An electron & a proton are each placed in an electric field

[amyc]

Homework Statement

An electron and a proton are each placed at rest in an electric field of 687 N/C. What is the velocity of the electron 56.5 ns after being released? Consider the direction parallel to the field to be positive. The fundamental charge is 1.602×10−19 C, the mass of a proton is 1.67267×10−27 kg and of an electron 9.109×10−31 kg. Answer in units of m/s.

Homework Equations

F=ma=Eq
v_f=v_i+a*t

The Attempt at a Solution

I did this using the electric field but I don't know how to incorporate the proton in here and the answer is wrong :(

This is what I got with the electric field:
E=687N/C
t=56.5*10^-9s
plug all the values into the force equations
F=(687)(1.602*10^-19=(9.109*10^-31)a
solve for acceleration and
a=1.208*10¹⁴
now to solve for final velocity plug into the kinematics equation
v_f=0+(1.208*10¹⁴)(56.5*10^-9) = 6.83*10⁶ m/s

Please help!
//

update! this was right it just had to be negative lol

 

[mfb]

The proton doesn't matter (it might become relevant later in the question?).

You should work with units consistently. Apart from that it looks fine.