10) Charge Q of mass m revolves around a point a charges q due to electrostatic attraction. Show that its period of revolution is given by $T^2=\frac{16{\mathrm{\pi }}^3 {\in }_0{\mathrm{mR}}^3}{Qq}.$

Dear Student,
Since the charge particle is moving along a circular path, a centripetal force , $\frac{m{v}^2}{R}$, must act on it. where $v = velocity of the particle = \frac{2\mathrm{\pi R}}{\mathrm{T}}, T is the time period$. The only force that acts is the electrostatic force of attraction between Q and q, which is $\frac{Qq}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\mathrm{R}}^2}$
Thus, for circular motion of the charged particle,

$$\begin{gathered} \frac{m{v}^2}{R}=\frac{Qq}{4\pi {\epsilon }_o{R}^2} \\ {v}^2=\frac{Qq}{4\pi {\epsilon }_{o}mR} \\ {\left(\frac{2\mathrm{\pi R}}{T}\right)}^2=\frac{Qq}{4\pi {\epsilon }_{o}mR} \\ {T}^2 = \frac{16{\mathrm{\pi }}^3{\mathrm{\epsilon }}_{\mathrm{o}}{\mathrm{mR}}^3}{\mathrm{Qq}} Ans \end{gathered}$$

Regards,