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:

$\frac{\sqrt{2} q^2}{4 \pi \varepsilon_{o} a}$ $\frac{\sqrt{2} q^2}{\pi \varepsilon_{o} a}$ $\frac{q^2}{\pi \varepsilon_{o} a}$ $\frac{q^2}{4 \pi \varepsilon_{o} a}$

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

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