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 the same mass is attached to 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 θ (say of about 5 degrees) with the field direction (see figure). Will be minimum time, needed for the rod to become parallel to the field after it is set free

• A. $t = 2\pi\sqrt{\frac{mL}{2pE}}$
• B. $t = \frac{\pi}{2}\sqrt{\frac{mL}{2qE}}$
• C. $t = \frac{3\pi}{2}\sqrt{\frac{mL}{2pE}}$
• D. $t = \pi\sqrt{\frac{2mL}{qE}}$

Answer 1

Answer: (B)

$t = \frac{\pi}{2}\sqrt{\frac{mL}{2qE}}$

Answer 2

In the given situation system oscillate in electric field with maximum angular displacement θ.

It’s time period of oscillation (similar to dipole)

$T = 2\pi\sqrt{\frac{I}{pE}}$ where I = moment of inertia of the

system and $p = qL$

Hence the minimum time needed for the rod becomes parallel to the field is $t = \frac{T}{4} = \frac{\pi}{2}\sqrt{\frac{I}{pE}}$

Here $I = M\left( \frac{L}{2} \right)^{2} + M\left( \frac{L}{2} \right)^{2} = \frac{ML^{2}}{2}$

⇒ $t = \frac{\pi}{2}\sqrt{\frac{ML^{2}}{2 \times qL \times E}} = \frac{\pi}{2}\sqrt{\frac{ML}{2qE}}$