The electric field $\mathbf{E}(\mathbf{x})$ due to a static charge distribution with density $\rho(\mathbf{x})$ satisfies

$$\mathbf{E}=-\boldsymbol{\nabla} \phi, \quad \boldsymbol{\nabla} \cdot \mathbf{E}=\frac{\rho}{\varepsilon_{0}},$$

where $\phi(\mathbf{x})$ is the corresponding electrostatic potential and $\varepsilon_{0}$ is a constant.

(a) Show that the total charge $Q$ contained within a closed surface $S$ is given by Gauss' Law

$$Q=\varepsilon_{0} \int_{S} \mathbf{E} \cdot \mathrm{d} \mathbf{S} .$$

Assuming spherical symmetry, deduce the electric field and potential due to a point charge $q$ at the origin i.e. for $\rho(\mathbf{x})=q \delta(\mathbf{x})$.

(b) Let $\mathbf{E}_{1}$ and $\mathbf{E}_{2}$, with potentials $\phi_{1}$ and $\phi_{2}$ respectively, be the solutions to (1) arising from two different charge distributions with densities $\rho_{1}$ and $\rho_{2}$. Show that

$$\frac{1}{\varepsilon_{0}} \int_{V} \phi_{1} \rho_{2} \mathrm{~d} V+\int_{\partial V} \phi_{1} \nabla \phi_{2} \cdot \mathrm{d} \mathbf{S}=\frac{1}{\varepsilon_{0}} \int_{V} \phi_{2} \rho_{1} \mathrm{~d} V+\int_{\partial V} \phi_{2} \nabla \phi_{1} \cdot \mathrm{d} \mathbf{S}$$

for any region $V$ with boundary $\partial V$, where $\mathrm{d} \mathbf{S}$ points out of $V$.

(c) Suppose that $\rho_{1}(\mathbf{x})=0$ for $|\mathbf{x}| \leqslant a$ and that $\phi_{1}(\mathbf{x})=\Phi$, a constant, on $|\mathbf{x}|=a$. Use the results of (a) and (b) to show that

$$\Phi=\frac{1}{4 \pi \varepsilon_{0}} \int_{r>a} \frac{\rho_{1}(\mathbf{x})}{r} \mathrm{~d} V$$

[You may assume that $\phi_{1} \rightarrow 0$ as $|\mathbf{x}| \rightarrow \infty$ sufficiently rapidly that any integrals over the 'sphere at infinity' in (2) are zero.]