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:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{\sqrt{2}\,q^{2}}{\pi\varepsilon_0 a} \)
The work done by an external agent in moving a charge slowly from one point to another equals the change in its potential energy, \( W = q_0(V_{\infty} - V_O) \), with \( V_{\infty} = 0 \).
Potential at the centre. Each corner is at a distance \( r = \frac{a}{\sqrt{2}} \) from O (half the diagonal), and potentials add as scalars.
\( \displaystyle V_O = 4\times\frac{1}{4\pi\varepsilon_0}\frac{q}{a/\sqrt{2}} = \frac{4\sqrt{2}\,q}{4\pi\varepsilon_0 a} = \frac{\sqrt{2}\,q}{\pi\varepsilon_0 a} \)
Work done on the charge \( -q \)
\( \displaystyle W = (-q)(0 - V_O) = qV_O = \frac{\sqrt{2}\,q^{2}}{\pi\varepsilon_0 a} \)