Answer:
The initial charges on the spheres are [tex]6.796\ 10^{-6}\ c[/tex] and [tex]-3.898\ 10^{-6}\ c[/tex]
Explanation:
Electrostatic Force
Two charges q1 and q2 separated a distance d exert a force on each other which magnitude is computed by the known Coulomb's formula
[tex]\displaystyle F=\frac{K\ q_1\ q_2}{d^2}[/tex]
We are given the distance between two unknown charges d=50 cm = 0.5 m and the attractive force of -0.9537 N. This means both charges are opposite signs.
With these conditions we set the equation
[tex]\displaystyle F_1=\frac{K\ q_1\ q_2}{0.5^2}=-0.9537[/tex]
Rearranging
[tex]\displaystyle q_1\ q_2=\frac{-0.9537(0.5)^2}{k}[/tex]
Solving for q1.q2
[tex]\displaystyle q_1\ q_2=-2.6492.10^{-11}\ c^2\ \ ......[1][/tex]
The second part of the problem states the spheres are later connected by a conducting wire which is removed, and then, the spheres repel each other with an electrostatic force of 0.0756 N.
The conducting wire makes the charges on both spheres to balance, i.e. free electrons of the negative charge pass to the positive charge and they finally have the same charge:
[tex]\displaystyle q=\frac{q_1+q_2}{2}[/tex]
Using this second condition:
[tex]\displaystyle F_2=\frac{K\ q^2}{0.5^2}=\frac{K(q_1+q_2)^2}{(4)0.5^2}=0.0756[/tex]
[tex]\displaystyle q_1+q_2=2.8983\ 10^{-6}\ C[/tex]
Solving for q2
[tex]\displaystyle q_2=2.8983\ 10^{-6}\ C-q_1[/tex]
Replacing in [1]
[tex]\displaystyle q_1(2.8983\ 10^{-6}-q_1)=-2.64917.10^{-11}[/tex]
Rearranging, we have a second-degree equation for q1.
[tex]\displaystyle q_1^2-2.8983.10^{-6}q_1-2.64917.10^{-11}=0[/tex]
Solving, we have two possible solutions
[tex]\displaystyle q_1=6,796.10^{-6}\ c[/tex]
[tex]\displaystyle q_1=-3.898.10^{-6}\ c[/tex]
Which yields to two solutions for q2
[tex]\displaystyle q_2=-3.898.10^{-6}\ c[/tex]
[tex]\displaystyle q_2=6.796.10^{-6}\ c[/tex]
Regardless of their order, the initial charges on the spheres are [tex]6.796\ 10^{-6}\ c[/tex] and [tex]-3.898\ 10^{-6}\ c[/tex]
