# Velocity of a Proton in a Capacitor

#### joshszman09

(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)

Last edited:

#### rock.freak667

Homework Helper
If you are given the voltage, and you know the charge on the proton, then you should have the energy that the field is supplying to the proton. This energy is converted into the kinetic energy (1/2)mv2.

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