Question

The time period of revolution of a charge $q_1$ and of mass m moving in a circular path of radius r due to Coulomb force of attraction with another charge $q_2$ at its centre is

• A. $\sqrt{\frac{16\pi {\epsilon }_{0}m{r}^3}{q_1{q}_2}}$
• B. $\sqrt{\frac{8{\pi }^2{\epsilon }_{0}m{r}^3}{q_1{q}_2}}$
• C. $\sqrt{\frac{{\epsilon }_{0}m{r}^3}{16q_1{q}_2}}$
• D. $\sqrt{\frac{16{\pi }^3{\epsilon }_{0}m{r}^3}{q_1{q}_2}}$
• E. $\sqrt{\frac{{\pi }^2{\epsilon }_{0}m{r}^3}{8q_1{q}_2}}$

Answer 1

The correct option is D $\sqrt{\frac{16{\pi }^3{\epsilon }_{0}m{r}^3}{q_1{q}_2}}$

Centripetal force $F=\frac{m{v}^2}{4}$ ..(i)
and electrostatic force
$F=K\cdot \frac{q_1\cdot q_2}{r^2}$ ...(ii)
Equating both equation
$\frac{m{v}^2}{r}=K\cdot \frac{q_1\cdot q_2}{r^2}$
$v=\sqrt{\frac{K}{m}\cdot \frac{q_1\cdot q_2}{r}}$
The time period of revolution
$T=\frac{2\pi r}{v}$
$=2\pi r\sqrt{\frac{m\cdot r}{K\cdot q_1{q}_2}}$
$=\sqrt{\frac{4{\pi }^2{r}^{2}mr}{K\cdot q_1{q}_2}}(\because K=\frac{1}{4\pi {\epsilon }_0})$
$=\sqrt{\frac{4{\pi }^2{r}^{2}mr\cdot 4\pi {\epsilon }_0}{q_1{q}_2}}$
$=\sqrt{\frac{16{\pi }^2{r}^{2}mr\cdot 4\pi {\epsilon }_0}{q_1{q}_2}}$