Question

Two point masses, $m$ each carrying charges $-q$ and $+q$ are attached to the ends of a massless rigid non-conducting wire of length $L$ . When this arrangement is placed in a uniform electric field, then it deflects through an angle $\theta $ . The minimum time needed by the rod to align itself along the field is

• A. $2\pi \sqrt{\frac{m L}{q E}}$
• B. $\frac{\pi }{2}\sqrt{\frac{m L}{2 q E}}$
• C. $\pi \sqrt{\frac{2 m L}{q E}}$
• D. $2\pi \sqrt{\frac{3 m L}{q E}}$

Answer 1

Answer: (B)

When the wire is brought in a uniform field E, then the torque is given by
$\tau=qELsin \theta $
$=qEL\theta $ $\left[\because \, \, \, \theta \, \, \, i s \, v e r y \, s m a l l\right]$
The moment of inertia of rod AB about O is
$I=m\left(\frac{L}{2}\right)^{2}+m\left(\frac{L}{2}\right)^{2}=\frac{m L^{2}}{2}$
As $\tau=I\alpha $

So, $\alpha =\frac{\tau}{I}=\frac{2q E L \theta }{m L^{2}}$
$\Rightarrow \, \, \, \omega ^{2}\theta =\frac{2 q E L \theta }{m L^{2}}$ $\left[\because \, \, \alpha = \omega ^{2} \theta \right]$
$\Rightarrow \, \, \, \omega ^{2}=\frac{2 q E}{m L}$
The time period of the wire is
$T=\frac{2 \pi }{\omega }=2\pi \sqrt{\frac{m L^{2}}{2 q E}}$
The rod will become parallel to the field in time $\frac{T}{4}$
So, $t=\frac{T}{4}=\frac{\pi }{2}\sqrt{\frac{m L}{2 q E}}$