# Potential electric engery

1. Jul 14, 2009

### talaroue

1. The problem statement, all variables and given/known data

What is the electric potential energy of the electron? The protons are fixed and cannot move, a=0.94 nm, s=1.24 nm.

2. Relevant equations

U=Vq=KQq/d

3. The attempt at a solution

U=Vq= 2KQq/d

K=8.99 x 10^9
Q=charge of proton 1.6 x 10^-19
q= charge of electron (-1.6)x10^-19
d= sqrt(a^2+s^2)

Also is my final anwser going to be negative? or should it be a positive?

2. Jul 14, 2009

### talaroue

oh and you multiply by 2 because there are 2 protons acting on the electron.

3. Jul 14, 2009

### talaroue

still can't figure it out

4. Jul 14, 2009

### talaroue

still nothing.

5. Jul 15, 2009

### turin

Looks to me like you did it correctly. Sorry I can't be of any help. Maybe you just didn't convert your nanometers properly into SI? Or maybe they ask for non-SI unit of energy?

6. Jul 16, 2009

### Redbelly98

Staff Emeritus
Just noticed this question.

Opposite charges will result in a negative potential, since U=2KQq/d will be negative for opposite charges: Qq is a positive times a negative value, hence a negative answer.

Yes, correct.

You have things set up correctly and just need to plug in the numbers. As you said, you use

U=2KQq/d

7. Jul 16, 2009

### talaroue

Oh makes since, so what you are saying is since they are the same charge U always equal negaive because it goes in the negative direction (aka being pushed away)

8. Jul 16, 2009

### Redbelly98

Staff Emeritus
Uh, no, I said U is negative because the charges are opposite. The electron is not pushed away from the protons, it is attracted to them.

9. Jul 17, 2009

### turin

In spite of the issue of negative or positive, do you care to post the absolute value of your answer (with units)? That will tell us immediately if you just screwed up your units.