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Two conducting sphere attract?
Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of -0.5092 N when separated by 50 cm, center-to-center. 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 0.1101 N.
1. What were the initial charges on the spheres?(Two answers)
1 Answer
- nyphdinmdLv 75 years ago
Initially you have
F1 = -0.5092 N = kq1*q2/r^2 where r = 50 cm and k = 9 x 10^9 N-m^2/C^2
Also conservation of charge gives Q = q1 + q2
Now after the spheres are connected
F2 = 0.1101 N = k q3^2/r^2 --> both spheres achieve the same charge when the wire connects them
q3 = r*sqrt(F2/k) = 1.75x10^-6 C
Hence q3 = Q = q1 + q2. Let's take F1 and substitute q2 = q3 - q1
F1 = kq1*q2/r^2 = kq1*(q3 - q1)/r^2 = (k/r^2)*(q1*q3 - q1^2) --> q1^2 - q1*q3 + (r^2/k)F = 0
You can use the quadratic formula to solve for q1
q1 = -q3/2 +/- (1/2) sqrt(q3^2 -4(r^2/k)F) --> this will generate two roots, q1+ and q1- you will get two corresponding values of q2 --
q2- = q3 - q1-
q2+ = q3 - q1+
