Answer:
x₃ = 0.397 m
Explanation:
For this exercise we use the vector sum of the forces, where the force is electric given by the Coulomb equation
F = [tex]k \frac{q_1q_2}{r^2}[/tex]
We also use that charges of the same sign repel and charges of a different sign attract.
In this case the force between load 1 and 3 has one direction and the force between 2 and 3 has the opposite direction, in the exercise they ask that the force on load 3 be zero
∑ F = F₁₃ - F₂₃ = 0
F₁₃ = F₂₃
we write the expressions of the forces, the distances are
r₁₃ = (x₁-x₃) ²
r₂₃ = (x₂ - x₃) ²
[tex]k \frac{q_1 \ q_3}{(x_1-x_3)^2} = k \frac{q_2 \ q_3}{(x_2-x_3)^2 }[/tex]
x₁ = 0 m
[tex]\frac{q_1}{x_3^2 } = \frac{q_2 }{(x_2 - x_3)^2 }[/tex]
q₁ (x₂ - x₃) ² = q₂ x₃²
[tex]\sqrt{\frac{q_1}{q_2} }[/tex] (x₂ -x₃) = x₃
we substitute the values
[tex]\sqrt\frac{9}{7} }[/tex] (0.4 -x₃) = x₃
x₃ (1 + [tex]\sqrt{9/7}[/tex]) = [tex]\sqrt{9/7}[/tex] 0.4
x₃ (1.13389) = 0.453557
x₃ = 0.453557 / 1.13389
x₃ = 0.397 m
