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An electron is released from rest in a uniform electric field and accelerates to the east at a rate of 4.0 X 10^6 m/s^2. What is the magnitude and direction of the electric field?
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An electron is released from rest in a uniform electric field and accelerates to the east at a rate of 4.0 X 10^6 m/s^2. What is the magnitude and direction of the electric field

F=ma=eE

F: force.

m: mass of electron.

a: acceleration.

e: charge of electron.

E=ma/e

E=(9. 11*10-31)*(4*106)/(1.6*10-19)=2.2775*10-5V/m in West. 

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