A) zero
B) \[\frac{Qq}{4\pi G{{\varepsilon }_{0}}r}\]
C) \[\frac{Qq}{2\pi G{{\varepsilon }_{0}}r}\]
D) None of these
Correct Answer: A
Solution :
Potential due to charge\[(q)\]at point\[(r)\]is given by \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q}{r}\] Since, charge\[Q\]is rotated in a circle of radius\[r\], hence its potential remains same at all points on the path, hence\[\Delta V=0\]. Also, work done\[=q\Delta V\] where\[q\]is charge and\[\Delta V=0\]. \[\therefore \]Work done = 0.You need to login to perform this action.
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