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1 November, 10:42

What is the magnitude of electrostatic force of attraction between two alpha particles separated

at a distance 5.3x10-11m? (1 apha particle = change of proton)

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Answers (1)
  1. 1 November, 14:09
    0
    Answer: F = 3.28 * 10^-7 N

    Explanation:

    Alpha particle charge (q) = 2e

    e = 1.6 * 10^-19

    q = 2e = 2 * (1.6 * 10^-19) = 3.2 * 10^-19

    1/4πEo = 9 * 10^9

    Distance (r) = 5.3 * 10^-11m

    The force of attraction between the two particles is given by:

    F = (1/4πEo) (q1q2 / r^2)

    F = [ (9 * 10^9) (3.2 * 10^-19) (3.2 * 10^-19) ] / (5.3 * 10^-11) ^2

    F = (92.16 * 10^ (9 - 19 - 19)) / 28.09 * 10^-22

    F = (92.16 * 10^-29) / 28.09 * 10^-22

    F = 3.28 * 10^ (-29 + 22)

    F = 3.28 * 10^-7 N
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