# Electron released into electric field

1. Jan 14, 2006

### 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.

2. Jan 14, 2006

### 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?

3. Jan 14, 2006

### 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.

4. Jan 14, 2006

### 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 !

Last edited: Jan 14, 2006
5. Jan 14, 2006

### that1grrl

I think I got it. THANKS!