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15 March, 12:05

An electron is released from rest in a uniform electric field and accelerates to the north at a rate of 115 m/s squared. What are the magnitude and direction of the electric field?

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  1. 15 March, 12:34
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    The direction of the electric field would be south.

    qE/m = 115

    E = 115*m/q

    = 115 * 9.1 * 10^ (-31) / 1.67*10^ (-19)

    = 762.87 * 10^ (-12)

    = 6.27 x 10^-10 N/C

    Hope this answers the question. Have a nice day. Feel free to ask more questions.
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