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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.

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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.

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  • $\begingroup$ 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
    Commented 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
    Commented Jan 25, 2023 at 1:06

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