 Solution

For the conductor in electrostatic equilibrium, the interior electric field is zero and the electrostatic potential is constant. Earthing fixes that constant to zero. Thus the conductor boundary conditions at $x=d$ areThe normal electric field need not vanish; it is supported by a surface charge density. The outward normal from this conductor into the exterior is $\mathbf{n}=-{\mathbf{e}}_x$, so the normal jump condition isThis sign convention will determine the sign of the induced surface charge density below.