Two charges q ₁=+6 q and q ₂=-3 q are placed as shown in figure. A proton is placed on x -axis away from q ₂ . To remain proton in equilibrium, the distance between q ₁ and proton is

Step-by-step solution

Equilibrium requires zero net electrostatic force on the proton. With charges q₁ = +6q at x=0 and q₂ = -3q at x=L , the equilibrium position must lie to the right of q₂ ( x > L ) since |q₂| Considering a proton at position x > L , the repulsive force from q₁ is F₁ = k 6qe x^2 and the attractive force from q₂ is F₂ = k 3qe (x-L)^2 . Setting F₁ = F₂ and canceling common factors k , q , and e : 6 x^2 = 3 (x-L)^2 Simplifying yields 2 x^2 = 1 (x-L)^2 . Taking square roots (positive since x > L ): 2 x = 1 x-L Solving for