In the figure three identical conducting spheres initially have the following ch

Question

In the figure three identical conducting spheres initially have the following charges: sphere A, 5Q; sphere B, -7Q; 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, sphere C first touches sphere A, and they divided up their total charge (5Q plus 0) equally between them. Thus, sphere A and sphere C each acquired charge 2.5Q. Then, sphere C touches B and those spheres split up their total charge (2.5Q –7Q) so that B ends up with charge equal to -2.5Q. The force of repulsion between A and B is therefore

F1 = k ((2.5Q)(2.5Q)) / d2

at the end of experiment 1. Now, in experiment 2, sphere C first touches B which leaves each of them with charge -3.5Q. When C next touches A, sphere A is left with charge Q/2. Consequently, the force of repulsion between A and B is

F2 = k ((3.5Q)((1/2)Q)) / d2

at the end of experiment 2. The ratio is

F2/ F1 = (3.5)(1/2) / (2.5)(2.5)

                = 0.28

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