Question

A simple pendulum of length $l$ has a bob of mass $m$, with a charge $q$ on it. A vertical sheet of charge, with surface charge density $\sigma$ passes through the point of suspension. At equilibrium, the spring makes an angle $\theta$ with the vertical. Its time period of oscillations is $T$ in this position. Then

• A. $\tan \theta =\frac{\sigma q}{2{\epsilon }_{0}mg}$
• B. $\tan \theta =\frac{\sigma q}{{\epsilon }_{0}mg}$
• C. $T>2\pi \sqrt{\frac{1}{g}}$
• D. $T=2\pi \sqrt{\frac{1}{g}}$

Answer 1

Answer: (A)

Electric intensity at $B$ due to sheet of charge,
$E=\frac{1}{2}\frac{\sigma q}{{\epsilon }_0}$
Force on the bob due to sheet of charge
$F=qE=\frac{1}{2}\frac{\sigma q}{{\epsilon }_0}$
As, the bob is in equilibrium, so
$\therefore \frac{mg}{OC}=\frac{F}{CB}=\frac{T}{BO}$
$\therefore \tan \theta =\frac{CB}{OC}=\frac{F}{mg}$
$=\frac{\frac{1}{2}\sigma q/{\epsilon }_0}{mg}=\frac{\sigma q}{2{\epsilon }_{0}mg}$