JEE Main 2024 Physics question
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
Answer: (C)
Solution
F₂/F₁ = [1/(K(r/5)²)]/[1/r²] = 25/K = 25/5 = 5, so the new force is 5F.