A) zero
B) \[\frac{\sqrt{2}{{q}^{2}}}{\pi {{\varepsilon }_{0}}a}\]
C) \[\frac{\sqrt{2}q}{\pi {{\varepsilon }_{0}}a}\]
D) \[\frac{{{q}^{2}}}{\pi {{\varepsilon }_{0}}a}\]
Correct Answer: B
Solution :
(b) \[\frac{\sqrt{2}{{q}^{2}}}{\pi {{\varepsilon }_{0}}a}\] Potential at the centre of the square due to four equal charges q at four corners \[V=\frac{4q}{4\pi {{\varepsilon }_{0}}\left( a\sqrt{2} \right)/2}=\frac{\sqrt{2}q}{\pi {{\varepsilon }_{0}}a}\] \[{{W}_{0\,\to \,\infty }}=-{{W}_{\infty \,\to \,0}}=-\left( -q \right)V=\frac{\sqrt{2}{{q}^{2}}}{\pi {{\varepsilon }_{0}}a}\]
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