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

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: $k = 2$.