Answer:
F = 2,8289 i ^ + 1,0909 j ^) 10⁻² N
F = 3.0226 10⁻² N , θ = 21.16º
Step-by-step explanation:
For this exercise we use Coulomb's law
F = k q₁q₂ / r₁₂²
We also use that the force is a vector magnitude, so we must calculate each component of the force , see the adjoint for the direction of the vectors
X axis
Fₓ = -F₁₄ + F₁₃ₓ
Y axis
= F₁₂ -F_{13y}
let's look for the expression for each force
where the side of the square is a = 1.25 m
F₁₂ = k Q₁Q₂ / a²
F₁₄ = k Q₁Q₄ / a²
the distance between 1 and 3 is
d = √(a² + a²) = a √2
F₁₃ = k Q₁Q₃ / d²
let's use trigonometry to find the components
cos 45 = F₁₃ₓ / F₁₃
F₁₃ₓ = F₁₃ cos 45
F₁₃ₓ = k Q₁Q₃ / 2a²
sin 45 = F_{13y} / F₁₃
F_{13y} = F₁₃ sin 45
F_{13y} = k Q₁Q₃ / 2a² sin 45
Taking all terms, we substitute in the force for each axis
X axis
Fₓ = - k Q₁Q₄ / a² + k Q₁Q₃ / 2a₂ cos 45
Fₓ = k Q₁ / a² ( -Q₄ + Q₃ /2 cos 45)
Fₓ = 9 10⁹ 1.5 10⁻⁶ / 1.25² (- 4.5 10⁻⁶ + 3.5/2 cos 45 10⁻⁶)
Fₓ = 8.64 10³ (3.2626 10⁻⁶)
Fₓ = 2.8189 10⁻² N
Y axis
F_{y} = k Q₁Q₂ / a² - k Q₁Q₃ /2a² sin 45
F_{y} = k Q₁ / a² (Q₂ - Q₃ /2 sin45)
F_{y} = 9 10⁹ 1.5 10⁻⁶/ 1.25² (2.5 10⁻⁶ - 3.5/2 sin 45 10⁻⁶)
F_{y} = 8.64 10³ (1.26256 10⁻⁶)
F_{y} = 1.0909 10⁻² N
The result can be given in two ways
1) F = Fₓ i ^ + F_{y} j ^
F = 2,8289 i ^ + 1,0909 j ^) 10⁻² N
2) in the form of a module and an angle, for which we use the Pythagorean theorem and trigonometry
F = √ (Fₓ² + F_{y}²)
F = 10⁻² √ (2,8189² + 1,0909²)
F = 3.0226 10⁻² N
tan θ = F_{y} / Fx
θ = tan⁻¹ (F_{y} / Fₓ)
θ = tan⁻¹ (1.0909 / 2.8189)
θ = 21.16º