An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling 4.50 m in the first 3.0

Question

An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling 4.50 m in the first 3.00μs3.00μs after it is released. (a) What are the magnitude and direction of the electric field? (b) Are we justified in ignoring the effects of gravity? Justify your answer quantitatively.

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Tài Đức 6 months 2021-07-28T03:12:17+00:00 1 Answers 26 views 0

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    2021-07-28T03:14:13+00:00

    Answer:

    Explanation:

    A. Using

    E= ma/q

    E=m/q(2s/t²)

    So

    E= 9.1×10^-31/1.6*10^-19( 2*4.5/ 3*10-12)

    E=5.7NC

    The electric field has to be downward since the force is positive that is upward

    B.

    The electron acceleration is of the order of 10^11 times greater so for practical purposes we neglect the effect of gravity

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