(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A proton is released from rest at the positive plate of a parallel plate capacitor. It crosses the capacitor and reaches the negative plate with a speed of 50,000 m/s. What will be the final speed of an electron released from rest at the negative plate?

2. Relevant equations

KE= 1/2 mv^2

U= qV

3. The attempt at a solution

I tried using conservation of energy, the left side is the proton and right side is the electron:

KE1 + U1 = KE2 + U2

1/2mv^2 + qV = 1/2mv^2 + qV

I do not understand what to plug in for V? I have tried searching for similar threads but I can't find a similar one. Any guidance would be appreciated, thank you.

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# Homework Help: Determining speed of electron in a parallel plate capacitor

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