Question

on rotating a point charge having a charge q around Q in a circle radius r. the work done will be

$$\begin{gathered} \mathbf{2}\mathbf{.}\mathbf{}OnrotatingapointhavingachargeqaroundachargeQinacircleofradiusr.Theworkdonewillbe \\ \left(1\right)q\times 2\mathrm{\pi r}\left(2\right)\frac{\mathrm{q}\times 2\mathrm{\pi Q}}{\mathrm{r}}\left(3\right)\mathrm{Zero}\left(4\right)\frac{\mathrm{Q}}{2{\mathrm{\epsilon }}_0\mathrm{r}} \end{gathered}$$

Answer 1

dear student


Force acting on charge during rotation is coulomb's force acting towards centre. it is behaving as centripital force (always perpendicular to velocitty.

workdone by centripital force is always zero.