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.)

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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the shielding can reduce the coupling of radio waves, electromagnetic fields and electrostatic fields. a conductive enclosure used to block electrostatic fields is also known as a faraday cage. the amount of reduction depends very much upon the material used, its thickness, the size of the shielded volume and the frequency of the fields of interest and the size, shape and orientation of apertures in a shield to an incident electromagnetic

from wikipedia