Four equal charges Q are placed at the four corners of a square of side 'a'. Work done in removing a charge -Q from its centre to infinity is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{\sqrt{2}\,Q^2}{\pi\epsilon_{o}\,a} \)
The work done by an external agent in moving a charge \( q \) slowly from a point to infinity is \( W = q\,(V_\infty - V) = -qV \), since \( V_\infty = 0 \).
Each corner is at a distance of half the diagonal from the centre.
\( \displaystyle d = \frac{\sqrt{2}\,a}{2} = \frac{a}{\sqrt{2}} \)
The potential at the centre is the sum of the four equal contributions.
\( \displaystyle V = 4 \times \frac{Q}{4\pi\epsilon_0\,(a/\sqrt{2})} = \frac{\sqrt{2}\,Q}{\pi\epsilon_0\,a} \)
Putting \( q = -Q \):
\( \displaystyle W = -(-Q)\,V = \frac{\sqrt{2}\,Q^2}{\pi\epsilon_0\,a} \)
The work is positive, because the negative charge is attracted by the four positive charges and has to be pulled away.