question_answer

Four equal charges \[Q\] are placed at the four corners of a square of each side is \['a'\]. Work done in removing a charge ? Q from its centre to infinity is                                                   [AIIMS 1995]

A)                    0                                        

B)                    \[\frac{\sqrt{2}{{Q}^{2}}}{4\pi {{\varepsilon }_{0}}a}\]

C)                                                               \[\frac{\sqrt{2}{{Q}^{2}}}{\pi {{\varepsilon }_{0}}a}\]

D)                                                 \[\frac{{{Q}^{2}}}{2\pi {{\varepsilon }_{0}}a}\]

Correct Answer: C

Solution :

                     Potential at centre O of the square \[{{V}_{O}}=4\,\left( \frac{Q}{4\pi {{\varepsilon }_{0}}(a/\sqrt{2})} \right)\] Work done in shifting (? Q) charge from centre to infinity \[W=-\,Q({{V}_{\infty }}-{{V}_{O}})=Q{{V}_{0}}\]\[=\frac{4\sqrt{2}\,{{Q}^{2}}}{4\pi {{\varepsilon }_{0}}a}\]\[=\frac{\sqrt{2}{{Q}^{2}}}{\pi {{\varepsilon }_{0}}a}\]