Calculating Electric Potential Energy of an Electron in a Fixed Proton System

[talaroue]

Homework Statement

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

Homework Equations

U=Vq=KQq/d

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?

 

[talaroue]

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

 

[talaroue]

still can't figure it out

 

[talaroue]

still nothing.

 

[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?

 

[Redbelly98]

Just noticed this question.

Homework Equations

U=Vq=KQq/d

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?

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.

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

Yes, correct.

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

U=2KQq/d​

 

[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)

 

[Redbelly98]

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.

 

[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.