Question

Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.154 N when their center-to-center separation is 42.2 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 0.0430 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.)

Answer 1

Suppose initial charge on Sphere 1 = q1

Suppose initial charge on Sphere 2 = q2

r = distance between both charges = 42.2 cm = 0.422 m

Force between both charges = 0.154 N

Now electrostatic force is given by:

F = k*q1*q2/r^2

q1*q2 = F*r^2/k

Using given values

q1*q2 = 0.154*0.422^2/(9*10^9)

q1*q2 = -3.047*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 = 42.2 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 (0.0430*0.422^2/(9*10^9))

Q = 0.922*10^-6 C

So,

(q1 + q2)/2 = 0.922*10^-6 C

q1 + q2 = 1.844*10^-6 C

We know that

q1*q2 = -3.047*10^-12 C

q1*(1.844*10^-6 - q1) = -3.047*10^-12

Solving above quadratic equation using scientific calculator

q1 = -1.052*10^-6 C

q2 = 2.896*10^-6 C

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