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.
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