A pith ball of mass 'm' gram and charge 'Q' is suspended using a mass less silk thread near a large charged conducting metal sheet of area 'A' and surface charged density '$\sigma$'. If the silk thread makes an angle $\Theta$ with the metal sheet, then:

The electric field near a large charged conducting metal sheet is given by $E = \dfrac{\sigma}{\epsilon_0}$.

The electric force acting on the pith ball is $F_e = QE = \dfrac{Q\sigma}{\epsilon_0}$.

The gravitational force acting downwards on the pith ball is $W = mg$.

Let $T$ be the tension in the silk thread. In equilibrium, the vertical and horizontal components of the tension balance the weight and the electric force respectively.

$T \cos\theta = mg$

$T \sin\theta = \dfrac{Q\sigma}{\epsilon_0}$

Dividing the second equation by the first equation, we get:

$\tan\theta = \dfrac{Q\sigma}{mg\epsilon_0}$

From this expression, it is evident that $\tan\theta \propto \sigma$.

Answer: $\tan\theta \propto \sigma$