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

Explanation

The torque on rod AB is given by
`tau = qE(l sin theta)`
`= q El theta`
The moment of inertia of the rod AB about `theta` is given by
`I = m((l)/(2))^(2) + m((l)/(2))^(2) = (ml^(2))/(2)`
We know that,
`tau = I alpha` or `alpha = (tau)/(I)`
`:. alpha = (q E l theta)/((ml^(2) //2)) = (2qE theta)/(ml) = omega^(2) theta` where `omega^(2) = (2qE)/(ml)`
An acceleration is directly propotional to `theta`, hence the motion of rod is SHM. The time period T is given by
`T = (2pi)/(omega) = 2pi sqrt((ml)/(2eE))`
The rod will become parallel to E in time
`t = (T)/(4) = (2pi)/(4)sqrt(((ml)/(2eE))) = (pi)/(2) sqrt(((ml)/(2qE)))`