**Answer:**

F_total = 0.275 / L² N, θ’= 156.6 º

**Explanation:**

For this exercise we use Coulomb's law

F = k q₁ q₂ / r₁₂²

In this case the force F₁₃ is attractive because the charge has a different sign and the outside F₂₃ is repulsive, we find each force and then add the vector

Let us find the distance between the charges as the triangle is equilateral the distance is the side of the triangle L, calculate the force

F₁₃ = 8.99 10⁹ 3 10⁻⁶ 7 10⁻⁶ / L²

F₂₃ = 8.99 10⁹ 5 10⁻⁶ 7 10⁻⁶ / L²

F₁₃ = 1.89 10⁻¹ / L²

F₂₃ = 3.15 10⁻¹ / L²

As the force is vectors, the easiest method to find it resulting is with the components, for local we use trigonometry, remember that the angles of an equilateral triangle are 60º

Sin 30 = F₁₃ₓ / F13

Cos 30 = F₁₃y / F13

F₁₃ₓ = F₁₃ sin 30

F₁₃y = F₁₃ cos 30

F₁₃ₓ = 1.89 10⁻¹ / L² sin 30

F₁₃y = 1.89 10⁻¹ / L² cos 30

F₁₃ₓ = 0.945 10⁻¹ / L²

F₁₃y = 1,637 10⁻¹ / L²

We do the same for force F₂₃

Let's take the angle of the horizontal 60º

Cos 60 = F₂₃ₓ / F₂₃

Sin 60 = F₂₃y / F₂₃

F₂₃ₓ = F₂₃ cos 60

F₂₃y = F₂₃ sin 60

F₂₃ₓ = 3.15 10⁻¹ / L² cos 60

F₂₃y = 3.15 10⁻¹ / L² sin 60

F₂₃ₓ = 1.575 10⁻¹ / L²

F₂₃y = 2.728 10⁻¹ / L²

Now we can find the components of the total force

F_total = F_totalx i + F_totaly

F_totalx = -F₁₃ₓ - F₂₃ₓ

F_totalx = - (0.945 +1.575) 10⁻¹ / L²

F_totalx = -0.252 / L²

F_totaly = -F₁₃y + F₂₃y

F_totaly = (- 1,637 + 2,728) 10⁻¹ / L²

F_totaly = 0.1091 / L²

The total force is

F_total = (-0.252i +0.1091j ) / L²

To give the result in the form of a module and angle, let's use the Pythagorean theorem

F_total = √ (Fₓ² + Fy²)

F_total = 1 / L² √ (0.252² + 0.1091²)

F_total = 0.275 / L² N

For the angle let's use trigonometry

tan θ = Fy / Fₓ

θ = tan⁻¹ (0.1091 / 0.252)

θ = 23.4º

To measure this angle from the positive side of the x axis

θ’= 180 - 23.4

θ’= 156.6 º

For a specific value we must know the distance from the side of the triangle