A block of mass `m` having charge `q` is attached to a spring of spring constant `k`. This arrangement is placed in uniform electric field `E` on smooth horizontal surface as shown in the figure. Initially spring in unstretched. Find the extension of spring in equilibrium position and maximum extension of spring.

Explanation

Due to electric force `qE` acting on the block , spring extends. Let at some instant extension of spring be `x`.
Net force on block , `F_(n e t) = qE - kx`
In equilibrium , `F_(n e t) = qE - kx = 0`
` x = x_(0) = (qE)/(k)`
In equilibrium , extension of spring `x_(0) = (qE)/(k)`
Maximum extension `= 2x_(0) = (2qE)/(k)`
OR
where `x' ` : maximum extension .
`(1)/(2) kx'^(2) = qEx' rArr x' = (2qE)/(k)`
In fact , the block is executing `S.H.M.` of time period
`T = 2pi sqrt((m)/(k))` , amplitude of oscillation `= (qE)/(k)`