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 vectical charged conductinb 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 of finding the surface charge density of a vertical charged conducting plate that keeps a small sphere of mass \( m \) and charge \( q \) in equilibrium at an angle \( \theta \), we can follow these steps:

### Step 1: Understand the Forces Acting on the Sphere
The sphere experiences three main forces:
1. The gravitational force \( \vec{F_g} = mg \) acting downward.
2. The tension force \( \vec{T} \) in the thread, which acts at an angle \( \theta \) to the vertical.
3. The electric force \( \vec{F_e} = qE \) due to the electric field \( E \) created by the charged plate.

### Step 2: Resolve the Tension Force
The tension \( T \) can be resolved into two components:
- The vertical component: \( T \cos \theta = mg \)
- The horizontal component: \( T \sin \theta = qE \)

### Step 3: Calculate the Electric Field
For an infinite charged conducting plate, the electric field \( E \) due to the surface charge density \( \sigma \) is given by:
\[ E = \frac{\sigma}{2\epsilon_0} \]
where \( \epsilon_0 \) is the permittivity of free space.

### Step 4: Substitute Electric Field into the Horizontal Force Equation
From the horizontal component of the tension, we have:
\[ T \sin \theta = qE \]
Substituting for \( E \):
\[ T \sin \theta = q \left(\frac{\sigma}{2\epsilon_0}\right) \]

### Step 5: Substitute Tension from the Vertical Force Equation
From the vertical component of the tension, we have:
\[ T = \frac{mg}{\cos \theta} \]
Substituting this into the horizontal force equation gives:
\[ \frac{mg}{\cos \theta} \sin \theta = q \left(\frac{\sigma}{2\epsilon_0}\right) \]

### Step 6: Simplify the Equation
This simplifies to:
\[ mg \tan \theta = \frac{q \sigma}{2\epsilon_0} \]

### Step 7: Solve for Surface Charge Density \( \sigma \)
Rearranging the equation to solve for \( \sigma \):
\[ \sigma = \frac{2\epsilon_0 mg \tan \theta}{q} \]

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