A uniform, upward-pointing electric field E of magnitude 3.50×10 3 N/C has been

Question

A uniform, upward-pointing electric field E of magnitude 3.50×103N/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 = 5 cm and separation d = 2.00 cm. Electrons are shot between the plates from the left edge of the lower plate.
The first electron has the initial velocity v0, which makes an angle =45° with the lower plate and has a magnitude of 7.35×106m/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.

Another electron has an initial velocity which has the angle =45° with the lower plate and has a magnitude of 6.13×106m/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.

Explanation / Answer

The horizontal velocity of the electron is constant and equal to 9.35 x 10^6 m/s x cos45 which is also the initial vertical component of velocity, vy =   9.35 x 10^6 m/s x cos45 =6.61*e^6

Now find the vertical acceleration of the electron ay

F = Eq =ma so ay = Eq/m for the electron. Look up q,m and calculate ay. q=1.6 x 10^-19 m = 9.1 x 10^-31 I think.

Now find the time it takes the electron to reach a height equal to the top plate.
Use deltay = 0.02 m = vyt - 1/2 ay t^2. It's a quadratic, but use your calculator (I'm sure you have a quadratic solver) to find t. Take the smaller real solution.

Now find the horizontal distance travelled using deltaX = vx t If this is bigger than 0.05 m, the particle doesn't hit the top plate.

For the first one I got that the electron hits the top plate 2.70 cm in from the left. For the second, there are no roots so it never gets that high. The time then to go across the plates is 0.05/vx = 7.39 x 10^-9 s

In that time deltaY = 0.016m above the top plate.

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