Posted by electric field on .
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.

Physics 
Jason,
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