Question

Two point charges placed at a distance r in air exert a force on each other . the value of distance r at which they experience force 4f when placed in a medium of dielectric constant k=16

Answer 1

Let two point charges q₁ and q₂ are separated r distance form each other.
Then, Force experienced between them , F = q₁q₂/4π∈₀r²

Now, both the charges are placed in a medium of dielectric constant , K = 16
Then, force experienced between them = 4F = q₁q₂/K(4π∈₀r'²)
Put F = q₁q₂/4π∈₀r²
∴ 4q₁q₂/4π∈₀r² = q₁q₂/K(4π∈₀r'²)
⇒ 1/r² = 1/4r'²K
⇒1/r² = 1/4 ×r'² × 16
⇒ 1/r² = 1/(8)²r'²
taking square root both sides,
⇒1/r = 1/8r'
∴r' = r/8

Hence, distance between them in medium = r/8

Answer 2

Answer:r/8

Explanation:see the below attachment