- #1
mitchy16
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Homework Statement
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?
Homework Equations
KE= 1/2 mv^2
U= qV
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.