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I Path orientation for calculating electric potential

  1. Sep 9, 2016 #1
    For line integrals in vector calculus,

    [tex]\int^a_b F \cdot dl[/tex]

    I almost always see the path oriented from a to b.

    But my text book has the following (look at the first equation for V(r):

    lapJZ5b.jpg

    Since the integral's limits are from O to r, I would have expected dl to also be pointing in the direction from O to r (i.e. pointing in the radially inward (minus r hat) direction), but the math in the textbook seems to imply that dl points radially outward (positive r hat direction, from r to O). I say this because the result of E dot dl has no minus sign in front of it.

    How do you know which direction to orient dl?
     
  2. jcsd
  3. Sep 10, 2016 #2

    vanhees71

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    It doesn't matter at which point you start your integration path as long as it is not at the singularity at ##\vec{r}=0## of the Coulomb potential (I don't know what's discussed in your book, because you didn't tell us; maybe it's a spherical surface carrying some charge?). Changing the starting point of your path just adds a constant to the potential, but the only thing interesting is the field ##\vec{E}=-\vec{\nabla} V##.

    Here they used a path starting from ##r=|\vec{r}| \rightarrow \infty##, making the potential ##0## at infinity, which is a convenient standard choice. Since the field is radial always, the only contribution is from the part going from infinity radially in, and you get the integral solved in your book.
     
  4. Sep 10, 2016 #3
    The path goes from b to a. If you reverse the order the modulus stay the same but the sign changes.
     
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