- Home
- Standard 12
- Physics
Positive charge $Q$ is distributed uniformly over a circular ring of radius $R$. A point particle having a mass $(m)$ and a negative charge $-q$ is placed on its axis at a distance $x$ from the centre. Assuming $x < R,$ find the time period of oscillation of the particle, if it is released from there [neglect gravity].
$\left[\frac{16 \pi^{3} \varepsilon_{0} R^{3} m}{Q q}\right]^{1 / 2}$
$\left[\frac{8 \pi^{2} \varepsilon_{0} R^{3}}{q}\right]^{1 / 2}$
$\left[\frac{2 \pi^{3} \varepsilon_{0} R^{3}}{3 q}\right]^{1 / 2}$
None of these
Solution
When the negative charge is shifted at a distance $x$ from the centre of the ring along its axis, then force acting on the point charge due to the ring.
$F=q E$ (towards centre)
$=q \cdot \frac{k Q x}{\left(R^{2}+x^{2}\right)^{3 / 2}}$
If $R \gg x,$ then $R^{2}+x^{2} \approx R^{2}$
and $F=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q q x}{R^{3}}$ (towards centre)
$\Rightarrow a=\frac{F}{m}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q q x}{m R^{3}}$
since, restoring force $F_{E} \propto x,$ therefore motion of charge particle will be $SHM.$
Time period of $SHM,$
$T=\frac{2 \pi}{\omega}=\left[\frac{16 \pi^{3} \varepsilon_{0} R^{3} m}{Q q}\right]^{1 / 2}\left[\because \omega^{2}=\frac{Q q}{4 \pi \varepsilon_{0} R^{3}}\right]$
