Question

If two charges \( q_1 \) and \( q_2 \) are separated with distance 'd' and placed in a medium of dielectric constant K. What will be the equivalent distance between charges in air for the same electrostatic force?

• A. d√K
• B. K√d
• C. 1.5d√K
• D. 2d√K

Hint:

The electrostatic force in a medium is \( \frac{1}{K} \) times the force in vacuum for the same charges and distance. To maintain the same force, increase the distance in vacuum by \( \sqrt{K} \) times the distance in the medium.

Answer 1

The Correct Option is A

The electrostatic force between two charges in a medium with dielectric constant K is given by: \[ F_{medium} = \frac{1}{4\pi\epsilon_0 K} \frac{|q_1 q_2|}{d^2} \] where \( \epsilon_0 \) is the permittivity of free space. The electrostatic force between the same two charges in air (or vacuum, with K=1) at a distance \( d' \) is given by: \[ F_{air} = \frac{1}{4\pi\epsilon_0} \frac{|q_1 q_2|}{d'^2} \] For the forces to be equal, \( F_{medium} = F_{air} \): \[ \frac{1}{4\pi\epsilon_0 K} \frac{|q_1 q_2|}{d^2} = \frac{1}{4\pi\epsilon_0} \frac{|q_1 q_2|}{d'^2} \] \[ \frac{1}{K d^2} = \frac{1}{d'^2} \] \[ d'^2 = K d^2 \] \[ d' = \sqrt{K d^2} = d\sqrt{K} \] Thus, the equivalent distance in air is \( d\sqrt{K} \).