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6 February, 09:20

A 2.0 g ball with a + 0.05 C charge is moved from the negative plate to the positive plate, and then is released. The potential difference between the plates is 12.0V. When the ball strikes the (-) plate its velocity is most nearly equal to

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  1. 6 February, 09:26
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    24.5 m/s

    Explanation:

    Since the work done by the electric field on the charge equals the kinetic energy of the ball, then

    qV = 1/2mv²

    and v = √ (2qV/m)

    where q = charge = + 0.05 C, V = potential difference = 12.0 V, m = mass of ball = 2.0 g = 0.002 kg

    Substituting the values into v, we have

    v = √ (2 * + 0.05 C * 12.0 V/0.002 kg)

    = √ (1.2 CV/.002 kg)

    = √ (600 CV/kg)

    = 24.5 m/s
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