Explanation

(d) Electric field due to a charged conducting sheet of surface charge density $\sigma$ is given by
$E = \frac{\sigma}{\varepsilon_{0}\varepsilon_{r}}$
Where, $\varepsilon_{0}$=permittivity in vacuum and $\varepsilon_{r}$relative permittivity of medium.
Here, electrostatics force on $B$
$\mathcal{Q}E = \frac{\mathcal{Q}\sigma}{\varepsilon_{0}\varepsilon_{r}}$
FBD of $B$ is shown in figure.
[image:1_image407.emf]
In equilibrium, $T\ \cos\ \theta = mg$
And $T\ \sin\ \theta = \frac{\mathcal{Q}\sigma}{\varepsilon_{0}\varepsilon_{r}}$
Thus, $\tan{\theta = \frac{\mathcal{Q}\sigma}{\varepsilon_{0}\varepsilon_{r}mg}}$
$\therefore\ \ \tan{\theta \propto \sigma}\ $
$or\ \ \ \sigma \propto \tan\theta$