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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Physics

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6 February, 18:45

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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Zayne Reilly · Answer 1

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

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