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


From the data given in the question, it is clear that, rod is rotation about its center of mass

Thus, finding the Torque acting on the dipole due to external field about center of mass.

Using, $\overrightarrow{\tau }=\overrightarrow{P}\times \overrightarrow{E}$

$\Rightarrow \tau =PE\sin \theta$

$\Rightarrow \tau =qLE\sin \theta$

$\Rightarrow qLE\sin \theta =I\alpha$

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

$t=\frac{T}{4}=\frac{\pi }{2}\sqrt{\frac{ML}{2qE}}$

Hence, option (c) is the correct answer.