Question

A simple pendulum of length $l$ has a bob of mass $m$ with a charge $q$ on it. A vertical sheet with surface charge density $\sigma$ passes through the point of suspension $O$. At equilibrium, the string makes an angle $\theta$ with the vertical, then -

• A. $\tan \theta =\frac{\sigma q}{2{ϵ}_{0}mg}$
• B. $\tan \theta =\frac{\sigma q}{3{ϵ}_{0}mg}$
• C. $\tan \theta =\frac{2\sigma q}{{ϵ}_{0}mg}$
• D. $\tan \theta =\frac{3\sigma q}{{ϵ}_{0}mg}$

Answer 1

The correct option is A $\tan \theta =\frac{\sigma q}{2{ϵ}_{0}mg}$


We know that electric field due to an infinite sheet $\left(E\right)=\frac{\sigma }{2{ϵ}_0}$.

From the equilibrium,

$T\sin \theta =qE...\left(1\right)$

$T\cos \theta =mg...\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$,

$\tan \theta =\frac{qE}{mg}$

$\Rightarrow \tan \theta =\frac{q\left(\frac{\sigma }{2{ϵ}_0}\right)}{mg}$

$\Rightarrow \tan \theta =\frac{\sigma q}{2{ϵ}_{0}mg}$