An electron is released from rest in a uniform electric field and accelerates to the north at a rate of 115 m/s2. What are the magnitude and direction of the electric field.

Answer 1

Assuming I did it correctly:

F=ma

mass = the mass of the electron = 9.11 * 10^-31 kg

a= 115 m/s

plug in your known variables
F will equal 1.05 * 10^-28 Newtons (N)

Next, F=qE

q equals the charge of the electron, or "point charge".
q=1.6 * 10^-19 C

Solve for E, so E=F/q

plug in your values, and E will equal 6.56 *10^-10 N/C

Answer 2

E=(ma/q)
Mass of electron=9.109e-31kg
Charge of electron=-1.602e-19C
so
E=(9.109e-31kg*115m/s^2)/-1.602e-19C
E=-6.539e-10
formated answer
E=6.539e-10 N/C South