Question

A point particle of mass $M$ is attached to one end of a massless rigid non-conducting rod of length $L$. Another point particle of same mass is attached to the other end of the rod. The two particles carry charges $+q$ and $-q$ repectively. This arrangement is held in a region of uniform electric field $E$ such that the rod makes a small angle $(\theta <5^{\circ })$ with the field direction. The minimum time needed for the rod to become parallel to the field after it is set free. (rod rotates about centre of mass)

• A. $2\pi \sqrt{\frac{ML}{2qE}}$
• B. $\pi \sqrt{\frac{ML}{2qE}}$
• C. $\frac{\pi }{2}\sqrt{\frac{ML}{2qE}}$
• D. $4\pi \sqrt{\frac{ML}{2qE}}$

Answer 1

The correct option is C $\frac{\pi }{2}\sqrt{\frac{ML}{2qE}}$

Consider the system as an electric dipole
$\overrightarrow{\tau }=\overrightarrow{P}\times \overrightarrow{E}$
$\Rightarrow \tau =PE\sin \theta$
$\Rightarrow \tau =qLE\sin \theta$
$\Rightarrow qLE\sin \theta =I\alpha$
Moment of inertia of rod about its COM
$I=M{\left(\frac{L}{2}\right)}^2+M{\left(\frac{L}{2}\right)}^2=\frac{M{L}^2}{2}$
Therefore,
$qLE\sin \theta =\frac{1}{2}M{L}^2\alpha$
$\alpha =\frac{2qE\sin \theta }{ML}$

$\Rightarrow \frac{\theta }{\alpha }=\frac{ML}{2qE}$ $(\because \sin \theta \approx \theta )$ for small angle
Now, time period
$T=2\pi \sqrt{\frac{ML}{2qE}}$
Rotating in clock wise direction, minimum time taken by rod to align itself field is time taken to complete ${\frac{1}{4}}^{th}$ oscillation
$t=\frac{T}{4}=\frac{\pi }{2}\sqrt{\frac{ML}{2qE}}$