Recall that the second uniqueness theorem states that in a volume V surrounded by conductors
and containing a specified charge density
ฯ , the electric field is uniquely determined if the total
charge on each conductor is known. This is true regardless of whether the region as a whole is
bound by another conductor or is unbounded. The proof provided by Griffiths went as follows:
Suppose there are two fields satisfying the conditions of the problem. Both obey Gaussโs law in
the space between the conductors,
โ
โโ
โE1=
ฯ
ฮต0
โ
โโ
โE2=
ฯ
ฮต0
}โ
{โฎ โ E1โ
d โ S=Qi
ฮต0
โฎ โ E2โ
d โ S=Qi
ฮต0
where the surface integrals are over the ith conducting surface. Similarly, the outer boundary
satisfies
โฎ โ E1โ
d โ S= Qtotal
ฮต0
โฎ โ E2โ
d โ S= Qtotal
ฮต0
where the surface integrals are over the outer boundary. Define
โ
E3= โ E1 โ โ E2 ,
which obeys โ โโ
โE3 =0 in the region between the conductors and โฎ โ E3โ
d โ S over each boundary
surface. Although we donโt know how the charge Qi distributes itself over the ith conductor, we
do know that each conductor is an equipotential, and hence
ฯ3 is a constant over each
conducting surface (although not necessarily the same constant). We can then write
โ
โโ
(
ฯ3 โ E3)=
ฯ3 ( โ โโ
โE3 )+ โ E3โ
( โ โ
ฯ3)=โ(E3)2 .
Integrating this over V and applying the divergence theorem to the left side,
โซ
V
โ
โโ
(
ฯ3 โ E3 )dV =โฎ
S
ฯ3 โ E3โ
d โ S=โโซ
V
(E3)2 dV .
Since
ฯ3 is constant over each surface, it comes outside each integral, and what remains is zero
since โ โโ
โE3 =0 . Therefore
โซ
V
(E3)2 dV =0 .
But, this integrand is never negative so the only way the integral can vanish is if E3=0
everywhere. Therefore โ E1= โ E2 , thus proving the theorem.
A more elegant proof uses Greenโs identity,
โซ
V
[ T โ 2 U +( โ โ T )โ
( โ โ U ) ] dV =โฎ
S
(T โ โ U )โ
d โ S ,
with T =U=
ฯ3 . Fill in the details of this proof