The electric field strength is 1.60×104 N/C inside a parallel-plate capacitor with a 1.50 mm spacing. An electron is released from rest at t

Question

The electric field strength is 1.60×104 N/C inside a parallel-plate capacitor with a 1.50 mm spacing. An electron is released from rest at the negative plate. You may want to review (Pages 688 – 690) . Part A What is the electron’s speed when it reaches the positive plate? Express your answer with the appropriate units. nothing nothing

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Thái Dương 6 months 2021-08-19T13:23:39+00:00 1 Answers 0 views 0

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    2021-08-19T13:25:29+00:00

    Answer:

    2903446.22825 m/s

    Explanation:

    m = Mass of electron = 9.11\times 10^{-31}\ kg

    q = Charge of electron = 1.6\times 10^{-19}\ C

    E = Electric field = 1.6\times 10^{4}\ N/C

    t = Time taken

    u = Initial velocity = 0

    v = Final velocity

    s = Displacement = 1.5 mm

    Acceleration is given by

    a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{1.6\times 10^{-19}\times 1.6\times 10^{4}}{9.11\times 10^{-31}}\\\Rightarrow a=2.81\times 10^{15}\ m/s^2

    Displacement is given by

    s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0+\dfrac{1}{2}at^2\\\Rightarrow t=\sqrt{\dfrac{2s}{a}}\\\Rightarrow t=\sqrt{\dfrac{2\times 1.5\times 10^{-3}}{2.81\times 10^{15}}}\\\Rightarrow t=1.0332548855\times 10^{-9}\ s

    Velocity is given by

    v=u+at\\\Rightarrow v=0+2.81\times 10^{15}\times 1.0332548855\times 10^{-9}\\\Rightarrow v=2903446.22825\ m/s

    The velocity of the electron is 2903446.22825 m/s

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