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 0.151 N when their center-to-center separation is 59.9 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 0.0272 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

Electrostatic force F = k q1 q2 / r^2
where k is constant and eual to 9x109, q1 and q2 are magnitude of charges, r is distance between the charges or distance between the center of spheres.
Initially force F1 = 0.151 = k q1 q2 / (r)2 .....1
When connected with wire, charge on the identical spheres is same and half of the total charge. Since initial force was force of attraction, sign of the charges are opposite. So final charge on both spheres is (q1 - q2)/2
Final force F2 = 0.0272 = k (q1-q2)2 / 4 r2  ........2
writing (q1-q2)2 = (q1+q2)2 - 4 q1q2
we get (q1 +q2)2 = 4(F2+F1) r2 /k ....3
from 2 and 3 we get
q1 = r / k1/2 ( (F1+ F2)1/2 + F21/2 ) = 117.2 micro Col
q2 = r / k1/2 ( (F1 +F2)1/2 - F21/2 ) = 51.3 micro Col.
q2 being smaller is -ve charge

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