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Gauss law for electrostatics

  1. Jan 22, 2009 #1
    I've "derived" the first Maxwell equation of the divergence of the E-field starting at the Coulomb force of a point charge using Gauss law and even the Dirac delta function to justify the divergence at the origin.

    Now I'm wondering: when you state the law in differential form, i.e. not specifying a volume over which the relation is to hold, I take it for being valid as a first order approximation in the neighborhood of a point. Reasonable?

    Also, does it make sense to think of the divergence at that point as independent of the charge density outside the local neighborhood? Just as Gauss law says. If so, given the electric field in a volume of space, the charge density at the same point is

    [tex]\rho(\vec{r}) \propto \nabla \cdot \vec{E} [/tex].

    Or am I missing something here? (Apart from the permitivity constant)
  2. jcsd
  3. Jan 24, 2009 #2
    You are missing something here. =P

    The differential forms of the Maxwell equations are point form equation; i.e. their validity prevails at all points in space with no reference to bounds. The differential form of Gauss' law does NOT say "the divergence at that point as independent of the charge density outside the local neighborhood." It does, however, say exactly the equation you have written.

    On the left side, you have a charge (density) distribution throughout space. On the right side is a spatial derivative. In one dimension, the derivative depends on points ahead and behind of the specific point in question; this generalizes to three dimensions. Therefore, the right side depends not just on the charge at a specific point but rather on the charge distribution throughout space.

    As a simple example, think of the field distribution of an isolated charge and its associated divergence. Now place a second charge at an arbitrary distance away from the first and think qualitatively of how the field distribution changes. The second charge, being arbitrarily placed, could be considered "outside the local neighborhood," but in general will affect the divergence.
  4. Jan 24, 2009 #3


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    No, it wont.

    Gauss' law says that the divergence of the electric field at any given point is proportional to the charge density at that point. It doesn't matter if another charge is placed somewhere else or not. In general, the electric field will change, but the divergence will not. The divergence of E at any point depends only on the charge density at that point and nothing else.
    Last edited: Jan 24, 2009
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