Question

Charge $q_2$ of mass m revolves around a stationary charge $q_1$ in a circular orbit of radius r. The orbital periodic time of $q_2$ would be.

• A. ${\left[\frac{4{\pi }^{2}m{r}^3}{k{q}_1{q}_2}\right]}^{1/2}$
• B. ${\left[\frac{k{q}_1{q}_2}{4{\pi }^{2}m{r}^3}\right]}^{1/2}$
• C. ${\left[\frac{4{\pi }^{2}m{r}^4}{k{q}_1{q}_2}\right]}^{1/2}$
• D. ${\left[\frac{4{\pi }^{2}m{r}^2}{k{q}_1{q}_2}\right]}^{1/2}$

Answer 1

The correct option is A ${\left[\frac{4{\pi }^{2}m{r}^3}{k{q}_1{q}_2}\right]}^{1/2}$

The force acting on the particle can be balanced by equating the electrostatic force by the centripetal force on the charge.

$\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_1}{r^2}=mr{\omega }^2=\frac{4{\pi }^{2}mr}{T^2}$

$T^2=\frac{\left(4\pi {\epsilon }_0\right)r^2\left(4{\pi }^{2}mr\right)}{q_1{q}_2}$

$T={\left[\frac{4{\pi }^{2}m{r}^3}{k{q}_1{q}_2}\right]}^{1/2}$.