Answer:
q₂ = 11.9988 C
, q₁ = 0.0012 C
Explanation:
For this exercise we will use Coulomb's law
F = K q₁q₂ / r²
They indicate that the sum of the charges is
q₁ + q₂ = 12
the force between them for r = 4 m is F = 8 10⁻³ N
q₁q₂ = F r² / K
q₁q₂ = 8 10⁻³ 4² / 8.99 10⁹
q₁q₂ = 14.24 10⁻¹²
substitute we write our system of equations
q₁ + q₂ = 12
q₁q₂ = 14.24 10⁻¹²
This system has two equations and two unknowns so it can be solved, let's solve in the first equation and substitute in the second
(12- q₂) q₂ = 14.24 10⁻¹²
12q₂ - q₂² - 14.24 10⁻¹² = 0
q₂² - 12 q₂ +14.24 10⁻¹² = 0
we solve the quadratic equation
q₂ = [12 ±√ (12² - 4 14,24 10⁻¹²)] / 2
q₂ = [12 ± 11,9999999] / 2
the two solutions are
q2 = 11,9999999997 C
q2 = 0.000000003
we substitute in the other equation to find the other charge
q₁ + q₂ = 12
q₁ = 12- q₂
q₁ = 0.000000003 C
this value is very small, I think the value of the force is wrong, if the force is F = 8 10⁶ N
q₁q₂ = 14.24 10⁻³
the quadratic equation is
q₂² - 12 q₂ + 14.24 10⁻³ = 0
the solution is
q₂ = [12 ±√ (12² - 4 14,24 10⁻³)] / 2
q2 = [12 ±√ (143,940)] / 2
q2 = [12 ± 11,9975] / 2
the two results are
q₂ = 11.9988 C
q₂ = 0.0025 C
as in the problem it indicates that q1 is smaller the correct answer is
q₂ = 11.9988 C
q₁ = 12-q₂
q₁ = 0.0012 C
