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Point charge(Q) has a divergence zero

  1. Aug 20, 2010 #1
    The D field due to a point charge(Q) has a divergence zero

    D = (Q/(4*pi*r2)) ar (ar --- unit vector)

    if we calculate the divergence we get zero.

    but guass law says that the divergence is a finite value not zero because a charge is present in there.

    is it that in calculating divergence we also take a particular surface?

    where im wrong? im confused of divergence..i tried wikipedia but could understand it upto my extent..i think somebody could help me
     
  2. jcsd
  3. Aug 20, 2010 #2

    arildno

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    Re: divergence

    1.Note the singularity of your vector field at r=0.

    2. You cannot make a regular volume integral over a region that includes this singularity, but you can perfectly well form a regular surface integral over a region containing that point in its interior.

    3. In order for the premises of Gauss' law to be strictly fulfilled, your function must be defined at ALL points in the region; since this is not the case here, you cannot really apply Gauss' law.

    4. It IS possible, however, to construct an irregular volume integral in this case, though, by representing the divergence of your vector field as a Dirac delta "function".
     
  4. Aug 20, 2010 #3
    Re: divergence

    so the divergence at a point shows the sink or source at a given point...so if at r=0 we are able to calculate the divergence then it would be equal to guass law result..is that what you are saying?
    if that is so, then divergence for every field except at the source of the field, every where it should be zero then..is it true?
     
  5. Aug 20, 2010 #4

    arildno

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    Re: divergence

    We CAN'T compute it at r=0 because the vector field isn't defined there. Therefore, rigorously, it has no meaning trying to apply Gauss' law, and thus, Gauss' law isn't "invalidated" by this case since the vector vield falls outside the class of vector fields Gauss' law is speaking about.

    Yes.
     
  6. Aug 20, 2010 #5
    Re: divergence

    so divergence is calculated at a particular point only...then in case we calculate for a surface then we get all the sources and sinks which is nothing but the total charge...and this is what gauss law says ..im correct??
     
  7. Aug 20, 2010 #6

    gabbagabbahey

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    Re: divergence

    I'd have to disagree with you there. Many quantities in physics are loosley called "functions", but are really distributions. This is true of charge density. For some idealized continuous charge distributions, you can represent the charge density as a function, but most of the time it can only be represented as a distribution. So, since Gauss' Law (for [itex]\textbf{E}[/itex] or [itex]\textbf{D}[/itex]) involves charge density (total or free), which is a distribution, it should be understood that the divergence of [itex]\textbf{D}[/itex] should be treated as a distribution and calculated accordingly.

    @bharath423

    You need to be careful when calculating the divergence near point, line, or surface charges. The easiest method to find the correct distribution is to calculate the divergence normally everyhwere except near you singularities (in this case, everywhere except a small region around your point charge), and then calculate the flux through any closed surface surrounding a single singularity, and using the divergence theorem, deduce what the divergence must be in the visciunity of that singularity.

    For your case, you should take a look at Griffith's Introduction to Electrodynamics where the Author calculates the divergence of [tex]\frac{\hat{\mathbf{r}}}{r^2}[/tex]. IIRC it is near the very end of the 1st chapter.
     
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