Question

A particle of mass $1\text{kg}$ and charge $-{10}^{-2}\text{C}$ is constrained to move along the axis of a ring of radius $\sqrt{2}\text{m}$. The ring carries a uniform charge density $+40\text{nC/m}$ along its length. Initially, the particle is in the centre of the ring where the force on it is zero. The period of oscillation of the particle when it is displaced slightly from its equilibrium position is given by

• A. $0.25\pi \text{s}$
• B. $0.45\pi \text{s}$
• C. $0.65\pi \text{s}$
• D. $0.85\pi \text{s}$

Answer 1

The correct option is C $0.65\pi \text{s}$

Given that,
Mass of particle, $m=1\text{kg}$
Charge of particle, $Q=-{10}^{-2}\text{C}$
Radius of ring, $R=\sqrt{2}\text{m}$
Charge density, $\lambda =+40\text{nC/m}$

Let the mass is at $r$ distance away from the centre of the ring, along the axis of the ring.

The distance of the particle from the circumference of the ring,
$R_1=\sqrt{R^2+r^2}$

The electric field at a distance $r$ from the centre of the ring is given by,
$E_{net}=\frac{2\pi k\lambda Rr}{(\sqrt{R^2+r^2}{)}^{\frac{3}{2}}}$
$\Rightarrow E_{net}=\frac{2\pi k\lambda Rr}{R^3}$ (As r<<R)
$\Rightarrow E_{net}=\frac{2\pi k\lambda r}{R^2}$
or $\overrightarrow{F}=-Q{E}_{net}=-\frac{2\pi kQ\lambda r}{R^2}$ (toward centre of ring)
This force is restoring in nature thus,
$m{\omega }^{2}r=\frac{2\pi kQ\lambda r}{R^2}$
$\omega =\sqrt{\frac{2\pi kQ\lambda }{m{R}^2}}$
so, time period, $T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m{R}^2}{2\pi kQ\lambda }}$
$\Rightarrow T=2\pi \sqrt{\frac{1\times (\sqrt{2}{)}^2}{2\times 3.14\times 9\times {10}^9\times {10}^{-2}\times 40\times {10}^{-9}}}$
$\Rightarrow T=0.65\pi \text{s}$