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?

E = 8.26*10⁻¹⁰ N/C, due south.

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:

       

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

       

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.