Continuity of the electric potential due to a surface charge

Question

The Electric potential due to a charge distribution on a surface is: $$\Phi \left ( x \right )=\int \frac{\sigma \left ( {x^{}}' \right )dx{}'}{\left \| x-x{}' \right \|}da.$$ I want to show that it's a continuous function everywhere.

Answer 1

The answer is not in general. There exist pathological charge densities for which the electric potential has discontinuities. Fortunately, these can always be ruled out on physical grounds: they usually have finite amounts of opposite charge infinitely close to each other, which allows for arbitrarily high electric fields, and therefore divergent energy densities and infinite configuration energies.

How does this work? Take two conducting plates and put their edges in contact. I'm quite fond of using the north and south hemispheres of a given sphere, separated at the equator (which has the advantage of not introducing any extraneous infinities), but for simplicity let's use two semi-infinite planes, in the $x,z$ plane and separated by the $x$ axis.

Now, connect both plates to a voltage source and set one to a potential $+V_0$ and the other to $-V_0$. The potential at the $x$ axis is (of course!) discontinuous, by construction.

If that makes you uneasy, consider spreading uniform but opposite surface charges $\pm \sigma_0$ on the $\pm z$ plates. Fix some positive coordinate $z_0$. Then if you zoom in more and more towards the plates (i.e. as you take $x\rightarrow 0$) the plates loom larger and larger, and when $x\ll z_0$ the $-$ plate looks so far away that the local potential has to be positive, and bounded away from zero. Similarly, if $z_0<0$ the potential close enough to the plates must be negative. Thus you have points arbitrarily near the $x$ axis which have potentials bounded away from zero, both positive and negative, so the potential is discontinuous there.

So why is this not a problem? In these pathological examples you are putting a lot of negative charge, in conducting surfaces, next to a lot of negative charge. This assumes that you have separated them with an insulator of zero thickness which can withstand large (albeit finite) potential differences (so infinite electric fields). This is just not physical: any dielectric will break down or spark as you approach these conditions. As soon as you introduce a finite separation, the discontinuity goes away.

More generally the electrostatic potential can have discontinuities. These are most usually considered as surface discontinuities, in which case they are known as dipole layers, which you can think of as infinitely thin capacitors set at constant potential. (Therefore, they must have two layers of opposite but infinite charge, which is again not physical.)

A real dipole layer would be something like a lot of polar molecules, all pointing the same way normally to the surface. We know that the potential inside is not discontinuous and that the electric fields inside are large but finite, but if you're outside then all you see are two plane layers of charge which look for all the world like they are setting up a discontinuous potential, so you might as well put that into your math and assume it is infinitely thin.

Answer 2

Ok so let $\mathbf x_0$ be some point on the surface, and let $\mathbf n$ denote a normal to the surface at $\mathbf x_0$. Consider the difference \begin{align} \Phi(\mathbf x_0 + \epsilon\mathbf n) - \Phi(\mathbf x_0 - \epsilon\mathbf n) &= \frac{1}{4\pi\epsilon_0}\int_S \sigma(\mathbf x')\left(\frac{1}{|\mathbf x_0+\epsilon\mathbf n-\mathbf x'|}-\frac{1}{|\mathbf x_0-\epsilon\mathbf n-\mathbf x'|}\right) da' \end{align} But recall the Taylor expansion $$ \frac{1}{|\mathbf x +\epsilon\mathbf a|} = \frac{1}{\mathbf x} + \epsilon\frac{\mathbf a\cdot\mathbf x}{|\mathbf x|^3} + \mathcal O(\epsilon^2) $$ from which it follows that \begin{align} \frac{1}{|\mathbf x_0+\epsilon\mathbf n-\mathbf x'|}-\frac{1}{|\mathbf x_0-\epsilon\mathbf n-\mathbf x'|} &= 2\epsilon \frac{\mathbf n\cdot(\mathbf x_0-\mathbf x')}{|\mathbf x_0 - \mathbf x'|^3} +\mathcal O(\epsilon^2) \end{align} and therefore $$ \Phi(\mathbf x_0 + \epsilon\mathbf n) - \Phi(\mathbf x_0 - \epsilon\mathbf n) =\epsilon\left(\frac{1}{2\pi\epsilon_0}\int_S \sigma(\mathbf x')\frac{\mathbf n\cdot(\mathbf x_0-\mathbf x')}{|\mathbf x_0 - \mathbf x'|^3} da'\right) + \mathcal O(\epsilon^2) $$ from which it follows that $$ \lim_{\epsilon\to 0}\Big(\Phi(\mathbf x_0+\epsilon\mathbf n)-\Phi(\mathbf x_0-\epsilon\mathbf n)\Big)=0 $$ So $\Phi$ is continuous at the surface charge.

Disclaimer. This is not a mathematically rigorous proof, but it's something that I think would satisfy most physicists. I propose that if more rigor is desired then the question should be migrated to math.SE whose possibility was first suggested by user Qmechanic.

Answer 3

An alternative, perhaps more physicsy argument is:

Since:

1. The electric field ought to be finite for a surface charge distribution e.g. because it is locally planar.
2. The potential is the line integral of the electric field.
3. The line integral of a function without singularities will be continuous.

the potential ought to be continuous. You may find the first claim to be lacking in rigor.

You are trying to solve this starting with the Green's function for the Poisson equation, which itself has a singularity. This is probably not the most natural method. As an alternative to the foregoing argument, you could look up the conditions that are necessary and/or sufficient for solutions of the Poisson equation to be continuous.

As an aside, I would note that what you are trying to prove in fact does not hold if you consider a point charge to be a special case of a surface charge distribution.