Paper 3, Section II, C

Vector Calculus | Part IA, 2011

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

E=ϕ,E=ρε0,\mathbf{E}=-\boldsymbol{\nabla} \phi, \quad \boldsymbol{\nabla} \cdot \mathbf{E}=\frac{\rho}{\varepsilon_{0}},

where ϕ(x)\phi(\mathbf{x}) is the corresponding electrostatic potential and ε0\varepsilon_{0} is a constant.

(a) Show that the total charge QQ contained within a closed surface SS is given by Gauss' Law

Q=ε0SEdS.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 qq at the origin i.e. for ρ(x)=qδ(x)\rho(\mathbf{x})=q \delta(\mathbf{x}).

(b) Let E1\mathbf{E}_{1} and E2\mathbf{E}_{2}, with potentials ϕ1\phi_{1} and ϕ2\phi_{2} respectively, be the solutions to (1) arising from two different charge distributions with densities ρ1\rho_{1} and ρ2\rho_{2}. Show that

1ε0Vϕ1ρ2 dV+Vϕ1ϕ2dS=1ε0Vϕ2ρ1 dV+Vϕ2ϕ1dS\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 VV with boundary V\partial V, where dS\mathrm{d} \mathbf{S} points out of VV.

(c) Suppose that ρ1(x)=0\rho_{1}(\mathbf{x})=0 for xa|\mathbf{x}| \leqslant a and that ϕ1(x)=Φ\phi_{1}(\mathbf{x})=\Phi, a constant, on x=a|\mathbf{x}|=a. Use the results of (a) and (b) to show that

Φ=14πε0r>aρ1(x)r dV\Phi=\frac{1}{4 \pi \varepsilon_{0}} \int_{r>a} \frac{\rho_{1}(\mathbf{x})}{r} \mathrm{~d} V

[You may assume that ϕ10\phi_{1} \rightarrow 0 as x|\mathbf{x}| \rightarrow \infty sufficiently rapidly that any integrals over the 'sphere at infinity' in (2) are zero.]

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