In the figure, three identical conducting spheres initially have the following c
ID: 2043497 • Letter: I
Question
In the figure, three identical conducting spheres initially have the following charges: sphere A, 10Q; 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 magnitude of the electrostatic force between A and B at the end of experiment 2 to the force magnitude at the end of experiment 1?
Explanation / Answer
Experiment #1:When sphere A touches sphere C the charge is conserved. The net charge Qc + Qa = 10Q. After they are separated both will have Qa= Qc= 5Q. The charge is equally distributed. Now sphere C is brought in contact with sphere B. Now the net charge is Qc+ Qb = 5Q - 6Q = - Q. This is distributed among them. Qc= Qb = -0.5Q The net force between the sphere A and B is F1 = k (5Q)(0.5Q) / d2 = k (2.5) 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 = -6Q. After they are separated both will have Qb= Qc= -3Q. The charge is equally distributed Now sphere C is brought in contact with sphere A. Now the net charge is Qc+ Qa = -3Q + 10Q = 7 Q. This is distributed among them. Qc= Qa = 3.5Q The net force between the sphere A and B is F2 = k (-3Q)(3.5Q) / d2 = k (10.5) Q / d2 ( attractive as the force is negative) The raio of the force is F2 / F1 = [k (10.5) Q / d2 ] / [ k (2.5) Q / d2 ] = 4.2 The charge is equally distributed Now sphere C is brought in contact with sphere A. Now the net charge is Qc+ Qa = -3Q + 10Q = 7 Q. This is distributed among them. Qc= Qa = 3.5Q The net force between the sphere A and B is F2 = k (-3Q)(3.5Q) / d2 = k (10.5) Q / d2 ( attractive as the force is negative) The raio of the force is F2 / F1 = [k (10.5) Q / d2 ] / [ k (2.5) Q / d2 ] = 4.2 Now sphere C is brought in contact with sphere A. Now the net charge is Qc+ Qa = -3Q + 10Q = 7 Q. This is distributed among them. Qc= Qa = 3.5Q The net force between the sphere A and B is F2 = k (-3Q)(3.5Q) / d2 = k (10.5) Q / d2 ( attractive as the force is negative) The raio of the force is F2 / F1 = [k (10.5) Q / d2 ] / [ k (2.5) Q / d2 ] = 4.2 = k (10.5) Q / d2 ( attractive as the force is negative) The raio of the force is F2 / F1 = [k (10.5) Q / d2 ] / [ k (2.5) Q / d2 ] = 4.2
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