Two charges q₁ and q₂ are separated with distance 'd' and placed in a medium of dielectric constant 'k'. The equivalent distance between charges in air for the same electrostatic force is

Step-by-step solution

Force between two charges in a medium of dielectric constant k is given by F_ m = 1 4 ₀ k q₁ q₂ d² Force between the same charges in air at an equivalent distance d' is given by F_ a = 1 4 ₀ q₁ q₂ d'² Given that the electrostatic force remains the same, F_ m = F_ a 1 4 ₀ k q₁ q₂ d² = 1 4 ₀ q₁ q₂ d'² 1 k d² = 1 d'² d'² = k d² d' = d k Answer: d k