A) \[zero\]
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: A
Solution :
Potential at centre 0 of the square \[{{V}_{0}}=4\left( \frac{Q}{4\pi {{\varepsilon }_{0}}(a/\sqrt{2})} \right)\] Work done in shifting charge from centre to infinity \[W=Q({{V}_{\infty }}-{{V}_{0}})\] \[W=Q{{V}_{0}}\] \[=\frac{4\sqrt{2}{{Q}^{2}}}{4\pi {{\varepsilon }_{0}}a}\] \[W=\frac{\sqrt{2}{{Q}^{2}}}{\pi {{\varepsilon }_{0}}a}\]You need to login to perform this action.
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