Question

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Two identical non-conducting solid spheres of same mass and charge are suspended in air from a common point by two non-conducting, massless strings of same length. At equilibrium, the angle between the strings is $\alpha$. The spheres are now immersed in a dielectric liquid of density $800{\text{kg m}}^{-3}$ and dielectric constant $21$. If the angle between the strings remains the same after the immersion, then

• A. electric force between the spheres changes
• B. mass density of the spheres is $8400{\text{kg m}}^{-3}$
• C. mass density of the spheres is $840{\text{kg m}}^{-3}$
• D. the tension in the strings holding the spheres remains unchanged

Answer 1

The correct option is C mass density of the spheres is $840{\text{kg m}}^{-3}$

By Coulomb's law the force due to one sphere on another remains the same in magnitude.
At equilibrium,
$T\cos \frac{\alpha }{2}=mg$
and $T\sin \frac{\alpha }{2}=F$
After being immersed in a dielectric liquid, given that there is no change in angle $\alpha$.
So $T\cos \frac{\alpha }{2}=mg-V\rho g$
Here $V\rho g$ is the buoyant force acting on body in upward direction, when it is immersed in liquid.
when $\rho =800{\text{kg/m}}^3$
and $T\sin \frac{\alpha }{2}=\frac{F}{K}$
The electrostatic force will get reduced by a factor of $K$ when body immersed in liquid.
$\Rightarrow \frac{mg}{F}=\frac{mg-V\rho g}{\frac{F}{K}}$
$\text{or}\frac{1}{K}=1-\frac{\rho }{\sigma }$
$\sigma =$density of sphere
$\Rightarrow \frac{1}{21}=1-\frac{800}{\sigma }$
$\sigma =840{\text{kg/m}}^3$