Two identical balls each having a density `rho` are suspended from as 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.

Explanation

Let v is th volume of each ball, then mass
of each ball is `m=rhov`. When balls are in air
`Tcos theta=m g, T sin theta=F`
`F=mg tan theta = rho v g tan theta`……..1
When balls are suspended in liquid. The coulumbic force is reduced to `F^(1)=F/K` and apparent weight = weight - upthrust `W^(1)=rho vt - sigma v g `
According to the problem, angle `theta` is uncharged- Therefore
`F^(1)=W^(1) tan tehta =(rho v g - sigma v g) tan theta `.........2
From 1 and 2 , `F/(F^(1))=K=(rho)/(rho-sigma)`