A thin fixed ring of radius 1 meter has a positive charge $1\times {10}^{-5}\mathrm{C}$ uniformly distributed over it. A particle of mass 0.9 g and having a negative charge of 1 x 10^-6 coulomb is placed on the axis at a distance of 1 cm from the centre of the ring. If the motion of the negatively charged particle is found to be approximately simple harmonic with the time period of oscillations is $\frac{\mathrm{\pi }}{\mathrm{n}}\mathrm{s}$ find the value of n. 

Let us first find the force on a -q charge placed at a distance x from centre of ring along its axis.

In this case force on particle P, ${\mathrm{F}}_{\mathrm{P}}=-\mathrm{qE}$

$\Rightarrow {\mathrm{F}}_{\mathrm{P}}=-\mathrm{q}\cdot \frac{\mathrm{Kqx}}{{\left({\mathrm{x}}^2+{\mathrm{R}}^2\right)}^{3/2}}$

For small $\mathrm{x}, \mathrm{x}<<\mathrm{R},$ we can neglect x, compared to R, we have

$\mathrm{F}=-\frac{\mathrm{KqQx}}{{\mathrm{R}}^3}$

Acceleration of particle, $\mathrm{a}=-\frac{\mathrm{KqQ}}{{\mathrm{mR}}^3}\mathrm{x}$

[Here we have x = 1 cm and R = 1 m hence $\mathrm{x}<<\mathrm{R}$ can be used]

This shows that particle P executes SHM, now comparing this acceleration with $\mathrm{a}=-{\mathrm{\omega }}^2\mathrm{x}$

We get $\mathrm{\omega }=\sqrt{\frac{\mathrm{KqQ}}{{\mathrm{mR}}^3}}$

Thus time period of SHM, $\mathrm{T}=\frac{2\mathrm{\pi }}{\mathrm{\omega }}=2\mathrm{\pi }\sqrt{\frac{{\mathrm{mR}}^3}{\mathrm{KqQ}}}$

$\Rightarrow \mathrm{T}=2\mathrm{\pi }\sqrt{\frac{0.9\times {10}^{-3}\times (1{)}^3}{9\times {10}^9\times {10}^{-5}\times {10}^{-6}}}=\frac{\mathrm{\pi }}{5}\mathrm{s}$