Question

A small particle of mass $m$ and charge $-q$ is placed at point P at a distance $x$ from the centre on the axis of a ring of radius $R$ containing charge $Q$ and released. If $R>>x$, the particle will undergo oscillations along the axis of symmetry with an angular frequency

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

Answer 1

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

Electric field at point P due to the charge element $dQ$ of the ring is $dE$.

We have $dE=\frac{dQ}{4\pi {ϵ}_o{r}^2}$ where $r=\sqrt{R^2+x^2}$

$⟹$ $dE=\frac{dQ}{4\pi {ϵ}_o(R^2+x^2)}$

By resolving $dE$ into horizontal and vertical components, we get net electric field at P only in horizontal direction as vertical components cancel out each other.

Net electric field at P $E_p=\int dE\cos \theta$

From figure, $\cos \theta =\frac{x}{\sqrt{R^2+x^2}}$

$⟹$ $E_p=\int \frac{xdQ}{4\pi {ϵ}_o(R^2+x^2{)}^{3/2}}$

As $R>>x$, we get $E_p=\int \frac{xdQ}{4\pi {ϵ}_o{R}^3}$

Or $E_p=\frac{x}{4\pi {ϵ}_o{R}^3}\int dQ$

Or $E_p=\frac{Qx}{4\pi {ϵ}_o{R}^3}$

Force on the charge $-q$ at P, $F=-q{E}_p$

$\therefore$ $ma=-\frac{Qqx}{4\pi {ϵ}_o{R}^3}$

Or $a=-\frac{Qq}{4\pi {ϵ}_{o}m{R}^3}x$

Or $a=-w^{2}x$

Thus angular frequency of oscillation $w=\sqrt{\frac{Qq}{4\pi {ϵ}_{o}m{R}^3}}$