In the figure, a uniform, upward-pointing electric field E of magnitude 5.00E3 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 (0.04 m) and separation d = 2.00 cm (0.02 m). 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 θ=45° with the lower plate and has a magnitude of 9.07E6 m/s.
a) 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 θ=45° with the lower plate and has a magnitude of 7.28×106 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).
v[initial, y]^2 + (2*a*deltaY) = 0
acceleration = [qE]/[m]
v[initial,y] + (a*t) = 0
The Attempt at a Solution
find the acceleration
qE=ma (mass and q are actually constant numbers of 9.11E-31 and 1.6E-19 respectively)
then you find the velocity components which is vo(sin theta)
solve for time for X and Y
Tx being: L= (velocity from above)(Tx)
Ty being: d= (.5(acceleration))Ty^2 + (velocity found above)Ty
0.02 m=((0.5)(8.78E14))Ty^2 + (6.41E6)Ty
(4.39E14)Ty^2 + (6.41E6)Ty - 0.02 = 0
whichever time is smaller, you solve for the height using time*acceleration
**This question has been eluding me for a while, and that's probably because I can't wrap my head around the question. The attempt I have up there was given to me by one of my classmates and I've attempted to follow along as you can see, but I don't understand it. Is the attempt, in your opinion, good? Did I do those parts correctly? What does it even mean? Tx and Ty? Please, help me.