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
