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Two expressions for work - which is correct?

  1. May 11, 2008 #1
    When I want to find the energy in a system containing two points charges q and -q, I can use the following formula:

    [tex]W = -Q\int_{\bf{a}}^{\bf{b}} {\bf{E}} \cdot d{\bf{l}} = Q(V(\mb b) - V(\mb a))[/tex]

    If my reference point is set to infinity, I can rewrite the last expression to:

    [tex]W=QV(\mb r)[/tex]

    The two point charges are seperated by a distance of 2d, so the energy in the system calculated from the first formula will be (when bringing in the charge q in from infinity):

    W = -K*q^2/(4d).

    If I use the last expression, I get

    W = -K*q^2/(2d).

    Ehh, as far as I can see, I haven't made an obvious mistake. So how can this be?
     
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  3. May 11, 2008 #2

    Hootenanny

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    Why is this 4d of the charges are separated by a distance of 2d?
     
  4. May 11, 2008 #3
    The electric field at the point x=d is E = -K*q/(4d^2).

    This I integrate, and it still gives me 4d.
     
  5. May 11, 2008 #4

    Hootenanny

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    Let's step back a bit. Suppose our first charge is located at the origin, simply because it ismore conveinient. The the electric field due to this point charge is,

    [tex]\mathbf{E} = k\frac{q}{R^2}\hat{\mathbf{r}}[/tex]

    Where R is the distance from the origin and [itex]\hat{\mathbf{r}}[/itex] is the radial unit vector. Agree?

    Now let's bring in the second charge,

    [tex]W = -Q\int^{2r}_{\infty}\mathbf{E}\cdot{d\mathbf{\ell}}[/tex]

    Now setting Q = -q and integrating along the (negative) radial direction,

    [tex]W = qk\int^{2r}_{\infty}\frac{dR}{R^2}\left(-\hat{\mathbf{r}}\cdot\hat{\mathbf{r}}\right)[/tex]

    [tex]W = -qk\int^{2r}_{\infty}\frac{dR}{R^2}[/tex]

    Do you agree?
     
  6. May 11, 2008 #5
    Yes, I do agree.
     
  7. May 11, 2008 #6

    Hootenanny

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    Hence integrating,

    [tex]W = qk\left[\frac{1}{R}\right]^{2r}_{\infty}[/tex]

    [tex]W = qk\left(\frac{1}{2r}-\lim_{R\to\infty}\frac{1}{R}\right)[/tex]

    [tex]W = \frac{kq}{2r}[/tex]
     
  8. May 11, 2008 #7
    I agree with what you have done, but I also still agree on what I have done. What is wrong with my method?
     
  9. May 11, 2008 #8

    Hootenanny

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    In your method, where have you placed your initial charge?
     
  10. May 11, 2008 #9
    I have my initial charge in x=-d and the other is brought in from infinity to x=d.
     
  11. May 11, 2008 #10

    Hootenanny

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    Simply integrating this will not give you the potential energy at this point, you need to integrate before you substitute the limits in.
     
  12. May 11, 2008 #11
    Ok, this is how I've done:

    We have that [tex]\[
    W = - Q\int_\infty ^d {Edl = QV(d)}
    \]
    [/tex]

    Approach 1, where [tex]\[
    E = \frac{{ - kq}}{{\left( {2d'} \right)^2 }}
    \]
    [/tex]:

    [tex]\[
    W = - q\int_\infty ^d {\frac{{ - kq}}{{\left( {2d'} \right)^2 }}dd' = } \frac{{ - kq^2 }}{{4d}}
    \]
    [/tex]

    Approach 2, where [tex]\[
    V(d) = \frac{{ - kq}}{{2d}}
    \]
    [/tex]:

    [tex]\[
    W = q \cdot \frac{{ - kq}}{{2d}} = \frac{{ - kq^2 }}{{2d}}
    \]
    [/tex]

    Can you point out the mistake here?
     
  13. May 11, 2008 #12

    Hootenanny

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    Sorry Niles, I have no idea what your doing with you're first approach.
     
  14. May 11, 2008 #13
    Don't be sorry, sadly I do not explain things very well to others.

    I found my error now. Instead of using an arbitrary distance in the electric field, I just inserted 2d, which is of course wrong, since I have to integrate it. This is of course also what you wrote in #10.

    You did it again.
     
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