In the figure three identical conducting spheres initially have the following ch
ID: 1836714 • Letter: I
Question
In the figure three identical conducting spheres initially have the following charges: sphere A, 3Q; 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 experiment are conducted. In experiment 1, sphere C is touched to sphere A and then (separately) to sphere 3, 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
Charge of sphere A is q = 3Q
Charge of sphere B is q ' = -7Q
Charge of sphere C is q " = 0
Experiment 1 :
Charge on sphere C and A after touch A is qa= (q +q " ) / 2
= 3Q / 2
= 1.5 Q
Charge on sphere C and B after touch B is qb = (qa +q ') / 2
=(1.5Q -7Q) /2
= -2.75 Q
Force between A and B at the end of experiment 1 is F = K(qa)(qb) /d 2
= K(1.5Q)(2.75Q)/d 2
Experiment 2 :
Charge on sphere C and B after touch B is qa= (q ' +q " ) / 2
= -7Q / 2
= -3.5 Q
Charge on sphere C and A after touch A is qb = (qa +q ) / 2
=(-3.5Q +3Q) /2
= -0.25 Q
Force between A and B at the end of experiment 2 is F ' = K(qa)(qb) /d 2
= K(3.5Q)(0.25Q)/d 2
Required answer is F ' / F = [ K(3.5Q)(0.25Q)/d 2]/[K(1.5Q)(2.75Q)/d 2]
= 0.2121
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