Question

Electrostatic force between two point charges placed in a medium of dielectric constant \( k \) is \( F_1 \). On changing the medium, electrostatic force between the charges becomes \( F_2 \). Dielectric constant of the medium will be:

• A. \( \dfrac{F_1}{kF_2} \)
• B. \( \dfrac{F_2}{kF_1} \)
• C. \( \dfrac{kF_1}{F_2} \)
• D. \( \dfrac{F_1}{F_2}k \)

Hint:

The electrostatic force is inversely proportional to the dielectric constant of the medium. If the medium changes, the force will change accordingly. The relationship is \( F = \dfrac{F_1}{k} \).

Answer 1

The Correct Option is C

The electrostatic force between two point charges in a vacuum is given by Coulomb's Law: \[ F_0 = \dfrac{k_e q_1 q_2}{r^2} \] where: - \( F_0 \) is the electrostatic force in a vacuum,
- \( k_e \) is Coulomb's constant,
- \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and
- \( r \) is the distance between the charges.
When a dielectric medium with dielectric constant \( k \) is introduced, the force between the charges decreases due to the presence of the medium. The force \( F \) between the charges in the dielectric medium is: \[ F = \dfrac{F_0}{k} \] This shows that the electrostatic force in the dielectric medium is reduced by a factor of \( k \). Now, if the initial force between the charges in the medium is \( F_1 \) and the new force is \( F_2 \), the relationship between the two forces is: \[ F_2 = \dfrac{F_1}{k} \] Rearranging this equation to find \( k \), we get: \[ k = \dfrac{F_1}{F_2} \] Thus, the dielectric constant \( k \) of the medium is \( \dfrac{kF_1}{F_2} \), and the correct answer is option (C).