Question

Two spherical bobs of same mass and radius having equal charges are suspended from the same point by strings of same length. The bobs are immersed in a liquid of relative permittivity ${\epsilon }_r$ and density ${\rho }_0$. The density $\sigma$ of the bob for which the angle of divergence of the strings is same in the air & in the liquid is $\frac{{x\epsilon }_r{\rho }_0}{{\epsilon }_r-1}$. Find $x$.

Answer 1

Let $F_1$ and $F_2$ be the electrostatic forces between the bobs in the two cases respectively.
now, $F_1={\epsilon }_r{F}_2$
When in air, $W_1=mg$ and $\tan \theta =\frac{F_1}{W_1}=\frac{F_1}{mg}.....\left(1\right)$
When immersed in the liquid, $W_2=mg(1-\frac{{\rho }_0}{\sigma })$ and $\tan \theta =\frac{F_2}{W_2}=\frac{F_2}{mg(1-\frac{{\rho }_0}{\sigma })}...\left(2\right)$
From (1) and (2), $\frac{F_1}{mg}=\frac{F_2}{mg(1-\frac{{\rho }_0}{\sigma })}$
or $\frac{{\epsilon }_r{F}_2}{mg}=\frac{F_2}{mg(1-\frac{{\rho }_0}{\sigma })}$
or ${\epsilon }_r-\frac{{\epsilon }_r{\rho }_0}{\sigma }=1$
or $\frac{{\epsilon }_r{\rho }_0}{\sigma }={\epsilon }_r-1$
$\therefore \sigma =\frac{{\epsilon }_r{\rho }_0}{{\epsilon }_r-1}$