Question

A fixed ring of radius R has uniformly distributed charge Q. A particle (m, –q) is at centre of ring. It is displaced slightly along the axis of ring and released. Time period of oscillation is

Answer 1

Given:

A fixed ring of radius R has uniformly distributed charge Q. A particle (m, –q) is at centre of ring. It is displaced slightly along the axis of ring and released.

To find:

Time period of SHM?

Calculation:

Now, when the charge is displaced slightly by a displacement of 'x' , it will experience a force as follows:

$F = E \times ( - q)$

$\implies F = \dfrac{kQx}{ {R}^{3} } \times ( - q)$

$\implies F = - \dfrac{kQq}{ {R}^{3} } x$

$\implies F \propto - x$

• SO, IT IS AN SHM.

So, time period will be :

$\implies T = 2\pi \sqrt{ \dfrac{m}{( \dfrac{kQq}{ {R}^{3} } )}}$

$\implies T = 2\pi \sqrt{ \dfrac{m {R}^{3} }{kQq}}$

$\implies T = 2\pi \sqrt{ \dfrac{4\pi \epsilon_{0} m {R}^{3} }{Qq}}$

So, final answer is:

$\boxed{ \bf T = 2\pi \sqrt{ \dfrac{4\pi \epsilon_{0} m {R}^{3} }{Qq}}}$