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 0.126 N when their center-to-center separation is 46.9 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 0.0499 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

Suppose initial charge on Sphere 1 = q1

Suppose initial charge on Sphere 2 = q2

r = distance between both charges = 46.9 cm = 0.469 m

Force between both charges = 0.126 N

Now electrostatic force is given by:

F = k*q1*q2/r^2

q1*q2 = F*r^2/k

Using given values

q1*q2 = 0.126*0.469^2/(9*10^9)

q1*q2 = -3.08*10^-12 C

[See that q1*q2 will be negative as force is attractive which means one of them is negative and other one is positive].

Now when both spheres are brought into contact, after that charge will be equally distributed. Now charge on each sphere will be Q, where

Q = (q1 + q2)/2

Now when returned to distance r = 46.9 cm, force will be repulsive because both charge will have same sign either positive or negative, So

F1 = k*Q*Q/r^2

Q^2 = F1*r^2/k

Q = sqrt (0.0499*0.469^2/(9*10^9))

Q = 1.10*10^-6 C

So,

(q1 + q2)/2 = 1.10*10^-6 C

q1 + q2 = 2.20*10^-6 C

We know that

q1*q2 = -3.08*10^-12 C

q1*(2.20*10^-6 - q1) = -3.08*10^-12

Solving above equation

q1 = -0.97*10^-6 C

q2 = 3.17*10^-6 C

Please Upvote. Let me know if you have any doubt.

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