Two identical balls each having a density `rho` are suspended from as common point by two insulating strings of equal length. Both the balls have equal mass and charge. In equilibrium each string makes an angle `theta` with vertical. Now, both the balls are immersed in a liquid. As a result the angle `theta` does not change. The density of the liquid is `sigma`. Find the dielectric constant of the liquid.

Explanation

To solve the problem, we need to analyze the forces acting on the two identical balls in both scenarios: when they are suspended in air and when they are immersed in a liquid.

### Step-by-Step Solution:

1. **Understanding the Initial Setup**:
- We have two identical balls, each with mass \( m \), density \( \rho \), and charge \( Q \).
- The balls are suspended by insulating strings of length \( l \), making an angle \( \theta \) with the vertical.
- In equilibrium, the forces acting on each ball must balance.

2. **Forces in Air**:
- The forces acting on each ball are:
- Weight \( W = mg = \rho V g \) (where \( V \) is the volume of the ball).
- Tension \( T \) in the string.
- Electrostatic force \( F_e \) due to the repulsion between the two charged balls.
- The vertical component of the tension balances the weight:
\[ T \cos \theta = \rho V g \quad \text{(1)} \]
- The horizontal component of the tension balances the electrostatic force:
\[ T \sin \theta = F_e \quad \text{(2)} \]

3. **Dividing the Forces**:
- Dividing equation (2) by equation (1):
\[ \frac{T \sin \theta}{T \cos \theta} = \frac{F_e}{\rho V g} \]
- This simplifies to:
\[ \tan \theta = \frac{F_e}{\rho V g} \quad \text{(3)} \]

4. **Forces in the Liquid**:
- When the balls are immersed in a liquid of density \( \sigma \), the buoyant force \( F_B \) acts upwards:
\[ F_B = \sigma V g \]
- The new tension in the string is \( T' \).
- The vertical component of the tension plus the buoyant force balances the weight:
\[ T' \cos \theta + \sigma V g = \rho V g \quad \text{(4)} \]
- The horizontal component of the tension still balances the electrostatic force, but now it is modified by the dielectric constant \( E_r \):
\[ T' \sin \theta = \frac{F_e}{E_r} \quad \text{(5)} \]

5. **Dividing the Forces in Liquid**:
- Dividing equation (5) by equation (4):
\[ \frac{T' \sin \theta}{T' \cos \theta + \sigma V g} = \frac{F_e}{E_r (\rho V g - \sigma V g)} \]
- This simplifies to:
\[ \tan \theta = \frac{F_e}{E_r (\rho - \sigma) g} \quad \text{(6)} \]

6. **Equating the Two Scenarios**:
- From equations (3) and (6):
\[ \frac{F_e}{\rho V g} = \frac{F_e}{E_r (\rho - \sigma) g} \]
- Canceling \( F_e \) and \( g \) from both sides:
\[ \frac{1}{\rho} = \frac{1}{E_r (\rho - \sigma)} \]
- Rearranging gives:
\[ E_r = \frac{\rho}{\rho - \sigma} \]

### Final Answer:
The dielectric constant of the liquid is given by:
\[ E_r = \frac{\rho}{\rho - \sigma} \]