Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 3 / 11A (Vector Calculus)

For two solutions let $u$ be their difference. Green’s identity and ${\partial }_{n}u=0$ give $0={\int }_{V}u({\nabla }^{2}u-m^{2}u)=-{\int }_V(|\nabla u{|}^2+m^2{u}^2)$, so $u=0$ when $m\ne 0$. At $m=0$, solutions are unique only up to an additive constant.

On a simply connected region, $\nabla \times E=0$ gives $E=-\nabla \varphi$. Then Gauss's law gives ${\nabla }^2\varphi =-\rho /{\epsilon }_0$.

Writing $r=|x|$ and imposing continuity gives

Gauss's law gives $E=Q\hat{r}/(4\pi {\epsilon }_0{r}^2)$ outside any spherically symmetric charge distribution. The shell in (ii) has $Q=2\pi {\rho }_0(b^2-a^2)$, and differentiating its exterior potential gives exactly this field.