Poisson's equation in electrostatics

Remarks

It is Poisson's equation $\Delta u = f$ with $f = -\rho / \varepsilon_0$. In a region without charge it becomes Laplace's equation $\Delta \varphi = 0$. If $\rho$ is twice continuously differentiable with compact support, then the Coulomb electric potential of $\rho$ is a classical solution on $\R^3$. Example: a ball of radius $R$ with constant density $\rho_0$ has, inside the ball, the potential $\varphi(x) = \rho_0 (3R^2 - \abs{x}^2) / (6 \varepsilon_0)$. The check: $\Delta \abs{x}^2 = 6$ in $\R^3$, so $\Delta \varphi = -\rho_0 / \varepsilon_0$. Non-example: $\varphi(x) = \rho_0 \abs{x}^2 / (6 \varepsilon_0)$ has $\Delta \varphi = +\rho_0 / \varepsilon_0$, so it has the wrong sign. In Gaussian units the equation reads $\Delta \varphi = -4 \pi \rho$.