Solution
= Solution
For $z\gt 0$, $\phi=q(4\pi\varepsilon_0)^{-1}(|x-d\hat z|^{-1}-|x+d\hat z|^{-1})$ and $E=-\nabla\phi$. The normal field just above the plane is $-qd/[2\pi\varepsilon_0(r^2+d^2)^{3/2}]$, so $\sigma=\varepsilon_0E_z$ is the stated expression. Integrating $2\pi r\sigma(r)dr$ from zero to infinity gives total induced charge $-q$.