Two identical balls each having a density $\rho$ are suspended 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. The density of the liquid is $\sigma$. Find the dielectric constant of the liquid.

 Each ball is in equilibrium under the following three forces :

(i) tension,

(ii) electric force and

(iii) weight

So, Lami's theorem can be applied,

In the liquid,

$F_e^{\text{'}}=\frac{F_e}{K}$

where, $K=$ dielectric constant of liquid and $W^{\text{'}}=W$ – upthrust

Applying Lami's theorem in vacuum

$\frac{W}{\sin \left(90°+\theta \right)}=\frac{F_e}{\sin \left(180°-\theta \right)}$

$\frac{W}{\cos \theta }=\frac{F_e}{\sin \theta }$            ……(i)

Similarly in liquid, $\frac{W^{\text{'}}}{\cos \theta }=\frac{F_e^{\text{'}}}{\sin \theta }$          ……(ii)

Dividing Eq. (i) by Eq. (ii), we get

$\frac{W}{W^{\text{'}}}=\frac{F_e}{F_e^{\text{'}}}$

or   $K=\frac{W}{W-\text{ upthrust }}$           $\left(as\frac{F_e}{F_e^{\text{'}}}=K\right)$

$=\frac{V\rho g}{V\rho g-V\sigma g}$        ($V=volume of ball$)

or   $K=\frac{\rho }{\rho -\sigma }$

Note: In the liquid $F_e$ and $W$ have changed. Therefore, $T$ will also change.