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8 August, 23:45

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.)

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  1. 8 August, 23:50
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    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
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