Correct formula to express the potential generated by a single layer charge distribution

Assume that the closed surface $S$ encircles a volume $V$, and that a surface charge with density $\sigma$ ("single layer") is distributed over $S$. My question regards the electrostatic potential $\phi$ generated inside the volume $V$ by this charge distribution: on the one hand, by using the superposition principle and Coulomb's law, I get $$\tag{1} \phi(x)=\frac{1}{4\pi \epsilon_0} \int_S \frac{\sigma(x')\, dS'}{\lvert x-x'\rvert}.$$ On the other hand, I know that $\phi$ solves Laplace's equation on $V$ with Neumann boundary conditions on $S$: $$\begin{cases} -\nabla^2 \phi =0 & \text{on }V\\ \frac{\partial \phi}{\partial \nu} \propto \sigma &\text{on }S \end{cases}$$ so that it may be expressed in an integral form by means of a suitable Green function (see Jackson, 3rd ed., equation (1.46) pag.39): $$\tag{2} \phi(x)=\langle \phi\rangle_S + C\int_S \sigma(x') \frac{\partial G}{\partial \nu'}(x, x')\, dS(x').$$

Question. Formulas (1) and (2) do not agree in general, otherwise all volumes would share the same Green function, and that's not true. So one of the two must be wrong. Which one is wrong, and for what physical reason?

Your Neumann boundary value problem is not correct. You actually have a free-space problem (your domain is $\mathbb{R}^3$). In that space some surfaces are charged. You do not have something like ideal conducting surfaces or symmetry conditions which would allow you to reduce the problem to a bounded domain.

An example which shows that your Neumann boundary problem is not the right formulation for the problem:

Assume $V$ is a cube and set $\sigma(x)=0$ on almost all plane surfaces of the cube except one where you set $\sigma(x)=1 \rm \frac{C}{m^2}$. Your Neumann boundary condition would wrongly imply that the field strength on the surface opposite to the one with $\sigma(x)=1\rm \frac{C}{m^2}$ would be zero.

PS: Why does this reference not work:\eqref{1}?? It worked when I first edited the answer.

• What does "quader" mean? I guess that it is the German word for "cube", right? Mar 25 '14 at 18:43
• Yes, sorry. I corrected the answer. Is it necessary to add a picture or is the answer reasonable with this description? Mar 25 '14 at 18:47
• It's OK. Also, you counter-example is good and I understand it (+1). Could you suggest me how to fix the boundary value problem? Mar 25 '14 at 18:53
• Let us discuss it at chat.stackexchange.com/rooms/13770/…. Mar 25 '14 at 19:10