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Electron repulsion work problem.

  1. Oct 23, 2012 #1

    B3NR4Y

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    The problem: If two electrons are held at fixed points (5,0) and (-5,0), respectively, find the work done in moving a third electron from the origin to (3,0)

    All equations:
    Force of repulsion between two electrons:
    \begin{equation}
    \frac{1}{d^{2}}.
    \end{equation}
    \begin{equation}
    W = \int_{a}^{b} \, f(x) dx
    \end{equation}
    "D" is the distance between the electrons

    Attempt at a solution:
    The distance between the electron at (-5,0) and the third electron is, 5+x where x is the position of the electron at any point on its journey to (3,0). The distance between the electron at (5,0) and the third electron is (5-x). I found the work by adding the two work equations like this:
    \begin{equation}
    \begin{split}
    W&= \int_{0}^{3} \frac{1}{(5+x)^{2}} dx + \int_{0}^{3} \frac{1}{(5-x)^{2}} dx \\
    & = -[-\frac{1}{5-x} |^{3}_{0}] + [-\frac{1}{5+x} |^{3}_{0}\\
    &=\frac{3k}{8} ergs
    \end{split}
    \end{equation}
    Book says the answer is
    \begin{equation}
    \frac{9k}{40}\, ergs
    \end{equation}
     
  2. jcsd
  3. Oct 23, 2012 #2

    Dick

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    I agree with your answer.
     
  4. Oct 23, 2012 #3

    B3NR4Y

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    So probably just an error in the book?

    Good :3
     
  5. Oct 23, 2012 #4

    Dick

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    Yeah, I think so.
     
  6. Oct 24, 2012 #5

    Dick

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    Hope it's not to late to change my mind. There is a problem. The forces you are integrating don't point in the same direction. You need a minus sign on one of your forces. The book answer is correct. Sorry about that!
     
  7. Oct 24, 2012 #6

    B3NR4Y

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    No worries! Thank you. Makes sense, the answer is correct now. I feel honored you answered me.
     
  8. Oct 24, 2012 #7

    Dick

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    Glad you feel honored by a not very accurate response :). Hope that won't deter you from posting again.
     
  9. Oct 24, 2012 #8

    B3NR4Y

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    It most definitely will not. I'm 15 and self-teaching, so I will post here often. Thank you again.
     
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