A uniform, upward-pointing electric field E of magnitude 5.00×103 N/C has been s

Question

A uniform, upward-pointing electric field E of magnitude 5.00×103 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. 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 9.95×106 m/s.

a) 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. I got answer 0.028 meters which is WRONG from using other Cramster posts so if someone could please show me another way to get another answer.

b) Another electron has an initial velocity which has the angle ?=45° with the lower plate and has a magnitude of 8.26×106 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.

Explanation / Answer

Ignore the plates for a second PositionElectron=Vo*t+1/2 a t^2+ Position initial 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.

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