Question

Chapter 21, Problem 008 Your answer is partially correct. Try again. In the figure three identical conducting spheres initially have thelowing charges: sphere A, 8Q: 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? Number T0.875 No units Units the tolerance is +/-2%

Answer 1

008)

In experiment 1,

Now when C and A are touched, After separating
Charge on A = Charge on C = (qa + qc)/2 = ( 8Q + 0)/2 = 4Q
Now when C and B are touched, After separating
Charge on B = Charge on C = (qb + qc)/2 = ((4Q) + (-7Q))/2 = -3Q/2

In experiment 2,

Now when C and B are touched, After separating
Charge on B = Charge on C = (qb + qc)/2 = ( -7Q + 0)/2 = -7Q/2
Now when C and A are touched, After separating
Charge on A = Charge on C = (qa + qc)/2 = ((8Q) + (-7Q/2))/2 = 9Q/4
Now Since electrostatic force is given by,

Fe = k*q1*q2/r^2

where, k = 9*10^9

q1 & q2 = charges

r = distance between charges

So, in experiment 1

Fe1 = k*qa*qb/r^2 eq(1)

in experiment 2,

Fe2 = k*qa*qb/r^2 eq(2)

now dividing eq (2)/(1),

Fe2/Fe1 = (qa*qb)/(qa*qb)

Fe2/Fe1 = [(7Q/2)*(9Q/4)]/[(4Q)*(3Q/2)]

Fe2/Fe1 = (63/4)/(6) = 63/24

Fe1/Fe2 = 2.625 unitless
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