Total charge $-Q$ is uniformly spread along the length of a ring of radius $R$. $A$ small test charge $+q$ of mass $m$ is kept at the centre of the ring and is given a gentle push along the axis of the ring.
$(a)$ Show that the particle executes a simple harmonic oscillation.
$(b)$ Obtain its time period.

(N/A) The force acting on a point charge $q$ at point $P$ at distance $x$ from the centre of the ring of radius $R$ on its axis due to a charge element $(-dQ)$ at $A$ is:
$dF = k \frac{(-dQ)q}{R^2 + x^2}$
As shown in the figure,the components $dF \sin \theta$ are equal in magnitude but opposite in direction,so they cancel each other. The net force $F$ is the sum of the $dF \cos \theta$ components,which are directed towards the centre $O$:
$F = \oint dF \cos \theta = \oint -k \frac{(dQ)q}{R^2 + x^2} \cdot \frac{x}{\sqrt{R^2 + x^2}} = -\frac{kQqx}{(R^2 + x^2)^{3/2}}$
For $x \ll R$,$R^2 + x^2 \approx R^2$,so $F \approx -\frac{kQq}{R^3} x$. Since $F \propto -x$,the motion is simple harmonic.
Comparing with $F = -m \omega^2 x$,we get $\omega^2 = \frac{kQq}{mR^3} = \frac{Qq}{4 \pi \epsilon_0 m R^3}$.
The time period is $T = \frac{2 \pi}{\omega} = 2 \pi \sqrt{\frac{4 \pi \epsilon_0 m R^3}{Qq}}$.