Question

Force between two point charges \( q_1 \) and \( q_2 \) placed in vacuum at a distance \( r \, \text{cm} \) apart is \( F \). Force between them when placed in a medium having dielectric \( K = 5 \) at \( r/5 \, \text{cm} \) apart will be:

• A. \( F / 25 \)
• B. \( 5F \)
• C. \( F / 5 \)
• D. \( 25F \)

Hint:

When charges are placed in a medium, always account for the dielectric constant \( K \) and any change in the distance between the charges. Both factors influence the resultant force.

Answer 1

The Correct Option is B

The electrostatic force between two charges in a vacuum is given by Coulomb's Law: \[ F = \frac{1}{4\pi\epsilon_0} \cdot \frac{q_1 q_2}{r^2}. \] In a medium with a dielectric constant \( K \), the force becomes: \[ F' = \frac{1}{4\pi K \epsilon_0} \cdot \frac{q_1 q_2}{r'^2}. \] Here, \( K = 5 \) and \( r' = \frac{r}{5} \). Substituting these values: \[ F' = \frac{1}{4\pi (5\epsilon_0)} \cdot \frac{q_1 q_2}{\left(\frac{r}{5}\right)^2}. \] Simplify: \[ F' = \frac{25}{5} \cdot \frac{1}{4\pi\epsilon_0} \cdot \frac{q_1 q_2}{r^2}. \] Thus: \[ F' = 5F. \] Conclusion: The force between the charges in the medium is \( 5F \).