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Divergence of Electric Field

  1. Aug 12, 2004 #1

    Gza

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    I'm not sure why this question comes to mind now, since I haven't had an E&M class for a few months now, but nonetheless. Place some charge at the origin. Surround the charge with a spherical Gaussian surface and calculate the surface integral. You obviously get a non-zero result(Gauss's Law), but this seems to violate the divergence theorem, equating the divergence of a field within a volume to the surface integral of the field over the surface enclosing the volume. This violates it because the divergence of the electric field is zero. So if the divergence of the charge's electric field within the spherical volume is zero, how can the surface integral be non-zero??? :confused: I know i'm missing something big here, but if someone can clarify where I went wrong with my reasoning, that would be great. Thanks.
     
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  3. Aug 12, 2004 #2

    arildno

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    The electric field (induced by the point charge) is undefined at the origin (you've got a singularity there).

    By setting the divergence of the electric field equal to (a specific multiple of) the delta function rather than 0, Gauss' law holds.
     
  4. Aug 12, 2004 #3
    Hi,
    I have passed the exam of vector analysis time ago, so I hope I remember right:
    If we have a solenoidal field (field without source) then the divergence of this field is equal 0. If that's true then the flow through the closed surface is equal 0.
    If we have a charge inside the sphere, then we have a source, thus the divergence is not 0.
    Electric field is potential type, so its rotor is equal 0. You have probably changed rotor and divergence or divergence and gradient.
    I have also forgotten this a little bit.. If we integrate the field of potential field through a closed curve, then we get zero (here we can see the independence of the curve form), but what about surface?
    If we have a sphere and the charge in the center of the sphere, then the sphere connects all points with same potential, so if we integrate the electric field through the surface we get 0.
    Hope you understand what I have written, because some time has passed since I have learned these things.
    If anything is wrong, comments would be really helpful.
    Regards,
    Niko
     
  5. Aug 12, 2004 #4

    ZapperZ

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    You asked a very good question. arildno has given you a hint on the fact that you cannot ignore that obvious delta function sitting at the origin. I think a clear derivation of the divergence of the electric field for a point charge might be useful, so here it is:

    http://www.lyon.edu/webdata/users/s...vergence and Curl of Electrostatic Fields.doc

    Zz.
     
    Last edited: Aug 12, 2004
  6. Aug 12, 2004 #5

    arildno

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    I'd rather say it lurked there..:wink:
     
  7. Aug 12, 2004 #6

    Gza

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    I think my confusion stemmed from my understanding of what the "delta function" is. I'm assuming it has something to do with the charge density as a function of position. But regardless of this, having an electric field:

    [tex]\vec{E} = \frac{kq}{r^2}\hat{r} [/tex]

    by calculating its divergence, you will find it to be zero. But there is a charge q present, so how can the divergence of the field possibly be zero, since my knowledge of Gauss's law tells me that there will be a nonzero surface integral?
     
  8. Aug 12, 2004 #7

    Tom Mattson

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    Not at the origin, you won't--the divergence of E is undefined there. That's the whole point that is being made.
     
  9. Aug 12, 2004 #8

    Gza

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    So I assume that the charge must occupy a volume, which is why you must bring density into the whole thing, and then run the integration over the volume using the charge density. I understand it verbally now, but what would this electric field with a non-zero divergence look symbolically (This is all kind of new to me, i've only taken an intro to E&M).
     
  10. Aug 12, 2004 #9

    arildno

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    Gza:
    You are "over-thinking" a bit here.
    Let's first agree upon the actual value we get by calculating the flux over the surface:
    [tex]4\pi{k}q[/tex] (Agreed?)

    A very common choice is to set [tex]k=\frac{1}{4\pi\epsilon_{0}}[/tex]
    where [tex]\epsilon_{0}[/tex] is called the permittivity of vacuum/free space.

    The Dirac delta "function" is a handy tool to tackle the effects of intense, extremely local phenomona (among other stuff)

    One example is to predict the vibrations in a bar induced by giving it a sharp knock initially. Another is to handle the effects of a "point" concentration of charge, as is your case.

    We "define" the delta function (for a single variable)as follows:
    [tex]\delta(x)=0,x\neq{0}[/tex]
    [tex]\int_{V}\delta(x)dx=1[/tex]

    for any volume V with 0 included.
    (To get the multivariable version, just use [tex]\delta(\vec{x})[/tex] instead)

    Clearly, the delta "function" is not any regular type of function (we refer to it as either an example of a generalized function or distribution)

    Note therefore, that if we define the divergence as
    [tex]\nabla\cdot\vec{E}=4\pi{k}q\delta(\vec{x})[/tex]
    the volume integral yields the same result as the surface integral.
     
  11. Aug 12, 2004 #10
    Would anyone mind explaining to a beginner what the divergence theorem is?

    I'm completely beginner at physics at this level, me being too young to enter university as yet, but hopefully it's not to dreary a question.
     
  12. Aug 12, 2004 #11

    pervect

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  13. Aug 13, 2004 #12

    turin

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    Gza,
    Can you relax your "point charge" to a very tiny ball of radius, R0, of charge with some constant (for simplicity) charge density, ρ0? Then, you can see a nontrivial divergence inside the ball:

    DivE = 4πkρ0

    for r < R.
     
  14. Aug 13, 2004 #13

    Gza

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    Thanks everyone. I had some time to think about it, and it seems blatantly obvious to me now. You can't use a point charge approximation (which I was insisting on doing) to understand it.
     
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