Question

A dipole consists of a light rod of length l, having two point charges +q and -q and of same mass m, joined at ends A and B of the rod. A uniform electric field E is present in the region and perpendicular to the length of the rod. The rod is released from rest from this position. The angular velocity of the rod, when it becomes parallel to the
electric field for the first time is

• A. $\sqrt{\frac{qE}{ml}}$
• B. $2\sqrt{\frac{qE}{ml}}$
• C. $\sqrt{\frac{qE}{2ml}}$
• D. $\frac{1}{2}\sqrt{\frac{qE}{ml}}$

Answer 1

The correct option is B $2\sqrt{\frac{qE}{ml}}$

As $\Delta W=-\Delta U$
Now, $\Delta W=\frac{1}{2}l{\omega }^2$
$=\frac{1}{2}\times \frac{m{l}^2}{2}{\omega }^2=\frac{1}{4}m{l}^2{\omega }^2$
$\Delta U=-qlE$
$\therefore \frac{m{l}^2}{4}{\omega }^2=qlE\Rightarrow \omega =2\sqrt{\frac{qE}{ml}}$