What is the potential energy of an electron

1. Oct 22, 2008

Foxhound101

1. The problem statement, all variables and given/known data
The electric potential at point A is 1800V.

Part B -
What is the potential energy of an electron at point A in the figure?

Point A is .01m away from a charge of +2nC

2. Relevant equations
V=Uq

U = (Kq1q2)/r

3. The attempt at a solution
I think the problem is, is that I am confused about the difference between electric potential and potential energy. This problem shouldn't be difficult.

So I tried the second equation

U = (9.00*10^9)(1.6*10^-19)(2*10^-9)/.01
U = 2.88*10^-18

But that isn't the answer...I don't know what equation I should use or what I am doing wrong.

2. Oct 22, 2008

alphysicist

Hi Foxhound101,

There's not much detail about the situation here, but one thing I noticed is that you are not putting the sign of the electron in the potential energy equation. For the potential and potential energy the sign is required (unless of course they are just asking for the magnitude for some reason).

3. Oct 22, 2008

Foxhound101

This time I tried the other equation...

1800 = U / (2*10^-9)
3.6*10^-6 = U

still wrong.

4. Oct 22, 2008

Foxhound101

5. Oct 22, 2008

alphysicist

I think there are two things wrong here. The first is the sign error I mentioned. But also, this calculation from your first post:

It looks like a calculation error here; it should not be to the -18 power.

Do you get the right answer now?

6. Oct 22, 2008

Foxhound101

Alright, thanks for the help. I have the correct answer now. I have the rest of the problem worked out as well.

Man...I wasted so much time on this problem when the only major mistake I made was the negative sign.

Normally MasteringPhysics tells me if the sign is incorrect...

7. Oct 22, 2008

alphysicist

Glad to help! And I know it's discouraging to spend so much time on something that turns out to be a seemingly small matter. However, not including the sign is a very common and persistant mistake for potential/potential energy calculations, and it's better to get a warning about it in your mind now before any exams.