physics
posted by Andy on .
A. A uniform, upwardpointing electric field E of magnitude 3.00 10e3 N/C has been set up between two horizontal plates by charging the lower plate positively and the upper plate negatively. The plates have length L = 4 cm and separation d = 2.00 cm. An electron is then shot between the plates from the left edge of the lower plate. The initial velocity v0 of the electron makes an angle theta=45 with the lower plate and has a magnitude of 6.9310e6 m/s. Will the electron strike one of the plates? If so, what is the horizontal distance from the left edge? If not enter the vertical position at which the particle leaves the space between the plates ( m).
B.The next electron has an initial velocity which has the same angle theta=45 with the lower plate and has a magnitude of 5.83 10e6 m/s. Will this electron strike one of the plates? If so, what is the horizontal distance from the left edge? If not enter the vertical position at which the particle leaves the space between the plates ( m).
I fixed the characters in the question please help

Ignore the plates for a second
PositionElectron=Vo*t+1/2 a t^2+ Positioninitial
now, you know a is F/m= Eq/m
PositionElectron= Vo*t+1/2 q/m E t^2+positioninitial
Vo and E are vectors.
Now breake Vo into horizontal and vertical components, and you have a position equation.
To see if they hit either plate, Just work with the vertical part of the equation.
find t when position=2, and find t when postion=0
if no solution,it does not hit plates.
Then, work with horizontal, find time when position horizontal is 4.
then use that time to find the vertical position. 
Bobpursley thanks for the help I see that Eq/m=a is equal to F/m=a but what vaule do I use for m in this problem

What is the mass of an electron?

haha oh ok thanks