If a small sphere of mass m and charge q is hanged from a silk thread at an angle `theta` with the surface of a vertical charged conducting plate, then for equilibrium of sphere the surface charge density of the plate -

Options

A. `(epsilon_(0)mg tan theta)/(q)`
B. `(2epsilon_(0)mg tan theta)`/q
C. `epsilon_(0)mg q tan theta`
D. `(epsilon_(0)mg tan theta)/(3q)`

Answer

To solve the problem, we need to analyze the forces acting on the small sphere of mass \( m \) and charge \( q \) that is suspended from a silk thread at an angle \( \theta \) with respect to a vertical charged conducting plate. The goal is to find the surface charge density \( \sigma \) of the plate for the sphere to be in equilibrium.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Sphere:**
- The weight of the sphere, \( W = mg \), acting downward.
- The tension \( T \) in the thread, which can be resolved into two components:
- Vertical component: \( T \cos \theta \)
- Horizontal component: \( T \sin \theta \)
- The electric force \( F_e \) acting on the sphere due to the electric field \( E \) from the charged plate, given by \( F_e = qE \).

2. **Determine the Electric Field due to the Charged Plate:**
- For a charged conducting plate, the electric field \( E \) is given by:
\[ E = \frac{\sigma}{2 \epsilon_0} \]
where \( \sigma \) is the surface charge density and \( \epsilon_0 \) is the permittivity of free space.

3. **Set Up the Equilibrium Conditions:**
- For vertical forces:
\[ T \cos \theta = mg \quad \text{(1)} \]
- For horizontal forces:
\[ T \sin \theta = qE \quad \text{(2)} \]

4. **Substitute the Electric Field into the Horizontal Force Equation:**
- From equation (2), substituting \( E \):
\[ T \sin \theta = q \left(\frac{\sigma}{2 \epsilon_0}\right) \]
- Rearranging gives:
\[ T \sin \theta = \frac{q \sigma}{2 \epsilon_0} \quad \text{(3)} \]

5. **Divide Equation (3) by Equation (1):**
- Dividing equation (3) by equation (1):
\[ \frac{T \sin \theta}{T \cos \theta} = \frac{\frac{q \sigma}{2 \epsilon_0}}{mg} \]
- This simplifies to:
\[ \tan \theta = \frac{q \sigma}{2 \epsilon_0 mg} \]

6. **Solve for Surface Charge Density \( \sigma \):**
- Rearranging the equation gives:
\[ \sigma = \frac{2 \epsilon_0 mg \tan \theta}{q} \]

### Final Expression:
Thus, the surface charge density \( \sigma \) of the plate is given by:
\[ \sigma = \frac{2 \epsilon_0 mg \tan \theta}{q} \]