Two small identical conducting balls start with different amounts of charge on t

Question

Two small identical conducting balls start with different amounts of charge on them. Initially, when they are separated by 90.0 cm (this is the center-to-center distance), one ball exerts an attractive force of 2.50 N on the second ball. The balls are then touched together briefly (charge is conserved in this process), and then again separated by 90.0 cm. Now both balls have a positive charge, and the force that one ball exerts on the other is a repulsive force of 1.20 N. For this problem, use k = 9.0 * 10^9 N m^2/C^2, and assume that the radius of each ball is small compared to 90.0 cm. After the two balls have been touched together, what is the charge on each ball Before they were touched together, one ball had a positive charge. How much charge did that ball have Before they were touched together, one ball had a negative charge. How much charge did that ball have

Explanation / Answer

Force between the two balls is given by:

F = k*q1*q2/r^2, where k = 9*10^9 Nm^2/C^2

Let initial charge on the balls are q1 and q2
So from the first relation(the force is attractive hence one of them will be negative),
-2.5 = 9*10^9 *q1*q2/(.90^2)
or q1*q2 = -2.5*(.90^2)/(9*10^9)
or q1*q2 = -2.25*10^-10 ------(1)

and from the second relation ,as they are touched, both have same amount of charge = (q1+q2)/2

So,
k*((q1+q2)/2)^2/(.90^2) = 1.20 N
((q1+q2)/2)^2 = 1.20*(.90^2)/(9*10^9) = 1.08*10^-10
or q1+q2 = 2.078*10^-5 -----------(2)

Solving equation (1) and (2) we get :

q1*q2 = -2.25*10^-10

q1+q2 = 2.078*10^-5

q1 and q2 will be solution of the quadratic equation,

x^2 - 2.078*10^-5 x + 2.25*10^-10 = 0

hence q1 = 2.863*10^-5 C and q2 = -0.7856*10^-5 C

a)

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