A charged ball $$B$$ hangs from a silk thread $$S$$ which makes an angle $$\theta$$ with a large charged conducting sheet $$P$$, as show in the figure. The surface charge density $$\sigma$$ of the sheet is proportional to

Let the mass of the charged ball $$B$$ be $$m$$ and its charge be $$q$$.

The electric field $$E$$ produced by a large charged conducting sheet $$P$$ with surface charge density $$\sigma$$ is given by:

$$E = \frac{\sigma}{\epsilon_0}$$

The forces acting on the charged ball are:

1. The weight of the ball, $$mg$$, acting vertically downwards.

2. The electric force, $$F_e = qE$$, acting horizontally away from the sheet.

3. The tension $$T$$ in the silk thread $$S$$.

In equilibrium, the thread makes an angle $$\theta$$ with the vertical sheet. We can resolve the tension $$T$$ into horizontal and vertical components:

The vertical component balances the weight:

$$T \cos\theta = mg$$

The horizontal component balances the electric force:

$$T \sin\theta = qE$$

Dividing the horizontal equilibrium equation by the vertical equilibrium equation, we get:

$$\frac{T \sin\theta}{T \cos\theta} = \frac{qE}{mg}$$

$$\Rightarrow \quad \tan\theta = \frac{qE}{mg}$$

Now, substitute the value of the electric field $$E$$ into the equation:

$$\tan\theta = \frac{q \sigma}{\epsilon_0 mg}$$

Rearranging the equation to solve for $$\sigma$$, we find:

$$\sigma = \left( \frac{\epsilon_0 mg}{q} \right) \tan\theta$$

Since $$\epsilon_0$$, $$m$$, $$g$$, and $$q$$ are all constants for this system, we can conclude that the surface charge density $$\sigma$$ is directly proportional to $$\tan\theta$$:

$$\sigma \propto \tan\theta$$