1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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]


    [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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook