Question

Four equal charge Q are placed at the four corners of a square of side a. The work doen in re-moving a charge -Q from the centre of the square to infinity is

• A. Zero
• B. $\frac{\sqrt{2}{Q}^2}{4\pi {ϵ}_{0}a}$
• C. $\frac{\sqrt{2}{Q}^2}{\pi {ϵ}_{0}a}$
• D. $\frac{Q^2}{2\pi {ϵ}_{0}a}$

Answer 1

The correct option is C $\frac{\sqrt{2}{Q}^2}{\pi {ϵ}_{0}a}$

AO=BO=CO=DO =$r=\frac{a}{\sqrt{2}}$
Since electric potential is a scalar and since the charges at corners are equal, the magnitude of the electric potential at point O due to the four charges =4 times that due to a single charge. Thus
$V=4\times \frac{1}{4\pi {ϵ}_0}.\frac{Q}{r}=\frac{1}{\pi {ϵ}_0}.\frac{Q}{\frac{a}{\sqrt{2}}}=\frac{\sqrt{2}Q}{\pi {ϵ}_{0}a}$

$\therefore$ Work done $=QV=\frac{\sqrt{2}{Q}^2}{\pi {ϵ}_{0}a}$. Hence the correct choice is (c).