In the figure three Identical conducting spheres Initially have the following ch
ID: 1412711 • Letter: I
Question
In the figure three Identical conducting spheres Initially have the following charges: sphere A, 8Q; sphere B, -5Q; 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 I 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
ANSWER
Experiment #1:
When sphere A touches sphere C the charge is conserved. The net charge Qc + Qa = 8Q. After they are separated both will have Qa= Qc= 4Q.
The charge is equally distributed.
Now sphere C is brought in contact with sphere B. Now the net charge is Qc+ Qb = 4Q - 5Q = - Q. This is distributed among them.
Qc= Qb = -0.5Q
The net force between the sphere A and B is F1 = k (4Q)(0.5Q) / d2
= k (2) Q / d2 ( attractive as the force is negative)
Experiment #2:
When sphere B touches sphere C the charge is conserved. The net charge Qc + Qb = -5Q. After they are separated both will have Qb= Qc= -2.5Q.
The charge is equally distributed
Now sphere C is brought in contact with sphere A. Now the net charge is Qc+ Qa = -2.5Q + 8Q = 5.5 Q. This is distributed among them.
Qc= Qa = 2.75Q
The net force between the sphere A and B is F2 = k (-2.5Q)(2.75Q) / d2
= k (6.875) Q / d2 ( attractive as the force is negative)
The ratio of the force is
F2 / F1 = [k (6.875) Q / d2 ] / [ k (2) Q / d2 ]
= 3.4375 ANS.
Regards!!!
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