Proton accelerated across 1000 V capacitor to reach opposite plate

[joshszman09]

(1). Homework Statement

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)

 

[gabbagabbahey]

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) ). If the caacitor is charged to 1000V, doesn't that mean the potential between the plates is 1000V?...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.