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 -0.4675 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 0.0655 N. What were the initial charges on the spheres? Since one is negative and you cannot tell which is positive or negative, there are two solutions. Take the absolute value of the charges and enter the smaller value here.
the larger value?

Explanation / Answer

i hav posted a similar question and its solution..plz make out ur answer from it..hope it will be helpful for u.rate me only if u find this helpful.:) Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.0911 N when their center-to-center separation is 43.3 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 0.0297 N. Of the initial charges on the spheres, with a positive net charge, what was (a) the negative charge in coulombs on one of them and (b) the positive charge in coulombs on the other This can be treated like a point-charge problem, since we are measuring over distances greater than the sphere radii. Then F1 = -kq1q2/r^2 = 0.0911 N, and F2 = kq3^2/r^2 = 0.0297 N, where 2q3 = q1+q2 due to conservation of charge. Solving for q3, q3 = sqrt(0.0297r^2/k) = 7.8658E-7 C Then solving for q1 we have F1 = -kq1(2*7.8658E-7-q1)/r^2 = 0.0911 which yields a quadratic -q1^2 + 2*7.8658E-7q1 + 0.0911r^2/k = 0 resulting in q1 = 2.37293E-6 C and q2 = -7.99772E-7 C, which sum to 2*7.8658E-7 C.

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