Consider the system shown below. If the charge is slightly displaced perpendicular to the wire from its equilibrium position then find out the time period of SHM.

Explanation

: At equilbrium position weight of the particle is balanced by the electric force
`implies mg = qE`
`mg=q(2klamda)/(d)" "...(1)`
Now if the particle is slightly displaced by a distance `x_lamda` (where `x ltlt d` ) net force on the body,
`F_("net")(2klamdaq)/(d+x)-mg`
from (1)
`F_("net")=(2klamdaq)/(d+x)-(2klamdaq)/(d)=(-2klamdaqx)/(d(d+x))`
`xltltd F_("net")~~(-2klamdaqx)/(d^(2))`
`impliesa=-(-2klamdaqx)/(md^2)`
for SHM
`a =-omega^2x`
`:. -omega^2=(2klamdaq)/(md^2)`
`impliesomega=sqrt((2klamdaq)/(md^(2))`
`T=(2pi)/(omega)=2pisqrt((md^2)/(2klamdaq))`