Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 3 / 11C / c

Use the auxiliary point charge of strength $q\ne 0$ at the origin, with potential ${\varphi }_2=q/(4\pi {\epsilon }_{0}r)$ from part (a). Apply the electrostatic reciprocity identity of part (b) to the exterior region $a<r<R$, then let $R\to \infty$. Its inner boundary has outward normal $-{\mathbf{e}}_r$, and ${\rho }_2=0$ everywhere in this exterior region. The outer boundary terms vanish under the stipulated decay assumption.

On $r=a$, ${\varphi }_1=\Phi$ andThus the surviving boundary term on the left is $\Phi q/{\epsilon }_0$. On the right, ${\varphi }_2$ is constant on the inner sphere, so its boundary term isIndeed, $\nabla {\varphi }_1=-{\mathbf{E}}_1$, and this integral is the outward electric flux from the interior ball, whose enclosed charge is zero by the hypothesis and Gauss's law. No symmetry of the exterior charge distribution is needed.

The electrostatic reciprocity identity now reduces toCanceling $q/{\epsilon }_0$ givesThe inward normal on the inner boundary is essential to the positive sign in this result.