A) \[\frac{Q}{4\pi {{\varepsilon }_{0}}r}\]
B) \[\frac{{{Q}^{2}}al}{4\pi {{\varepsilon }_{0}}r}\]
C) zero
D) \[\frac{Qal}{2r}\]
Correct Answer: C
Solution :
There will be same potential on all the points of the charged circle and hence work done by charge in moving on circular path will be zero.You need to login to perform this action.
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