Question

An electric dipole with dipole moment p and moment of inertia / is released from rest in a uniform external electric field E with angle between bar(p) and bar(E) equal to (pi)/(2) .The angular velocity of the dipole when bar(p) and bar(E) become parallel to each other is​

Answer 1

Explanation:

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Answer 2

Answer:

Hence, the period of oscillation is $T=\frac{2\PI}{\omega}=2\Pi (\frac{I}{pE})$

Explanation:

The dipole experiences a torque p Esinθ tending to bring itself back in the direction of field.

​Therefore, on being released (i.e. rotated) the dipole oscillates about an axis through its centre of mass and perpendicular to the field. If I is the moment of inertia of the dipole about the axis of rotation, then the equation of motion is

$I \times \frac{d^2 \theta }{dt^2}=-pEsin\theta$

As we know that,

For small amplitude $sin\theta \approx \theta$

Thus,

$\frac{d^2\theta}{dt^2}==-(\frac{pE}{I})\\\theta =-\omega ^2\theta$

where $\omega = (\frac{pE}{I})$

This is a S.H.M., whose period of oscillation is $T=\frac{2\PI}{\omega}=2\Pi (\frac{I}{pE})$