# Homework Help: What is the change in the proton's electric potential energy?

1. Apr 12, 2013

### RedLego

1. The problem statement, all variables and given/known data
A proton moves in a constant electric field E from point A to point B. the magnitude of the electric field is 6.4x10^4 N/C. The direction of electric field is opposite to the motion of the proton.

If the distance from point A to point B is 0.50m, what is the change in the proton's electric potential energy, EPEb-EPEa?

E=6.4x10^4 N/C
Δd= 0.50
ΔV=?

2. Relevant equations
ΔV=Δd*E
V=W/q
W=qEcosθ

3. The attempt at a solution
ΔV=Δd*E=(6.4x10^4)(0.50) = 32000J
Clearly this is wrong, as the answer is supposed to be 5.1x10^-15J.
Apparently you need to multiply it by the proton's charge, 1.602x10^-19C, and you would get the answer. but why? i dont see any formulas that would suggest me to multiply my answer by the fundamental charge.

2. Apr 12, 2013

### Dick

One of your equations is V=W/q. That means W=qV. For a proton q is not 1 coulomb. It's charge of a proton. And the units of Δd*E aren't joules. It's joules/coulomb.

3. Apr 12, 2013

### RedLego

:) Arn't you fantastic!!! Thank you sir!!!