1. Electric field due to a large charged conducting sheet

   For a large charged conducting sheet with surface charge density $\sigma$, the electric field just outside the sheet is $E=\frac{\sigma}{\varepsilon_0}$ and it is perpendicular to the sheet.

2. Forces on the charged ball

   Let the charge on the ball be $q$ and its mass be $m$.

   The forces acting on the ball are:

   • Weight: $mg$ vertically downward
   • Electric force: $qE$ perpendicular to the sheet
   • Tension: $T$ along the string

   Since the sheet is vertical in the figure, the electric field is horizontal. The string makes angle $\theta$ with the sheet, so it also makes angle $\theta$ with the vertical.

3. Resolve forces

   In equilibrium:

   • Horizontal balance: $T\sin\theta = qE$

   • Vertical balance: $T\cos\theta = mg$

4. Take ratio

   Dividing the two equations, $\frac{T\sin\theta}{T\cos\theta} = \frac{qE}{mg}$ $\tan\theta = \frac{qE}{mg}$

   Hence, $E = \frac{mg}{q}\tan\theta$

5. Relate $\sigma$ to $\theta$

   Since $E=\frac{\sigma}{\varepsilon_0},$ we get $\sigma = \varepsilon_0 E = \varepsilon_0\frac{mg}{q}\tan\theta$

   Therefore, $\sigma \propto \tan\theta$

6. Check options

   • A: $\cot\theta$ ❌
   • B: $\cos\theta$ ❌
   • C: $\tan\theta$ ✅
   • D: $\sin\theta$ ❌

Final Answer: Option C