KVPY 2020-21 SA Exam Previous Papers

The selection can be solved through image method in electromagnetism. Here A, B and C represent electric images of q
${\mathit{F}}_2=\frac{\mathit{K}{\mathit{q}}^2}{4{\mathit{d}}^2}$ (attractive)
${\mathit{F}}_3=\frac{\mathit{K}{\mathit{q}}^2}{4{\mathit{d}}^2}$ (attractive)
${\mathit{F}}_1=\frac{\mathit{K}{\mathit{q}}^2}{\left(2\sqrt{2}\mathit{d}\right)}$ (repulsion)
$\therefore {\mathit{F}}_2={\mathit{F}}_3=\mathit{F}$

$=\left(2\sqrt{2}-1\right)\frac{\mathit{K}{\mathit{q}}^2}{8{\mathit{d}}^2}$
$=\frac{{\mathit{q}}^2\left(2\sqrt{2}-1\right)}{32\mathrm{\pi }{\mathrm{\epsilon }}_0{\mathit{d}}^2}$, towards $\mathit{O}$