Answer:
The initial charges of the spheres were q₁=6.7712×10⁻⁶C and q₂=-4.4350×10⁻⁶C.
Explanation:
As the spheres attract each other, the charges of the spheres are opposite.
The atracction force is given by:
F=[tex]-\frac{Kq_{1}q_{2}}{r^{2}}[/tex]
where:
K: Coulomb constant
q₁: charge of sphere 1
q₂: charge of sphere 2
r: distance between both charges
The electrostatic atraction force is 0.108 N so:
0.108N=-8.99×10⁹[tex]\frac{N*m^2}{C^2}[/tex] [tex]\frac{q_{1}q_{2}}{(0.5m)^2}[/tex]
q₁·q₂=[tex]-\frac{0.108N*(0.5m)^2}{8.99*10^9\frac{N*m^2}{C^2}}[/tex]
q₁·q₂=-3.003×10⁻¹² C²
When the wire is connected the charges are equally distributed as the spheres are identical. Hence, the final charge is of each sphere is [tex]\frac{q_1q_2}{2}[/tex]
The repel force is 0.360 N and it is given by:
F=[tex]\frac{K(\frac{q_1+q_2}{2})(\frac{q_1+q_2}{2})}{r^{2}}[/tex]
F=[tex]\frac{K(q_1+q_2)^2}{4r^{2}}[/tex]
Then, we get a secong equation:
(q₁+q₂)²=[tex]\frac{0.360N*4*(0.5m)^2}{8.99*10^9\frac{N*m^2}{C^2}}[/tex]
(q₁+q₂)=√4.004×10⁻¹¹ C²
q₁+q₂=6.3277×10⁻⁶ C
We solve the equation system:
[tex]\left \{ {{q_1=\frac{-3.003*10^{-12}C^2}{q_2} } \atop {q_1+q_2=6.3277*10^{-6}C}} \right.[/tex]
We replace q₁ in the second equation:
[tex]\frac{-3.003*10^{-12}C^2}{q_2} +q_2=6.3277*10^{-6}C[/tex]
[tex]-3.003*10^{-12}C^2+(q_2)^2=6.3277*10^{-6}C*q_2[/tex]
[tex](q_2)^2-6.3277*10^{-6}C*q_2-3.003*10^{-12}C^2=0[/tex]
The solutions are:
q₁=6.7712×10⁻⁶C
q₂=-4.4350×10⁻⁶C
