= Solution
Set the grounded plane at $z=0$ and the real <point charge> at $(0,0,h)$. The <image charge> $-q$ at $(0,0,-h)$ makes the <electrostatic potential> vanish throughout the plane. The <uniqueness theorem for the electrostatic potential> then identifies the physical solution in $z>0$:
$$
\Phi(a,z)=\frac{q}{4\pi\epsilon_0}\left[\frac1{\sqrt{a^2+(z-h)^2}}-\frac1{\sqrt{a^2+(z+h)^2}}\right].
$$
The normal <electric field> is $E_z=-\partial_z\Phi$. Differentiation gives
$$
E_z(a,z)=\frac{q}{4\pi\epsilon_0}\left[\frac{z-h}{[a^2+(z-h)^2]^{3/2}}-\frac{z+h}{[a^2+(z+h)^2]^{3/2}}\right],
$$
and therefore
$$
\boxed{E_z(a,0^+)=-\frac{q}{2\pi\epsilon_0}\frac{h}{(a^2+h^2)^{3/2}}.}
$$
For $q>0$ the <electric field> points into the <electrical conductor>, consistently with negative induced <surface charge density>.
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