Respuesta :
Answer:
F_total = -65.5 N
Explanation:
Coulomb's law is
F = [tex]k \frac{q_1 q_2}{r^2}[/tex]k q1q2 / r2
In this exercise the force on the charge q3 is asked, for this we use that the force is a vector
F_total = F₁₃ + F₂₃
let's look for every force
F₁₃ = [tex]k \frac{q_1 q_3}{r_{13}^2}[/tex]
it is indicated that charge 1 is equal to q₁ = -11.5 nC = -11.5 10⁻⁹ C, located at x₁ = -1.675 mm = -1.675 10⁻³ m and charge q₃ = 48.0 nC = 48.0 10⁻⁹ C located at x₃ this position is not written in the exercise, suppose the position x₃ = -0.5 mm = 0.5 10⁻³ m
the distance is
r₁₃ = [tex]\sqrt{(x_3 - x_1 )^2}[/tex]
r₁₃ = [tex]\sqrt{(0.5-1.675)^2} \ 10^{-3}[/tex]
r₁₃ = 1.175 10⁻³ m
we calculate
F₁₃ = [tex]\frac{9 \ 10^{9} 11.5\ 10^{-9} 48.0\ 10^{-9} }{(1.175 10^{-3})^2 }[/tex]
F₁₃ = 3.598 N
as the charge q₁ is negative and the charge q₃ is positive the force is attractive directed to the right
we look for F₂₃, where q₂ = 40 nC = 40.0 10⁻⁹ C located at x₂ = 0
F₂₃ = [tex]k \frac{q_2q_3}{r_{23}^2}[/tex]
r₂₃ = [tex]\sqrt{(x_3-x_2)^2}[/tex]
r₂₃ = [tex]\sqrt{(0.5 -0)} \ 10^{-3}[/tex]RA (-0.5 0) 2 103
r₂₃ = 0.5 10⁻³ m
F₂₃ = [tex]\frac{9 \ 10^{9}\ 40\ 10^{-9} \ 48.0\ 10^{-9} x}{(0.5 \ 10^{-3})^2 }[/tex]
F₂₃ = 6.912 10¹ N
F₂₃ = 69.12 N
as the two charges are of the same sign, the force is repulsive, therefore it is directed to the left
the total force is
F = total = 3.598 - 69.12
F_total = -65.5 N
the negative sign indicates that the force is to the left
