Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of 37^ with each othe

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/cm^3, the angle remains the…

Accepted Answer

Given: Two identical charged spheres are suspended symmetrically. The angle between the strings remains the same in air and in a liquid. Density of sphere material is $\rho_s = 1.4 \, \text{g/cm}^3$ and density of liquid is $\rho_l = 0.7 \, \text{g/cm}^3$ .

Find: The dielectric constant $K$ of the liquid.

Let the angle each string makes with the vertical be $\theta$ . Since the angle between the strings is $37^\circ$ , the same $\theta$ is maintained in both situations.

For equilibrium in air:

T \cos\theta = mg

T \sin\theta = F_e

Dividing,

\tan\theta = \frac{F_e}{mg}

For equilibrium in the liquid, buoyancy reduces the effective weight and the electrostatic force becomes $F_e/K$ :

T' \cos\theta = mg - F_b

T' \sin\theta = \frac{F_e}{K}

So,

\tan\theta = \frac{F_e/K}{mg - F_b}

Since the angle remains the same, the two expressions for $\tan\theta$ are equal:

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

Cancelling $F_e$ ,

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

Hence,

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

Now write weight and buoyant force in terms of densities:

mg = V\rho_s g

F_b = V\rho_l g

Substituting,

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

Using the given values:

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

Therefore, the dielectric constant of the liquid is $2$ . The correct option is B.

Given: The angular separation does not change when the spheres are immersed in the liquid.

Find: The value of dielectric constant $K$ .

The unchanged angle means the ratio of horizontal force to vertical force remains unchanged.

In air:

\frac{\text{electrostatic force}}{\text{weight}} = \frac{F_e}{mg}

In liquid:

\frac{\text{electrostatic force in medium}}{\text{apparent weight}} = \frac{F_e/K}{mg - F_b}

Equating these because the angle is the same:

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

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

Now,

mg = V\rho_s g, \qquad F_b = V\rho_l g

Therefore,

K = \frac{V\rho_s g}{V(\rho_s - \rho_l)g} = \frac{\rho_s}{\rho_s - \rho_l}

Substitute $\rho_s = 1.4 \, \text{g/cm}^3$ and $\rho_l = 0.7 \, \text{g/cm}^3$ :

K = \frac{1.4}{0.7} = 2

Hence, the correct option is B.

Answer

Step-by-step solution

Common mistakes

Practice more Coulomb's Law & Superposition Principle questions

Related questions