How Does an Electron Behave Between Parallel Conducting Plates?

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An electron is fired between two parallel conducting plates with a speed of 5.6 × 10^6 m/s at a -45° angle, with a potential difference of 100 V and a distance of 2.0 mm between the plates. For part (a), the user correctly applied energy conservation principles to find that the closest distance to the bottom plate is approximately 1.79 mm. For part (b), it is suggested to use projectile motion equations, substituting gravitational acceleration with the acceleration derived from the electric field. This approach will help determine where the electron strikes the top plate. Understanding the relationship between electric fields and motion is crucial for solving these types of problems.
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Homework Statement


An electron is fired at a speed of v0 = 5.6 × 106 m/s and at an angle of θ0 = – 45° between two parallel conducting plates that are D = 2.0 mm apart, as in Figure P16.66. If the potential difference between the plates is ΔV = 100 V, determine (a) how close d the electron will get to the bottom plate and (b) where the electron will strike the top plate.
That is the exact question from the textbook. there is a diagram that shows the two plates and the electron starting in the middle of the two plates.


Homework Equations


I'm assuming you'll need Ei=Ef or some sort of projectile motion. I really don't even know that.


The Attempt at a Solution



I haven't the foggiest idea for part b, but for part a I attempted to do it like this and got an answer. I did Ei=Ef for the first part and set KE=electrical potential energy. Then substituted (delta)v over D for E and was able to solve for E and get 50000 v/m and then solved the equation for d and got 1.79 x 10^-3 m. Is that ok? and please any help. I really need it badly.
 
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Your method for part a) is correct.

For b), use the standard equations for projectile motions, but instead of using the acceleration due to gravity, use the acceleration due to the electric field.
 
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