0
$\begingroup$

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.

$\endgroup$

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – Leonardo
    Jan 24, 2023 at 22:31
  • $\begingroup$ 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 $4\pi R^2 \sigma$. $\endgroup$
    – Woe
    Jan 25, 2023 at 1:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.