Question

Two identical balls each having a density $\rho$ are suspended in vacuum from a common point by two insulating strings of equal length. Both the balls have equal mass and charge. In equilibrium each string makes an angle $\theta$ with vertical. Now both the balls are immersed in a liquid, as a result the angle $\theta$ does not change. If the density of the liquid is $\sigma$, then find the dielectric constant of the liquid.

• A. $\frac{\sigma }{\rho -\sigma }$
• B. $\frac{\rho }{\rho -\sigma }$
• C. $\frac{\sigma }{\sigma +\rho }$
• D. $\frac{2\sigma }{\rho -\sigma }$

Answer 1

The correct option is B $\frac{\rho }{\rho -\sigma }$

The balls are in equilibrium due to balanced condition of forces acting on each ball, i.e. Tension $\left(T\right)$, weight $\left(W\right)$ and electrostatic force of repulsion $\left(F_e\right)$


In liquid the electrostatic force $\left(F_e^{{}^{\prime }}\right)$ gets reduced,
$F_e^{{}^{\prime }}=\frac{F_e}{k}\dots \dots \left(i\right)$
Let $V$ be the volume of each ball, hence mass of ball
$m=\rho V$
When balls are in vacuum:
$T\cos \theta =mg$
$T\sin \theta =F_e$
$\Rightarrow \tan \theta =\frac{F_e}{mg}$
or, $F_e=mg\tan \theta =\rho Vg\tan \theta \dots \dots \left(ii\right)$
When balls are immersed in liquid,
Apparent weight $W^{{}^{\prime }}=W-$ upthrust
$\Rightarrow W^{{}^{\prime }}=W-V\sigma g=V\rho g-V\sigma g$
($\because$ upthrust is buoyant force)
Thus, $\tan \theta =\frac{F_e^{{}^{\prime }}}{W^{{}^{\prime }}}$
or, $F_e^{{}^{\prime }}=(V\rho g-V\sigma g)\tan \theta \dots \dots \left(iii\right)$
From eq. (i), (ii), (iii):
$Vg(\rho -\sigma )\tan \theta =\frac{\rho Vg\tan \theta }{k}$
$\Rightarrow k=\frac{\rho }{\rho -\sigma }$

$\begin{array}{c}\text{Tip: Draw the FBD of balls in vacuum and in liquid.}\\ \text{Always keep in mind that "presence of liquid will}\\ \text{decrease the effective weight and Coulombic force}\left(F_e^{{}^{\prime }}\right)\end{array}$