Two small identical balls are suspened from a common point by two identical strings. When they are given identical charges, they move apar and the strings make with the vertical an angle of q. Now the system is immersed in a liquid and as a result the angle q does not change. If r is the density of the material of the balls and s is the density of the liquid, the dielectric constant of the liquid is

Options

A. `p/(p-sigma)`
B. `(p-sigma)/p`
C. `(sigma)/(p-sigma)`
D. `(p-sigma)/(sigma)`

Explanation

To solve the problem, we need to analyze the forces acting on the charged balls both in air and when submerged in a liquid. Here’s a step-by-step solution:

### Step 1: Analyze the forces acting on the balls in air
When the balls are charged and suspended, they repel each other and make an angle \( \alpha \) with the vertical. The forces acting on each ball are:
- The tension \( T \) in the string.
- The gravitational force \( mg \) acting downwards.
- The electrostatic force \( F_e \) acting horizontally due to the repulsion between the charged balls.

From the equilibrium of forces, we can write:
1. In the vertical direction:
\[ T \cos \alpha = mg \]
2. In the horizontal direction:
\[ T \sin \alpha = F_e = \frac{k q^2}{r^2} \]

### Step 2: Relate the forces to the angle
From the two equations above, we can express \( T \) in terms of \( mg \) and \( F_e \):
- From the vertical equation:
\[ T = \frac{mg}{\cos \alpha} \]
- From the horizontal equation:
\[ T = \frac{F_e}{\sin \alpha} = \frac{k q^2}{r^2 \sin \alpha} \]

### Step 3: Set the two expressions for tension equal
Equating the two expressions for tension \( T \):
\[ \frac{mg}{\cos \alpha} = \frac{k q^2}{r^2 \sin \alpha} \]
Rearranging gives:
\[ \tan \alpha = \frac{k q^2}{m g r^2} \]

### Step 4: Analyze the forces when submerged in the liquid
When the system is submerged in a liquid of density \( \sigma \), the forces acting on the balls change slightly due to the buoyant force. The new equilibrium conditions are:
1. In the vertical direction:
\[ T' \cos \alpha = mg - V \sigma g \]
2. In the horizontal direction:
\[ T' \sin \alpha = \frac{k q^2}{k' r^2} \]

### Step 5: Set the new expressions for tension equal
From the vertical direction:
\[ T' = \frac{mg - V \sigma g}{\cos \alpha} \]
From the horizontal direction:
\[ T' = \frac{k q^2}{k' r^2 \sin \alpha} \]
Setting these equal gives:
\[ \frac{mg - V \sigma g}{\cos \alpha} = \frac{k q^2}{k' r^2 \sin \alpha} \]

### Step 6: Solve for the dielectric constant \( k' \)
Equating the two expressions for \( T' \) and simplifying:
\[ mg - V \sigma g = \frac{k q^2}{k' r^2} \tan \alpha \]
Substituting \( V = \frac{m}{\rho} \) (volume in terms of density) gives:
\[ mg - \frac{m \sigma g}{\rho} = \frac{k q^2}{k' r^2} \tan \alpha \]
Rearranging leads to:
\[ \frac{1}{k'} = \frac{1}{\rho} - \frac{\sigma}{\rho} \]
Thus:
\[ k' = \frac{\rho}{\rho - \sigma} \]

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