= Solution
Let $x\ge0$ denote the radial distance from the axis in the plane, and write $r_h=(x^2+h^2)^{1/2}$. A charge $q$ at height $h$ and its image $-q$ give a combined normal <electric field> $-2k_eqh/r_h^3$ at $z=0^+$; the tangential components cancel. Superposing the two pairs yields
$$
\boxed{\mathbf E(x,0,0^+)=\frac{Qd}{2\pi\epsilon_0}\left[\frac4{(x^2+4d^2)^{3/2}}-\frac1{(x^2+d^2)^{3/2}}\right]\hat{\mathbf z}}.
$$
The conductor boundary condition gives $\sigma=\epsilon_0E_z$, so
$$
\boxed{\sigma(x)=\frac{Qd}{2\pi}\left[\frac4{(x^2+4d^2)^{3/2}}-\frac1{(x^2+d^2)^{3/2}}\right]}.
$$
To integrate the induced <surface charge>, use concentric annuli of area $2\pi x\,dx$. Since $\int_0^\infty x(x^2+h^2)^{-3/2}\,dx=1/h$,
$$
Q_{\rm sheet}=Qd\left(\frac4{2d}-\frac1d\right)=\boxed{+Q}.
$$
The local <surface charge density> is negative near the origin but positive farther out; integrating only the negative patch would miss the total. The total real charge is $-Q$, so the induced $+Q$ also agrees with the sum of the two image charges. An isolated finite plate constrained to zero net charge would require a different global boundary-value problem; the formula is the infinite-plane solution just specified.
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