Question

Two charges q1 and q2 have a total charge of 12 C. When they are separated by 4 m, the force exerted by one charge on the other has a magnitude of 8 mN. Find q1 and q2 if both are positive so that they repel each other, and q1 is the smaller of the two. (For the universal constant k use the value 8.99 109 N m2/C2.)

Answer 1

q₂ = 11.9988 C , q₁ = 0.0012 C

Step-by-step 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