At a separation distance of 0.500-m, two like-charged balloons experience a repulsive force of 0.320 N. If the distance is is decreased by a

Question

At a separation distance of 0.500-m, two like-charged balloons experience a repulsive force of 0.320 N. If the distance is is decreased by a factor of 3 and the charge on one of the balloons is doubled, then the repulsive force would be?

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Hưng Khoa 6 months 2021-07-26T08:56:04+00:00 1 Answers 15 views 0

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    2021-07-26T08:57:58+00:00

    Answer:

    2.9 N

    Explanation:

    When the separation distance, r, is 0.5 m, the electrostatic force is 0.32 N. Electrostatic force is given as:

    F = (k * q1 * q2) / r²

    Where F = force acting on the balloons

    k = Coulombs constant

    Therefore:

    0.32 = (k * q1 * q2) / 0.5²

    => k * q1 * q2 = 0.32 * 0.5² ————(1)

    When the distance is decreased by 3, that is r = r/3 = 0.5/3

    F = (k * q1 * q2) / (0.5/3)² ————(2)

    Putting (1) into (2):

    => F = (0.32 * 0.5²) / (0.5/3)²

    F = (0.32 * 0.5² * 3²) / 0.5²

    F = 2.9 N

    Therefore, the force would be 2.9 N

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )