Question

Two identical charged spheres are suspended by strings of equal length and the strings make certain angle with each other. When suspended in a liquid of density $400kg/m^3$ , the angle between the threads remains the same. If the density of the material of the sphere is $1600kg/m^3$ , the dielectric constant of the liquid:

• A. $1.33$
• B. $2$
• C. $3.12$
• D. $5$

Answer 1

The correct option is A $1.33$

Initially when the balls are in air let the angle be $\theta$ between them
$\Rightarrow \frac{K{q}^2}{4l^{2}si{n}^2\left(\frac{\theta }{2}\right)}=Tsin\left(\frac{\theta }{2}\right)andTcos\left(\frac{\theta }{2}\right)=mg$
$\Rightarrow tan\frac{\theta }{2}=\frac{K{q}^2}{4l^{2}si{n}^2\left(\frac{\theta }{2}\right)mg}$
$\Rightarrow tan\frac{\theta }{2}si{n}^2\frac{\theta }{2}=\frac{1}{4\pi {ϵ}_o}\frac{q^2}{4l^2{\rho }_{b}Vg}$
Now when it is inserted in water we get electric $K{\epsilon }_0$ and buoyant force $={\rho }_{l}Vg$
$\therefore Weight,\frac{1}{4\pi {ϵ}_o}\frac{q^2}{4l^2{\rho }_{b}Vg}=\frac{q^2}{4\pi \left(K{ϵ}_o\right)4l^2({\rho }_b-{\rho }_l)Vg}$
$\Rightarrow K({\rho }_b-{\rho }_l)={\rho }_b\Rightarrow K=\frac{1600}{1600-400}=1.33$