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. Here’s a step-by-step solution:

### Step 1: Analyze the forces acting on the spheres in air
When the two identical charged spheres are suspended in air, the forces acting on each sphere are:
- The gravitational force (weight) acting downwards: \( mg \)
- The tension in the string acting at an angle \( \theta/2 \)
- The electrostatic repulsive force between the two charged spheres

From the equilibrium of forces, we can write:
1. The vertical component of the tension balances the weight:
\[ T \cos\left(\frac{\theta}{2}\right) = mg \]
2. The horizontal component of the tension balances the electrostatic force:
\[ T \sin\left(\frac{\theta}{2}\right) = F_e \]

Where \( F_e \) is the electrostatic force given by Coulomb's law:
\[ F_e = k \frac{q^2}{r^2} \]

### Step 2: Relate the forces using tangent
From the above equations, we can derive:
\[ \tan\left(\frac{\theta}{2}\right) = \frac{F_e}{mg} = \frac{k \frac{q^2}{r^2}}{mg} \]

### Step 3: Analyze the forces acting on the spheres in the liquid
When the spheres are submerged in a liquid of density \( \sigma \), the forces acting on each sphere are:
- The gravitational force (weight) acting downwards: \( mg \)
- The buoyant force acting upwards: \( F_b = \sigma V g \) (where \( V \) is the volume of the sphere)
- The tension in the string acting at an angle \( \theta/2 \)

The equilibrium of forces gives us:
1. The vertical component of the tension:
\[ T \cos\left(\frac{\theta}{2}\right) = mg - F_b \]
2. The horizontal component of the tension still balances the electrostatic force:
\[ T \sin\left(\frac{\theta}{2}\right) = F_e' \]

Where \( F_e' \) is the electrostatic force in the liquid:
\[ F_e' = \frac{1}{K} k \frac{q^2}{r^2} \]

### Step 4: Set up the equations for the liquid case
From the equations for the liquid case, we can write:
\[ \tan\left(\frac{\theta}{2}\right) = \frac{F_e'}{mg - F_b} \]
Substituting \( F_b = \sigma V g \) (where \( V = \frac{4}{3} \pi r^3 \) for a sphere) gives:
\[ \tan\left(\frac{\theta}{2}\right) = \frac{\frac{1}{K} k \frac{q^2}{r^2}}{mg - \sigma V g} \]

### Step 5: Equate the two expressions for tangent
Since the angle \( \theta \) remains the same in both cases, we can equate the two expressions for \( \tan\left(\frac{\theta}{2}\right) \):
\[ \frac{k \frac{q^2}{r^2}}{mg} = \frac{\frac{1}{K} k \frac{q^2}{r^2}}{mg - \sigma V g} \]

### Step 6: Simplify and solve for \( K \)
Cross-multiplying and simplifying gives:
\[ mg - \sigma V g = \frac{mg K}{1} \]
Rearranging gives:
\[ K = \frac{mg}{mg - \sigma V g} \]

### Step 7: Substitute volume of the sphere
Substituting \( V = \frac{4}{3} \pi r^3 \):
\[ K = \frac{\rho}{\rho - \sigma} \]

Where \( \rho \) is the density of the material of the spheres.

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