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 charge density $\sigma$ . The angle made by the thread with the vertical in equilibrium is:

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

Answer 1

The correct option is D

$ta{n}^{-1}\left(\frac{\sigma q}{2{ϵ}_{0}mg}\right)$

Tension in the string is along it.

Resolving, we get,

$Tsin\theta =qE=\frac{q\sigma }{2{ϵ}_0}$

$Tcos\theta =mg$

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