Q.
A charged ball $B$ hangs from a silk thread $S$, which makes an angle $\theta$ with a large charged conducting sheet $P$, as shown in the figure. The surface charge density $\sigma$ of the sheet is proportional to
Chhattisgarh PMTChhattisgarh PMT 2006
Solution:
In equilibrium
$T \cos \theta=m g$ ...(i) and $T \sin \theta=F=E q$
$T \sin \theta=F =E q$ ...(ii)
$\therefore \frac{T \sin \theta}{T \cos \theta}=\frac{\sigma q}{2 \varepsilon_{0} \times m g}$
$\tan \theta=\frac{\sigma q}{2 \varepsilon_{0} m g}$
or $\sigma \propto \tan \theta$
