Two point charges q ₁ and q ₂ are ' l ' distance apart. If one of the charges is doubled and distance between them is halved. The magnitude of force becomes n times, where n is

Step-by-step solution

Let the initial charges be q₁ and q₂ , separated by a distance l . The initial force magnitude is F₁ = k |q₁ q₂| l^2 . When one charge is doubled ( q₁' = 2q₁ ) and the distance halved ( l' = l/2 ), the new force becomes: F₂ = k |q₁' q₂'| (l')^2 = k |(2q₁) q₂| (l/2)^2 Simplifying: F₂ = k 2|q₁ q₂| l^2/4 = k 2|q₁ q₂| l^2 4 = 8 (k |q₁ q₂| l^2 ) = 8F₁ The final force magnitude is 8 times the original force. The correct option is C.