= Solution
The <electric field> of a signed point <electric charge> in vacuum follows <Coulomb's law>:
$$
E(r)=\frac{Q}{4\pi\epsilon_0}\frac{r}{|r|^3}.
$$
For the <conducting plane>, the <method of images> replaces each charge $Q_j$ at height $h_j$ by an <image charge> $-Q_j$ at height $-h_j$. The real and image <electric fields> have cancelling tangential components and equal normal components at the surface. Thus the <electric field at a conducting plane from axial point charges>, just above the point directly below them, is
$$
E_z(0^+)=-\frac{2}{4\pi\epsilon_0}\sum_j\frac{Q_j}{h_j^2}
=\frac{60}{4\pi\epsilon_0}\left(\frac1{5000^2}-\frac1{8000^2}\right).
$$
Numerically,
$$
\boxed{E(0^+)\simeq1.31\times10^4\ \mathrm{V\,m^{-1}}\,\hat z,\quad\text{upwards}.}
$$
The surface value means the air-side limit. Inside the ideal <conductor> the electrostatic <electric field> is zero; its discontinuity across the surface is supported by induced <surface charge>. Omitting the <image charges> would give half the external answer.
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