 Solution

At electrostatic equilibrium the electric field in the metal is zero. The empty cavity also has zero electric field, by the uniqueness theorem for the electrostatic potential. The electric charge lies on the outer surface. For the isolated conducting sphere, spherical symmetry and Gauss's law giveTaking the electric potential to vanish at infinity and using $E_r=-d\varphi /dr$ givesThe electric potential is continuous at the surface, while the normal electric field jumps by $\sigma /{ϵ}_0$, with surface charge density $\sigma =Q/(4\pi R^2)$.