Question

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

• A. $\frac{\pi }{2},\sqrt{\frac{ML}{2qE}}$
• B. $\frac{\pi }{2},\sqrt{\frac{ML}{qE}}$
• C. $\pi ,\sqrt{\frac{ML}{qE}}$
• D. None of these

Answer 1

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

The force acting at both the charged constitute a couple

$Z=$ Force $\times$ perpendicular distance

$Z=qE\times AB\sin \theta =qEl\sin \theta$

Since, $\theta =$ small

$\therefore$ $\sin \theta =0$

$\Rightarrow Z=qEl\theta$

restoring torque $Z=-PEl\theta$

If $\alpha$ is angular acceleration

$Z=J\alpha$

$\Rightarrow Z=\frac{{Ml}^2}{2}\alpha$

$\Rightarrow \alpha =-W^2\theta$

Now, $\alpha =-W^2\theta$

$\Rightarrow W^2=\frac{2qE}{Ml}$

$W={\left(\frac{2qE}{Ml}\right)}^{1/2}$

Time period of oscillation $=\frac{2\pi }{W}$

$\Rightarrow T=2\pi {\left(\frac{MI}{2qE}\right)}^{1/2}$

Rotating in clock wise direction, minimum time taken by rod to align itself field is time taken to complete $1/4th$ oscillation

$\Rightarrow t=\frac{T}{4}=\frac{\pi }{2}{\left(\frac{MI}{2qE}\right)}^{1/2}$