Past exam of the electromagnetism course of the University of Oxford / 2023 / 1 / a / ii / Solution

Let $\mathbf{n}=\mathbf{r}/a$ be the unit normal from the electrical conductor into the vacuum. The electrostatic boundary conditions at a conductor, together with the vanishing interior electric field, giveWrite $u=\cos \theta$ in spherical coordinates, $D=|\mathbf{r}-{\mathbf{r}}_0|$ on $r=a$, and $z^{\star }=a^2/r_0$. Differentiating the electric potential at fixed angles givesHere the distance identity in the preceding answer was used before simplifying. Hence the induced surface charge density isFor positive $q$ this surface charge density is negative everywhere, with largest magnitude nearest the point charge.