Two identical conducting spheres, fixed in place,attract each other with an elec
ID: 1664440 • Letter: T
Question
Two identical conducting spheres, fixed in place,attract each other with an electrostatic force of -0.4013N when separated by 50 cm, center-to-center. Thespheres are then connected by a thin conducting wire. When the wireis removed, the spheres repel each other with an electrostaticforce of 0.1740 N. What were the initial charges on thespheres? Since one is negative and you cannot tell which ispositive or negative, there are two solutions. Take the absolutevalue of the charges and enter the smaller value hereExplanation / Answer
repel force F1 = 0.7250 N, so2 charges are different signs F2 = 0.1740 N. r = 0.50 m initial charges (absolute values) q1 and q2(assume q1 > q2) F1 =kq1q2/r2q1q2= F1r2/k (1) final charges: q = (q1 - q2)/2 for eachsphere F2 =kq2/r2q2 =F2r2/k [(q1 - q2)/2]2 =F2r2/k (q1 - q2)2 =4F2r2/k (2) (2) + 4*(1): (q1 + q2)2 = 4(F1 +F2)r2/k (3) from (3) q1 + q2 = 2r[(F1 +F2)/k] (4) from (2) q1 - q2 = 2r(F2/k) (5) solve from (4) and (5), q1 = r*{[(F1 + F2)/k] +(F2/k)] = 7.20*10-6 C q2 = r*{[(F1 + F2)/k] -(F2/k)] = 2.80*10-6 C
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