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Finding the E field direction of integration

  1. Jul 3, 2010 #1
    If you have a line of charge with charge density [tex]\lambda=\frac{dq}{dl}[/tex] and you want to find the electric field at a perpendicular distance z from the midpoint, you get

    [tex]dE = \frac{1}{4\pi\epsilon_0}\frac{\lambda}{r^2}dl[/tex]

    Then you integrate [tex]dE[/tex] from one end of the line of charge to the other. (e.g. [tex]\int_{-L}^L ... dl[/tex])

    Obviously if you reverse the integral terminals you get the negative of your original answer, but physically, why should reversing integral terminals even matter? (i.e. [tex]\int_L^{-L}...dl[/tex])

    After all, the physical interpretation of the integral is just summing up the little [tex]dq[/tex]s over the line, what does it matter which direction you do it in? And importantly, how do you know which is the correct direction to sum up the [tex]dq[/tex]s?

    Thanks
     
  2. jcsd
  3. Jul 3, 2010 #2

    Hurkyl

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    Because one way has positive length, and the other way has negative length.

    The real line is oriented so that displacements towards [itex]+\infty[/itex] are positive.


    Why do the signs creep in? Because you are using an oriented notion of integration -- a parametrized curve (from -L to L) with respect to a differential form ([itex]d\ell[/itex]).

    There are unsigned notions of integration. If you have a measure (say, [itex]\mu[/itex]), you can define integrals over sets -- e.g.
    [tex]\int_{S} \ldots d\mu[/tex]​
    Of course, it turns out that integrals of the standard length measure on R over intervals can be computed in the oriented way:
    [tex]\int_{[-L,L]} f \, d\mu = \int_{-L}^L f(x) \, dx = -\int_L^{-L} f(x) \, dx[/tex]​
    (I'm assuming L>0 in the above)
     
  4. Jul 4, 2010 #3
    Thanks hurkyl
     
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