In the figure three identical conducting spheres initially have the following ch
ID: 1571867 • Letter: I
Question
In the figure three identical conducting spheres initially have the following charges: sphere A, 6Q; 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
First take Experiment - 1:
Here sphere A touches sphere C the charge is conserved.
Then the net charge Qc + Qa = 6Q.
And after they are separated both will have Qa= Qc= 3Q.
The charge is equally distributed.
Now sphere C is brought in contact with sphere B.
So the net charge is Qc+ Qb = 3Q - 7Q = -4 Q. This is distributed among them.
So, Qc= Qb = -2.0Q
The net force between the sphere A and B is F1 = k (3Q)(2.0Q) / d2
= - k (6.0) Q^2 / d2 ( attractive as the force is negative)
Again for Experiment - 2:
When sphere B touches sphere C the charge is conserved. The net charge Qc + Qb = -7Q. After they are separated both will have Qb= Qc= -3.5Q.
The charge is equally distributed
Now sphere C is brought in contact with sphere A. Now the net charge is Qc+ Qa = -3.5Q + 6Q = 2.5 Q. This is distributed among them.
Qc= Qa = 1.25Q
The net force between the sphere A and B is F2 = k (-3.5Q)(1.25Q) / d2
= - k (4.375) Q^2 / d2 ( attractive as the force is negative)
The raio of the force is
F2 / F1 = [k (4.375) Q^2 / d2 ] / [ k (6.0) Q^2 / d2 ]
= 0.73
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