PHY 5346
HW Set 3 Solutions – Kimel
3. 2.5
a)
W = ∫
r
∞
|F|dy =
q2a
4π 0
∫
r
∞
dy
y3 1 − a2
y2
2 =
q2a
8π 0r2 − a2
Let us compare this to disassemble the charges
− W′ = − 1
8π 0
∑
i≠j
q iq j
|x⃗i−x⃗j |
=
1
4π 0
aq2
r
1
r 1 − a2
r2
=
q2a
4π 0r2 − a2
> W
The reason for this difference is that in the first expression W, the image charge is moving and
changing size, whereas in the second, whereas in the second, they don’t.
b) In this case
W = ∫
r
∞
|F|dy =
q
4π 0
∫
r
∞ Qdy
y2
− qa3 ∫
r
∞ 2y2 − a2
yy2 − a2 2
dy
Using standard integrals, this gives
W =
1
4π 0
q2a
2r2 − a2
− q
2a
2r2
− qQr
On the other hand
− W′ =
1
4π 0
aq2
r2 − a2
− qQ +
a
r q
r
=
1
4π 0
aq2
r2 − a2
− q
2a
r2
− qQr
The first two terms are larger than those found in W for the same reason as found in a), whereas the
last term is the same, because Q is fixed on the sphere.