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Two questions on electrostatics

  1. Jun 19, 2006 #1
    I'm a bit stuck with some things in electrostatics.

    My first problem:

    in my textbook, when they try to derivate the formula for the potential of a point charge: [tex]V(b) = - \int E.d\mathbf{l} = -\frac{q}{4 \pi \varepsilon_0} \int_\infty^b \frac{1}{r^3} \mathbf{r}.d \mathbf{l} [/tex]

    they say that [tex] \mathbf{r}.d \mathbf{l} = r.dr [/tex]
    There's also a picture which looks like this:

    http://img232.imageshack.us/img232/6044/charge3zj.jpg [Broken]

    My question is: Why isn't [tex] d \mathbf{r} = d \mathbf{l} [/tex] ? Why does [tex] d \mathbf{r} [/tex] have the same direction as [tex] \mathbf{r} [/tex] ?

    My second problem:

    There is some law or theorem which says that there is no electric field inside a conductor. Can this be proved, or is it just an empirical law?
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Jun 19, 2006 #2


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    The dl is a line element of the path you are taking to evaluate the integral (it doesn't matter what path you take), just some path from infinity to b.
    [tex]\vec r[/tex] is a vector pointing in the direction of the electric field (if the charge is positive), thus in the radial direction.
    The line element in spherical coordinates is something like [tex]d\vec l = dr\hat r+rd\theta \hat \theta+r\sin \theta d\phi \hat \phi[/itex], so [tex]\vec r \cdot d\vec l=rdr[/tex].
  4. Jun 19, 2006 #3


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    Well, you need to be a bit more precise. The electric field is zero in a region completely surrounded by a conductor, provided there is no charge enclosed in that region.

    It can be proved easily using the first uniqueness theorem.
  5. Jun 19, 2006 #4
    great, thank you very much for your help.
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