Question

Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of $30^{\circ }$ with each other. When suspended in a liquid of density $0.8gc{m}^{-3}$, the angle remains the same. If density of the material of the sphere is $1.6gc{m}^{-3}$, the dielectric constant of the liquid is

• A. 4
• B. 3
• C. 2
• D. 1

Answer 1

Answer: (C)

From fig, $\tan \theta =qE/mg$
$\tan 15^{\circ }=\frac{k{q}^2}{d^{2}mg}$
$\tan 15^{\circ }=\frac{k{q}^2}{1.6Vg{d}^2}\dots ..[V$ is the volume $]$.
When system is suspended in liquid,
$\begin{array}{l}\tan 15^{\circ }=\frac{k{q}^2}{K(mg-\rho gg)d^2}\\ \tan 15^{\circ }=\frac{k{q}^2}{K(1.6-0.8)Vg{g}^2}\dots \dots .(2)\end{array}$
from (1) and (2) we get,
$\frac{k{q}^2}{K(1.6-0.8)Vg{d}^2}=\frac{k{q}^2}{1.6Vg{d}^2}$
$\therefore K=2=$ Dielectric constant of liquid.