Respuesta :
Answer:
x₁ = 0.62 m
Explanation:
In this exercise the force is electric, given by Coulomb's law
F =[tex]k \frac{q_1q_2}{r^2}[/tex]
This force is a vector, since the three charges are in a line we can reduce the vector sum to a scalar sum.
For the sense of force let us use that charges of the same sign repel and charges of the opposite sign attract.
∑ F = F₁₂ - F₂₃
They ask us to find the point where the summaries of the force is zero.
F₁₂ - F₂₃ = 0
F₁₂ = F₂₃
let's fix a reference system located in the first charge (more to the left), the distance between the two charges is d = 1.5 m and x is the distance to the location of the second sphere
k q₁q₂ / x² = k q₂q₃ / (d-x) ²
q₁ (d-x) ² = q₃ x²
let's solve
d² - 2 x d + x² = [tex]\frac{q_3}{q_1}[/tex] x²
x² (1 - [tex]\frac{q_3}{q_1}[/tex]) - 2x d + d² = 0
we substitute the values
x² (1- 4/2) - 2 1.5 x + 1.5² = 0
x² (-1) - 3.0 x + 2.25 = 0
x² + 3 x - 2.25 = 0
let's solve the quadratic equation
x = [-3 ± [tex]\sqrt{ 3^2 + 4 \ 2.25}[/tex]] / 2
x = [-3 ± 4.24] / 2
x₁ = 0.62 m
x₂ = 3.62 m
since it indicates that the charge q₂ e places between the spheres, the correct solution is
x₁ = 0.62 m
