Question

Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of -0.2532 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 0.2525 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. Enter the larger value here.

Answer 1

Suppose initial charge on Sphere 1 = q1

Suppose initial charge on Sphere 2 = q2

r = distance between both charges = 50 cm = 0.5 m

Force between both charges = -0.2532 N

Now electrostatic force is given by:

F = k*q1*q2/r^2

q1*q2 = F*r^2/k

Using given values

q1*q2 = -0.2532*0.5^2/(9*10^9)

q1*q2 = -7.03*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 = 50 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 (0.2525*0.5^2/(9*10^9))

Q = 2.65*10^-6 C

So,

(q1 + q2)/2 = 2.65*10^-6 C

q1 + q2 = 5.3*10^-6 C

We know that

q1*q2 = -7.03*10^-12 C

q1*(5.3*10^-6 - q1) = -7.03*10^-12

Solving above quadratic equation, using Scientific calculator

q1 = -1.10*10^-6 C & q2 = 6.40*10^-6 C

|q1| = 1.10*10^-6 C

|q2| = 6.40*10^-6 C

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