# Speed of a proton through potential difference

## Homework Statement

What is the speed of a proton that has been accelerated from rest through a potential difference of -1000V?

## Homework Equations

ΔV=potential difference
V=U/q
Conservataion of energy
U=qEd (d is distance the proton is moving...i think)

## The Attempt at a Solution

Here is my thought process with this problem:
ΔV=-1000V this means the change in electric potential is (-), doesnt tell us what V(initial) or V(final) is, but we know the difference between them is 1000
I know that V=U/q
For the proton going through this potential difference it follows conservation of energy:
Ki+Ui=Kf+Uf→0+qEd=1/2mpv2+0→qEd=1/2mpv2
here is where I get stuck...
it looks to me like we have one equation and two unknowns
the equation being:qEd=1/2mpv2
the unknowns: E and v
And I guess Im not even sure if Uf=0.

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berkeman
Mentor
Are you familiar with the conversion between eV and Joules? And in problems like this, usually you would assume the particle starts at rest, unless stated otherwise.

Orodruin
Staff Emeritus
Homework Helper
Gold Member
You do not need E or d, you were given the potential difference.

Are you familiar with the conversion between eV and Joules? And in problems like this, usually you would assume the particle starts at rest, unless stated otherwise.
I am not familiar with this.

You do not need E or d, you were given the potential difference.
So we can then say ΔV=U/q=qEd/q=Ed? thats all I am seeing from this. There may be another equation I am not remembering that relates potential difference to potential energy...

ahh, are you saying that ΔV=U/q can be solved so U=qΔV and in this case everything is known?
then the equation:
Ki+Ui=Kf+Uf→0+Ui=(1/2)mv2+0→qΔV=(1/2)mv2 and then we can solve for v? That makes sense

Last edited:
Orodruin
Staff Emeritus