Two identical conducting spheres, fixed in place, attract each other with an ele

Question

Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.146 N when their center-to-center separation is 74.8 cm. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of 0.0372 N. Of the initial charges on the spheres, with a positive net charge, what was (a) the negative charge on one of them and (b) the positive charge on the other? (Assume the negative charge has smaller magnitude.)

Explanation / Answer

The problem can be solved like a point-charge problem, since we are measuring over distances greater than the sphere radii. Then
F1 = -kq1q2/r^2 = 0.146 N,

and in the second condition,
F2 = kq3^2/r^2 = 0.0372 N, where 2q3 = q1+q2 due to conservation of charge.
Solving for q3,
q3 = sqrt(0.0372r^2/k) = 1.52157E-6 C
Then solving for q1 we have
F1 = -kq1(2*1.52157E-6-q1)/r^2 = 0.146

Which yields a quadratic
-q1^2 + 2*1.52157E-6q1 + 0.146r^2/k = 0

-q1^2 + 2*1.52157E-6q1 + 9.086E-12 = 0

=> q1^2 - 2*1.52157E-6q1 - 9.086E-12 = 0

=> q1^2 - 2*1.52157E-6q1 + (1.52157E-6)2 = 9.086E-12 + (1.52157E-6)2

=> q1 - 1.52157E-6 = 3.37656E-6

=> q1 = 4.8981E-6 C

Hence, q2 = 2*1.52157E-6 - 4.8981E-6 = -1.85496E-6 C

(a)  -1.85496E-6 C

(b)  4.8981E-6 C

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