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Differential form of gauss's law.

  1. Oct 12, 2014 #1
    I don't understand what charge density is meant in the equation: div E = constant times charge density. I have the derivation in front of me and the last step follows from accepting that the rate of change of the integral of the field divergence per change in volume is the same as the rate of change of the integral of charge density per that same change in volume, and as such the divergence of an e field is one over permitivity times the... charge density at that point? What does this mean? How does it account for divergence of an e field away from point charges?
  2. jcsd
  3. Oct 13, 2014 #2


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    Homework Helper
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    Charge density is defined as the charge in a small volume divided by the volume. More exactly: the charge density at a point P is[tex]\rho=\lim_{\Delta V \rightarrow 0}\frac{\Delta Q}{\Delta V}[/tex]

    On the other hand, the divergence of a vector-vector function is defined as
    See http://en.wikipedia.org/wiki/Divergence
    So the divergence of E integrated for a volume V surrounding the point P is equal to the surface integral of the normal component of E enclosing the volume, which is equal to ##Q/\epsilon _0##, Q being the charge of the volume, according to Gauss' Law. At the limit of ## \Delta V \rightarrow 0##,
    ## div E = \rho/\epsilon_0##

  4. Oct 13, 2014 #3


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    Are you thinking about the spreading-out of electric field lines from a point charge? It might be instructive to calculate the divergence explicitly for the Coulomb's-law field of a point charge at the origin: [tex]\vec E = k \frac {\hat r}{|\vec r|^2} = k \frac {\vec r}{|\vec r|^3} = k \frac {x \hat i + y \hat j + z \hat k}{(x^2 + y^2 + z^2)^{3/2}}[/tex] Then consider what happens at ##\vec r = 0## and read up on the Dirac delta "function" if necessary (e.g. in Griffiths, section 1.5).
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