Answer:
q₁ =± 1.30 10⁻⁶ C and q₂ = ± 1.28 10⁻⁶ C
Explanation:
We will solve this problem with Coulomb's law
F = K q₁q₂ / r²
Where the Coulomb constant is value 8.99 10⁹ N m² / C²
Let's apply this equation to our problem
Case 1
F1 = k q₁ q₂ / r₁²
Where r₁ = 0.440 m and F1 = 0.0765 N
Case 2
The charges are the same
F2 = k q q / r₂²
With r₂ = r₁ = 0.440 m, the spheres are fixed and the force is F2 = 0.100 N
When the spheres are joined with the wire, the charge is distributed, distributed and matched in the two spheres
q₁ + q₂ = 2 q
Let's replace
F2 = k ½ (q₁ + q₂) / r²
Let's write the two equations and solve the system of equations
F1 = k q₁ q₂ / r²
F2 = ½ k (q₁ + q₂) / r²
F1 r² / k = (q₁ q₂)
F2 r² / k = (q₁ + q₂)/2
q₁ = 2F2 r² / k - q₂
We substitute in the other equation
F1 r² / k = (2F2 r² / k - q₂) q₂
0 = -F1 r² / k + (2F2 r² / k) q₂ - q₂²
Let's solve the second degree equation
F1 r² / K = 0.0765 0.440² / 8.99 10⁹
F1 r² / K = 1.65 10⁻¹²
(2F2 r² / k) =2 0.10 0.44² / 8.99 10⁹
(2F2 r2 / k) = 4.30 10⁻¹²
q₂² - 4.30 10⁻¹² q₂ + 1.65 10⁻¹² = 0
q₂ = ½ {4.30 10⁻¹² ± √ [(4.30 10⁻¹²)² - 4 1.65 10⁻¹²]}
q₂ = ½ {4.30 10⁻¹² ± 2,569 10⁻⁶}
q₂ = ± 1.2845 10⁻⁶ C
Now we calculate q1
F1 = k q₁ q₂ / r²
q₁ = F1 r² / (k q₂)
q₁ = 0.0765 0.440² / (8.99 10⁹ 1.2845 10⁻⁶)
q₁ = 1.30 10⁻⁶ C
