Question

A particle of mass m and charge -Q is constrained to move along the axis of a ring of radius a. The ring carries a uniform charge density +$\lambda$ along its length. Initially the particle is in the centre of the ring where the force on it is zero. Show that the period of oscillation of the particle when it is displaced slightly from its equilibrium position is given by
$T=2\pi \sqrt{\frac{2{\epsilon }_{0}m{a}^2}{\lambda Q}}$

Answer 1

Given that a particle of mass $m$ and charge $-Q$ is constrained to move along the axis of a ring of radius $a$ and carries a charge density $\lambda$. We have to show that the period of oscillation of particle when it is displaced slightly from its equilibrium position is given by:

$T=2\pi \sqrt{\frac{2{\epsilon }_{0}m{a}^2}{\lambda Q}}$

Electric potential on the axis of ring at a distance $x$ is $=\frac{kq}{(x^2+a^2{)}^{1/2}}$

and electric field on the axis of ring is

$E_x=\frac{-\delta V}{\delta x}\hat{i}$

$E_x=-\left(\frac{1}{2}\frac{2kqx}{(x^2+a^2{)}^{3/2}}\right)$

$E_x=\frac{kqx}{(x^2+a^2{)}^{3/2}}=\frac{\lambda ax}{2{\epsilon }_0(x^2+a^2{)}^{3/2}}$

Restoring force, $F_{res}=-Q{E}_x$

$=\frac{-Q\lambda ax}{2{\epsilon }_0(x^2+a^2{)}^{3/2}}$

Since, $x<<a$

$F_{res}\approx \frac{-Q\lambda ax}{2{\epsilon }_0{a}^3}$

$F_{res}\alpha -x$. Thus the particle perform simple harmonic motion.

As, $F_{res}=m\alpha$

$\alpha =\frac{F_{res}}{m}=\frac{-Q\lambda ax}{m{a}^{3}2{\epsilon }_0}=-{\omega }^{2}x$

where $\omega =\sqrt{\frac{Q\lambda a}{2{\epsilon }_{0}m{a}^3}}$

Time period, $T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{2{\epsilon }_{0}m{a}^2}{Q\lambda }}$

Hence Proved.