 2 (30)

Problem 2.36Two spherical cavities, of radii a and b, are hollowed out from the interior of a (neutral)conducting sphere of radius R (Fig. 2.49). At the center of each cavity a point chargeis placed – call these charges qa and qb.a) Find the surface charges σa, σb, and σR.The charge at the middle of each cavity will induce an equal and opposite charge density on thecavity surface. Let the radii of the cavities be a and b, respectively. Then

σa = −qa/4πa2, σb = −qb/4πb2 (31)

Now, since the sphere is overall neutral, the remaining charge must be on the outer surface R.Therefore,

σR = (qa + qb)/4πR2 (32)

b) What is the field outside the conductor?The field outside a conductor is always E = σ/ε0n. Therefore, outside of R:

E =qa + qb4πε0R2

r (33)

Note that this is consistent with Gauss’s Law:∫E · da = Qenc/ε0.

c) What is the field within each cavity?Once again, by Gauss’s law (or due to the boundary conditions, whichever you prefer),

Ea =1

4πε0

qar2

ra ∀ r < a (34)

and

Eb =1

4πε0

qbr2

rb ∀ r < b (35)

Note that these unit vectors are, in general, different from each other and the unit vector r inequation 33.d) What is the force on qa and qb?The electric field is due to the charges themselves, and the field due to charge a does not influencecharge b, and vice versa. Therefore the force on each charge is 0. 2

e) Which of these answers would change if a third charge, qc, were brought near theconductor?σR would be different, since the new charge would induce some more charge on the outer surface.Furthermore, it would affect ER, since the new charge contributes its own electric field. Note thatit would be difficult to apply Gauss’s law here, since we don’t have the required symmetry.

Problem 3.18The potential at the surface of a sphere (radius R) is given by V0 = kcos3θ, where k

5

is a constant. Find the potential inside and outside the sphere, as well as the surfacecharge density σ(θ) on the sphere. (Assume there’s no charge inside or outside thesphere.)This potential can be decomposed into a linear combination of Legendre polynomials:

V0 = kcos3θ = k[4cos3θ − 3cosθ] = k[xP3(cosθ) + yP1(cosθ)] (36)

I can tell immediately that it is some combination of P3 and P1 since the function is odd. So nowI have the equality (if I replace P3 and P1 with the explicit forms of the polynomials):

4cos3θ − 3cosθ = x5cos3θ − 3cosθ

2+ ycos(θ) =

5x

2cos3θ +

2y − 3x

2cosθ (37)

and therefore we have

5x

2= 4

2y − 3x

2= −3 ⇒ x =

8

5, y =

−3

5(38)

So our potential, written in terms of the Legendre polynomials, is

V0 =k

5[8P3(cosθ)− 3P1(cosθ)] (39)

Now the general solution to Laplace’s equation (in spherical coordinates) is

V (r, θ) =∞∑l=0

AlrlPl(cosθ) +

Blrl+1

Pl(cosθ) (40)

Inside the sphere, the Bl’s must be 0, or we will have problems at the origin, while outside thesphere, the Al’s must be 0, since otherwise we will have problems at infinity:

Vin =∞∑l=0

AlrlPl(cosθ), Vout =

∞∑l=0

Blrl+1

Pl(cosθ) (41)

Clearly the only l’s that contribute are l = 1 and l = 3:

Vin =∞∑l=0

AlrlPl(cosθ) =

k

5[8P3(cosθ)− 3P1(cosθ)] ⇒ A0 = 0, A2 = 0, Al = 0 ∀ l > 3 (42)

and so, at the surface of the sphere,

Vin =k

5[8P3(cosθ)−3P1(cosθ)] = A1RP1(cosθ)+A3R

3P3(cosθ) ⇒ A1 =−3k

5R, A3 =

8k

5R3(43)

while equating this with outside the surface yields (Bl 6=1,3 = 0):

Vout =k

5[8P3(cosθ)−3P1(cosθ)] =

B1

R2P1(cosθ)+

B3

R4P3(cosθ) ⇒ B1 =

−3kR2

5, B3 =

8kR4

5(44)

Thus, outside, we have, (plugging our A1, A3, B1, B3 into the general solution, equation 40):

Vin =−3k

5RrP1(cosθ) +

8k

5R3r3P3(cosθ) 2 (45)

Vout =−3kR2

5

1

r2P1(cosθ) +

8kR4

5

1

r4P3(cosθ) 2 (46)

To obtain σ, we use the fact that the discontinuity in the electric field at the boundary is equal tothe surface charge density:

σ

ε0= −(

∂V

∂n

∣∣∣∣out− ∂V

∂n

∣∣∣∣in

) =∂Vin∂r

∣∣∣∣R− ∂Vout

∂R

∣∣∣∣R

(47)

6

Cranking out the derivatives:

∂Vin∂r

=−3k

5RP1(cosθ) +

24kr2

5R3P3(cosθ)

∣∣∣∣R

=−3k

5RP1(cosθ) +

24k

5RP3(cosθ) (48)

∂Vout∂r

=6kR2

5r3P1(cosθ)−

32kR4

5r5P3(cosθ)

∣∣∣∣R

=6k

5RP1(cosθ)−

32k

5RP3(cosθ) (49)

Subtracting one from the other yields:

σ = ε0−9k

5RP1(cosθ) + ε0

56k

5RP3(cosθ) 2 (50)

Problem 3.25Charge density σ(φ) = asin5φ (where a is a constant) is glued over the surface of aninfinite cylinder of radius R. Find the potential inside and outside the cylinder. Usethe general solution to Laplace’s equation with cylindrical symmetry:

V (s, φ) = a0 + b0ln(s) +∞∑k=1

[sk(akcos(kφ) + bksin(kφ)) + s−k(ckcos(kφ) + dksin(kφ)] (51)

Inside the cylinder, we must have b0 = ck = dk = 0, since otherwise the potential blows up at s = 0.Outside we require b0 = ak = bk = 0, since otherwise the potential blows up at infinity. So,

Vin(s, φ) = a0 +∞∑k=1

sk(akcos(kφ) + bksin(kφ)) (52)

Vout(s, φ) = A0 +∞∑k=1

s−k(ckcos(kφ) + dksin(kφ)) (53)

Using the definitionσ

ε0= −(

∂V

∂n

∣∣∣∣out− ∂V

∂n

∣∣∣∣in

) =∂Vin∂s− ∂Vout

∂s

∣∣∣∣s=R

(54)

where σ is given by σ(φ) = asin5φ. Taking the derivatives of equations 52 and 53 and subtractingyields

asin5φ

ε0=∞∑k=1

kRk−1(akcos(kφ) + bksin(kφ)) + kR−k−1(ckcos(kφ) + dksin(kφ)) (55)

Clearly, ak = ck = 0, unless k = 5 to match the left hand side.

asin5φ

ε0= 5R4b5sin(5φ) + 5R−6d5sin(5φ) (56)

In order to calculate b5 and d5, we need a second equation. The second equation is the continuityof the potential inside and outside the cylinder Vin|R = Vout|R:

a0 +R5b5sin(5φ) = A0 +R−5d5sin(5φ) ⇒ a0 = A0, b5 =d5R10

(57)

We can choose a0 = A0 = 0, and plugging this into equation 56 to solve for d5

asin5φ

ε0= 5R4 d5

R10sin(5φ) + 5

d5R6

sin(5φ) ⇒ d5 =aR6

10ε0⇒ b5 =

a

10R4ε0(58)

Now we have everything we need to plug in to the potentials inside (equation 52) and outside(equation 53). If we note that all the coefficients vanish except b5 and d5:

Vin = s5a

10R4ε0sin(5φ) 2 (59)

7

Vout = s−5aR6

10ε0sin(5φ) 2 (60)

Problem 3.26A sphere of radius R, centered at the origin, carries charge density

ρ(r, θ) = kR

r2(R− 2r)sin(θ) (61)

where k is a constant, and r, θ are the usual spherical coordinates. Find the approxi-mate potential for points on the z axis, far from the sphere.For reference, the exact form of the multipole expansion potential is

V (r) =1

4πε0

∞∑n=0

1

rn+1

∫rnPn(cosθ)ρ(r)dV (62)

The monopole term in the potential is

Vmon(r) =1

4πε0

Q

r(63)

We need to find Q.

Q =

∫ρdV = 2π

∫ π

0

∫ R

0kR

r2(R− 2r)sin(θ)r2sin(θ)drdθ = kπ2

∫ R

0(R2 − 2r)dr = 0 (64)

Thus the total charge on the sphere is 0, so there is no monopole term, Vmon = 0.The dipole term is

Vdip(r) =1

4πε0

1

r2

∫rcos(θ)ρdV =

1

4πε0

2π

r2

∫ π

0

∫ R

0rcos(θ)k

R

r2(R− 2r)sin(θ)r2sin(θ)drdθ (65)

and we can see immediately that the θ integral vanishes, so the dipole term is also zero, Vdip = 0.The quadruple term won’t be zero:

Vquad =1

4πε0

1

r3

∫r2

3cos2θ − 1

2ρdV =

2πkR

8πε0r3

∫ π

0

∫ R

0r2(3cos2θ − 1)

1

r2(R− 2r)sin(θ)r2drdθ

(66)This is not a difficult integral, and it equals kπ2R5/48. Thus, the potential at a point on the z-axisis

V (z) =1

4πε0

kπ2R5

48z32 (67)

where I have let the original r → z.

Problem 3.27Four particles are placed as follows: charge q at (0, 0,−a), charges −2q at (0, a, 0) and(0,−a, 0), and charge 3q at (0, 0, a). Find a simple approximate formula for the potentialvalid at points far from the origin. Express your answer in spherical coordinates.To first order, (the monopole term) the potential vanishes, since the total charge is zero. Let’sexamine the dipole term:

Vdip =1

4πε0

p · rr2

(68)

But p =∑qidi, so the dipole moment is

p = (3q)(a)z + (q)(−a)z + (−2q)(−a)y + (−2q)(a)y + 0x = 2qaz (69)

So the dipole potential is

Vdip =1

4πε0

2qaz · rr2

=1

4πε0

2qacosθ

r22 (70)

8

Problem 3.30Two point charges, 3q and −q, are separated by a distance a. For each of the arrange-ments below, find (i) the monopole moment, (ii) the dipole moment, and (iii) the ap-proximate potential (in spherical coordinates) at large r (include both the monopoleand dipole contributions).a) 3q at (0, 0, a), −q at origin.The monopole moment is just the charge, Q = 3q + (−q) = 2q. The dipole moment is p =

∑qidi

p = (−q)(0) + (3q)(a)z = 3qaz 2 (71)

The approximate potential is the sum of the monopole and dipole terms:

V =1

4πε0(Q

r+

p · rr2

) =1

4πε0(2q

r+

3qacosθ

r2) 2 (72)

where I have replaced z · r with cosθ.b) 3q at origin, −q at (0, 0,−a).The monopole moment is that total charge, which is again 2q. The dipole moment is

p = (3q)(0) + (−q)(−a)z = qaz 2 (73)

The approximate potential is

V =1

4πε0(Q

r+

p · rr2

) =1

4πε0(2q

r+qacosθ

r2) 2 (74)

c) 3q at (0, a, 0), −q at origin.The monopole moment is the total charge 2q, the dipole moment is

p = (3q)(a) + (−q)(0)y = 3qay 2 (75)

meaning that the potent is

V =1

4πε0(Q

r+

p · rr2

) =1

4πε0(2q

r+

3qasinθsinφ

r2) 2 (76)

since y · r = sinθsinφ.

Problem 3.31A ”pure” dipole p is situated at the origin, pointing in the z direction.a) What is the force on a point charge q at (a, 0, 0) (Cartesian coordinates)?The potential on the x-axis due to a dipole is

V =1

4πε0

p · rr2

=1

4πε0

pcosθ

r2(77)

So the force is

F = −q∇V = −q∂V∂r

r− q

r

∂V

∂θθ =

q

4πε0

2pcosθ

r3r +

1

4πε0

qpsinθ

r3θ

∣∣∣∣r=a,θ=π/2

= − pq

4πε0a3z (78)

since for θ = π/2, the unit vector θ is in the -z direction. By the way, θ = π/2 since the point weare looking at is on the x axis, while the dipole moment is in the z direction.b) What is the force on q at (0, 0, a)?This time, θ = 0, so

F =pq

2πε0a3z (79)

Since the sinθ kills the second term in equation 78 above, and the r direction is in z.c) How much work does it take to move q from (a, 0, 0) to (0, 0, a)?

9

This is just the difference between the answers to parts a and b, multiplied by q, since W = qV .Thus,

W =pq

4πε0a2(80)

Problem 3.32Three point charges are located as shown in Fig. 3.38, each a distance a from theorigin. Find the approximate electric field at points far from the origin. Express youranswer in spherical coordinates, and include the two lowest orders in the multipoleexpansion.The first order term in the potential is V = −q/4πε0r2, since the net charge is -q. The dipole termis

V =1

4πε0

∫p · rr2

=1

4πε0

∫p · rr2

(81)

where the dipole moment is

p =

∫ρ(r)rdV =

∫−qδ(y + a) + qδ(z − a)− qδ(y − a)(xx + yy + zz)dxdydz = (82)

multiplying this out:

p =

∫qδ(z − a)zdzz− q(δ(y − a) + δ(y + a))ydyy = qaz− qay− q(−a)y = qaz (83)

Thus, the dipole potential is

V =1

4πε0

∫qaz · rr2

=1

4πε0

qa

r2cosθ (84)

Thus the total potential, monopole and dipole terms included, is:

V = − 1

4πε0

q

r2+

1

4πε0

qa

r2cosθ (85)

The electric field, therefore, is

E = −∇V =q

4πε0[−1

r2r +

a

r3(2cosθr + sinθθ)] 2 (86)

Problem 4.20A sphere of linear dielectric material has embedded in it a uniform free charge densityρ. Find the potential at the center of the sphere (relative to infinity), if its radius isR and its dielectric constant is εr.We need to find the electric field everywhere first. Inside the sphere,∫

D · da = D4πr2 = Qf =

∫ρdV = ρ

4

3πr3 ⇒ Din =

1

3ρrr (87)

while outside the sphere,∫D · da = D4πr2 = Qf =

∫ρdV = ρ

4

3πR3 ⇒ Dout =

1

3r2ρR3r (88)

Therefore, inside the sphere,

Ein =D

ε=

D

ε0εr=

ρr

3ε0εrr (89)

Outside the sphere, the electric field is

Eout =D

ε=

D

ε0=

ρR3

3ε0r2r (90)

10

From which we can find the potential:

V =

∫ ∞0

E · dl =

∫ R

0Ein · dl +

∫ ∞R

Eout · dl =

∫ R

0

ρr

3ε0εrdr +

∫ ∞R

ρR3

3ε0r2dr (91)

=ρR2

6ε0εr+ρR2

3ε0=ρR2

3ε0(1 +

1

2εr) 2 (92)

Problem 4.23Find the field inside a sphere of linear dielectric material in an otherwise uniformelectric field E0 by the following method of successive approximations: First pretendthe field inside is just E0, and use Eq. 4.30 to write down the resulting polarizationP0. This polarization generates a field of its own, E1 (Ex. 4.2), which in turn modifiesthe polarization by an amount P1, which further changes the field by an amount E2,and so on. The resulting field is E0 + E1 + E2 + .... Sum the series, and compare youranswer with Eq. 4.49.The polarization due to the field E0 is P0 = ε0χeE0. The field resulting from this polarization isgiven by equation 4.18: E = −P/3ε0. Thus,

E1 = −P0

3ε0= −ε0χeE0

3ε0= −χeE0

3(93)

Now the polarization due to this field is P1 = ε0χeE1, which results in the field E2 = −P1/3ε0:

E2 = −P1

3ε0= −ε0χeE1

3ε0= −χe

3(−χeE0

3) =

χ2e

9E0 (94)

Thus we obtain an infinite geometric series

Etot = E0

∞∑n=0

(−χe3

)n =E0

1 + χe/3(95)

Using χe = εr − 1, we obtain:

Etot =E0

2/3 + εr/3=

3

εr + 2E0 2 (96)

Problem 4.26A spherical conductor, of radius a, carries a charge Q. It is surrounded by a lineardielectric material of susceptibility χe, out to radius b. Find the energy of this config-uration.We will use equation 4.58 in Griffiths:

W =1

2

∫D ·Edτ (97)

It is evident that for r < a, both D and E are 0. Outside the conductor, using the fact thatε = ε0(1 + χe):

D =Q

4πr2r r > a (98)

E =Q

4πεr2r a < r < b (99)

E =Q

4πε0r2r r > b (100)

Thus, the energy is

W =1

2

∫ b

a

Q

4πr2r · Q

4πεr2r4πr2dr +

1

2

∫ ∞b

Q

4πr2r · Q

4πε0r2r4πr2dr (101)

11

where the integral from 0 to a contributes no energy to the system. Thus,

W =Q2

8πε

∫ b

a

dr

r2+

Q2

8πε0

∫ ∞b

dr

r2=

Q2

8πε(1

a+

1

b)− Q2

8πε0b(102)

Or, writing the answer in terms of the susceptibility,

W =Q2

8πε0(1 + χe)(1

a+

1

b)− Q2

8πε0b2 (103)

Problem 4.31A dielectric cube of side a, centered at the origin, carries a ”frozen-in” polarizationP = kr, where k is a constant. Find all the bound charges, and check that they addup to zero.The polarization can be rewritten in Cartesian coordinates,

P = k(xx + yy + zz) (104)

We also know that −∇ ·P = ρb and P · n = σb. Let’s find the bound volume charge first:

ρb = −∇ ·P = −k(1 + 1 + 1) = −3k (105)

Meanwhile, the bound surface charge on one side is

σb = P · n = k(xx + yy + zz) · z∣∣∣∣z=a/2

=ka

2(106)

Thus, since there are six sides to a cube, the total bound surface charge is σb = 3ka. Now we needto find the total charge:

Qtot =

∫ρbdV +

∫σbda = −3k(a3) + 3ka(a2) = 0 2 (107)

Problem 4.32A point charge q is imbedded at the center of a sphere of linear dielectric material(with susceptibility χe and radius R). Find the electric field, the polarization, andthe bound charge densities ρb and σb. What is the total bound charge on the surface?Where is the compensating negative bound charge located?The displacement field D is easily found to be D = q/4πr2r. The electric field, written in terms ofthe susceptibility, defined such that ε = ε0(1 + χe), is

E =q

4πε0(1 + χe)r2r (108)

The polarization is P = ε0χeE, so

P =χeq

4π(1 + χe)r2r (109)

The bound volume charge is

ρb = −∇ ·P = − χeq

4π(1 + χe)∇ · r

r2= − χeq

(1 + χe)δ3(r) (110)

where I have used the fact that ∇ · (r/r2) = 4πδ3(r). Meanwhile the bound surface charge is

σb = P · n∣∣∣∣R

=χeq

4π(1 + χe)R2(111)

Thus the total bound charge on the surface is

Qsurface =

∫σbda =

χeq

4π(1 + χe)R24πR2 =

χe1 + χe

q (112)

12

The compensating negative charge is located at the origin (in the ρb term):

Qcomp =

∫ρbdV = − χeq

(1 + χe)

∫δ3(r) = − χeq

(1 + χe)(113)

which exactly cancels the surface charge in equation 112.

Problem 5.4Suppose that the magnetic field in some region has the form B = kzx. (where k is aconstant). Find the force on a square loop (side a), lying in the yz plane and centeredat the origin, if it carries a current I, flowing counterclockwise, when you look downthe x axis.The force is F = Idl ×B. The force on the two horizontal segments in z will cancel out, since thefield flips sign.

F = I

∫dl×B = I

∫ a

0dzy× kzx =

1

2Ika2z (114)

But that was only for one edge of the loop. The force on both edges will conspire, so the total forceis

F = Ika2z 2 (115)

Problem 5.5A current I flows down a wire of radius a.a) If it is uniformly distributed over the surface, what is the surface current densityK?K = dI/dl⊥, so I =

∫Kdl⊥ = K2πa, since the length perpendicular to the flow is the circumference

(not the surface area, which has a component parallel to the flow). Thus, K = I/2πa 2

b) If it is distributed in such a way that the volume current density is inverselyproportional to the distance from the axis, what is J?We have that J ∝ 1/s, or more precisely J = α/s, with α some constant. Thus,

I =

∫JdS = α

∫ a

0

2πs

sds = 2πaα ⇒ α =

I

2πa(116)

and therefore

J =I

2πsa2 (117)

Problem 5.9Find the magnetic field at point P for each of the steady current configurations shownin Fig. 5.23We will try to use the Biot Savart law:a)

B =µ04π

∫Idl × r

r2=µ0I

4π

∫ π/2

0

adφφ× r

a2+µ0I

4π

∫ 0

π/2

bdφφ× r

b2(118)

where I have used that dl = rdφφ, for r = a and r = b. There is clearly no contribution from thestraight sides, since here dl × r = 0. Thus,

B =µ0I

4πa(z)

π

2+µ0I

4πb(z)(−π

2) =

µ0I

8(1

a− 1

b)z 2 (119)

b)Here, the Biot Savart integral becomes (for the semicircular side):

B =µ0I

4π

∫ π/2

−π/2

Rdφ(−φ)× r

R2= −µ0I

4Rz (120)

13

For the two straight lines, the Biot Savart integral is awkward and difficult to compute, so we willrefer to Griffiths example 5, in which he states that the magnetic field between these two wires isgoing to be B = µ0I/2πs, so the total field at s = R is:

B = (−µ0I4R

+µ0I

2πR)z (121)

Problem 5.12Suppose you have two infinite straight line charges λ, a distance d apart, movingalong at a constant speed v. How great would v have to be in order for the magneticattraction to balance the electric repulsion?A line charge λ traveling at velocity v constitutes a current I = λv. The magnetic field at a distanced due to the first (second) wire is

B =µ0

2πdI1(I2) (122)

There is an attractive force between the wires, since the current is in the same direction. Themagnitude of the force is F = ILB, where L is the length of the wire. Now, the total force isinfinite, but the force per unit length is f = IB, so the force per unit length between these twowires is, substituting I = λv,

fB =µ0I1I2

2πd=µ0λ

2v2

2πd(123)

Meanwhile, we need to find the electric field due to a wire of charge density λ:∫E · dl = E2πd =

λL

ε0⇒ E =

λL

2πε0d, (124)

meaning that the force per unit length between the wires is

fe =λE

L=

λ2

2πε0d(125)

Thus the electric force per unit length fe ≡ Fe/L can be set equal to the magnetic force per unitlength:

fB = fe ⇒ µ0λ2v2

2πd=

λ2

2πε0d⇒ v =

√1

ε0µ0= c 2 (126)

Thus you could never get the wires moving fast enough – the electric force will always dominateover the magnetic force, and the wires will repel.

Problem 5.15Two long coaxial solenoids each carry current I, but in opposite directions, as shownin Fig. 5.42. The inner solenoid (radius a) has n1 turns per unit length, and the outerone (radius b) has n2. Find B in each of the three regions: (i) inside the inner solenoid,(ii) between them, and (iii) outside both.The field inside a solenoid is B = µ0nI. Outside, the field is zero. Thus, in region (i), we havecontributions from both the larger and the smaller solenoid. The field in region (i) is

B(i) = µ0n1I z + µ0n2I(−z) = µ0I(n1 − n2)z 2 (127)

where the minus sign comes from the fact that the current is being carried in the opposite directionin the larger solenoid. In region (ii), the only contribution to the magnetic field is from the largersolenoid. The region is outside the smaller solenoid, so the magnetic field due to the smaller one iszero. Thus, the field in region (ii) is

B(ii) = µ0n2I(−z) = −µ0n2I z 2 (128)

14

Outside both solenoids, the field is zero. Thus, in region (iii),

B(iii) = 0 2 (129)

Problem 5.16A large parallel-plate capacitor with uniform surface charge σ on the upper plate and−σ on the lower is moving with a constant speed v, as shown in Fig. 5.43.a) Find the magnetic field between the plates and also above and below them.We consider each moving plate to be a surface current K, where K = σv. According to example5.8, the magnetic field due to such a surface current is B = ±µ0K/2, where the + is below theplane and the − is above the plane. How is this derived? Well, use Ampere’s law:

∫B · dl = 2BL =

µ0Ienc = µ0KL ⇒ B = µ0K/2. Thus, above and below the capacitor, the field is zero, since thesurface current K is pointed in opposite direction due to the opposite charge on the plates. Inbetween the capacitor plates, the field adds:

Bbetween =µ02σv − µ0

2(−σv) = µ0σv (in) 2 (130)

b) Find the magnetic force per unit area on the upper plate, including its direction.The force per unit area is f = K×B, so we obtain, for theupperplate

f = σvµ02σv =

µ0σ2v2

2(131)

Doing the right hand rule will convince you that the force is up.c) At what speed v would the magnetic force balance the electrical force?The electric field of the bottom plate is E = σ/2ε0, so the force on the upper plate is F = σ2A/2ε0.The force per unit area is, thus, f = σ2/2ε0. Equating this force with the force per unit area inpart (b) yields:

µ0σ2v2

2=

σ2

2ε0⇒ v =

1√µ0ε0

2 (132)

So the plates would have to move at the speed of light. Thus, the forces would never balance.

Problem 5.22Find the magnetic vector potential of a finite segment of a straight wire, carrying acurrent I. [Put the wire on the z axis, from z1 to z2, and use equation 5.64:

A =µ04π

∫Idl

|r− r′|(133)

Check that your answer is consistent with equation 5.35:

B =µ0I

4πs(sinθ2 − sinθ1)φ (134)

The distance between the wire and any point s off the wire is√z2 + s2, and, letting I = I z, and

dl = dz,

A =µ0I

4π

∫ z2

z1

zdz√z2 + s2

=µ0I

4πln

√s2 + z22 + z2√s2 + z21 + z1

z 2 (135)

To confirm whether this matches equation 5.35, we take the curl of A:

B = ∇×A = −∂Az∂s

φ (136)

15

If you take this derivative, you get

B = φµ0I

4πs[

z2√z22 + s2

− z21√z21 + s2

] (137)

But recall that sinθj = zj/√z2j + s2, so

B = φµ0I

4πs(sinθ2 − sinθ1) 2 (138)

Problem 6.1Calculate the torque exerted on the square loop shown in Fig. 6.6, due to the circularloop (assume r is much larger than a or b. If the square loop is free to rotate, whatwill its equilibrium orientation be?Torque is given by N = m × B. The dipole moment of the circular loop is m = Iπa2z. Themagnetic field due to the circular loop is given by everyone’s favorite formula for the magnetic fielddue to a dipole:

B =µ04π

1

r3[3(m · r)r−m] = −µ0

4π

Iπa2

r3z (139)

since m · r = 0. The m in the torque equation, however, refers to the dipole moment of the squareloop, which is m = Ib2x. Thus,

N = −Ib2x× µ04π

Iπa2

r3z = −I

2a2b2µ04r3

y (140)

The final orientation of the square loop will be when there is no torque on it. Since the magneticfield is in the z direction, the loop will rotate until it points in -z. Why minus? Since the torqueis in -y, the loop is spinning in the direction of negative z, so it will anti align with the other loop.

Problem 6.7An infinitely long circular cylinder carries a uniform magnetization M parallel to itsaxis. Find the magnetic field (due to M) inside and outside the cylinder.If the magnetization is parallel to the axis then we can write M = M z. Then,

Jb = ∇×M = 0, Kb = M× n = M z× s = Mφ (141)

So the cylinder acts like a solenoid, since it carries a surface current in the azimuthal direction.When that happens, the magnetic field is B = µ0nI, where n is the number of turns per unit length.But n = N/L, and I = KbL, thus the product nI = NKb, and, since N = 1 for this cylinder,

B = µ0Kbz = µ0M z = µ0M 2 (142)

Outside the cylinder, where the magnetization is zero, the field is also zero.

Problem 6.8A long circular cylinder of radius R carries a magnetization M = ks2φ, where k is aconstant, s is the distance from the axis, and φ is the usual azimuthal unit vector (Fig.6.13). Find the magnetic field due to M, for points inside and outside the cylinder.The easiest way to do this is to calculate the bound currents:

Jb = ∇×M =1

s

∂(sMφ)

∂sz = 3ksz, Kb = M× n = −ks2z

∣∣∣∣s=R

= −kR2z (143)

Now we can get the magnetic field inside the cylinder by using Ampere’s law. Since the surfacecurrent is... on the surface, the only contribution to Ienc is from Jb:∫

B · dl = B2πs = µ0Ienc = µ0

∫Jbda = µ0

∫ s

03ksz · sdφz = 2πµ0ks

3 ⇒ B = µ0ks2φ (144)

16

or, in terms of the magnetization,

Bin = µ0ks2φ = µ0M 2 (145)

Inside the cylinder, the magnetic field is zero since the total current enclosed by an Amperian loopis zero. Let’s check that:

Ienc =

∫Jbda+

∫Kbdl =

∫ R

03ksz · sdφz− kR2

∫dl = 2πkR3 − kR22πR = 0 (146)

Problem 6.12An infinitely long cylinder, of radius R, carries a frozen-in magnetization, parallel tothe axis, M = ksz, where k is constant and s is the distance from the axis; there is nofree current anywhere. Find the magnetic field inside and outside the cylinder by twodifferent methods:a) As in Sect. 6.2, locate all the bound currents. and calculate the field they produce.The bound volume current is Jb = ∇ ×M and the bound surface current is Kb = M × n. Let’scalculate these:

Jb = −dMz

dsφ = −kφ, Kb = M× s

∣∣∣∣s=R

= kRφ (147)

Thus, the total current through a square Amperian loop of side length l, placed a distance R − sinto the cylinder is

Ienc =

∫Jbda+

∫Kbdl = −k(R− s)l + kR(l) = ksl (148)

and therefore the magnetic field is∫B · dl = Bl = µ0Ienc = µ0ksl ⇒ Bin = µ0ksz 2 (149)

It is in the z direction, since the magnetization is in that direction and the surface and volumecurrents are in the φ directions. Meanwhile, outside, there is no current through the Amperianloop outside, so Bout = 0.b) Use Ampere’s law (in the form of Eq. 6.20) to find H, and the get B from Eq. 6.18.(Notice that the second method is much faster, and avoids any explicit reference tothe bound currents.)Ampere’s law is

∫H · dl = Ifree,enc. Since there is no free current anywhere, H = 0. Therefore,

since H = B/µ0 −M, we can immediately find the magnetic field. Outside, where M = 0, B = 0also. Inside, B = µ0M, so

B = µ0ksz 2 (150)

Problem 6.16A coaxial cable consists of two very long cylindrical tubes, separated by linear in-sulating material of magnetic susceptibility chim. A current I flows down the innerconductor and returns along the outer one; in each case the current distributes itselfuniformly over the surface (Fig. 6.24). Find the magnetic field in the region betweenthe tubes. As a check, calculate the magnetization and the bound currents, and con-firm that (together, of course, with the free currents) they generate the correct field.Using Ampere’s law:

∫H · dl = Ifree ⇒ H = Ifree/2πs. Therefore, since B = µH = µ0(1+χm)H,

we obtain

B = µ0(1 + χm)I

2πsφ (151)

Magnetization is given by M = χmH, so

M =Iχm2πs

φ (152)

17

Therefore,

J = ∇×M =1

s

∂(sMφ)

∂sz = 0 (153)

Ka = M× n

∣∣∣∣s=a

=Iχm2πs

φ× s

∣∣∣∣s=a

=Iχm2πa

z (154)

and

Ka = M× n

∣∣∣∣s=b

=Iχm2πs

φ× s

∣∣∣∣s=b

= −Iχm2πb

z (155)

Thus, the total current is

Itot = I +

∫Kdl = I +

∫Iχm2πa

z · dlz = I +Iχm2πa

2πa = I(1 + χm) (156)

Thus, ∫B · dl = µ0Ienc ⇒ B2πs = µ0I(1 + χm) ⇒ B =

µ0I(1 + χm)

2πsφ 2 (157)

Problem 6.17A current I flows down a long straight wire of radius a. If the wire is made oflinear material (copper, say, or aluminum) with susceptibility χm, and the current isdistributed uniformly, what is the magnetic field a distance s from the axis? Find allthe bound currents. What is the net bound current flowing down the wire.At a distance s < a from the center of the wire,

∫H · dl = 2πsH = Ifree. Since the current is

distributed uniformly, the current at that distance is Is2/a2. Thus,

Hin = Is

2πa2φ (158)

Meanwhile, outside the wire, the enclosed current is just I. Thus,

Hout =I

2πsφ (159)

Now, B = µH = µ0(1 + χm)H. Thus,

Bin = Iµ0(1 + χm)s

2πa2φ (160)

and

Bout =Iµ02πs

φ (161)

since the susceptibility of free space vanishes. The free current is Jf = I/πa2 and Jb = χmJf , so

Jb =I

πa2χmz (162)

while

Kb = M× n = χmH× n

∣∣∣∣s=a

=χmI

2πa(−z) (163)

The net bound current is

Ib = Jbπa2Kb2πa = Iχm − Iχm = 0 2 (164)

Problem 7.7A metal bar of mass m slides frictionlessly on two parallel conducting rails a distancel apart. A resistor R is connected across the rails and a uniform magnetic field B,pointing into the page, fills the entire region.a) If the bar moves to the right at speed v, what is the current in the resistor? In

18

what direction does it flow?If the bar is moving to the right, that means that the flux through the loop is increasing. The areaof the loop is lvt, so that the flux (and it’s time derivative) through the loop is

φB = lvtB ⇒ dφBdt

= Blv (165)

By Faraday’s law, this produces an EMF in the loop, whose magnitude is

EMF = −dφBdt

= −Blv = IR (166)

meaning that the current is

I = −BlvR

2 (167)

The negative sign means that the current is flowing counterclockwise (i.e. downward through theresistor), since the positive sense was defined by the direction of the magnetic field.b) What is the magnetic force on the bar? In what direction?The magnetic force is F = Idl×B. Since the current is up through the bar and the magnetic fieldis into the page, the force is to the left, as it must be. Otherwise, the bar would gain more andmore energy as it sped up, without a source of that energy. The bar must slow down. The force is

F = IlB = −B2l2v

R(168)

where the minus sign means to the left. 2.c) If the bar starts out with speed v0 at time t = 0, and is left to slide, what is its speedat a later time t?

mdv

dt= −B

2l2v

R⇒ v(t) = v0e

−B2l2t/Rm 2 (169)

d) The initial kinetic energy of the bar was, of course, mv20/2. Check the energydelivered to the resistor is exactly mv20/2.The power delivered to the resistor is P = I2R, and power is the time derivative of energy:P ≡ dW/dt:

P =dW

dt= I2R =

B2l2v2

R⇒ dW =

B2l2v2

Rdt (170)

Plugging in my result for v2 from equation 169:

W =B2l2

R

∫ ∞0

v20e−2B2l2t/Rmdt =

B2l2v20R

Rm

−2B2l2(e−2B

2l2t/Rm)

∣∣∣∣∞0

=1

2mv20 2 (171)

Problem 7.12A long solenoid, of radius a, is driven by an alternating current, so that the field insideis sinusoidal: B(t) = B0cos(ωt)z. A circular loop of wire, or radius a/2 and resistanceR, is placed inside the solenoid, and coaxial with it. Find the current induced in theloop, as a function of time.The flux through the loop is

φB = π(a

2)2B0cos(ωt) =

π

4a2B0cos(ωt) (172)

so the induced EMF is

EMF = −dφBdt

=πa2B0ω

4sin(ωt) (173)

19

meaning that the current I = EMF/R is

I =πa2B0ω

4Rsin(ωt) (174)

Problem 7.15A long solenoid with radius a and n turns per unit length carries a time-dependentcurrent I(t) in the φ direction. Find the electric field (magnitude and direction) ata distance s from the axis (both inside and outside the solenoid), in the quasi staticapproximation.The magnetic field inside a solenoid is B = µ0nI, and outside is 0. Thus, the flux through anAmperian loop of radius s inside the solenoid is

φB = µ0nI(t)πs2 (175)

The induced EMF is

EMF = −µ0nπs2dI

dt(176)

and the electric field can be calculated via

EMF =

∫E · dl = E2πs ⇒ Ein = −µ0ns

2

dI

dt2 (177)

The electric field circles around the solenoid, since the magnetic field is in the z direction throughthe solenoid. Meanwhile, the flux through an Amperian loop outside the solenoid is

φB = µ0nI(t)πa2 (178)

since the magnetic field outside the solenoid is 0, the contribution to∫

B · da vanishes outside.Thus, the electric field can be found through∫

E · dl = − d

dtµ0nI(t)πa2φ (179)

Since the loop of integration is taken around the point outside the solenoid, the integral becomes2πsE.

E2πs = −µ0nπa2dI

dt⇒ E = −µ0na

2

2s

dI

dtφ (180)

Problem 7.31A fat wire, radius a, carries a constant current I, uniformly distributed over its crosssection. A narrow gap in the wire, of width w � a, forms a parallel-plate capacitor, asshown in Fig. 7.43. Find the magnetic field in the gap, at a distance s < a from theaxis.This is a displacement current problem. The displacement current is given by the change of electricflux. The electric field due to the charge density on the edge of the wire is E = σ/ε0. Thus, theflux through an Amperian loop of radius s is

φE =σ

ε0πs2 ⇒ dφE

dt=πs2

ε0

dσ

dt=πs2

ε0

1

πa2dq

dt=

Is2

a2ε0(181)

Thus, we have ∫B · dl = ε0µ0

∂φE∂t

=ε0µ0Is

2

a2ε0⇒ B =

µ0Is2

a21

2πsφ =

µ0Is

2πa2φ 2 (182)

Problem 7.32The preceding problem was an artificial model for the charging capacitor, designed toavoid complications associated with the current spreading out over the surface of the

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plates. For a more realistic model, imagine thin wires that connect to the centers ofthe plates (Fig. 7.44a). Again, the current I is constant, the radius of the capacitor isa, and the separation of the plates is w � a. Assume that the current flows out overthe plates in such a way that the surface charge is uniform, at any given time, and iszero at t = 0.a) Find the electric field between the plates, as a function of t.Between the plates, we have the usual electric field in a capacitor, E = σ/ε0 = q/πε0a

2. Butq/t = I, so the electric field as a function of time is

E =It

πε0a2z 2 (183)

b) Find the displacement current through a circle of radius s in the plane midwaybetween the plates. Using this circle as your ”Amperian loop,” and the flat surfacethat spans it, find the magnetic field at a distance s from the axis.The flux through this circle is this electric field multiplied by the area of the circle, πs2:

φE =It

πε0a2πs2 (184)

Therefore, the displacement current is

Id = ε0dφEdt

= ε0I

πε0a2πs2 =

Is2

a22 (185)

From this we can obtain the magnetic field∫B · dl = µ0Id ⇒ B2πs =

µ0Is2

a2⇒ B =

µ0Is

2πa2φ 2 (186)

c) Repeat part (b), but this time use the cylindrical surface in Fig. 7.44b, which isopen at the right end and extends to the left through the plate and terminates outsidethe capacitor. Notice that the displacement current through this surface is zero, andthere are two contributions to Ienc. We know have a total enclosed current of (i) the currentin the wire with radius s, and (ii) the displacement current from the donut region around the wire.The displacement current will be the old displacement current in equation 185, minus the currentin the wire. So

Id,new = Idisp,old − Iwire = Iwires2

a2− Iwire (187)

To obtain the total current to be enclosed by our Amperian loop, we need to add the real currentflowing in the wire, I:

Itot = Id,new + Iwire = Iwires2

a2− Iwire + Iwire = Iwire

s2

a2(188)

Thus, we again find, as in equation 186, that

B =µ0Is

2πa2φ 2 (189)

Problem 8.2Consider the charging capacitor in Prob. 7.31.a) Find the electric and magnetic fields in the gap, as functions of the distance s fromthe axis and the time t. (Assume the charge is zero at t = 0.)We already found the magnetic field in problem 7.31 by using Maxwell’s law

∫B · dl = ε0µ0φE .

We found it to be

B =µ0Is

2πa2φ 2 (190)

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Meanwhile, the electric field in the gap is given by E = σ/ε0z. Using σ = q/A = It/A, we obtain

E =It

ε0πa2z 2 (191)

We could have also found this by using Maxwell’s law in differential form:

∇×B =1

s

∂(sBφ)

∂sz =

µ0I

πa2z = ε0µ0

∂E

∂t⇒ E =

It

ε0πa2z (192)

b) Find the energy density uem and the Poynting vector S in the gap. Note especiallythe direction of S. Check that Eq. 8.14 is satisfied.The energy density is

uem =ε02E2 +

B2

2µ0=

1

2

I2t2

ε0π2a4+

1

2

µ0I2s2

4π2a4=

µ0I2

2π2a4(c2t2 +

s2

4) (193)

Meanwhile, the Poynting vector is trivially computed:

S =1

µ0E×B = − I2st

2π2ε0a4s (194)

Eq. 8.14 is∂

∂t(umech + uem) = −∇ · S (195)

umech = 0 here, so it remains to take the time derivative of equation 193 and compare it to thedivergence of equation 194:

∂uem∂t

=µ0I

2

π2a4c2t = −∇ · S =

I2t

2π2ε0a4∇ · (ss) =

I2t

π2ε0a4(196)

(Recall that the divergence is taken in cylindrical coordinates). Since c2 = 1/µ0ε0, equation 8.14 isconfirmed.c) Determine the total energy in the gap, as a function of time. Calculate the totalpower flowing into the gap, by integrating the Poynting vector of the appropriatesurface. Check that the power input is equal to the rate of increase of energy in thegap (Eq. 8.9 – in this case W = 0, because there is no charge in the gap).The total energy is given by

Uem =

∫uemdV =

∫ b

0uemw2πsds =

µ0wI2b2

2πa4[c2t2 +

b2

16] (197)

for an arbitrary surface of radius b. w is the width of the gap. The total power is

P =

∫S · da =

I2t

2π2ε0a4[bs · (2πbws] =

I2wtb2

πε0a4(198)

Problem 10.3Find the fields, and the charge and current distributions, corresponding to V (r, t) =0, A(r, t) = −qt/4πε0r2r.This is a rather simple exercise:

E = −∇V − ∂A

∂t=

1

4πε0

q

r2r (199)

B = ∇×A = 0 (200)

It looks like this corresponds to a point charge at the origin:

∇ ·E =q

4πε0∇ · ( r

r2) =

q

ε0δ3(r) ⇒ ρ = qδ3(r) (201)

∇×B = 0 ⇒ J = 0 (202)

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