Q. A simple pendulum of length $l$ has a bob of mass $m,$ with a charge $q$ on it. A vertical sheet of change, 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) $\tan\ \theta = \frac{\sigma q}{2\varepsilon_{0}mg}$
B) $\tan\ \theta = \frac{\sigma q}{\varepsilon_{0}mg}$
C) $T > 2\pi\sqrt{\frac{1}{g}}$
D) $T = 2\pi\sqrt{\frac{1}{g}}$
Explanation
(a) Electric intensity at $B$ due to sheet of charge,
$E = \frac{1}{2}\frac{\sigma}{\varepsilon_{0}}$
force on the bob due to sheet of charge
$F = qE = \frac{1}{2}\frac{\sigma q}{\varepsilon_{0}}$
As the bob is in equilibrium, so
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$\frac{mg}{OC} = \frac{F}{CB} = \frac{T}{BO}$
$\tan{\theta = \frac{CB}{OC}} = \frac{F}{mg}$
$= \frac{\frac{1}{2}\sigma q/\varepsilon_{0}}{mg} = \frac{\sigma q}{{2\varepsilon}_{0}mg}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.