A particle of mass m and charge q is placed at rest in a uniform electric field E and then released. The kinetic energy attained by the particle after moving a distance y is :

Explanation

Let the charge q at the mid-point is displaced slightly to the left. The force on the displaced charge q due to charge q at A,
`F_(1)=1/(4piepsi_(0)) q^(2)/((l+x)^(2))`
The force on the displaced charge q due to charge at B,
`F_(2)=1/(4piepsi_(0)) q^(2)/((l-x)^(2))`

Net restoring force on the displaced charge q.
`F=F_(2)-F_(1)` or `F=1/(4pi epsi_(0)) q^(2)/((l-x)^(2))-1/(4piepsi_(0)) q^(2)/((l+x)^(2))`
or `F=q^(2)/(4piepsi_(0))[1/((l-x)^(2))-1/((l+x)^(2))]=q^(2)/(4piepsi_(0)) (4lx)/((l^(2)-x^(2))^(2))`
since `l gt gt x, :. F=(q^(2) lx)/(pi epsi_(0) l^(4))` or `F=(q^(2) x)/(pi epsi_(0) l)`
Hence we see that `F prop x` and it is opposite to the direction of displacement. Therefore, the motion is SHM.
`T=2pi sqrt(m/k), ("here "k=q^(2)/(pi in_(0) l^(3))) T=2pi sqrt((mpi in_(0) l^(3))/q^(2))`