A uniform rod AB of mass m and length `l` is hinged at its mid point C. The left half (AC) of the rod has linear charge density `-lamda` and the right half (CB) has `+ lamda` where `lamda` is constant. A large non conducting sheet of uniform surface charge density `sigma` is also present near the rod. Initially the rod is kept perpendicular to the sheet. The end A of the rod is initialy at a distance d. Now the rod is rotated by a small angle in the plane of the paper and released. Prove that the rod will perform SHM and find its time period.

Answer

`T=2 pi sqrt((2m in_(0))/(3 lamda sigma))`

Explanation

The sheet produces a uniform electric field `E = (sigma)/(2 in_(0))` towards right. The part AC and BC will experience electric force F as shown. They can be considered to be activing at the mid points of those parts respectively. The rod will experience torque about the point c in the anticlockwise direction

whose magnitude is `tau = F (l)/(2) sin theta = (Fl)/(2) theta`:
But `F = lambda (l)/(2). (sigma)/(2 in_(0)) = (lambda l sigma)/(4 in_(0))`
`tau = ((lambda l^(2) sigma)/(8 in_(0))) theta`
Now since `tau` is towards the mean position & `tau prop theta`
`:.` it will do SHM `rarr` Hence proved
& `tau = 1 prop = (lambda l^(2)sigma)/(8 in_(0))`
`rArr (ml^(2) sgma)/(12) alpha = (lambda l^(2) sigma)/(8 in_(0)) alpha`
`alpha = ((3 lambda sigma)/(2m in_(0))) theta`
`:. omega^(2) = (3 lambda sigma)/(2 m in_(0)) rArr ((2pi)/(T))^(2)`
`rArr T = 2pi sqrt(2m in_(0))/(3lambda sigma)`