I am going over the proof of second uniqueness theorem in International 4th Ed of Griffiths Electricity and Magnetism. In page 122, followed till the part where two fields $\vec{E_1}$ and $\vec{E_2}$ were taken such that the satisfy gauss's law on each conductor inside the volume of consideration, and then, $\vec{E_3}$, the difference, being taken such that it obeys the following results:

$$ \vec{E_3} = \vec{E}_1 - \vec{E}_2 \tag{1}$$ In region between conductors: $$ \nabla \cdot \vec{E_3} = 0 \tag{2}$$ Over each boundary surface: $$ \oint \vec{E_3} \cdot da = 0 \tag{3}$$

Now, after this Griffith mentions that there is one final piece of information to complete the proof

Although we do not know how the charge $Q_i$ distributes itself over the ith conductor, we do know that each conductor is an equipotential, and hence $V_3$ is a constant (not necessarily the same constant) over each conducting surface. (It need not be zero, for the potentials $V_1$ and $V_2$ may not be equal - all we know for sure is that both are constant over any given conductor.)

I went over this paragraph a few times but I still can't understand what role it played in the proof.. because it seems to me that the rest can be done without it even being considered. Here is how I think it would play out:

$$ \int_V \nabla \cdot (V_3 \vec{E}_3) d \tau = \int_V \underbrace{ V_3 ( \nabla \cdot \vec{E}_3) }+ \vec{E}_3 \cdot (\nabla V_3) dV=- \int_V (\vec{E_3})^2 dV$$

Underbraced term zero due to (2), and by the divergence theorem, the LHS becomes:

$$ \int_V V_3 \vec{E_3} \cdot dA = -\int_V (\vec{E_3})^2 dV$$

Now conside a sufficiently large surface containing our charge distribution such that $V=0$ at all points on boundary of this surface (potential at infinity is zero), then it must be that:

$$ \int_V \vec{E_3} \cdot \vec{E_3} dV=0$$

The only way the above equality can happen is if $\vec{E_3} = 0$ through all of spaces since the integrand is a strictly positive quantity.

Hence my question, what is the point of the quoted paragraph?


1 Answer 1


The volume $V$ consists of the free space between the outer boundary and the conductors; it does not include the conductors themselves. This means that the boundary of $V$ is not just the outer boundary, but also includes the surfaces of the conductors. Even if you define your potential such that $V_3 = 0$ on the outer boundary (we can always do this by adding or subtracting a constant), this does not immediately imply that $V_3 = 0$ on the other pieces of the boundary, namely the surfaces of the conductors. In fact, that's what you're trying to prove.

The fact that $V_3$ must be constant over the surface of each conductor comes into play because it implies that for the surface $S_i$ of each individual conductor, we have $$ \int_{S_i} V_3 \vec{E}_3 \cdot d\vec{a} = V_3 \int_{S_i} \vec{E}_3 \cdot d\vec{a}, $$ and each of these latter integrals vanishes by your Eq. (3). Without the constancy of $V_3$, we cannot argue that these surface integrals all vanish. From there, the proof proceeds as you have it.

Of course, you could redefine your volume $V$ so that it consists of everything inside the outer boundary, including the conductors. But in that case, you have charge inside $V$, so Laplace's equation doesn't hold in the first place and the whole proof falls apart.

  • $\begingroup$ ". This means that the boundary of V is not just the outer boundary, but also includes the surfaces of the conductors" I am really confused here, so to understand, I thought of the 'cheese' with holes. Are you saying that the surface of the holes in the cheese is also boundary of the cheese? picture $\endgroup$ Commented May 26, 2021 at 15:03
  • $\begingroup$ In the last sentece "you have charge inside V, so Poisson's equation doesn't hold in the first place and the whole proof falls apart." I don't think I have used poissons equation in writing any of my statements $\endgroup$ Commented May 26, 2021 at 15:05
  • $\begingroup$ @Buraian: Yes, that's correct. For a simpler example, if $V$ was a thick spherical shell $a< r < b$, then the boundary of $V$ would consists of the two spheres $r = a$ and $r = b$. $\endgroup$ Commented May 26, 2021 at 15:06
  • $\begingroup$ @Buraian: Sorry, I meant Laplace's equation, which is equivalent to saying that $\vec{\nabla} \cdot \vec{E}_3 = 0$. $\endgroup$ Commented May 26, 2021 at 15:07
  • 1
    $\begingroup$ @Buraian: We know that $\nabla \cdot \vec{E}_3 = 0$ in the space between the outer surface and the conductors, since we are assuming that $\rho$ is specified there and $\vec{E}_3$ is the difference there. But we don't know for sure that $\rho = 0$ everywhere for $\vec{E}_3$ on the surface of the conductors; while we know that the net charges on the conductors are the same in configurations #1 and #2, they could be distributed differently. So if we included the conductors in $V$, we would not necessarily have $\vec{\nabla} \cdot \vec{E}_3 = 0$ in $V$. $\endgroup$ Commented May 26, 2021 at 23:39

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