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Sagot :
Answer:
F_net = 1.125 10⁻⁶ N
positive direction of the x axis
Explanation:
The electric force is given by Coulomb's law
F =[tex]k \frac{q_1q_2}{r_{12}^2}[/tex]
also used charge of equal sign repel each other.
We appease this information to our problems
the charge q₁ is positive and the charge q₃ is positive for which the force is repulsive; ; a charge q₂ is negative so the force is attractive.
If we apply that the forces are vectors
F_net = - F₁₃ + F₂₃
F_net = [tex]-k \frac{q_1q_3}{r_{13}^2} + k \frac{q_2q_3}{r_{23}^2}[/tex]
F_net = k q₃ ( [tex]- \frac{q_1}{r_{13}^2} + \frac{q_2}{r_{23}^2}[/tex] )
let's look for the distances
r₁₃ = 2-0 = 2 cm = 2 10⁻² m
r₂₃ = 4-0 = 4 cm = 4 10⁻² m
let's substitute
F_net = 9 10⁹ 5 10⁻⁹ ( [tex]- \frac{1 \ 10^{-9}}{2 \ 10^{-2}} + \frac{3 \ 10^{-9}}{4 \ 10^{-2}}[/tex])
F_net = 45 (- 0.5 10⁻⁷ + 0.75 10⁻⁷)
F_net = 1.125 10⁻⁶ N
the positive sign indicates that the outside has a positive direction of the x axis, that is, it is to the right
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