Velocity of a Proton in a Capacitor (1 Viewer)

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(1). The problem statement, all variables and given/known data

Suppose a proton is fired from the negative plate of a capacitor charged up to 1000 Volts. How fast must it be traveling to reach the other side?

(2) Relevant equations

Okay, so I figured that this would be a conservation of energy problem and used: 1/2mv2 = (KeQq)/r
I think this would work, but it requires that you know both the charge on the capacitor and the distance between the plates which is not given. I am really stuck and could use some help. I just need to be pushed in the right direction and hinted towards what to do. I have literally tried everything I know about voltage and capacitors and just can't get it. If you need and more info/have questions just let me know. Thanks in advance!

3. Attempt at Solution

There isn't much to put here except a few of the equations I tried using.

I know that V = U/Q and that U= (KQq)/r and so I came up with V = (Kq)/r and tried solving for r(distance between the plates), but I got a really small number, r = 1.44E^-12. Even if that is right, I still can't think of a way to solve for the charge, Q, on the capacitor and plug the numbers into the equation I wrote in (2)


Homework Helper
Gold Member
I know that V = U/Q
This is true for a point charge Q in an external electrostatic potential V (like a proton in a capacitor) :wink: ). If the caacitor is charged to 1000V, doesn't that mean the potential between the plates is 1000V?:wink:...What does that make U? What do you get when you apply conservation of energy to that?

and that U= (KQq)/r
This is only true for a point charge Q subject to the potential of another point charge q a distance r away. You have a point charge and a capacitor in this problem, not two point charges.

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