Answer:
F_net = 9.87 10⁻⁴ N
Explanation:
Let's use that force is a vector magnitude
∑ F = F₁₃ + F₂₃
De bold arfe vectros. The force is the electric force, we use that charges of the same sign repel and when the charges are of a different sign they attract
the charges q1 and q2 are negative and the charge q3 is positive with the positions y1 = 38 m, y2 = -7m, y3 = 16 m
∑ F = F₁₃ - F₂₃
F_net = [tex]k \frac{q_1q_3}{r_{13}^2 } - k \frac{q_2q_3}{r_{23}^2 }[/tex]
in this case q₁ = q₂ = q
F_net = k q q₃ ( )
let's look for the distance
r₂₃ = y₂ - y₃
r₂₃ = -7 -16
r₂₃ = - 23 m
r₁₃ = 38 - 16
r₁₃ = 22 m
let's calculate
F_net = 9 10⁹ 24 26 10⁻¹² ( )
F_net = 5.616 ( 1.758 10⁻⁴ )
F_net = 9.87 10⁻⁴ N
