Question

A positive charge Q is distributed uniformly over a circular ring of radius R. A particle of mass m, and a negative charge q, is placed on its axis at a distance x from the centre. Find the force on the particle. Assuming x << R, find the time period of oscillation of the particle if it is released from there.

Answer 1

Consider an element of angular width $d\theta$ at a distance r from the charge q on the circular ring, as shown in the figure.


So, charge on the element,
$dq=\frac{Q}{2\pi R}Rd\mathrm{\theta }=\frac{Q}{2\mathrm{\pi }}d\mathrm{\theta }$
Electric field due to this charged element,
$$\begin{gathered} dE=\frac{dq}{4\mathrm{\pi }{\epsilon }_0}\frac{1}{r^2} \\ =\frac{{\displaystyle \frac{Q}{2\mathrm{\pi }}}d\mathrm{\theta }}{4\mathrm{\pi }{\epsilon }_0}\frac{1}{R^2+x^2} \end{gathered}$$
By the symmetry, the Esinθ component of all such elements on the ring will vanish.
So, net electric field,
$d{E}_{net}=dE\cos \theta =\frac{Qd\theta }{8{\pi }^2{\epsilon }_0{\left(R^2+x^2\right)}^{3/2}}$
Total force on the charged particle,
$$\begin{gathered} F=\int q\mathrm{d}{E}_{\mathrm{net}} \\ =\frac{q\mathrm{Q}}{8{\mathrm{\pi }}^2{\epsilon }_0}\frac{\mathrm{x}}{{\left(R^2+x^2\right)}^{3/2}}{\int }_0^{2\mathrm{\pi }}d\mathrm{\theta } \\ =\frac{xQq}{4\pi {\epsilon }_0{\left(R^2+x^2\right)}^{3/2}} \end{gathered}$$
According to the question,
$$\begin{gathered} x<<R \\ F=\frac{Qqx}{4\pi {\epsilon }_0{R}^3} \end{gathered}$$
Comparing this with the condition of simple harmonic motion, we get
$$\begin{gathered} F=m{\omega }^{2}x \\ \Rightarrow m{\omega }^2=\frac{Qq}{4\pi {\epsilon }_0{R}^3} \\ \Rightarrow m{\left(\frac{2\pi }{T}\right)}^2=\frac{Qq}{4\pi {\epsilon }_0{R}^3} \\ \Rightarrow T={\left[\frac{16{\pi }^3{\epsilon }_{0}m{R}^3}{Qq}\right]}^{1/2} \end{gathered}$$