Q. 744.5( 2 Votes )

Two particles, carrying –q and +q and having equal masses m each, are fixed at the ends of a light rod of length a to form a dipole. The rod is clamped at an end and is placed in a uniform electric field E with the axis of the dipole along the electric field. The rod is slightly tilted and then released. Neglecting gravity find the time period of small oscillations.

Answer :


Given:
Magnitude of charge on both the particles: q
Mass of both the particles: m
Electric field : E
The axis of the dipole is along the electric field.
Length of the dipole: l=a

Formula used:
We know that, a charge placed in an electric field E will experience electric force.
When a dipole with charge of magnitude q at its ends is suspended under a magnetic field and having one end clamped (-q), the free end (+q) will experience an electric force and it will start oscillating.
Electric force is given as:
F=qE

and
F = ma
qE = ma
Where a is the acceleration of the particle, q is the charge on the particle, m is the mass of the particle and E is the electric field.
Also, time period for Simple pendulum is:
Here, g=a(acceleration) as we are neglecting gravity.
Substituting we get,

Hence, time period of the small oscillations is .


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