Two point charges q₁ and q₂ are ' l ' distance apart. If one of the charges is doubled and the distance between them is halved. The magnitude of the force becomes 'n' times, where 'n' is

Step-by-step solution

Initial force between the charges is given by Coulomb's law: F = 1 4 ₀ q₁ q₂ l^2 When one of the charges is doubled, let q₁' = 2q₁ and q₂' = q₂ . When the distance between them is halved, let l' = l 2 . The new force is: F' = 1 4 ₀ q₁' q₂' l'^2 F' = 1 4 ₀ 2q₁ q₂ ( l 2 )^2 F' = 1 4 ₀ 2q₁ q₂ l^2 4 F' = 8 ( 1 4 ₀ q₁ q₂ l^2 ) F' = 8F Thus, the magnitude of the force becomes 8 times the initial force, so n = 8 . Answer: 8