Answer :
Answer:
q₂ = ± 1.306 10⁻⁶ C
q₂ = ± 3.073 10⁻⁶ C
Explanation:
We must solve this exercise using Coulomb's law, let's write the equilibrium equation for each case
first case
F₁ = k q₁ q₂ / r² (1)
where F1 is the force on sphere 1, with a value of F₁ = 0.110 N; r is the distance between them r = 0.573 m
second case.
The spheres were joined with a conductive wire, so the charge of the two is equal, then the wire is removed and
-F₂ = k q₃ q₃ / r² (2)
In this case the force is expelling and is equal to F₂ = 0.0489 N
when the spheres are joined with the wire the charge of the spheres is distributed evenly between them two
q₁ + q₂ = 2 q₃ (3)
we can see that there are three equations with three unknowns for which the system can be solved
let's substitute equation 3 in 2
- F₂ = k (q₁ + q₂)² / r²
we join this equation with equation 1
F₁ = k q₁ q₂ / r²
-F₂ = k (q₁ + q₂)²/4r²
q₁ = F₁ r² / k q₂
- F₂ = k (F₁ r² / k q₂ + q₂)²/4r²
Let's replace the values and work out
- 0.0489 = 9 109 (0.110 0.573² / (9 10⁹ q₂) + q₂)² / 0.573²
- 0.0489 4 0.573²/9 10⁹ = (0.110 0.573² / (9 10⁹ q₂) + q₂)²
- 7.1356 10⁻¹² = (4.013 10⁻¹² / q₂ + q₂)²
- 7.1356 10⁻¹² = 1 / q₂² (4.013 10⁻¹² + q₂² )²
- 7.1356 10⁻¹² q₂² = (16.104 10⁻²⁴ + 4.013 10⁻¹² q₂² + q₂⁴
q₂⁴ + q₂² (4,013 10⁻¹² + 7,1356 10⁻¹²) + 16,104 10⁻²⁴ = 0
we change variables
q’ = q₂²
q'² + q' 11.1486 10⁻¹² + 16.104 10⁻²⁴ = 0
we solve this quadratic equation
q '= [- 11.1486 10⁻¹² ±√ ((11.1486 10⁻¹² )² - 4 16.104 10⁻²⁴)] / 2
q '= [- 11.1486 10 12 ±√ (59.8753 10⁻²⁴)] / 2
q '= [- 11.1486 ± 7.7379] / 2 10⁻¹²
q'1 = -1.70535 10⁻¹²
q'2 = -9.44325 10⁻¹²
q' = q₂²
q₂ = √ q'
q₂ = ± 1.306 10⁻⁶ C
q₂ = ± 3.073 10⁻⁶ C
since q 'is squared of these charges they can be positive or negative giving the same result.