On a typical clear day, the atmospheric electric field points downward and has a magnitude of approximately 103 N/C. Compare the gravitation

Question

On a typical clear day, the atmospheric electric field points downward and has a magnitude of approximately 103 N/C. Compare the gravitational and electric forces on a small dust particle of mass 2.2 ✕ 10−15 g that carries a single electron charge. Fg FE = What is the acceleration (both magnitude and direction) of the dust particle? (Enter the magnitude in m/s2.) magnitude m/s2 direction —Select—

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Maris 3 years 2021-08-29T12:05:46+00:00 1 Answers 137 views 0

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  1. Edana
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    2021-08-29T12:07:26+00:00
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    Answer:

    a) FE = 0.764FG

    b) a = 2.30 m/s^2

    Explanation:

    a) To compare the gravitational and electric force over the particle you calculate the following ratio:

    \frac{F_E}{F_G}=\frac{qE}{mg}              (1)

    FE: electric force

    FG: gravitational force

    q: charge of the particle = 1.6*10^-19 C

    g: gravitational acceleration = 9.8 m/s^2

    E: electric field = 103N/C

    m: mass of the particle = 2.2*10^-15 g = 2.2*10^-18 kg

    You replace the values of all parameters in the equation (1):

    \frac{F_E}{F_G}=\frac{(1.6*10^{-19}C)(103N/C)}{(2.2*10^{-18}kg)(9.8m/s^2)}\\\\\frac{F_E}{F_G}=0.764

    Then, the gravitational force is 0.764 times the electric force on the particle

    b)

    The acceleration of the particle is obtained by using the second Newton law:

    F_E-F_G=ma\\\\a=\frac{qE-mg}{m}

    you replace the values of all variables:

    a=\frac{(1.6*10^{-19}C)(103N/C)-(2.2*10^{-18}kg)(9.8m/s^2)}{2.2*10^{-18}kg}\\\\a=-2.30\frac{m}{s^2}

    hence, the acceleration of the particle is 2.30m/s^2, the minus sign means that the particle moves downward.

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