 Solution

Let $x\ge 0$ 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 $-2k_{e}qh/r_h^3$ at $z=0^{+}$; the tangential components cancel. Superposing the two pairs yieldsThe conductor boundary condition gives $\sigma ={ϵ}_0{E}_z$, soTo integrate the induced surface charge, use concentric annuli of area $2\pi xdx$. Since ${\int }_0^{\infty }x(x^2+h^2{)}^{-3/2}dx=1/h$,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.