Question

An electron is released from rest in a unifor electric field and accelerates to the north at a rate of 145 m/s^2. What is the magnitude and direction of the electric field?

Answer 1

E = 8.26*10⁻¹⁰ N/C, due south.

Step-by-step explanation:

• Assuming no other forces acting on the electron than the electrostatic force due to the electric field, we can apply Newton's 2nd law as follows:

`F = -eE =ma (1)`

• Solving for E, we can find its magnitude as follows:

`E =(m*a)/(e) = (9.1e-31 kg*145m/s2)/(1.6e-19C) = 8.26e-10 N/C (1)`

• The direction of the electric field is by definition the one that would take a positive test charge, so if the electron is accelerated to the north, the electric field would exactly oppose to this direction, so it is directed due south.