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Meaning of p in Gauss's law

  1. Aug 3, 2010 #1
    So I'm reading Div, Grad, Curl and All That which is my only exposure to Maxwell's equations. The author started with the integral form of Gauss's law (which I get), where the surface integral of the electric field = the enclosed charge (q) divided by epsilon nought. But in deriving the differential form using the divergence theorem, q becomes the volume times the average charge density (p), but in the final equation (of the differential form) is says the divergence of the electric field equals the charge density (not average) divided by epsilon nought.

    Getting to my point, what is the meaning of the charge density in the differential form? If it's not an average, it must vary at the different points in the volume you're considering... so I'm not sure what the p means here. If I'm not being clear please let me know and I'll try to explain it better. Thanks
    Last edited by a moderator: Aug 4, 2010
  2. jcsd
  3. Aug 3, 2010 #2
    Does this help? :



    \vec{E}=\lim_{q\rightarrow 0}\frac{\vec{F_e}}{q}
    =\frac{1}{4\pi \epsilon_o}\int\frac{\rho (r)}{r^2}\hat{r}d\tau

    \oint \vec{E}\cdot d\vec{a} =\frac{Q_{enc}}{\epsilon_0}

    Qenc is the volume integral of the charge density ρ(r) in the volume enclosed by the surface integral on the left hand side

    Bob S
    Last edited: Aug 3, 2010
  4. Aug 3, 2010 #3
    I think you stated it correctly right here. The charge density is a property that is defined at every point, and it is not an average over a volume. You can think of it as similar to mass density. You can have a material where the mass density is not constant over a volume. Every point has a different value in general. It is similar to a scalar variable as a function of position, but strictly speaking it is not a scalar because it does not transform like a scalar with coordinate transformations. This last statement is more that you need and will likely confuse you a little, so ignore it for now, but store that information away for later use because it does become important later in your studies.

    In case it is still not clear, think of [tex]\rho[/tex] as a limit taken at a point. Consider a small volume element [tex] \Delta V[/tex] containing charge [tex]\Delta Q[/tex]. The charge density is then approximately [tex]\rho\approx{{\Delta Q}\over{\Delta V}}[/tex]. Now take the limit as volume goes to zero and you have charge density precisely.

    [tex]\rho=lim_{\Delta V \rightarrow 0}{{\Delta Q}\over{\Delta V}}[/tex]
  5. Aug 3, 2010 #4
    Okay so that part is making sense. So from skimming ahead it looks like actually finding the charge has something to do with the laplacian (which I haven't gotten to yet), but could you get the enclosed charge (q) by taking the triple integral of p dV? Or would you just use some other method?
  6. Aug 3, 2010 #5
    Yep, it sounds like you get it.
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