Question

Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle θ with each other. When suspended in water the angle remains the same. If density of the material of the sphere is 1.5 g/cc, the dielectric constant of water will be ________ (Take density of water = 1 g/cc)

Answer 1

Correct Answer: 3

Consider the forces acting on the spheres in air and water:

\(\tan \frac{\theta}{2} = \frac{F}{mg} = \frac{q^2}{4 \pi \epsilon_0 r^2 mg}\)

When suspended in water:

\(\tan \frac{\theta}{2} = \frac{F'}{mg'} = \frac{q^2}{4 \pi \epsilon_0 \epsilon_r r^2 m g_{\text{eff}}}\)

Equating both expressions and using the relation:

\(\epsilon_r g = \epsilon_0 \epsilon_r g \left[ 1 - \frac{1}{1.5} \right]\)

Simplifying gives:

\(\epsilon_r = 3\)

Answer 2

This problem involves analyzing the equilibrium of two charged spheres suspended by strings, first in air and then in water. The goal is to find the dielectric constant of water, given that the angle between the strings remains the same in both media.

Concept Used:

The solution involves the principle of static equilibrium, where the net force on each sphere is zero. We will analyze the forces acting on one of the spheres in both scenarios (air and water).

In Air: The forces are Tension (\(T\)), Weight (\(mg\)), and Electrostatic Force (\(F_e\)).

In Water: The forces are Tension (\(T'\)), Weight (\(mg\)), Buoyant Force (\(F_b\)), and the new Electrostatic Force (\(F_e'\)).

The electrostatic force in a medium with dielectric constant \(K\) is given by:

\[ F_e' = \frac{F_e}{K} \]

The buoyant force is given by Archimedes' principle:

\[ F_b = V \rho_{\text{water}} g \]

where \(V\) is the volume of the sphere and \(\rho_{\text{water}}\) is the density of water.

The weight of the sphere is:

\[ mg = V \rho_{\text{sphere}} g \]

In equilibrium, the horizontal and vertical components of the forces must balance. For a single sphere, if the string makes an angle \(\alpha\) with the vertical (here \(\alpha = \theta/2\)), the condition for equilibrium is:

\[ \tan(\alpha) = \frac{\text{Horizontal Force}}{\text{Vertical Force}} \]

Step-by-Step Solution:

Step 1: Analyze the equilibrium in air.

Let each string have length \(L\), and the angle between them be \(\theta\). The angle each string makes with the vertical is \(\alpha = \theta/2\). The forces acting on one sphere are:

• Weight \(mg\) acting vertically downwards.
• Electrostatic force \(F_e\) acting horizontally.
• Tension \(T\) acting along the string.

Resolving the tension into components, for equilibrium:

\[ T \cos(\theta/2) = mg \quad \cdots(1) \] \[ T \sin(\theta/2) = F_e \quad \cdots(2) \]

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

\[ \tan(\theta/2) = \frac{F_e}{mg} \quad \cdots(\text{A}) \]

Step 2: Analyze the equilibrium in water.

When the system is submerged in water, the angle \(\theta\) remains the same. The forces on one sphere are:

• Weight \(mg\) acting vertically downwards.
• Buoyant force \(F_b\) acting vertically upwards.
• Electrostatic force in water, \(F_e' = F_e/K\), acting horizontally.
• New tension \(T'\) acting along the string.

The net vertical force is the apparent weight, \(W_{\text{app}} = mg - F_b\).

For equilibrium in water:

\[ T' \cos(\theta/2) = mg - F_b \quad \cdots(3) \] \[ T' \sin(\theta/2) = F_e' \quad \cdots(4) \]

Dividing equation (4) by (3), we get:

\[ \tan(\theta/2) = \frac{F_e'}{mg - F_b} \quad \cdots(\text{B}) \]

Step 3: Equate the conditions for equilibrium.

Since the angle \(\theta\) is the same in both cases, \(\tan(\theta/2)\) is the same. From equations (A) and (B):

\[ \frac{F_e}{mg} = \frac{F_e'}{mg - F_b} \]

Substitute \(F_e' = F_e/K\):

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

Canceling \(F_e\) from both sides, we get:

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

Step 4: Express forces in terms of densities.

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

\[ mg = V \rho_s g \] \[ F_b = V \rho_w g \]

Substitute these into the expression for \(K\):

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

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

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

Final Computation & Result:

Step 5: Substitute the given numerical values.

Density of the material of the sphere, \(\rho_s = 1.5 \text{ g/cc}\).

Density of water, \(\rho_w = 1 \text{ g/cc}\).

\[ K = \frac{1.5}{1.5 - 1} = \frac{1.5}{0.5} \] \[ K = 3 \]

The dielectric constant of water is 3.