Question

A small particle of mass $m$ and charge $-q$ is placed at point $P$ on the axis of uniformly charged ring and released, if $R>>x$, the particle will undergo oscillations along the axis of symmetry with an angular frequency that is equal to :

• A. $\sqrt{\frac{qQ}{4\pi {\epsilon }_{0}m{R}^3}}$
• B. $\sqrt{\frac{qQx}{4\pi {\epsilon }_{0}m{R}^4}}$
• C. $\frac{qQ}{4\pi {\epsilon }_{0}m{R}^3}$
• D. $\frac{qQx}{4\pi {\epsilon }_{0}m{R}^4}$

Answer 1

The correct option is A $\sqrt{\frac{qQ}{4\pi {\epsilon }_{0}m{R}^3}}$

The electric field at point P due to the charged ring is $E=\frac{Q}{4\pi {ϵ}_0}\frac{x}{\sqrt{(x^2+R^2{)}^3}}$

when $R>>x,E\sim \frac{Q}{4\pi {ϵ}_0}\frac{x}{R^3}$

Now the equation of motion of charge -q is $m\frac{d^{2}x}{d{t}^2}=-qE=-\frac{qQ}{4\pi {ϵ}_0}\frac{x}{R^3}$

or,$\frac{d^{2}x}{d{t}^2}=-\frac{qQ}{4\pi {ϵ}_{0}m{R}^3}x$

therefore, angular frequency ,$\omega =\sqrt{\frac{qQ}{4\pi {ϵ}_{0}m{R}^3}}$