= Solution
Apply the <superposition principle> to the result of part (a). If $F(a,h)=-qh/[2\pi\epsilon_0(a^2+h^2)^{3/2}]$, then the exact normal <electric field> from the two real <point charges> and their <image charges> is
$$
E_z=F(a,h)-F(a,h+d)
=\frac{q}{2\pi\epsilon_0}\left[\frac{h+d}{[a^2+(h+d)^2]^{3/2}}-\frac{h}{(a^2+h^2)^{3/2}}\right].
$$
For $d\ll h$, <Taylor's theorem> gives $E_z=-d\partial_hF+O(d^2)$. Since
$$
\frac{d}{dh}\frac{h}{(a^2+h^2)^{3/2}}
=\frac{a^2-2h^2}{(a^2+h^2)^{5/2}},
$$
the leading <electric dipole> result is
$$
\boxed{E_z(a,0^+)=\frac{qd}{2\pi\epsilon_0}\frac{a^2-2h^2}{(a^2+h^2)^{5/2}}+O\!\left(\frac{qd^2}{\epsilon_0h^4}\right).}
$$
The <electric dipole moment> is $\mathbf p=-qd\hat{\mathbf z}$. The circle $a=\sqrt2h$ is the zero of the leading normal <electric field> and hence of the induced <surface charge density>. Inside it, induced <surface charge density> is negative; outside it, positive. The tangential <electric field> is zero everywhere on the grounded <electrical conductor>, so the leading full <electric field> vanishes on that circle. The corresponding <dipole-induced charge on a grounded plane> integrates to zero over the whole plane: the negative central region is balanced by the positive exterior.
For finite $d$, $a=\sqrt2h$ is only the leading-order zero. With $u=h^{2/3}$ and $v=(h+d)^{2/3}$, the exact zero satisfies $a^2=uv(u+v)$, giving $a=\sqrt2(h+d/2)+O(d^2/h)$. For example, inserting any $d>0$ in the exact expression at $a=\sqrt2h$ does not give zero. This is a controlled <electric dipole> approximation, not an exact finite-separation identity.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2022-dipole-induced-charge.png]
{title=Sign reversal and enclosed induced <electric charge> for a normal <electric dipole> above a grounded plane}
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