Electron released into electric field

[that1grrl]

An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling 4.50m in the first 3.00us after released. What are the mag and direction of electric field and are we justified in ignoring the effect of gravity (justify quantitatively).

Im not sure where to start. Do I look for finding acceleration to use a=-eE/m to get the E? I just need a start.

[lightgrav]

Yes. From kinematics, what's its average speed? what's its final speed?
What's its acceleration? What Force must act to provide that acceleration?

[that1grrl]

Ok so I may be grasping at straws but would I do this:
(using constant-acceleration formula)

v=4.5m/3X10^-6
v= 1.33x10^6m/s= (2a(4.50m))^1/2
a=1.98x10^11m/s^2

a=F/m
1.98x10^11m/s= F/9.11x10^-31kg

etc? Even close. Eventually my units don't work out so Im thinking this isnt the right track.

[lightgrav]

save the Kinetic Energy trweatment for next chapter...

v_average = 1.5E6 m/s , so v_final = 3E6 m/s , right?

(v_final)^2 = 2 a x ...

but a = v_final / t !

[that1grrl]

I think I got it. THANKS!