Four equal charges each equal to 'q' coulomb are placed at the four vertices of a square of side 'a'. The work done in displacing a charge -q from the centre 'O' of the square to infinity is given by:

$\frac{\sqrt{2} q^2}{\pi \varepsilon_{o} a}$

The correct answer is Option (2) → $\frac{\sqrt{2} q^2}{\pi \varepsilon_{o} a}$

The electric potential at a point due to charge q,

$V=\frac{kq}{r}$

Now,

At the center of the square (O), the distance of each vertex from the center -

$r=\frac{a}{\sqrt{2}}$

$∴V_{single}=\frac{kq}{a/\sqrt{2}}=\sqrt{2}\frac{kq}{a}$

$V_{total}=4×\sqrt{2}\frac{kq}{a}=\frac{4\sqrt{2}k}{a}$

∴ Work done, $W = q.(V_∞-V_{initial})$

$=-q\left(0-\frac{\sqrt{2}q}{πε_0a}\right)$

$=\frac{\sqrt{2}q^2}{πε_0a}$