Force between two point charges q₁ and q₂ placed in vacuum at 'r' cm apart is F. Force between them when placed in a medium having dielectric constant K=5 at 'r / 5' cm apart will be:

Solution

Physics sir · 3 steps

The medium divides the force by 5, the shorter distance multiplies it by 25, so the net change is 5 times.

1. In vacuum. Coulomb's law gives F = 1/4πε₀(q₁ q₂)/r².
2. In the medium. A medium of dielectric constant K weakens the force by a factor K, so F' = 1/(4πε₀ K)(q₁ q₂)/d². Here K = 5 and d = r/5.

   F'/F = 1/K(r/d)² = 1/5 × 5² = 5

3. So. The new force is F' = 5F, which is option (a).

The idea to keep. When both the medium and the distance change, write the force as a ratio: divide by K and multiply by the square of the distance ratio. The units cm or m cancel in the ratio.

Answer from the NTA final answer key. Solution by PrepWiser.