 Solution

The method of images places an image charge $-q$ at the reflected position below the plane. The candidate potential in the physical upper half-space isAt $z=0$ the two terms cancel, so $\Phi =0$ and $E_x=E_y=0$. Direct differentiation givesThus the upper-side field satisfies exactly the conductor boundary conditions from part (i). The fictitious pair does have a field below the plane, but that part of its solution is not used for the physical conductor.

In $z>0$, the potential has the correct real point source, the correct zero Dirichlet boundary condition, and the required decay at infinity. By Uniqueness of the Dirichlet problem, it is the physical potential there. This is why the artificial charge can represent the induced conductor response without being a second physical charge.