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3 June, 05:34

An electron with a speed of 1.1 * 107 m/s moves horizontally into a region where a constant vertical force of 3.7 * 10-16 N acts on it. The mass of the electron is 9.11 * 10-31 kg. Determine the vertical distance the electron is deflected during the time it has moved 22 mm horizontally.

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  1. 3 June, 07:09
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    GIven data:

    Speed V = 1.1*10⁷ m/s

    Force F = 3.7*10⁻¹⁶N

    Mass of electron = m = 9.11*10⁻³¹Kg

    Horizontal Distance X = 2.2*10⁻²m

    Vertical distance Y = ?

    Solution:

    As we know that,

    F = ma

    where a is the acceleration of electron.

    (3.7*10⁻¹⁶) = (9.11*10⁻³¹) a

    a = 4.06*10¹⁴m/s² upward.

    ∵opposing weight of electron is negligible.

    now we find how long it takes the electron to move 22mm horizontally.

    X = V*T

    T = X/V

    T = 2.2*10⁻²/1.1*10⁷

    = 2*10⁻⁹s

    now we can find out the vertical distance Y.

    Y = 0.5aT²

    = (0.5) (4.06*10¹⁴) (2*10⁻⁹) ²

    = 8.12*10⁻⁴m
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