= Solution
Find the potential from $\mathbf E=-\nabla\phi$ and $\nabla\cdot(\epsilon\nabla\phi)=-\rho_f$. One must specify boundary data: a potential on each Dirichlet boundary, a normal displacement on each Neumann boundary, or an appropriate mixture. Conductors are equipotentials; at <dielectric> interfaces $\phi$ is continuous and the normal displacement jumps by the specified free surface charge. With no singular dipole sheet, tangential <electric field> is continuous.
To establish <uniqueness of Poisson equation>, subtract two solutions with the same charge and boundary data, writing their difference as $u$. <Integration by parts> gives
$$
\int_V\epsilon|\nabla u|^2\,dV
=\int_{\partial V}u\epsilon\partial_nu\,dA,
$$
since $\nabla\cdot(\epsilon\nabla u)=0$. The right side vanishes for identical Dirichlet data, identical Neumann data, or their usual mixed combination. Interface terms cancel under the homogeneous difference conditions. With positive <permittivity>, $\nabla u=0$: \b[the field is unique], and the potential is unique if any suitable reference potential is fixed. For a pure Neumann problem an additive constant remains, and the prescribed outward displacement flux must equal the enclosed free charge. This theorem justifies image-charge solutions once their boundary data have been checked.
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