Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

Two point particles of same mass M are attached to two ends of a massless non-conducting rigid rod of length l. The charge on particle are +q and -q. The rod is held in a region of uniform electric field E making an angle $\theta $( very small) with E. The time t , after itself parallel to electric field after releasing it is :


$(A)\;2 \pi ( \frac{Ml}{2qE})^{1/2} \\ (B)\;2 \pi ( \frac{2Ml}{qE})^{1/2} \\ (C)\;2 \pi ( \frac{Ml}{qE})^{1/2} \\ (D)\;\frac{\pi}{2} ( \frac{Ml}{2qE})^{1/2} $

Can you answer this question?

1 Answer

0 votes
The force acting at both the charged constitute a couple.
z= Force $\times$ perpendicular disatnce
=>$z= qE \times AB \sin \theta$
$\qquad= qEl \sin \theta$
Since , $\theta$ is small
$\therefore \sin \theta =0$
=>$ z =qEl \theta$
Restoring torque $z= -qEL \theta$
If $\alpha $ is angular acceleration
$z=I \alpha$
$I= \large\frac{Ml^2}{2}$
=> $z= \large\frac{Ml^2}{2}$$ \alpha$
=> $\alpha =-w^2 \theta$
Now, $\alpha =-w^2 \theta$
=> $w^2=\bigg(\large\frac{2qE}{Ml}\bigg)$
=> $w=\bigg(\large\frac{2qE}{Ml}\bigg)^{1/2}$
Time period of oscillation $=\large\frac{2 \pi}{w}$
=> $ T= 2 \pi \bigg( \large\frac{Ml}{2qE}\bigg) ^{1/2}$
Rotating in clock wise direction, minimum time taken by rod to align itself parallel to electric field is time taken to complete $1/4\; th$ oscillation.
=> $ t= \large\frac{T}{4} =\frac{\pi}{2} \bigg( \large\frac{ Ml}{2qE}\bigg)^{1/2}$
Hence D is the correct answer.
answered Jan 15, 2014 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App