# If a charge q is rotated around charge Q in a circle of radius y by $90^{\circ}$, the work done will be

$(A)\;q \times 2 \pi r \\ (B)\;0 \\ (C)\; q \times \large\frac{\pi}{2} r \\ (D)\;q \times \large\frac{2 \pi Q}{r}$

Since q moves on equipotential surface therefore work done is zero.
=> $q \times \large\frac{2 \pi Q}{r}$
Hence D is the correct answer.