Electrostatics Question 254

Question: A small sphere carrying a charge -q? is hanging in between two parallel plates by a string of length L. Time period of pendulum is $ T _{0} $ . When parallel plates are charged, the time period changes to $ T $ . The ratio $ T/T _{0} $ is equal to [UPSEAT 2003]

Options:

A) $ {{( \frac{pE}{I} )}^{1/2}} $

B) $ {{( \frac{pE}{I} )}^{3/2}} $

C) $ {{( \frac{I}{pE} )}^{1/2}} $

D) $ {{( \frac{p}{IE} )}^{1/2}} $

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Answer:

Correct Answer: A

Solution:

When dipole is given a small angular displacement q about it’s equilibrium position, the restoring torque will be

$ \tau =-pE\sin \theta =-pE\theta $ (as sinq = q) or $ I\frac{d^{2}\theta }{dt^{2}}=-pE\theta $ (as $ \tau =I\alpha =I\frac{d^{2}\theta }{dt^{2}} $ )

or $ \frac{d^{2}\theta }{dt^{2}}=-{{\omega }^{2}}\theta $

with $ {{\omega }^{2}}=\frac{pE}{I} $

therefore $ \omega =\sqrt{\frac{pE}{I}} $



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