Two spherical conductors B and C having equal radii and carrying equal charges in them repel each other with a force F when kept apart at some distance. A third spherical conductor having same radius as that of B but uncharged, is brought in contact with B , then brought in contact with C and finally removed away from both. The new force of repulsion between B and C is

Step-by-step solution

Let the spherical conductors B and C have same charge as q . The electric force between them is F= 1 4 ₀ q^2 r^2 r , being the distance between them. When third uncharged conductor A is brought in contact with B , then charge on each conductor q_A=q_B= q_A+q_B 2 = 0+q 2 = q 2 When this conductor A is now brought in contact with C , then charge on each conductor q_A=q_C= q_A+q_C 2 = (q / 2)+q 2 = 3 q 4 Hence, electric force acting between B and C is aligned F^ = 1 4 ₀ q_B q_C r^2 & = 1 4 ₀ (q / 2)(3 q / 4) r^2 & = 3