A) \[\frac{qQ}{4\pi {{\varepsilon }_{0}}r}\]
B) \[\frac{qQ}{4\pi {{\varepsilon }_{0}}^{2}{{r}^{2}}}\]
C) \[\frac{qQ}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\]
D) none of these
Correct Answer: D
Solution :
The potential at each point on the circular path will be equal. So, work done \[=q\times \] potential difference \[=q\times 0\] = 0You need to login to perform this action.
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