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 infinityYou need to login to perform this action.
You will be redirected in
3 sec