= Solution
The <electric field> within a <conductor in electrostatic equilibrium> vanishes. The normal <electric field> boundary condition, with outward <normal vector> $+\hat{\mathbf z}$, therefore gives $\sigma=\epsilon_0E_z(0^+)$. Directly below the monopole, $a=0$:
$$
\boxed{\sigma_{\rm mono}(0)=-\frac{q}{2\pi h^2}.}
$$
For the <electric dipole>,
$$
\boxed{\sigma_{\rm dip}(0)=-\frac{qd}{\pi h^3}+O(qd^2/h^4).}
$$
The exact finite-separation answer is
$$
\sigma(0)=\frac{q}{2\pi}\left[\frac1{(h+d)^2}-\frac1{h^2}\right],
$$
whose <Taylor series> agrees with the boxed leading term. Although the real <point charges> have zero total <electric charge>, their unequal distances leave negative induced <surface charge density> immediately beneath them.
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