= Solution
First use the usual idealization of a grounded infinite <conducting plane> at $z=0$, with the real <point charge> $+q$ at $(0,0,d)$. The <method of images> replaces the metal, for purposes of calculating the exterior <electric field>, by an image charge $-q$ at $(0,0,-d)$. The exact <electric potential> in $z>0$ is
$$
V(\rho,z)=\frac{q}{4\pi\epsilon_0}\left[\frac1{\sqrt{\rho^2+(z-d)^2}}-\frac1{\sqrt{\rho^2+(z+d)^2}}\right].
$$
It vanishes on the plane, and the <uniqueness theorem for the electrostatic potential> identifies it with the physical exterior solution. The tangential <electric field> cancels on the plane, while the normal components add:
$$
\boxed{\mathbf E(\rho,0^+)=-\frac{qd}{2\pi\epsilon_0(\rho^2+d^2)^{3/2}}\hat{\mathbf z}.}
$$
The <electric field> inside the <conductor> is zero, so the <electrostatic boundary condition> gives $\sigma=\epsilon_0E_z(0^+)$. Integrating the <surface charge density> over the facing plane yields
$$
Q_{\rm face}=\int_0^\infty\frac{-qd}{2\pi(\rho^2+d^2)^{3/2}}2\pi\rho\,d\rho=-qd\left[-\frac1{\sqrt{\rho^2+d^2}}\right]_0^\infty,\qquad \boxed{Q_{\rm face}=-q.}
$$
The exact image expression is essential near the charge: using the far-field <electric dipole> approximation at $\rho$ comparable with $d$ would be invalid.
\b[Electrical boundary condition.] The source does not specify grounding or the initial total <electric charge> of the slab. The boxed total is the grounded infinite-plane answer and the large-plane approximation to the charge on its nearby face. An initially neutral, finite, electrically isolated slab retains zero \i[total] <electric charge>; an approximately $-q$ accumulation near the real charge is then balanced by approximately $+q$ on its remote face and edges. This gives a concrete counterexample to interpreting $-q$ as the unconditional total charge of every large metal slab. The local image formula assumes the infinite-plane limit with no applied background <electric field>.
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