Inf. Parallel Plate Capacitors and Potential Energy

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kevtimc25
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Hopefully no one minds that I post a couple questions in a row as long as I'm following the rules:

1. Homework Statement


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Two infinite parallel plates labelled W and X are separated by 5.2 cm. The electric potential between the plates is 150 V. An electron starts from rest at time tW and reaches plate X at time tX. The electron continues through the opening and reaches point P at time tP

c.) Find the electron’s kinetic energy and potential energy at point *P.
d.) Find the minimum speed of the electron at time tw needed for it to escape to infinity.


Homework Equations


U = qV , energy conserved

The Attempt at a Solution


We can calc. the KE X easily: Ux = 0, Ex = Uw, Uw = 150V * -1.60*10^-19 C = 1/2 m(e) v^2 so we have v and there is no electric field once it is outside of the plates. A previous given answer indicates that the speed is constant once we hit X (I don't understand why other than that no outside forces are acting on the electron if we exclude gravity), so how can we calculate U and K at *P?
 
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Do we know the initial position of the electron?
If the electron is accelerated through a potential difference of V then it's kinetic energy is V electron-volts.

Per your question: do you know how to calculate the potential due to a single plate?
Do you know the superposition principle?
 
Yes it is at rest at W.

I know the EF is σ/2e0 everywhere due to each plate and you can find V by integrating E through a distance. And I know you can also find V by adding the potentials from each plate. But we don't now the Surface density of the plates.
 
You also know the potential difference between the plates is 150V... this gives you the kinetic energy at X.
Per the charge density - just put it equal to ##+\sigma## for the +ve plate and ##-\sigma## for the -ve plate.
 
Right, outside of the PPC the EF is 0 everywhere, so the ΔV would have to be 0 right? That means the potential doesn't change so the KE can't change as well (at point X and beyond).