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 -0.9580 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 0.1298 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.

It is also asking for the larger value.

Explanation / Answer

By Coulomb’s law F1 = kq1q2/r12 -------------------(1)

-0.9580 N = (9*109* q1q2)/0.52

q1q2 = -2.66*10-11C ………………..(2)

Now when both charges are connected and separated new charge on each sphere = q3 = (q1+ q2)/2

Thus q1+ q2 = 2q3----------------(3)

F1 = kq32/r12

0.1298 N = (9*109*q32)/0.52

q3 = 1.9*10-6 C    or   q3 = -1.9*10-6 C   

Assume q3 is positive thus q3 = 1.9*10-6 C    ( you will get same answer if you choose 2nd value)

Plugging this value in equation (3) we get,

q1+ q2 = 2*1.9*10-6 C   = 3.8*10-6 C  

q2 = (3.8*10-6 C - q1 ) ---------------(4)

Plugging eqn (4) in eqn (1),

F1 = kq1q2/r12

-0.9580 N = (9*109* q1(3.8*10-6 C - q1 ))/0.52

q1 = -3.6*10-6C     or   q1 = 7.4*10-6C      

Assume it’s q1 = -3.6*10-6C and plug this in eqn (4) , we will get

q2 = -3.6*10-6C     or   q2 = 7.4*10-6C      

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