Question

Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of ${30}^0$ with each other. When suspended in a liquid of density 0.8 $gc{m}^{-3}$, the angle remains the same. lf density of the material of the sphere is 1.6 $gc{m}^{-3}$, the dielectric constant of the liquid is:

• A. 3
• B. 4
• C. 2
• D. 1

Answer 1

Answer: (C)

From fig,

$tan\theta =qE/mg$

$tan{15}^o=\frac{k{q}^2}{d^{2}mg}$

$tan{15}^o=\frac{k{q}^2}{1.6Vg{d}^2}$ .....[$V$ is the volume]...........(1)

When system is suspended in liquid,

$tan{15}^o=\frac{k{q}^2}{K(mg-\rho Vg)d^2}$

$tan{15}^o=\frac{k{q}^2}{K(1.6-0.8)Vg{d}^2}$ ......(2)

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.