Two small equal point charges of magnitude q are suspended from a common point on the ceiling by insulating mass less strings of equal lengths. They come to equilibrium with each string making angle `theta` from the vertical. If the mass of each charge is m, then the electrostatic potential at the centre of line joining the will be `(1/(4pi epsilion_(0))=k)`

Options

A. `2sqrt(kmgtan theta)`
B. `sqrt(k mg tan theta)`
C. `4sqrt(kmg//tan theta)`
D. `3sqrt(kmg//tan theta)`

Explanation

To solve the problem, we need to find the electrostatic potential at the center of the line joining two equal point charges suspended from a common point. Here’s a step-by-step solution:

### Step 1: Understand the Setup
We have two equal point charges \( q \) suspended from a common point by massless strings, making an angle \( \theta \) with the vertical. Each charge has a mass \( m \).

### Step 2: Analyze Forces Acting on the Charges
When the charges are in equilibrium, the forces acting on each charge include:
- The gravitational force \( mg \) acting downward.
- The tension \( T \) in the string acting along the string.

The tension can be resolved into two components:
- Vertical component: \( T \cos \theta \)
- Horizontal component: \( T \sin \theta \)

### Step 3: Set Up Equations for Equilibrium
For vertical equilibrium:
\[ T \cos \theta = mg \quad \text{(1)} \]

For horizontal equilibrium, the electrostatic force \( F \) between the two charges must balance the horizontal component of the tension:
\[ F = T \sin \theta \quad \text{(2)} \]

### Step 4: Calculate the Electrostatic Force
The electrostatic force \( F \) between the two charges \( q \) separated by a distance \( r \) is given by Coulomb's law:
\[ F = \frac{1}{4\pi \epsilon_0} \frac{q^2}{r^2} \]

### Step 5: Relate Distance \( r \) to Angle \( \theta \)
The distance \( r \) between the two charges can be expressed in terms of the angle \( \theta \) and the length of the strings \( L \):
\[ r = 2L \sin \theta \]

### Step 6: Substitute \( r \) into the Electrostatic Force Equation
Substituting \( r \) into the electrostatic force equation gives:
\[ F = \frac{1}{4\pi \epsilon_0} \frac{q^2}{(2L \sin \theta)^2} = \frac{1}{4\pi \epsilon_0} \frac{q^2}{4L^2 \sin^2 \theta} = \frac{q^2}{16\pi \epsilon_0 L^2 \sin^2 \theta} \]

### Step 7: Equate Forces from Equations (1) and (2)
From equations (1) and (2), we have:
\[ T \sin \theta = \frac{q^2}{16\pi \epsilon_0 L^2 \sin^2 \theta} \]
Substituting \( T \) from equation (1):
\[ \frac{mg}{\cos \theta} \sin \theta = \frac{q^2}{16\pi \epsilon_0 L^2 \sin^2 \theta} \]
This simplifies to:
\[ mg \tan \theta = \frac{q^2}{16\pi \epsilon_0 L^2 \sin^2 \theta} \]

### Step 8: Solve for \( q^2 \)
Rearranging gives:
\[ q^2 = 16\pi \epsilon_0 L^2 mg \tan \theta \sin^2 \theta \]

### Step 9: Calculate the Electrostatic Potential \( V \)
The electrostatic potential \( V \) at the center of the line joining the two charges is given by:
\[ V = k \left( \frac{q}{r_1} + \frac{q}{r_2} \right) \]
Since \( r_1 = r_2 = \frac{r}{2} \):
\[ V = 2k \frac{q}{\frac{r}{2}} = \frac{4kq}{r} \]

### Step 10: Substitute \( r \) and \( q \)
Substituting \( r = 2L \sin \theta \) and \( q^2 \) from earlier:
\[ V = \frac{4kq}{2L \sin \theta} = \frac{2kq}{L \sin \theta} \]

### Final Expression for Electrostatic Potential
Substituting \( q \) from the earlier expression:
\[ V = \frac{2k \sqrt{16\pi \epsilon_0 L^2 mg \tan \theta \sin^2 \theta}}{L \sin \theta} \]

### Conclusion
Thus, the electrostatic potential at the center of the line joining the two charges is given by:
\[ V = 4 \sqrt{km g \tan \theta} \]