Question

There is an insulator of length $L$ and of negligible mass with two small balls of mass $m$ and electric charge $q$ attached to its ends. The rod can rotate in the horizontal plane around a vertical axis crossing it at a distance $\frac{L}{4}$ from one of its ends. A uniform electric field $E$ exists in the region as shown in the figure. Find the time period for the small oscillations of the rod.

• A. $T=2\pi \sqrt{\frac{4qE}{5mL}}$
• B. $T=2\pi \sqrt{\frac{5mL}{4qE}}$
• C. $T=2\pi \sqrt{\frac{mL}{5qE}}$
• D. $T=2\pi \sqrt{\frac{3mL}{2qE}}$

Answer 1

The correct option is B $T=2\pi \sqrt{\frac{5mL}{4qE}}$

Let, the rod is displaced by an angle $\theta$ from its equlibrium position as shown in figure,


From figure, the net torque about hinge point,

${\tau }_{net}=+qE\frac{L}{4}\sin \theta -qE\left(\frac{3L}{4}\sin \theta \right)$

$\Rightarrow {\tau }_{net}=-\frac{qEL}{2}\sin \theta$

We know that, ${\tau }_{net}=I\alpha$ so,

$\Rightarrow I\alpha =-\frac{qEL}{2}\sin \theta ...\left(1\right)$

For small oscillations. $\sin \theta \approx \theta$

Moment of inertia, $I$ about hinge

$I=m{\left(\frac{L}{4}\right)}^2+m{\left(\frac{3L}{4}\right)}^2$

$\Rightarrow I=\frac{m{L}^2}{16}+\frac{9m{L}^2}{16}=\frac{5}{8}m{L}^2...\left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$ we have,

$\alpha =-\frac{\frac{qEL}{2}\theta }{I}$

$\Rightarrow \alpha =-\left(\frac{qEL}{2}\theta \right)\left(\frac{8}{5m{L}^2}\right)$

$\Rightarrow \alpha =-\frac{4qE}{5mL}\theta$

Now comparing with $\alpha =-{\omega }^2\theta$

$\omega =\sqrt{\frac{4qE}{5mL}}$

Now time period of oscillations,

$\Rightarrow T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{5ml}{4qE}}$

Hence, option (b) is correct answer.