Question

Positive charge Q is distributed uniformly over a circular ring of radius R. A particle having a 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

We know: Electric field 'E' at 'P' due to teh charged ring

$=\frac{KQx}{(R^2+x^2{)}^{3/2}}=\frac{KQx}{R^3}$

Force experienced ${}^{\prime }{F}^{\prime }=Q\times E=\frac{q\times K\times Qx}{R^3}$

Now, amplitude$=x$

So, $T=2\pi \sqrt{\frac{x}{KQqx/m{R}^3}}=2\pi \sqrt{\frac{m{R}^{3}x}{KQqx}}=2\pi \sqrt{\frac{4\pi {\epsilon }_{0}m{R}^3}{Qq}}=\sqrt{\frac{4{\pi }^2\times 4\pi {\epsilon }_{0}m{R}^3}{qQ}}$

$\Rightarrow T={\left[\frac{16{\pi }^3{\epsilon }_{0}m{R}^3}{qQ}\right]}^{1/2}$.