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6 April, 22:22

tres cargas eléctricas cuyos valores son q1=3uc q2=-5uc y q3=-7uc de encuentran distribuidas en un triángulo. cuál es la magnitud de las fuerzas resultante sobré la carga q3 y su ángulo respecto a su eje horizontal

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  1. 6 April, 23:55
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    F_total = 0.275 / L² N, θ' = 156.6 º

    Explanation:

    For this exercise we use Coulomb's law

    F = k q₁ q₂ / r₁₂²

    In this case the force F₁₃ is attractive because the charge has a different sign and the outside F₂₃ is repulsive, we find each force and then add the vector

    Let us find the distance between the charges as the triangle is equilateral the distance is the side of the triangle L, calculate the force

    F₁₃ = 8.99 10⁹ 3 10⁻⁶ 7 10⁻⁶ / L²

    F₂₃ = 8.99 10⁹ 5 10⁻⁶ 7 10⁻⁶ / L²

    F₁₃ = 1.89 10⁻¹ / L²

    F₂₃ = 3.15 10⁻¹ / L²

    As the force is vectors, the easiest method to find it resulting is with the components, for local we use trigonometry, remember that the angles of an equilateral triangle are 60º

    Sin 30 = F₁₃ₓ / F13

    Cos 30 = F₁₃y / F13

    F₁₃ₓ = F₁₃ sin 30

    F₁₃y = F₁₃ cos 30

    F₁₃ₓ = 1.89 10⁻¹ / L² sin 30

    F₁₃y = 1.89 10⁻¹ / L² cos 30

    F₁₃ₓ = 0.945 10⁻¹ / L²

    F₁₃y = 1,637 10⁻¹ / L²

    We do the same for force F₂₃

    Let's take the angle of the horizontal 60º

    Cos 60 = F₂₃ₓ / F₂₃

    Sin 60 = F₂₃y / F₂₃

    F₂₃ₓ = F₂₃ cos 60

    F₂₃y = F₂₃ sin 60

    F₂₃ₓ = 3.15 10⁻¹ / L² cos 60

    F₂₃y = 3.15 10⁻¹ / L² sin 60

    F₂₃ₓ = 1.575 10⁻¹ / L²

    F₂₃y = 2.728 10⁻¹ / L²

    Now we can find the components of the total force

    F_total = F_totalx i + F_totaly

    F_totalx = - F₁₃ₓ - F₂₃ₓ

    F_totalx = - (0.945 + 1.575) 10⁻¹ / L²

    F_totalx = - 0.252 / L²

    F_totaly = - F₁₃y + F₂₃y

    F_totaly = ( - 1,637 + 2,728) 10⁻¹ / L²

    F_totaly = 0.1091 / L²

    The total force is

    F_total = (-0.252i + 0.1091j) / L²

    To give the result in the form of a module and angle, let's use the Pythagorean theorem

    F_total = √ (Fₓ² + Fy²)

    F_total = 1 / L² √ (0.252² + 0.1091²)

    F_total = 0.275 / L² N

    For the angle let's use trigonometry

    tan θ = Fy / Fₓ

    θ = tan⁻¹ (0.1091 / 0.252)

    θ = 23.4º

    To measure this angle from the positive side of the x axis

    θ' = 180 - 23.4

    θ' = 156.6 º

    For a specific value we must know the distance from the side of the triangle
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