8c21p7) Two identical conducting spheres, fixed in place, attract each other wit

Question

8c21p7) Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of -0.8515 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 0.1586 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. What is the smaller value? What is the larger value?

Explanation / Answer

two charges: +q1 and -q2

F = 0..8515 N, d = 50 cm.

F = kq1q2/d2, so q1q2 = Fd2/k = 23.67*10-12 C2 -------(1)
Now say X charge move to q2 from q1
q1 - x = q2+x
x = (q1 - q2)/2
New charge on q1 = q1+ (q1 - q2)/2  = (q1 - q2)/2
Similarly on q2 new charge will be = (q1 - q2)/2
Say new charge = q
q = (q1 - q2)/2, F' = 0.1586 N

F' = kq2/d2

q2 = F'd2/k

(q1 - q2)2 = 4F'd2/k

q1 - q2 = 2d(F'/k) = 4.2*10-6 C

q1 = q2 + 4.2*10-6 C

plug it into (1)

q22 + 4.2*10-6 q2 - 23.67*10-12 C2 = 0

solve it and get

q2 =  3.199*10-6 C

q1 = 7.399*10-6 C

answer: (a) the negative charge = - 3.199*10-6C; (b) the positive charge = 7.399*10-6C

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