An electron with velocity v = 1.0 ´ 106 m/s is sent between the plates of a capacitor where the electric field is E = 500 V/m. If the distan

Question

An electron with velocity v = 1.0 ´ 106 m/s is sent between the plates of a capacitor where the electric field is E = 500 V/m. If the distance the electron travels through the field is 1.0 cm, how far is it deviated (Y) in its path when it emerges from the electric field? (me = 9.1 ´ 10 – 31 kg, e = 1.6 ´ 10 – 19 C)

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Huyền Thanh 2 months 2021-08-02T18:44:48+00:00 1 Answers 3 views 0

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    2021-08-02T18:46:22+00:00

    Answer:

    The deviation in path is 4.39 \times 10^{-3}

    Explanation:

    Given:

    Velocity v = 1 \times 10^{6} \frac{m}{s}

    Electric field E = 500 \frac{V}{m}

    Distance x = 1 \times 10^{-2} m

    Mass of electron m = 9.1 \times 10^{-31} kg

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

    Time taken to travel distance,

        t = \frac{x}{v}

        t = \frac{1 \times 10^{-2} }{1 \times 10^{6} }

        t = 10^{-8} sec

    Acceleration is given by,

      F = qE

    ma = qE

       a = \frac{qE}{m}

       a = \frac{1.6 \times 10^{-19} \times 500}{9.1 \times 10^{-31} }

       a = 8.77 \times 10^{13} \frac{m}{s^{2} }

    For finding the distance, we use kinematics equations.

       y = vt + \frac{1}{2}  at^{2}

    Where v = 0 because here initial velocity zero

       y = \frac{1}{2} at^{2}

       y = \frac{1}{2} \times  8.77 \times 10^{13 } \times (10^{-2} )^{2}

       y = 4.39 \times 10^{-3} m

    Therefore, the deviation in path is 4.39 \times 10^{-3}

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