jack 4b

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Solutions to Problems in Jackson,

Classical Electrodynamics, Third Edition

Homer Reid

October 8, 2000

Chapter 4: Problems 8-13

Problem 4.8

A very long, right circular, cylindrical shell of dielectric constant /0 and inner andouter radii a and b, respectively, is placed in a previously uniform electric field E0with its axis perpendicular to the field. The medium inside and outside the cylinderhas a dielectric constant of unity.

(a) Determine the potential and electric fields in the three regions, neglecting endeffects.

(b) Sketch the lines of force for a typical case of b 2a.

(c) Discuss the limiting forms of your solution appropriate for a solid dielectric

cylinder in a uniform field, and a cylindrical cavity in a uniform dielectric.

We will take the axis of the cylinder to be the z axis and the electric field tobe aligned with the x axis: E0 = E0i. Since the cylinder is very long and weretold to neglect end effects, we can ignore the z direction altogether and treatthis as a two-dimensional problem.

(a) The general solution of the Laplace equation in two dimensional polar co-ordinates is

(r, ) =

[Anrn + Bnr

n][Cn sin(n) + Dn cos(n)]

For the region inside the shell (r < a), the B coefficients must vanish to keepthe potential from blowing up at the origin. Also, in the region outside the shell

1

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Homer Reids Solutions to Jackson Problems: Chapter 4 2

(r > b), the only positive power of r in the sum must be that which gives riseto the external electric field, i.e. E0r cos with An = 0 for n > 1. With theseobservations we may write expressions for the potential in the three regions:

(r, ) =

rn[An sin n + Bn cos n], r < a

rn[Cn sin n + Dn cos n] + rn[En sin n + Fn cos n], a < r < b

E0r cos +

rn[Gn sin n + Hn cos ], r > b

The normal boundary condition at r = a is

0

r

x=a

=

r

x=a+

or

0

nan1[An sin n + Bn cos n] =

na

n1

[Cn sin n + Dn cos n] na

(n+1)

[En sin n + Fn cos n]

From this we obtain two equations:

0

An = Cn Ena2n (1)

0

Bn = Dn Fna2n (2)

Next, the tangential boundary condition at r = a is

x=a+

=

x=a

or

nan[An cos n Bn sin n] =

nan[Cn cos n Dn sin n] + na

n[En cos n Fn sin n]

from which we obtain two more equations:

An = Cn + Ena2n (3)

Bn = Dn + Fna2n (4)

Similarly, from the normal boundary condition at r = b we obtain

0

E0 cos 0

nb(n+1)[Gn sin n + Hn cos ] =

nbn1[Cn sin n + Dn cos n] nb(n+1)[En sin n + Fn cos ]

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which leads to

0

Gn = Cnb2n En (5)

0

b2E0n1 0

Hn = Dnb2n Fn (6)

Finally, we have the tangential boundary condition at r = b:

bE0 sin +

nbn[Gn cos n Hn sin n] =nbn[Cn cos n Dn sin n] + nb

n[En cos n Fn sin n]

giving

Gn = Cnb2n + En (7)

b2E0n1 + Hn = Dnb2n + Fn. (8)

The four equations (1), (3), (5), and (7) specify a degenerate system of linearequations, which can only be satisfied by taking An = Cn = En = Gn = 0 for

all n. Next, for n = 1, the system of equations (2), (4), (6), and (8) specify thesame degenerate system of equations, so Bn = Dn = Fn = Gn = 0 for n = 0.However, for n = 1, we have

0

B1 = D1 F1a2 D1 =

1

2

1 +

0

B1

B1 = D1 + F1a2 F1 =

1

2a2

1 0

B1.

and

H1 = b2E0 +

0D1b

2

0F1

H1 = b

2

E0 + D1b

2

+ F1

0 = 2b2E0 + b2

1 +

0

D1 +

1

0

F1

Substituting from above,

4b2E0 =1

0

b2( + 0)

2 a2( 0)2

B1

or

B1 =40b

2

b2( + 0)2 a2( 0)2E0.

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Then

D1 =20( + 0)b

2

b2( + 0)2 a2( 0)2E0

F1 =20( 0)a

2b2

b2( + 0)2 a2( 0)2E0

H1 =b2(b2 a2)(20 2)

b2( + 0)2 a2( 0)2E0.

The potential is

(r, ) =

40b2

b2( + 0)2 a2( 0)2 E0rcos , r < a

20b2

b2( + 0)2 a2( 0)2

( + 0)r + ( 0)

a2

r

E0cos , a < r < b

(b2 a2)(20 2)

b2( + 0)2 a2( 0)2

b2

rE0cos E0rcos , b < r.

As 0, E0r cos in all three regions, which is reassuring.The electric field is

E(r, ) =

40b2

b2( + 0)2 a2( 0)2E0 [cos r sin ] , r < a

20b2

b2( + 0)2 a2( 0)2

( + 0) ( 0)

a2

r2

E0 cos r

( + 0) + ( 0)

a2

r2

E0 sin

, a < r < b

(b2 a2)(20

2)

b2( + 0)2 a2( 0)2

b

r

2E0 [cos r + sin ]

+E0 [cos r sin ] , b < r.

(b) In Figure 4.1 Ive plotted the field lines for b = 2a, = 50. Also, as anappendix to this document Ive included the C program I wrote to generate thisplot.

(c) For a solid dielectric cylinder in a uniform field, we would have a 0. Inthat case the field would look like

E(r, ) =

20 + 0

E0i, r < b

E0i (20

2)

( + 0)2

b

r

2E0[cos r + sin ], r > b

On the other hand, a cylindrical cavity in a uniform dielectric corresponds to

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Figure 1: Field lines in Problem 4.8 for b = 2a, = 50.

b , in which case the field becomes

E(r, ) =

40( + 0)2

E0i, r < a

20( + 0)E0i 20( 0)( + 0)2

ar2

E0[cos r + sin ], r > a.

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Problem 4.9

A point charge q is located in free space a distance d away from the center of adielectric sphere of radius a (a < d) and dielectric constant /0.

(a) Find the potential at all points in space as an expansion in spherical harmonics.(b) Calculate the rectangular components of the electric field near the center of

the sphere.

(c) Verify that, in the limit /0 , your result is the same as that for theconducting sphere.

We will take the origin of coordinates at the center of the sphere, and putthe point charge on the z axis at z = +h. Then the problem has azimuthalsymmetry.

(a) Since there is no free charge within the sphere, D = 0 there. But since thepermittivity is uniform within the sphere, we may also write (D/) = E = 0

there. This means that polarization charge only exists on the surface of thesphere, so within the sphere the potential satisfies the normal Laplace equation,whence

(r, ) =l

AlrlPl(cos ) (r < a).

Now, in the region r > a, the potential may be written as the sum of twocomponents 1 and 2, where 1 comes from the polarization charge on thesurface of the sphere, while 2 comes from the external point charge. Since1 satisfies the Laplace equation for r > a, we may expand it in Legendrepolynomials:

1(r, ) =l

Blr(l+1)Pl(cos ) (r > a).

On the other hand, 2 is just the potential due to a point charge at z = d:

2(r, ) =

q

40

rldl+1

Pl(cos ), r < d

q

40

dlrl+1

Pl(cos ), r > d.

(9)

Putting this all together we may write the potential in the three regions as

(r, ) =

Alr

lPl(cos ), r < a

Blr

(l+1) +q

40

rl

dl+1

Pl(cos ), a < r < d

Bl + qd

l

40

r(l+1)Pl(cos ), r > d.

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The normal boundary condition at r = a is

r

r=a

= 0

r

r=a+

0lAla

l1 = (l + 1)Bla(l+2) +

lqal1

40dl+1

Al =0

(l + 1)

lBla

(2l+1) +q

40dl+1

(10)

The tangential boundary condition at r = a is

r=a

=

r=a+

Alal = Bla

(l+1) +q

40

al

d(l+1)

Bl = Ala2l+1

q

40

a2l+1

dl+1(11)

Combining (10) and (11), we obtain

Al =1

0

+ l+1l

2l + 1

l

q

40dl+1

Bl =1

0

+ l+1l

1

0

qa2l+1

40dl+1

In particular, as /0 we have

Al 0

as must happen, since the field within a conducting sphere vanishes; and

Bl qa2l+1

40dl+1. (12)

With the coefficients (12), the potential outside the sphere due to the polariza-tion charge at the sphere boundary is

1(r, ) =1

40

qa

d

a2d

l1

rl+1Pl(cos ).

Comparing with (9) we see that this is just the potential of a charge qa/d onthe z axis at z = a2/d. This is just the size and position of the image charge

we found in Chapter 2 for a point charge outside a conducting sphere.

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(b) Near the origin, we have

(r, ) = A1rP1(cos ) + A2r2P2(cos ) +

=q

40

30

d2( + 20)z +

1

2

50

d3(2 + 30)

(z2 x2 y2) +

so the field components are

Ex =q

40d2

502 + 30

xd

+

Ey =q

40d2

502 + 30

yd

+

Ez = q

40d2

30

+ 20+

502 + 30

zd

+

Problem 4.10

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 hemi-spherical shell of dielectric (of dielectric constant /0), as shown in the figure.

(a) Find the electric field everywhere between the spheres.

(b) Calculate the surface-charge distribution on the inner sphere.

(c) Calculate the polarization-charge density induced on the surface of the dielectricat r = a.

Well orient this problem such that the boundary between the dielectric-

filled space and the empty space is the xy plane. Then the region occupiedby the dielectric is the region a < r < b, 0 < < /2, and the problem hasazimuthal symmetry.

(a) Since the dielectric has uniform permittivity, all the polarization chargeexists on the boundary of the dielectric, so within its body we may take thepotential to be a solution of the normal Laplace equation. The potential in theregion between the spheres may then be written

(r, ) =

[Alr

l + Blr(l+1)]Pl(cos ), 0

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