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 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 degree) with the field direction, fig. Find an expression for the minimum time needed for the rod to become parrallel to the field after it is set free.

Explanation

As shown in Fig, AB = l
Tonque on the rod AB, `tau = qE (l sin theta)`
When `theta` is small, `tau = q E l theta` …(i)
Moment of inertia of the rod about O.
`l = m(l//2)^(2) + m(l//2)^(2) = (ml^(2))/(2)`
As `tau = l alpha`
`:. alpha = (tau)/(l) = (qE t theta)/(ml^(2)//2) = (2q E theta)/(ml) = omega^(2) theta`
Clearly, `alpha` is directly proportional to `theta`, and torque is trying to decreses `theta`.
`:.` The motion of the rod si S.H.M., with `omega^(2) = (2qE)/(m l), omega = sqrt((2qE)/(ml)) = (2pi)/(T) :. T = 2pi sqrt((ml)/(2qE))`
The rod will become parallel to `vec(E)` from a position , `theta = 90^(@)` in a time `t = (T)/(4) = (2pi)/(4) sqrt((ml)/(2q E)) = (pi)/(2) sqrt((ml)/(2qE))`