Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 4 / 7B (Electromagnetism)

The differential Gauss law is $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$. On a large ball $B_R$, the product rule and $\nabla \varphi =-\mathbf{E}$ giveThe divergence theorem consequently expresses the electrostatic energy asFor a localized charge distribution with the electric potential chosen to vanish at infinity, $\varphi =O(R^{-1})$ and $\mathbf{E}=O(R^{-2})$, so the surface term is $O(R^{-1})$ and tends to zero. More generally, it suffices to assume that this boundary integral vanishes. Passing to all space givesThe physical decay and potential reference are relevant: $\mathbf{E}\to 0$ by itself does not specify the additive constant in $\varphi$. Adding a constant $C$ changes the charge-based expression by $CQ/2$ when the total charge is $Q$, while leaving the field unchanged. The zero-at-infinity potential is the usual reference implicit in this all-space electrostatic energy formula.

Spherical symmetry and the integral Gauss law give $4\pi r^2{E}_r(r)=Q/{ϵ}_0$ for $r>R$. Together with the stipulated zero interior field,The electric field jumps across the ideal charged surface; its value exactly on that surface is not needed for the volume integral. The electrostatic energy of a uniformly charged spherical shell isAs a check, the shell has constant electric potential $Q/(4\pi {ϵ}_{0}R)$, with zero reference at infinity, so $U=Q\varphi (R)/2$ gives the same result.