Two identical spheres of same mass and specific gravity (which is the ratio of density of a substance and density of water) 2.4 have different charges of Q and -3Q. They are suspended from two stings of a same length l fixed to points at the same horizontal level, but distant l from each other. When the entire set up is transferred inside a liquid of specific gravity 0.8, it is observed that the inclination of each string in equilibrium remains unchanged. Then the dielectric constant of the liquid is

Answer

To solve the problem step by step, we will analyze the forces acting on the two spheres and how they behave when submerged in a liquid.

### Step 1: Understand the Given Information
- Two identical spheres have the same mass and specific gravity of 2.4.
- The charges on the spheres are \( Q \) and \( -3Q \).
- The specific gravity of the liquid is 0.8.
- The inclination of the strings remains unchanged when submerged in the liquid.

### Step 2: Calculate the Density of the Spheres
Specific gravity is defined as the ratio of the density of a substance to the density of water.
- Density of the sphere \( \rho_s = 2.4 \cdot \rho_w \) (where \( \rho_w \) is the density of water).

### Step 3: Identify the Forces Acting on the Spheres
When the spheres are in equilibrium, the forces acting on them are:
1. Gravitational force \( F_g = mg \)
2. Electric force \( F_e \) due to the charges
3. Tension in the string \( T \)

### Step 4: Establish Equations for Forces in Air
For sphere 1 (charge \( Q \)):
- Vertical component: \( T \cos \theta = mg \) (1)
- Horizontal component: \( T \sin \theta = F_e \) (2)

For sphere 2 (charge \( -3Q \)):
- The forces are similar, but the electric force will be in the opposite direction.

### Step 5: Relate the Electric Force to Charges
The electric force \( F_e \) between the two spheres can be expressed as:
\[ F_e = k \frac{|Q \cdot (-3Q)|}{r^2} = k \frac{3Q^2}{r^2} \]
where \( k \) is Coulomb's constant and \( r \) is the distance between the spheres.

### Step 6: Establish the Ratio of Forces
From equations (1) and (2):
\[ \tan \theta = \frac{F_e}{mg} \]
This is our first equation (A).

### Step 7: Analyze the Forces When Submerged in Liquid
When submerged in the liquid, the buoyant force \( F_b \) acts on the spheres:
\[ F_b = V \cdot \rho_l \cdot g \]
where \( \rho_l \) is the density of the liquid.

### Step 8: Establish New Equations for Forces in Liquid
For the spheres in the liquid:
- Vertical component: \( T' \cos \theta = mg - F_b \) (3)
- Horizontal component: \( T' \sin \theta = \frac{F_e}{\epsilon_r} \) (4)

### Step 9: Relate the Two Equations
From equations (3) and (4):
\[ \tan \theta = \frac{F_e / \epsilon_r}{mg - F_b} \]
This is our second equation (B).

### Step 10: Set Equations A and B Equal
Since the angle \( \theta \) remains unchanged, we can set the two expressions for \( \tan \theta \) equal:
\[ \frac{F_e}{mg} = \frac{F_e / \epsilon_r}{mg - F_b} \]

### Step 11: Simplify the Equation
Cancelling \( F_e \) from both sides (assuming \( F_e \neq 0 \)):
\[ \frac{1}{mg} = \frac{1 / \epsilon_r}{mg - F_b} \]

### Step 12: Substitute for Buoyant Force
Substituting \( F_b = V \cdot \rho_l \cdot g \) and using the volume of the sphere \( V = \frac{4}{3} \pi r^3 \):
\[ F_b = \frac{4}{3} \pi r^3 \cdot \rho_l \cdot g \]

### Step 13: Solve for the Dielectric Constant
After substituting and simplifying, we find:
\[ 1 = \epsilon_r \left(1 - \frac{\rho_l}{\rho_s}\right) \]
Substituting the specific gravities:
\[ 1 = \epsilon_r \left(1 - \frac{0.8 \rho_w}{2.4 \rho_w}\right) \]
This simplifies to:
\[ 1 = \epsilon_r \left(1 - \frac{1}{3}\right) = \epsilon_r \cdot \frac{2}{3} \]
Thus:
\[ \epsilon_r = \frac{3}{2} = 1.5 \]

### Final Answer
The dielectric constant of the liquid is \( \epsilon_r = 1.5 \).

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