Past exam of the mathematics course of the University of Cambridge / 2025 / ib / Paper 1 / 15B / Solution

Using $\rho ={\epsilon }_0\nabla \cdot E$, $E=-\nabla \varphi$, integration by parts, and $\varphi =0$ on the boundary givesPut $C=q/(4\pi {\epsilon }_0)$. Gauss's law gives the radial fieldTaking zero potential outside,The charge formula givesThe field formula gives the same result after integrating $4\pi r^2{E}_r^2$ over the two nonzero annuli.