- #1

cookiemnstr510510

- 162

- 14

## 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 (-), doesn't 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:

K

_{i}+U

_{i}=K

_{f}+U

_{f}→0+qEd=1/2m

_{p}v

^{2}+0→qEd=1/2m

_{p}v

^{2}

here is where I get stuck...

it looks to me like we have one equation and two unknowns

the equation being:qEd=1/2m

_{p}v

^{2}

the unknowns: E and v

And I guess I am not even sure if U

_{f}=0.