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1 August, 15:35

Two tiny spheres of mass 5.80 mg carry charges of equal magnitude, 77.0 nC, but opposite sign. They are tied to the same ceiling hook by light strings of length 0.530 m. When a horizontal uniform electric field E that is directed to the left is turned on, the spheres hang at rest with the angle θ between the strings equal to 58.0∘

What is the magnitude E of the field?

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  1. 1 August, 15:38
    0
    Mass m = 5.80 mg = 5.80 x 10⁻⁶ kg

    Charge q = 77 nC = 77 x 10⁻⁹ C

    String length L = 0.530 m

    angle θ = 58.0°

    Here we can force on one mass

    gravitational force F1 = mg is acting in downward direction

    F1 = (5.80 x 10⁻⁶ kg) (9.81 m/s²)

    F1 = 57 x 10⁻⁶ N

    Second force is electrostatic force due to another charge

    F2 = kqq/r₂

    Where r is the distance between two charges

    r = 2 LSinθ = 2 (0.53) Sin58

    r = 0.9 m

    F2 = (9x 10⁹) (77 x 10⁻⁹) (77 x 10⁻⁹) / (0.9²)

    F2 = 65.9 x 10⁻⁶ N

    Third force is F3 = qE

    Fourth force is tension in the string

    In equilibrium sum of forces along horizontal will be zero

    and sum of forces along vertical will be zero

    First we will use sum of vertical forces equal to zero

    TCosθ - F1 = 0

    T Cos58 - (57 x 10⁻⁶) = 0

    T = 107.56 x 10⁻⁶ N

    Now Sum of horizontal forces = 0

    F3 - F2 - TSinθ = 0

    qE - (65.9 x 10⁻⁶) - (107.56 x 10⁻⁶) = 0

    (77 x 10⁻⁹) E = 173.46 x 10⁻⁶

    E = 2253 N/C
  2. 1 August, 15:53
    -2
    Look it up on google
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