Two identical charged spheres of material density `rho`, suspended from the same point by inextensible strings of equal length make an angle `theta` between the string. When suspended in a liquid of density `sigma` the angle `theta` remains the same. The dielectric constant K of the liquid is

Options

A. `(rho)/(rho - sigma)`
B. `(rho - sigma)/(rho)`
C. `(rho)/(rho + sigma)`
D. `(rho + sigma)/(rho)`

Answer

To solve the problem, we need to analyze the forces acting on the charged spheres when they are suspended in air and then in a liquid.

### Step 1: Analyze the forces acting on the spheres in air
When the two charged spheres are suspended in air, they experience:
- Gravitational force acting downwards: \( F_g = mg = V \rho g \)
- Electrostatic force acting horizontally due to repulsion between the two charged spheres: \( F_e = \frac{k \cdot q^2}{r^2} \)

Where:
- \( m \) is the mass of the sphere,
- \( V \) is the volume of the sphere,
- \( \rho \) is the density of the sphere,
- \( g \) is the acceleration due to gravity,
- \( q \) is the charge on each sphere,
- \( r \) is the distance between the centers of the spheres.

### Step 2: Set up the equilibrium condition in air
In equilibrium, the horizontal and vertical forces must balance. The vertical component of the tension \( T \) in the strings must balance the weight of the spheres, and the horizontal component must balance the electrostatic force.

Let \( T \) be the tension in the strings. The angle \( \theta \) gives us:
- Vertical component: \( T \cos(\theta) = mg \)
- Horizontal component: \( T \sin(\theta) = F_e \)

### Step 3: Analyze the forces acting on the spheres in the liquid
When the spheres are submerged in a liquid of density \( \sigma \), they experience:
- Gravitational force acting downwards: \( F_g' = mg - V \sigma g \) (the buoyant force reduces the effective weight)
- The electrostatic force remains the same as in air.

### Step 4: Set up the equilibrium condition in the liquid
In the liquid, the equilibrium conditions are similar:
- Vertical component: \( T' \cos(\theta) = mg - V \sigma g \)
- Horizontal component: \( T' \sin(\theta) = F_e \)

### Step 5: Relate the two cases
Since the angle \( \theta \) remains the same in both cases, we can equate the ratios of the forces:
\[ \frac{T \sin(\theta)}{T \cos(\theta)} = \frac{F_e}{mg} \quad \text{(in air)} \]
\[ \frac{T' \sin(\theta)}{T' \cos(\theta)} = \frac{F_e}{mg - V \sigma g} \quad \text{(in liquid)} \]

### Step 6: Simplify the equations
From the above equations, we can derive:
\[ \frac{F_e}{mg} = \frac{F_e}{mg - V \sigma g} \]

### Step 7: Solve for the dielectric constant \( K \)
From the balance of forces, we can express the dielectric constant \( K \) as:
\[ K = \frac{\rho}{\rho - \sigma} \]

### Final Answer
Thus, the dielectric constant \( K \) of the liquid is given by:
\[ K = \frac{\rho}{\rho - \sigma} \]