Question

A small charged particle of mass $m$ and charge $q$ is suspended by an insulated thread in front of a very large conducting charged sheet of uniform surface density of charge $\sigma$. The angle made by the thread with the vertical in equilibrium is

• A. ${\tan }^{-1}\left(\frac{\sigma q}{2{\epsilon }_{0}mg}\right)$
• B. ${\tan }^{-1}\left(\frac{\sigma }{q{\epsilon }_{0}mg}\right)$
• C. ${\tan }^{-1}\left(\frac{q}{2\sigma {\epsilon }_{0}mg}\right)$
• D. zero

Answer 1

The correct option is A ${\tan }^{-1}\left(\frac{\sigma q}{2{\epsilon }_{0}mg}\right)$

X-axis:- $T\sin \theta =qE=\frac{q\sigma }{2{ϵ}_0}$--------------1 [Equilibrium $\Rightarrow forwardforce=backwardforce$]

Y-axis:- $T\cos \theta =mg$--------------2

Dividing Equation 1 by Equation 2

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

$\theta ={\tan }^{-1}\left(\frac{q\sigma }{2{ϵ}_{0}mg}\right)$