GATE Electrical Engineering (EE) 2018 Solved Paper

Due to infinite conducting plane, we have to consider an image charge symmetrically below plane. Since one charge is placed at (0, 0, 0.2), the image charge will be at (0, 0, -0.2)
Electric field at P will be sum of Field due to $\mathit{Q}$ \& -Q
Net electric field at point due to a charge $\mathit{Q}$ is given by
$\mathit{E}=\frac{\mathit{Q}}{4\mathit{\pi }{\mathit{\epsilon }}_0{\mathit{r}}^2}$
So, electric field a point P is${\mathit{E}}_{\mathit{P}}=\frac{\mathit{Q}}{4\mathit{\pi }{\mathit{\epsilon }}_0{\left(0.1\right)}^2}+\frac{\mathit{Q}}{4\mathit{\pi }{\mathit{\epsilon }}_0{\left(0.3\right)}^2}$
Since, it is in downward direction so, ${\mathit{E}}_{\mathit{P}}=-999.09\mathit{V}/\mathit{m}$