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13 August, 21:16

A positively charged particle initially at rest on the ground accelerates upward to 140 m/s in 2.10 s. The particle has a charge-to-mass ratio of 0.100 C/kg and the electric field in this region is constant and uniform. What are the magnitude and direction of the electric field?

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  1. 13 August, 22:02
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    764.7 N/C, the direction is upward

    Explanation:

    E electric field of the electron = F / charge

    F force of the electron = Eq

    acceleration of the body = change in velocity / time = 140 m / 2.10 s = 66.67 m/s² since the electron is from rest.

    F upward = F of electron - force of gravity

    F upward = Eq - mg

    ma = Eq - mg

    divide through with m

    a = E (q/m) - g

    (a + g) / (q/m) = E

    (66.67 + 9.8) / 0.100 = 764.7 N/C, the direction is upward
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