jack9a

Solutions to Problems in Jackson,

Classical Electrodynamics, Third Edition

Homer Reid

June 27, 2003

Chapter 9: Problems 1-8

Problem 9.2

We place the square in the xy plane, centered at the origin, and aligned such thatat time t = 0 the q charges are on the x axis at x = R (where R = a/2).Then the charge density in cylindrical coordinates is

(r, , z, t) =q

R(r R)(z)

{( t) ( pi t)

}. (1)

The Fourier series expansion of is

(x, ) =

n=1

{An(x) cosnt+Bn(x) sinnt

}(2)

Multiplying by cosnt and sinnt and integrating from t = 0 to t = 2pi/, wefind

An =

2pi

2pi/0

(x, t) cosnt dt

Bn =

2pi

2pi/0

(x, t) sinnt dt.

1

Homer Reids Solutions to Jackson Problems: Chapter 9 2

Inserting (1), we find

An =q

2piR(r R)(z)

2pi/0

{( t) ( pi t)

}cosnt dt

=q

2piR(r R)(z)

{cosn cosn( pi)

}

=

{qpiR(r R)(z) cosn, n odd0, n even.

Similarly,

Bn =

{qpiR(r R)(z) sinn, n odd0, n even.

Plugging into (2),

(r, , z, t) =q

piR(r R)(z)

n odd

{cos(nt) cos(n) + sin(nt) sin(n)

}

=q

piR(r R)(z)

n odd

cos[n(t )]. (3)

To find the associated current density, we use

J = t

Clearly there is only a component of the current density, so this is

1

R

J

= +q

piR(r R)(z)

n odd

n sin[n(t )]

from which evidently

J = qpi(r R)(z)

n odd

cos[n(t )].

Evidently the space integrals of both J and r J(= RJ) vanish, so there is noelectric or magnetic dipole component to the radiation. We define the constantQn, for n positive and negative, to be the quadupole moment associatedwith the component of the charge density varying as eint. They are given by(Jackson equation (9.41))

Qn =

(3xx r2

)(x)d3x

=q

2piR

(3xx r2

)(r R)(z)ein dx.


Since the charge is confined to the plane z = 0, any moment with , = zvanishes. The remaining three moments are

Qxxn =qR2

2pi

2pi0

(3 cos2 1)eind

Qyyn =qR2

2pi

2pi0

(3 sin2 1)eind

Qxyn =3qR2

2pi

2pi0

sin coseind.

These are only nonvanishing if n = 2, and in that case we have

Qxx,2 =3qR2

4

Qyy,2 = 3qR2

4Qxy,2 = Qyx,2 = 0.

The vector defined by Jacksons equation (9.43) is

Q =3qR2

4

[nxi ny j

]and we then have

nQ = 3qR2

4

[nzny i+ nznxj 2nxnyk

](nQ

) n = 3qR

2

4

[nx(n2z + 2n2y) i+ ny(n2z + 2n2x) j+ nz(n2x n2y)k

]

|(nQ

) n|2 =

(3qR2

4

)2n2z(n

2x + n

2y).

Plugging into Jacksons equation (9.44) and switching to spherical coordinates,

H = icqR2k3

32pi

eikr

r

[cos sin sin i + cos sin cos j 2 sin2 sin cos k.

]dP

d=

c2q2R4Z0k6

2048pi2cos2 sin2

Integrating over solid angles, we find the total radiated power:

P =c2q2R4Z20k

6

2048pi2

pi0

2pi0

cos2 sin3 dd

=c2q2R4Z20k

6

7680pi.


Problem 9.3

The components of the vector potential are determined by the wave equation(2 + 1

c22

t2

)A(r, t) = 0.

The solution is

A(r, t) =04pi

dr

|r r|J(r, t |r r

|c

)

For sources varying in time as eit, we have

=04pi

eit

dr

|r r|J(r)eik|rr|. (4)

In the radiation zone we approximate

|r r| r n r1

|r r| 1

r

(1 +

n rr

)(5)

eik|rr| eikr

(1 ikn r

). (6)

Inserting (5) and (6) into (4) and keeping only terms of zeroth and first orderin r/r, we have

A(r, t) =04pi

ei(krt)

r

dr

{J(r) +

(1

r ik

)(n r)J(r) +

}. (7)

In the present case the lowest-order term in this expansion is nonvanishing, soin what follows we drop the higher-order terms. We then need to compute thevolume integral of J :

J(r) dr =

r( J) dr

= i

r(r) dr

= ip. (8)

Then (7) becomes

A(r, t) = i04pi

ei(krt)

rp. (9)

In this case we know the potential difference between the upper and lowerhemispheres is

V = 2V cost.


Hence the total charge on one hemisphere is

Q = CV = 2CV cost

where C is the capacitance of the spherical shell, which we will work out later.This charge is spread out uniformly over the 2piR2 surface area of the hemi-sphere. Then the amplitude of the oscillating charge density is

(r, , ) =

2CV

2piR2(r R), 0 < < pi

2

2CV2piR2

(r R), pi2< < pi.

Only the z component of the dipole moment is nonvanishing:

pz =

z(r)dr

=2CV

2piR2

0

2pi0

pi/20

(r R)r2 sin d d dr

2CV2piR2

0

2pi0

pipi/2

(r R)r2 sin d d dr

= 2CV R

[ pi/20

sin d pipi/2

sin d

]

= 4CV R.

The vector potential is

A =i0

pi

eikr

rCV Rk.

The fields in the radiation zone are

H =ck2

pi

eikr

rCV R

(n k

)E = Z0 ck

2

pi

eikr

rCV R

[n

(n k

)].

The angular distribution of power is

dP

d=

c2Z08pi2

k4(CV R)2 sin2

and the total power radiated is

P =4c2Z0k

4

3pi(CV R)2.

Finally, for completeness, lets calculate the capacitance C. Suppose we havetotal chargesQ on the top hemisphere and Q on the bottom hemisphere. Thenthe surface charge density is

= Q2piR2

.


The electric potential at the south pole of the shell is

=1

4pi0

[upper hemisphere

dA

|RR| lower hemisphere

dA

|RR|].

The distance from the south pole to a point on the surface with angles , isjust R

2(1 + cos ), so

=Q

42pi0R

[ 10

du

(1 + u)1/2 01

du

(1 + u)1/2

]

=Q

22pi0R

[ 1 + u101 + u

10]

=Q

22pi0R

[2 2

].

The electric potential at the north pole is the negative of this, so the potentialdifference is

V =Q

2pi0R

[2 2

].

Then the capacitance of the shell is

C =Q

V= pi0R

[1 +

2].

Is this right?

jack9a

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