Question

A point charge $-q$ having mass $m$ is placed at the centre of a uniformly charged ring of charge $Q$ and radius $R$. Find the time period of oscillation if the point charge is slightly displaced and released.

• A. $T=2\pi \sqrt{\frac{m{R}^3}{2KQq}}$
• B. $T=2\pi \sqrt{\frac{m{R}^3}{KQq}}$
• C. None of the above
• D. $T=2\pi \sqrt{\frac{m{R}^2}{KQq}}$

Answer 1

The correct option is B $T=2\pi \sqrt{\frac{m{R}^3}{KQq}}$


Let us displace the particle by an amount $x$ from its mean point.

Restoring force, $\overrightarrow{F}=q\overrightarrow{E}...\left(1\right)$

Electric field $E$ due to circular ring at displaced position is

$E=\frac{KQx}{(R^2+x^2{)}^{\frac{3}{2}}}...\left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$ , we have

$F=-\frac{KQqx}{(R^2+x^2{)}^{\frac{3}{2}}}...\left(3\right)$

For slight displacement
$R>>x$

$R^2+x^2\approx R^2$

From eaqaution $\left(3\right)$,

$F=-\frac{KQqx}{R^3}$

Comparing with general equation of $\text{SHM}$, we get

$F=-kx$
We have,
$k=\frac{KQq}{R^3}$

We know that, time period, $T$

$T=2\pi \sqrt{\frac{m}{k}}$

On substituting the value of $k$, we get

$T=2\pi \sqrt{\frac{m{R}^3}{KQq}}$

Hence, option (a) is correct answer.