Question

A particle of mass $m$ carrying charge $-q_1$ is moving at a constant angular speed around a fixed charge $+q_2$ at the centre of a circular path of radius $r$. Find the period of revolution of charge $-q_1$.

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

Answer 1

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

Particle of charge $-q_1$ and mass $m$ is performing uniform circular motion. Therefore, the centripetal force is provided by the electrostatic force of attraction between $-q_1$ and $+q_2$.


$\therefore |F_e|=F_{centripetal}$

$\Rightarrow |F_e|=m{\omega }^{2}r$

$\Rightarrow \frac{q_1{q}_2}{4\pi {ϵ}_0{r}^2}=m{\omega }^{2}r$

$\Rightarrow \omega ={\left(\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{m{r}^3}\right)}^{\frac{1}{2}}$

Time period $T$ of revolution will be

$T=\frac{2\pi }{\omega }=\frac{2\pi }{{\left(\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{m{r}^3}\right)}^{\frac{1}{2}}}$

$\Rightarrow T=2\pi \sqrt{\frac{4\pi {ϵ}_{0}m{r}^3}{q_1{q}_2}}$

$\Rightarrow T=\sqrt{\frac{4\pi {ϵ}_{0}m{r}^3}{q_1{q}_2}\left(4{\pi }^2\right)}$

$\therefore T=\sqrt{\frac{16{\pi }^3{ϵ}_{0}m{r}^3}{q_1{q}_2}}$

Hence, option $\left(a\right)$ is the correct answer.

$\begin{array}{c}\text{Why this question ?}\\ \text{Tip: Electrostatic force is always directed along line}\\ \text{joining of two charges, thus at each point on path}\overrightarrow{F_e}\perp \overrightarrow{v}\\ \text{Bottomline:}\overrightarrow{F_e}\text{will only provide}\text{centripetal force, thus charge revolves at constant speed.}\end{array}$