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Using Gauss' Law (differential form) to get Coulomb's law.

  1. Feb 1, 2010 #1
    I don't really know much about serious business electrostatics. I've only taken the AP Physics C E&M exam (using the integral form), but I was looking at wikipedia and I was curious about the differential form of Gauss' Law.

    I don't understand what I'm doing wrong with this. I'm trying to derive coulomb's law (just so I get a better understanding of how to use the differential form) from the differential form of Gauss' law.

    So, you have

    [tex]\rho (r) = \frac{Q}{V(r)}[/tex]

    and I guess for a point charge, you'd use a spherical volume, so,

    [tex]V(r) = \frac{4}{3}\pi r^3[/tex]

    Therefore,

    [tex]\rho (r) = \frac{3Q}{4\pi r^3}[/tex]

    Then plug that into Gauss' law:

    [tex]\nabla \cdot E = \frac{3Q}{4\pi \epsilon_0 r^3}[/tex]

    Because there's spherical symmetry, I'm just going to ignore the rest of the divergence operator and just say

    [tex]\nabla \cdot E = \frac{dE}{dr}[/tex]

    Perhaps this is where I went wrong, and the divergence is way more complicated than that for this situation, but this makes sense to me intuitively.

    Then you'd have after separating variables and ignoring a negative sign:

    [tex]E = \int_\infty^r \frac{3Q}{4\pi\epsilon_0 r^3} dr[/tex]

    [tex]E = \frac{3Q}{8\pi\epsilon_0 r^2}[/tex]

    [tex]\frac{3Q}{8\pi\epsilon_0 r^2} \neq \frac{Q}{4\pi\epsilon_0 r^2}[/tex]

    lol wat? There's a 2 in the denominator when there should be a 3 to make the 3 in the top cancel out and be happy. What am I doing wrong? Or am I just making utter non-sense?
     
  2. jcsd
  3. Feb 2, 2010 #2
    Lol, I was doing the divergence wrong, as I thought. I was being stupid.

    [tex]\nabla \cdot E = \frac{1}{r^2} \frac{d(r^2 E)}{dr}[/tex]

    :/

    integrate and you get:

    [tex]E = \frac{\rho r}{3 \epsilon_0}[/tex]

    replace rho

    [tex]E = \frac{Q}{4 \pi \epsilon_0 r^2}[/tex]

    k, good. I'm not crazy, just doing it wrong.
     
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