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

(B) electric force between the spheres reduces; (C) mass density of the spheres is $840 \text{ kg m}^{-3}$

(A) electric force between the spheres remains unchanged

(D) the tension in the strings holding the spheres remains unchanged

JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Electrostatics

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

α

. The spheres are now immersed in a dielectric liquid of density

800 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 remains unchanged
(B)electric force between the spheres reduces✓ Correct
(C)mass density of the spheres is $840 kg m^{- 3}$ ✓ Correct
(D)the tension in the strings holding the spheres remains unchanged

Answer: (B), (C)

Solution

In air, at equilibrium:

T cos \frac{α}{2} = m g

and

T sin \frac{α}{2} = F

where F is the electric force.

In dielectric liquid:

T^{'} cos \frac{α}{2} = m g - V ρ g

(buoyant force acts upward)

T^{'} sin \frac{α}{2} = \frac{F}{ε _{r}}

(electric force reduced by dielectric constant)

Since angle

α

remains the same:

\frac{m g}{F} = \frac{m g - V ρ g}{F / ε _{r}}

\frac{1}{ε _{r}} = 1 - \frac{ρ}{d}

where d is the density of the sphere.

\frac{1}{21} = 1 - \frac{800}{d}

\frac{20}{21} = \frac{800}{d}

d = \frac{800 \times 21}{20} = 840 kg m^{- 3}

✓ (Option C)

The electric force between spheres reduces by factor

ε_{r} = 21

. ✓ (Option B)

JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · One or more correct options

Practise more: all JEE Advanced PYQs on Electrostatics →

JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Electrostatics

Solution

More JEE Advanced questions on Electrostatics