Question

Two like charges kept in air medium experience a force \( F \), when they are separated by a certain distance \( r \). When the same charges are kept in a dielectric medium at the same distance of the separation, the force between them is 0.5F. The dielectric constant of the medium is

• A. 5
• B. \( \frac{3}{2} \)
• C. \( \frac{5}{2} \)
• D. 2
• E. \( \frac{2}{5} \)

Hint:

In the presence of a dielectric, the force between charges decreases by a factor equal to the dielectric constant.

Answer 1

The Correct Option is D

The force between two charges in a medium is given by the formula: \[ F = \frac{1}{4 \pi \epsilon} \frac{q_1 q_2}{r^2} \] where: - \( \epsilon \) is the permittivity of the medium, and
- \( r \) is the separation between the charges. 
In air, the force is \( F = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2} \). In the dielectric medium, the force is reduced by a factor of the dielectric constant \( K \), so the force becomes: \[ F' = \frac{1}{K} \cdot F \] Given that the force in the dielectric medium is 0.5F, we have: \[ \frac{F'}{F} = \frac{1}{K} = 0.5 \] Solving for \( K \): \[ K = 2 \] Hence, the dielectric constant of the medium is 2. 
Therefore, the correct answer is (D).