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. $\tan \theta=\frac{\sigma q}{2 \varepsilon_{0} m g}$
• B. $\tan \theta=\frac{\sigma q}{\varepsilon_{0} m g}$
• 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,
$E=\frac{1}{2} \frac{\sigma q}{\varepsilon_{0}}$
Force on the bob due to sheet of charge
$F=q E=\frac{1}{2} \frac{\sigma q}{\varepsilon_{0}}$
As, the bob is in equilibrium, so
$\therefore \frac{m g}{O C}=\frac{F}{C B}=\frac{T}{B O} $
$\therefore \tan \theta=\frac{C B}{O C}=\frac{F}{m g}$
$=\frac{\frac{1}{2} \sigma q / \varepsilon_{0}}{m g}=\frac{\sigma q}{2 \varepsilon_{0} m g}$