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Homework Help: Divergence Form of gauss's law

  1. Sep 14, 2009 #1
    Im really having troubles understanding the divergence form of gauss's law. I have done research on it and am still not able to understand it. it sates that E=[tex]\rho/\epsilon[/tex] or E=rho/epslom, so does that mean that the upside down triangle has no significance, ie does that mean i can simply solve and get rho=E*epslom. Also must the charge density be constant or can it vary with the distance from the centre as the electric field varies??

    Thankx in advance for any help provided
    Last edited: Sep 14, 2009
  2. jcsd
  3. Sep 14, 2009 #2


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    The "upside down triangle" is the "nabla" symbol used to express Divergence, Gradient, and Curl operations in differential calculus.

    [tex] \nabla \bullet \mathbf{E} = \rho/\epsilon[/tex]
    The divergence of the E field is the charge density over epsilon.

    Understanding first that the E field has nothing to do with fluid flow we can still use the analogue of fluid flow to understand the divergence operation.

    If you have an incompressible fluid then the only way fluid can flow away from a point in all directions is if fluid is being generated at this point. This is called a source.

    The divergence measures the extent to which a vector field seems to be spreading out i.e. if the vector field is a velocity field for a fluid it measure the extent to which that fluid has a source (negative divergence means a sink = negative source sucking up fluid).

    In electromagnetism the charge density is the analogue of a source and the E field the analogue of a flow (flux). Even though we don't consider the E field to correspond to any physical flow we still use the terms "flux" and "source" and say the charge density is the source of the E field.

    By the same token the curl of a fluid's velocity field expresses the vorticity. You can use this analogy to help visualize the differential forms of Ampere and Maxwell-Faraday laws.

    I suggest you get a really clear picture of Gauss' Law first.
    (The integral of a vector field's component normal to a closed surface is equal to the integral of its divergence inside the region.)
    [tex] \iint_{\partial \Omega} \mathbf{F}\bullet \mathbf{dS}=\iiint_\Omega \nabla \bullet \mathbf{F} dV [/tex]
    Imagine Gauss' Law applied to an arbitrary small spherical region at a point and you'll have basically the divergence at that point. The divergence equals the flux through the surface which must equal the total source inside that region.
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