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Moving a charge from inf

  1. Sep 21, 2007 #1
    Suppose there was a charge +Q located at x=-a and a +Q charge at x=a. If i wanted to bring in a charge of -Q form infinity to x=0, what is the work that I have to do?

    just wondering if i have the right expression:
    [tex]V=-\int _{\inf} ^a \frac{1}{4 \pi \epsilon_o } \frac{Q}{r^2}dr + -\int_{\inf}^{-a} \frac{1}{4 \pi \epsilon_o} \frac{Q}{r^2}dr[/tex]

    if i wanted to move the charge at the origin to x=+2a, the work i would have to do:

    [tex]V=-\int _{a} ^{3a} \frac{1}{4 \pi \epsilon_o } \frac{Q}{r^2}dr + -\int_{-a}^{a} \frac{1}{4 \pi \epsilon_o} \frac{Q}{r^2}dr[/tex]

    I was also wondering where the charge of the charge i'm moving comes into play, it doesnt seem like it matters for these equations?

    also, why is calculating the potential the same as finding the work?
    Last edited: Sep 21, 2007
  2. jcsd
  3. Sep 21, 2007 #2


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    Yes, those look right to me. But that gives the potential difference between infinity and that point... to get the work... you need to multiply the result by -Q.
  4. Sep 21, 2007 #3


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    But have you studied potential and energy due to point charges?

    You can write the result immediately without integrals...
  5. Sep 21, 2007 #4
    how do you do that?
  6. Sep 21, 2007 #5


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    The energy of the charge at infinity is 0...

    The energy of the charge at x = 0, is just:

    [tex]\frac{kQ(-Q)}{a} + \frac{kQ(-Q)}{a} = \frac{-2kQ^2}{a}[/tex]

    I'm using the formula... electric potential energy between two point charges is:


    So the change in energy is negative. So the charge loses energy... ie: you do negative work on the object...
    Last edited: Sep 21, 2007
  7. Sep 27, 2007 #6
    something weird happened and i'm not sure if it's supposed to:
    [tex]V=-\int _{\inf} ^a \frac{1}{4 \pi \epsilon_o } \frac{Q}{r^2}dr + -\int_{\inf}^{-a} \frac{1}{4 \pi \epsilon_o} \frac{Q}{r^2}dr =0[/tex]

    I get:

    [tex]V=-\frac{kQ}{a} + \frac{kQ}{a} =0[/tex]

    so it takes no work to move the charge there?
  8. Sep 27, 2007 #7


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    No... you have to be careful about the integrals... if you are integrating along the x-axis...

    let's examine the charge at x = a, and the voltage due to that charge. (here r = x - a, where x is the coordinate of the charge being brought in... a is the coordinate of the stationary charge)

    [tex]-\int_{\inf}^{-a} \vec{E}\cdot\vec{dr}[/tex]

    this equals:

    [tex]-(\int_{\inf}^{a} \vec{E}\cdot\vec{dr} + \int_{a}^{0} \vec{E}\cdot\vec{dr}+\int_{0}^{-a} \vec{E}\cdot\vec{dr})[/tex]

    [tex]-(\int_{\inf}^{a} \frac{1}{4 \pi \epsilon_o} \frac{Q}{r^2}dr +\int_{a}^{0} \frac{1}{4 \pi \epsilon_o} \frac{Q}{r^2}dr +\int_{0}^{-a} \frac{-1}{4 \pi \epsilon_o} \frac{Q}{r^2}dr)[/tex]

    see, the last section of that integral... the field is in the opposite direction, so a minus needs to be put in... also the a to 0... and 0 to -a parts cancel each other out... so we end up with the same integral as with the other stationary charge...

    I think maybe this integral is problematic because there's a discontinuity when r = 0... but it works out anyway.
  9. Sep 27, 2007 #8
    so the potential at x=-a is the same as the potential at x=a?
    Last edited: Sep 27, 2007
  10. Sep 27, 2007 #9
    [tex]-(\int_{\inf}^{a} \frac{1}{4 \pi \epsilon_o} \frac{Q}{r^2}dr -\int_{a}^{-a} \frac{1}{4 \pi \epsilon_o} \frac{Q}{r^2}dr)[/tex]
    [tex]=-(kQ (-1/a)-kq(2/a))= kQ 3/a [/tex]

    maybe i'm just doing the integrals wrong
  11. Sep 27, 2007 #10


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    The integral from a to 0 has kQ/r^2 inside the integral, but the integral from 0 to -a should have -kQ/r^2 (with the minus sign) inside the integral... the direction of the field has changed from rightward to leftward.

    So you can't put it all under the same integral...
  12. Sep 27, 2007 #11
    why couldnt i do this:

    [tex]-(\int_{\inf}^{0} \frac{1}{4 \pi \epsilon_o} \frac{Q}{r^2}dr +\int_{0}^{-a} \frac{-1}{4 \pi \epsilon_o} \frac{Q}{r^2}dr)[/tex]

    since the field from infinity to zero is the same?
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