Three identical conducting spheres initially have the following charges: sphere

Question

Three identical conducting spheres initially have the following charges: sphere A, 4Q; sphere B, -6Q; and sphere C, 0. Spheres A and B are fixed in place, with a center-to-center separation that is much larger than the spheres. Two experiments are conducted. In experiment 1, sphere C is touched to sphere A and then (separately) to sphere B, and then it is removed. In experiment 2, starting with the same initial states, the procedure is reversed: Sphere C is touched to sphere B and then (separately) to sphere A, and then it is removed. What is the ratio of the electrostatic force between A and B at the end of experiment 2 to that at the end of experiment 1?

Explanation / Answer

in experiment 1)

given that in the problem sphere is having initial charge 0,then when C touches with A,C gets charged on par with the charge A,now new charges on A and C are same i.e 2Q

charge on A = charge on C = (0+4Q)/2

now C touches with B then

charge on C = charge on B =(2Q-6Q)/2 = -2Q

Force between A and B is F1 = k*QA*QB/d^2 = k*(2Q)*(-2Q)/d^2..........(1)

d is the distance of seperation between A and B

In experiment 2)

after touching C with B,charge on C = charge on B = (0-6Q)/2 = -3Q

now C touches with A,then charge on A = charge on C = (-3Q+4Q)/2 = Q/2

Now force between A and B is F2 = k*QA*QB/d^2 = k*(Q/2)*(-3Q)/d^2.........(2)

required ratio is F2/F1 =(-3/2) / (-4) = 3/8

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