Question

Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of 37° with each other. When suspended in a liquid of density 0.7 g/cm³, the angle remains the same. If the density of the material of the sphere is 1.4 g/cm³, the dielectric constant of the liquid is ___________. (tan 37° =\( \frac{3}{4}\))

Answer 1

Correct Answer: 2

The gravitational force acting on the spheres is given by:

\( F_g = \rho_B V g \),

where \( \rho_B \) is the density of the spheres, \( V \) is the volume of the sphere, and \( g \) is the acceleration due to gravity.

The electric force acting between the spheres is:

\( F_e = \frac{Q^2}{4 \pi \epsilon_0 d^2} \),

where \( Q \) is the charge on the spheres, \( d \) is the distance between the centers of the spheres, and \( \epsilon_0 \) is the permittivity of free space.

The buoyant force acting on the spheres in the liquid is:

\( F_b = \rho_L V g \),

where \( \rho_L \) is the density of the liquid.

In equilibrium, the tension in the strings balances both the horizontal and vertical forces. The vertical and horizontal components of the tension are:

\( T \cos \theta = F_g - F_b \),

\( T \sin \theta = F_e \).

The ratio of these equations gives:

\( \tan \theta = \frac{F_e}{F_g - F_b} \).

For the spheres in the liquid, substituting the expressions for \( F_g \) and \( F_b \):

\( \tan \theta = \frac{F_e}{(\rho_B - \rho_L) V g} \).

When the spheres are in air, the buoyant force is negligible, and the force balance becomes:

\( \tan \theta = \frac{F_e}{\rho_B V g} \).

Equating the two expressions for \( \tan \theta \), we get:

\( \frac{F_e}{\rho_B V g} = \frac{F_e}{(\rho_B - \rho_L) V g} \).

Simplifying:

\( \rho_B V g = (\rho_B - \rho_L) V g \).

Cancelling \( V g \) from both sides:

\( \rho_B = \rho_B - \rho_L \Rightarrow \rho_B - \rho_L = k \cdot \rho_B \).

Substituting the given values:

\( \rho_B = 1.4 \, g/cm^3 \), \( \rho_L = 0.7 \, g/cm^3 \).

From the ratio:

\( k = \frac{\rho_B}{\rho_L} = \frac{1.4}{0.7} = 2 \).

Final Answer: The dielectric constant of the liquid is: \( k = 2 \).

Answer 2

The problem describes a classic electrostatic equilibrium setup with two identical charged spheres suspended by strings. Initially, they are in equilibrium in air, and then they are submerged in a liquid. The angle between the strings remains unchanged. We are asked to find the dielectric constant of the liquid, given the densities of the liquid and the sphere's material.

Concept Used:

The solution involves analyzing the forces acting on one of the charged spheres in two different scenarios: in air and in the liquid. The sphere is in static equilibrium in both cases. The key concepts are:

1. Condition for Equilibrium: The net force on the sphere is zero. We resolve the forces into horizontal and vertical components and set their sums to zero.

2. Forces in Air: - Tension (T) - Gravitational force (Weight, \( W = mg \)) - Electrostatic force (\( F_e = \frac{1}{4\pi\epsilon_0} \frac{q^2}{r^2} \))

3. Forces in Liquid: - Tension (T') - Gravitational force (\( W = mg \)) - Buoyant force (\( F_b = V\rho_{l}g \), where \( V \) is the volume of the sphere and \( \rho_{l} \) is the density of the liquid). The net downward force is the apparent weight \( W_{app} = W - F_b \). - Electrostatic force in a dielectric medium (\( F'_e = \frac{F_e}{K} \), where K is the dielectric constant of the liquid).

By analyzing the equilibrium in both cases, we can establish a relationship to find the dielectric constant K.

Step-by-Step Solution:

Step 1: Analyze the equilibrium of a sphere in air.

Let \( 2\theta \) be the angle between the strings, so each string makes an angle \( \theta \) with the vertical. The forces acting on one sphere are Tension (T), Weight (\(mg\)), and electrostatic repulsion (\(F_e\)).

For vertical equilibrium:

\[ T \cos\theta = mg \quad \cdots(1) \]

For horizontal equilibrium:

\[ T \sin\theta = F_e \quad \cdots(2) \]

Dividing equation (2) by (1), we get:

\[ \tan\theta = \frac{F_e}{mg} \quad \cdots(3) \]

Step 2: Analyze the equilibrium of the sphere in the liquid.

The angle \( \theta \) remains the same. The new forces are Tension (T'), Weight (\(mg\)), Buoyant force (\(F_b\)), and the new electrostatic force (\(F'_e\)).

The buoyant force acts upwards, opposing the weight. The net vertical force balanced by the tension component is the apparent weight, \( W_{app} = mg - F_b \).

For vertical equilibrium:

\[ T' \cos\theta = mg - F_b \quad \cdots(4) \]

The electrostatic force in the liquid is reduced by the dielectric constant K, so \( F'_e = F_e/K \).

For horizontal equilibrium:

\[ T' \sin\theta = F'_e = \frac{F_e}{K} \quad \cdots(5) \]

Dividing equation (5) by (4), we get:

\[ \tan\theta = \frac{F_e/K}{mg - F_b} \quad \cdots(6) \]

Step 3: Equate the expressions for \( \tan\theta \) from both cases.

Since the angle \( \theta \) is the same in air and in the liquid, we can equate equations (3) and (6):

\[ \frac{F_e}{mg} = \frac{F_e/K}{mg - F_b} \]

Canceling \( F_e \) from both sides (since \( F_e \neq 0 \)):

\[ \frac{1}{mg} = \frac{1}{K(mg - F_b)} \]

Rearranging to solve for K:

\[ K = \frac{mg}{mg - F_b} \]

Step 4: Express weight and buoyant force in terms of densities.

Let V be the volume of the sphere, \( \rho_s \) be the density of the sphere's material, and \( \rho_l \) be the density of the liquid.

Weight of the sphere: \( mg = V \rho_s g \)

Buoyant force: \( F_b = V \rho_l g \)

Substitute these expressions into the equation for K:

\[ K = \frac{V \rho_s g}{V \rho_s g - V \rho_l g} \]

Cancel the common term \( Vg \) from the numerator and denominator:

\[ K = \frac{\rho_s}{\rho_s - \rho_l} \]

Step 5: Substitute the given numerical values to find K.

Given:

Density of the sphere's material, \( \rho_s = 1.4 \, \text{g/cm}^3 \)

Density of the liquid, \( \rho_l = 0.7 \, \text{g/cm}^3 \)

\[ K = \frac{1.4}{1.4 - 0.7} = \frac{1.4}{0.7} \] \[ K = 2 \]

The dielectric constant of the liquid is 2.