Question

Two like charges are separated in air by a certain distance. The air is replaced with a dielectric medium of constant \( K \). Then the repulsive force between them:

• A. Decrease by \( K \) times
• B. Increase by \( K \) times
• C. No change
• D. \( K^2 \) times decreases

Hint:

When charges are surrounded by a dielectric medium, the repulsive force between them decreases by a factor equal to the dielectric constant \( K \).

Answer 1

The Correct Option is A

According to Coulomb's law, the force between two charges in air is given by:

\[ F = k_e \frac{q_1 q_2}{r^2} \]

Where:
- \( F \) is the force,
- \( k_e \) is Coulomb's constant,
- \( q_1 \) and \( q_2 \) are the magnitudes of the charges,
- \( r \) is the distance between the charges.

When a dielectric medium of constant \( K \) is introduced, the force is reduced by a factor of \( K \):

\[ F' = \frac{F}{K} \]

Thus, the force between the charges decreases by \( K \) times when a dielectric medium with constant \( K \) is placed between the charges.