Two charges placed in air repel each other by | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The force between two charges in air is given by $F_a = \frac{1}{4\pi\epsilon_0} \frac{q_1q_2}{r^2} = 10^{-4} \ N$.When a medium (oil) with dielec

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) The force between two charges in air is given by $F_a = \frac{1}{4\pi\epsilon_0} \frac{q_1q_2}{r^2} = 10^{-4} \ N$.When a medium (oil) with dielectric constant $K$ is introduced,the force becomes $F_m = \frac{1}{4\pi\epsilon_0 K} \frac{q_1q_2}{r^2} = 2.5 	imes 10^{-5} \ N$.The ratio of the force in air to the force in the medium is given by $K = \frac{F_a}{F_m}$.Substituting the values: $K = \frac{10^{-4}}{2.5 	imes 10^{-5}} = \frac{10}{2.5} = 4$.Therefore,the dielectric constant of oil is $4$.

Two charges placed in air repel each other by a force of $10^{-4} \ N$. When oil is introduced between the charges,the force becomes $2.5 \times 10^{-5} \ N$. The dielectric constant of oil is:

Similar Questions

What is polarisation?

Assertion : In the absence of an external electric field,the dipole moment per unit volume of a polar dielectric is zero.
Reason : The dipoles of a polar dielectric are randomly oriented.

Which of the following is the dimensional formula for electric polarisation?

The dielectric constant of a metal is ........ .

What is a dielectric?

Vedclass Test Series

Exam Paper Generator

Online Exam Module