Charge q_{2} is at the centre of a circular p | vedclass

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(B) The electric potential $V$ at any point on a circular path of radius $r$ due to a charge $q_{2}$ at the center is given by $V = \frac{1}{4 \pi \va

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zero

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(B) The electric potential $V$ at any point on a circular path of radius $r$ due to a charge $q_{2}$ at the center is given by $V = \frac{1}{4 \pi \varepsilon_{0}} \frac{q_{2}}{r}$.Since $q_{2}$ and $r$ are constant,the potential $V$ is constant at all points on the circular path,making it an equipotential surface.The work done $W$ in moving a charge $q_{1}$ between two points $A$ and $B$ is given by $W = q_{1}(V_{B} - V_{A})$.Since the path is equipotential,the potential at the starting point and the ending point (after one full rotation) is the same,i.e.,$V_{A} = V_{B}$.Therefore,the work done $W = q_{1}(V - V) = 0$.

Charge $q_{2}$ is at the centre of a circular path with radius $r$. Work done in carrying charge $q_{1}$,once around this equipotential path,would be

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