# Poisson equation and surface charge distribution

Poisson equation is given by $$\nabla^2V=\frac{\rho}{\epsilon_0}$$. Here $$\rho$$ indicates a volumic charge distribution, which is known in the region $$\Omega$$ where we solve the Poisson equation.

Is it correct using instead a known surface charge distribution $$\sigma$$ given in $$\Omega$$?

I think that the answer is yes, but only in a distributional sense (Dirac delta). In fact the electric field presents a discontinuity near a surface with charge density $$\sigma$$, then over the surface is not valid the local Gauss law $$\vec{\nabla} \cdot \vec{E}=\frac{\rho}{\epsilon_0}$$ (unless we interpret $$\rho$$ as a distribution $$\delta$$).

Thank you.

## 1 Answer

Every differential equation must have its associated boundary conditions to provide a particular solution. This is no different for Poisson's equation for the electric potential. In the case you described, the surface charge density $$\sigma$$ would be one boundary condition for your problem. The other boundary condition (as Poisson's equation is a second-order PDE) could, for example, be the continuity of the potential $$V$$ across the surface.

"In fact the electric field presents a discontinuity near a surface with charge density $$\sigma$$, then over the surface is not valid the local Gauss law..."

Actually, the discontinuity of $$\mathbf{E}$$ at the surface in terms of $$\sigma$$ is derived exactly from Gauss' law -- therefore, the law is valid everywhere, provided you treat the discontinuities as appropriate boundary conditions. You could, alternatively, define a surface charge density in terms of a volumetric charge density $$\rho$$ and a Dirac delta distribution at the boundary, but this would be just a mathematical convenience (or not) to deal with your problem, having no impact in the Physics at all.

• Thank you for your answer. Let's suppose for example that we want to solve Poisson equation in all $\mathbb{R}^3$, with Dirichlet boundary conditions given by $V=0$ at infinity. Moreover, let's suppose the only charge that we have in $\mathbb{R}^3$ is a surface charge density $\sigma$ (that we know). In this case can we solve in $\mathbb{R}^3$ the Poisson equation $\nabla^2 V=\frac{\sigma}{\epsilon_0}$, obtaining a unique solution $V$? Thank you for your time. Jan 24, 2023 at 22:31
• Is this case you would need to solve $\nabla^2 V=0$ for all space, where the surface region with charge density $\sigma$ will be one boundary condition and the other one will be $V \to 0$ for distances tending to infinity. Try for example solving the classic situation of a spherical shell of radius $R$ and surface charge density $\sigma$ -- the field inside the shell should be zero and outside it should be equal to a point particle with charge $4\pi R^2 \sigma$.
– Woe
Jan 25, 2023 at 1:06