= 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 \b[the conductor boundary conditions] at $x=d$ are
$$
\boxed{\phi(d,y,z)=0,\qquad E_y(d^-,y,z)=E_z(d^-,y,z)=0.}
$$
The 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 is
$$
\mathbf E(d^-)\cdot\mathbf n=\frac\sigma{\varepsilon_0},\qquad E_x(d^-)=-\frac\sigma{\varepsilon_0}.
$$
This sign convention will determine the sign of the induced <surface charge density> below.
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