A small spherical insulator of mass 3.70 10-2 kg and charge +0.600 µC is hung by a thin wire of negligible mass. A charge of −0.900 µC is he

Question

A small spherical insulator of mass 3.70 10-2 kg and charge +0.600 µC is hung by a thin wire of negligible mass. A charge of −0.900 µC is held 0.150 m away from the sphere and directly to the right of it, so the wire makes an angle θ with the vertical.
A) Find the angle.
B) Find the tension in the wire.

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Nguyệt Ánh 2 days 2021-07-21T22:38:29+00:00 1 Answers 0 views 0

Answers ( )

    0
    2021-07-21T22:39:40+00:00

    Answer:

    A) θ = 36.27°

    B) T = 0.4327N

    Explanation:

    A) If we imagine the free body diagram of this problem , and taking moments about the x-axis and equate to zero, we have;

    ΣFx = F_el – T_x = 0

    Now, From coulombs law, we know that F = k•q1•q2/r²

    Thus, applying to this question, we have;

    F_el – T_x gives

    (k•q_bob•q_free)/d² – T_x = 0

    Now, due to the angle θ, T_x when resolved, is T•sinθ

    Thus, we have;

    (k•q_bob•q_free)/d² – T•sinθ = 0

    And so;

    (k•q_bob•q_free)/d² = T•sinθ – – – (eq1)

    Now, resolving about the y-axis and equating to zero gives;

    ΣFy = T_y – m_bob•g = 0

    When T_y is resolved, it gives T•cosθ

    Thus;

    T•cosθ – m_bob•g = 0

    T•cosθ = m_bob•g

    T = m_bob•g/cosθ – – – – (eq2)

    Let’s put m_bob•g/ cosθ for T in eq 1 to get;

    (k•q_bob•q_free)/d² = (m_bob•g/ cosθ)•sinθ

    This gives;

    (k•q_bob•q_free)/d² =m_bob•g tanθ

    Making tanθ the subject ;

    tanθ = (k•q_bob•q_free)/(d².m_bob•g)

    We are given;

    q_bob = +0.600 µC = 0.6 x 10^(9)C

    q_free = −0.900 µC = −0.9 x 10^(9)C (we will use positive in calculation as the negative sign shows that electric field points radially towards the charge.)

    m_bob = 3.6 x 10^(-2) kg

    d = 0.15m

    k is coulombs constant with a standard value of 8.99 x 10^(9) N.m²/C²

    Plugging in relevant values ;

    tanθ = (8.99 x 10^(9)•0.6 x 10^(9)•0.9 x 10^(9))/(0.15²•3.6 x 10^(-2)•9.8)

    tanθ = 0.7339

    θ = tan^(-1)0.7339

    θ = 36.27°

    B) Let’s put 36.27° for θ in eq 2.

    Thus;

    T = (3.6 x 10^(-2)•9.8)/(cos 36.27)

    T = 0.4327N

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