Answer the question regarding the following experiment:

q₂ = -4.80 10⁻⁴  C  = - 0.48 mC, charge is negative

Explanation:

Let's use coulomb's law

           F =

and the sum of forces, remember that charges of the same sign repel and of different sign attract

           ∑ F = F₁₃ + F₂₃           (1)

Let's start by fixing a reference system located at charge 1 with the positive direction to the right.  In the problem it indicates that the net force on charge 3 is F = - 20.0 N, the negative sign indicates that the force is towards the left

let's look for every force, the charge q₁ = 12 10-⁻³ C and q₃ = 15 10⁻³ C

           F₁₃ =

           F₁₃ = 9 10⁹ 12.0 15.0 10⁻⁶ / (5-0)²

           F₁₃ = 64.8 10 3 N

This force is repulsive, that is, it is directed to the right

          F₂₃ = k \frac{q_2 q_3}{x_{23}^2}

          F₂₃ = 9 10⁹ q₂ 15.0 10⁻³ / (5-4)²

          F₂₃ = 135 10⁶ q₂  N

we substitute in equation 1

          -20.0 = 64.8 10³ + 135 10⁶ q₂

           q₂ = (-20 - 64.8 10³) / 135 10⁶

           q₂ = -4.80 10⁻⁴  C

the sign indicates that the charge is negative