Electrical Potential Energy problem

1. Jul 23, 2007

sirdeity

Hello. I'm just stuck on a problem. A point charge Q=+4.60x10^(-9)C is held fixed at the origin. A second point charge q=+1.20x10^(-9)C with mass of 2.80x10^(-4)kg is placed on the x-axis, 0.250m from the origin.

The problem asks, "What is the electrical potential energy U of the pair of charges?"

Okay, so I tried using the formulas:
-(Ub-Ua)=-deltaU and U=((1)/(4*pi*Eo))*((q1*qo)/(r))

Where I'm confused is that q1 and qo appear the same in both Ua and Ub, thus wouldn't Ub-Ua cancel and equal zero? Here's my obviously incorrect work:

-(((1)/(4*pi*Epsilono))*(((1.2x10^(-9))(4.6x10^(-9)))/(.25m)) = -1.98x10^(-7)

However, the correct answer according to the back of the book is "0.198J." Can anyone with patience please explain what I'm doing wrong? This material is new to me. I'm in a 5 week summer calculus 2 based engineering physics 2 course (PHY 122). We only just started talking about this material yesterday.

2. Jul 24, 2007

G01

You do not need the delta U equation. That formula applies when you change from one situation at one potential energy, to another situation with another potential energy.

Here you do not have a change in potential energy. The potential energy is staying the same, and they want you to find what it is. So, the only relationship you'll need is:

$$U = \frac{1}{4\pi\epsilon_o}\frac{q_1q_2}{r}$$

I am getting the same answer you are, but positive. Your extra negative sign comes in because you have a negative sign because that negative sign you have in the potential energy formula shouldn't be there. Unlike the gravitational potential energy expression, the electrical potential energy expression does not have that extra negative sign. Just make sure if the charges are negative, you use put their negative signs in the "U" formula!

Other than that, I think your work is correct. You are off by seven factors of ten. I think you may want to check to make sure that you are actually using the correct units, just in case you misread them. Also, if check your unit conversions if you did any. Then, if that isn't it, check the accuracy of the books answer with someone else like your professor.

Last edited: Jul 24, 2007
3. Jul 24, 2007

sirdeity

Thank you very much for your help. I actually figured out the problem last night, after about 3 hours of trying various things. The thing that was messing me up was that I assumed micro was ^-9 when it is actually ^-6.

It was a silly mistake on my part. Thanks again.

4. Jul 24, 2007

G01

No problem. I'm glad you figured it out!