A charged particle of mass `m` having charge `q` remains in equilibrium at height `d` above a fixed charge `Q`. Now the charged particle is slightly displaced along the line joining the charges, show that it will executes `S.H.M.` and find the time period of oscillation.

Explanation

In equilibrium
`mg = (1)/( 4 pi in_(0)) (Qq)/(d^(2))` ….(i)
If the charged particle is slightly displaced by `y` along line `AB`.
Restoring force `F = -[mg - (1)/(4 pi in_(0)) (Qq)/((d + y)^(2))]`
` = - [mg - (1)/(4 pi in_(0)) (Qq)/(d^(2) (1 + y //d^(2)))]`
`= -[ mg - mg (1 + y//d)^(-2)] , y lt lt d`
` = mg [ 1 - (1 - (2y)/(d))] = - ( 2mg)/(d) y`
Acceleration `a = (F)/(m) = -(2g)/(d) y`
`a alpha - y` (Case of `S.H.M.`)
Time period `T = 2 pi sqrt((d)/(2g))`