An electron is initially moving at 1.4 x 107 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude 120 N/C. What is

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An electron is initially moving at 1.4 x 107 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude 120 N/C. What is the kinetic energy of the electron at the end of the motion

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RI SƠ 5 years 2021-08-26T19:35:08+00:00 1 Answers 10 views 0

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  1. Gerda
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    2021-08-26T19:36:18+00:00
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    Answer:

    K.E = 15.57 x 10⁻¹⁷ J

    Explanation:

    First, we find the acceleration of the electron by using the formula of electric field:

    E = F/q

    F = Eq

    but, from Newton’s 2nd Law:

    F = ma

    Comparing both equations, we get:

    ma = Eq

    a = Eq/m

    where,

    E = electric field intensity = 120 N/C

    q = charge of electron = 1.6 x 10⁻¹⁹ C

    m = Mass of electron = 9.1 x 10⁻³¹ kg

    Therefore,

    a = (120 N/C)(1.6 x 10⁻¹⁹ C)/(9.1 x 10⁻³¹ kg)

    a = 2.11 x 10¹³ m/s²

    Now, we need to find the final velocity of the electron. Using 3rd equation of motion:

    2as = Vf² – Vi²

    where,

    Vf = Final Velocity = ?

    Vi = Initial Velocity = 1.4 x 10⁷ m/s

    s = distance = 3.5 m

    Therefore,

    (2)(2.11 x 10¹³ m/s²)(3.5 m) = Vf² – (1.4 x 10⁷)²

    Vf = √(1.477 x 10¹⁴ m²/s² + 1.96 x 10¹⁴ m²/s²)

    Vf = 1.85 x 10⁷ m/s

    Now, we find the kinetic energy of electron at the end of the motion:

    K.E = (0.5)(m)(Vf)²

    K.E = (0.5)(9.1 x 10⁻³¹ kg)(1.85 x 10⁷ m/s)²

    K.E = 15.57 x 10⁻¹⁷ J

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