A) \[\frac{qQ}{4\pi {{\varepsilon }_{0}}a}\]
B) \[\frac{qQ}{4\pi {{\varepsilon }_{0}}{{a}^{2}}}\]
C) \[\frac{q}{4\pi {{\varepsilon }_{0}}a}\]
D) zero
Correct Answer: D
Solution :
The electrical potential at any point on circle of radius a due to charge Q at its centre is\[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{a}\]. It is an equipotential surface. Hence, work done in carrying a charge q round the circle is zero.
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