An electric dipole formed by two particles fixed at the ends of a light rod of length l. The mass of each particle is m and the charges are -q and +q . The system is placed in such a way that the dipole axis is parallel to a uniform electric field E that exist in the region. The dipole is slightly rotated abut its centre and released. Show that for small angular displacement, the motion is angular simple harmonic and find its time period.

Explanation

Suppose, the dipole axis makes an angle `theta` with
the electric field at an instant. The magnitude of the
torque on it is
`|tau| =| vecp xx vecE|`
`=qlE sin theta.`
This torque will tend to rotate the dipole back towards
the electric field. Also, for small angular displacement
`sin theta =~~ theta so that`
`tau = -qlEtheta.`
The moment of inertia of the system about the axis of
rotation is
`I = 2 xx m ((l)/(2))^2 = (ml^2)/(2).`
Thus the angular acceleration is
`alpha = (tau)/(I) = - (2qE)/(ml) theta = - omega ^2theta`
. where `omega^2 = (2qE)/(ml)`
Thus, the motion is angular simple harmonic and the
time period is `T = 2 pi sqrt (ml)/(2qE)`.