Correct Answer - `2 pi sqrt((mh^(2))/(2kq lambda))`
At equilibrium `mg=qE`
`mg=(q2k lambda)/(h)`
when particle is displaced to distance x in upwar direction. Then net force in upward direction is
`F=q((2k lambda)/(h+x))-mg`
`=(2kq lambda)/(h)(1+x/h)^(-1)-mg=(2kq lambda)/(h)(1+x/h)-mg`
`=-((2kq lambda)/(h^(2)))x=-K_(0)x`
`:. F prop-x` (Motion is SHM)
Time period of motion is `T=2pisqrt((m)/(K_(0)))=2 pi sqrt((mh^(2))/(2kq lambda))`.