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 \[W=-\,Q({{V}_{\infty }}-{{V}_{O}})=Q{{V}_{0}}\]\[=\frac{4\sqrt{2}\,{{Q}^{2}}}{4\pi {{\varepsilon }_{0}}a}\]\[=\frac{\sqrt{2}{{Q}^{2}}}{\pi {{\varepsilon }_{0}}a}\]You need to login to perform this action.
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