Solutions for J. D. Jackson, Classical Electrodynamics, 3rd ed.
1.12.
This result follows from Green’s second identity, as shown on p. 36:
Z
I ∂Φ0
∂Φ
(Φ∇2 Φ0 − Φ0 ∇2 Φ)d3 x =
Φ 0 − Φ0 0 dA.
∂n
∂n
V
S
We can clearly identify the Poisson equation for the potential inside the volume:
∇2 Φ = −ρ/0
∇2 Φ0 = −ρ0 /0 .
(1)
(2)
Moreover, the components of the gradients of Φ and Φ0 normal to the surface are identified
with the surface charge density. This follows from the result on p. 36 that, for a point x
which lies outside of a volume V:
Z
I 1
ρ(x0 ) 3 0
1
∂Φ
1
1
∂
dA0 = 0
d
x
+
−
Φ
4π0 V |x − x0 |
4π S |x − x0 | ∂n0
∂n0 |x − x0 |
This tells us that outside of V, we may view the volume integral of ρ(x0 )/|x − x0 | in it as an
integral over the surface S only. If there are no other charges than those in V, then outside
of V:
Z
I 1
ρ(x0 ) 3 0
1
∂
1
1
∂Φ
Φ(x) =
d x =
Φ 0
−
dA0 .
4π0 V |x − x0 |
4π S
∂n |x − x0 |
|x − x0 | ∂n0
Comparing the surface integral terms to those found for potentials from a dipole layer and
a surface charge density shows that the effect of ρ on the potential is equivalently described
by:
∂
1
D = −0 Φ 0
∂n |x − x0 |
and
∂Φ
σ = −0 0
(3)
∂n
on the surface. The second result is vital here, as it allows us to write:
∂Φ
∂Φ0
= −σ/0
= −σ 0 /0 ,
∂n
∂n
and hence directly from eq. 1 and eq. 2:
Z
I
Z
Z
ρΦ0 d3 x +
σΦ0 dA =
ρ0 Φd3 x +
σ 0 ΦdA.
V
S
V
S
Note that explicitly showing eq. 3 is not strictly speaking required by the question. If we
assume the surface S to be conductive, we can immediately state:
∂Φ
= σ/0
∂n
where the second equality follows from the surface being equipotential and the last one is a
well-known result for conductors provable by Gauss’ law.
Hence the theorem is valid irrespective of whether the surface is conductive or not.
E = −∇Φ = −
J. Kosata, December 2016