Electromagnetism 70006 Answers to Problem Set 3 Spring 2006

1. Jackson Prob. 2.1: Charge above a grounded plane

(a) Surface charge density
· ¸
∂φ q (z − d) (z + d)
Ez (ρ, z) = − = − 2
∂z 4π²0 [ρ2 + (z − d)2 ]3/2 [ρ + (z + d)2 ]3/2
Evaluate Ez at z = 0 and multiply by ²0 to find σ.
qd 1
σ(ρ) = −
2π [ρ + d2 ]3/2
2

Plot of σ with d = 1 and q = 2π (Actually we show -σ)

1 2

0.5 1
0
-2 0
-1
0 -1
1
2 -2

(b) The direction of the force on the plane is along z axis. Its magnitude
is
Z ∞ Z
1 qσ(ρ) d q 2 d2 ∞ ρdρ q2 1
Fz = 2πρdρ 2 2
p = 2 2 3
= .
0 4π²0 ρ + d ρ2 + d2 4π²0 0 (ρ + d ) 4π²0 4d2

This is precisely the force on the image charge predicted by Coulomb’s

law.
(c) The force obtained by integrating σ 2 /(2²0 ) is
Z ∞ Z
1 q 2 d2 2πρdρ q 2 d2 ∞ ρdρ q2 1
Fz = 2 2 2 3
= 2 2 3
= .
2²0 (2π) 0 (ρ + d ) 4π²0 0 (ρ + d ) 4π²0 4d2

(d) Work done to move charge to infinity.

Z ∞ 2
q dz q2 1
W = 2
=
d 4π²0 4z 4π²0 4d

(e) Potential energy of charge and image.

q2 1
V =− .
4π²0 2d

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 This is twice the value naively expected. The work-energy relation
for a pair of charges discussed in Sec. 1.11 assumed that one charge
was fixed! In this case the both charges move as work is done on the
system.
(f) W for an electron at d= 1Å from surface.

q2 1
W =
4π²0 4d
(1.60217653 × 10−19 )2 1
=
4 × π × 8.854187817 × 10−12 4 × 10−10
= 5.767693 × 10−19 J
= 3.6eV

2. Jackson Prob 2.2:

(a) Assuming that the charge is on the z axis at distance d from the
origin, the potential at points inside the sphere is
½ ¾
1 q q0
Φ(r, θ) = √ −√ ,
4π²0 r2 − 2rd cos θ + d2 r2 − 2rd0 cos θ + d02
where q 0 = qa/d and d0 = a2 /d.
(b) Induced charge density: First, we determine the radial electric field
∂Φ
Er = −
∂r ½ ¾
1 q(r − d cos θ) q 0 (r − d0 cos θ)
= − 2
4π²0 [r2 − 2rd cos θ + d2 ]3/2 [r − 2rd0 cos θ + d02 ]3/2

The surface charge density σ is −²0 Er , evaluated at r = a. (In this

case, the normal points inward!). After simplification, this becomes

q 1 − d2 /a2
σ(θ) = − 2
4πa [1 − 2(d/a) cos θ + d2 /a2 ]3/2

(c) Magnitude and direction of force on q:

Z 1
q 2 |σ(µ)| (aµ − d)
Fz = 2πa dµ 2
4π²0 −1 [a − 2adµ + d2 ]3/2
2 2 2 Z 1
q a(a − d ) (aµ − d)
= dµ 2
4π²0 2 −1 [a − 2adµ + d2 ]3
q2 ad
= .
4π²0 (a − d2 )2
2

The force on q is in the +z direction. This is precisely the force

exerted on q by the image charge q 0 .

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 (d) Changes in the solution:
i. Sphere at a fixed potential V . In this case,

Φ(r, θ) → Φ(r, θ) + V, r<a

aV
→ , r≥a
r
There is an additional uniformly distributed charge Q = q +
4π²0 V on the sphere. Thus
q + 4π²0 V
σ→σ+ .
4πa2
The uniformly distributed charge exerts no additional force of q
since Z 1
(aµ − d)
dµ 2 =0
−1 [a − 2adµ + d2 ]3/2
ii. Sphere has a fixed charge Q. In this case, an additional charge
Q + q is again uniformly distributed over the surface. Therefore
Q+q
Φ(r, θ) → Φ(r, θ) + , r<a
a
1 Q+q
→ , r≥a
4π²0 r
The additional uniformly distributed charge is
Q+q
σ→σ+ ,
4πa2
and, as above, there is no added force on q.

3. Jackson Prob 2.5:

(a) Quasistatic force needed to balance charge q above a grounded sphere.

q2 ay
Fy =
4π²0 y − a2
2

Work done to remove charge to infinity

Z ∞
q2 a qq 0 1
W = Fy dy = 2 − a2
=
d 8π² 0 d 8π² 0 d − d0

This is 1/2 of the (negative of) the potential energy of the charge
and its image. Here again the image is not fixed as the charge moves
out, so the work-energy theorem, in its usual form, is not valid.

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 (b) Quasistatic force needed to balance charge q above an isolated sphere
carrying charge Q.
· ¸
q Q qa3 (2y 2 − a2 )
Fy = − −
4π²0 y 2 y 3 (y 2 − a2 )2
Work needed charge to remove charge to infinity
Z ∞ · ¸
1 Qq q 2 a q2 a
W = Fy dy = − − 2 + 2
d 4π²0 d d d − a2
· 0 0
¸
1 q(Q + q ) qq
= − +
4π²0 d 2(d − d0 )
The first term is the negative of the potential energy of the added
charge Q + q 0 and the charge q. This term has the correct sign as the
uniformly distributed added charge charge is effectively at the origin.
The second term is the negative of the charge-image potential with
the factor of 1/2 associated with the fact that the image moves along
with the original charge.

Addendum on the work-energy theorem:

In the plane and spherical image problems worked out above, we found that
the work needed to bring the charge q in from infinity was 1/2 the potential
energy of the charge and its image. To explain this factor 1/2, let us examine
the general expression for energy of a charge distribution
Z Z
1 ρ(r1 )ρ(r2 ) 1
W = dτ1 dτ2 = Φ(r)ρ(r) dτ
8π²0 |r1 − r2 | 2
• For two fixed charges, q and qi , we have Φ = Φq + Φi and ρ = qδ(r − rq ) +
qi δ(r − ri ). Here,
q 1
Φq (r) = ,
4π²0 |r − rq |
with a similar expression for Φi . One finds
1 1 1 1
W = q Φq (rq ) + q Φi (rq ) + qi Φq (ri ) + qi Φi (ri )
2 2 2 2
The first and fourth terms are (infinite) “self-energy” terms and must be
excluded from the sum. The second and third terms have identical values
and lead to the well-known expression for the interaction energy between
two charges
qqi 1
W = ,
4π²0 |ri − rq |
• For a charge q and a surface distribution σ such as we have in the present
case, the energy expression becomes
Z Z
1 1 1
W = q Φi (rq ) + Φq (r)σ(r)da + Φi (r)σ(r)da,
2 2 S 2 S

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where we have omitted the self-energy of q. Since the two contributions
to the potential Φq and Φi precisely cancel on the surface, the second and
third terms above cancel and we are left with
1 qqi 1
W = q Φi (rq ) = .
2 8π²0 |ri − rq |

This is, as expected, just 1/2 of the charge-image interaction energy.