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Gauss's law proof

  1. Jan 21, 2005 #1


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    See the "Gauss's Law Sketch Proof" paragraph here. Griffiths uses mainly the same argument to prove Gauss's law, but I don't really see the connection between the fact that the flux is Q/epsilon for a sphere and the conclusion that it is worth this quantity too for any closed surface.

    Can someone point to a more mathematical proof? Or simply unveil what I'm missing in this argument.

  2. jcsd
  3. Jan 21, 2005 #2
    what kind of prove you wanna see? The surface integral of E field = q/epsilon? if yes, why don't you try the simpliest case... a point charge at the center and integrate the sphere with radii R surround it...
  4. Jan 21, 2005 #3

    Doc Al

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    You might find this helpful: http://www.rpi.edu/dept/phys/Dept2/phys2/activities/gauss/proof.html
    It doesn't give a rigorous proof, but it does expand the argument a bit more than the site you gave. The key is that the flux depends on the solid angle, not on the radius (see how the r^2 terms cancel).
  5. Jan 21, 2005 #4
    My book starts with the differential form first and then applies the divergence theorm. If you understand how to get to the differential form then I think that's a better way to go.
  6. Jan 21, 2005 #5
    differential form?
    you mean divE=charge density?
    how could you PROVE that w/o guass law? I really wanna know
    if you cant, how could you PROVE guass law by the differential form?
    are you doing a circular logic here?

    div E= charge density is the direct result of guass law...... you can PROVE it by the definition of "div" & gauss' law,
    Last edited: Jan 21, 2005
  7. Jan 21, 2005 #6

    E(r)=\frac{1}{4\pi\epsilon_0} \iiint \frac{r-r'}{|r-r'|^3}\rho(r') d^3r'


    [tex]r = x\hat{i} + y\hat{j}+z\hat{k}[/tex] (region where you want to find the E-field)

    [tex]r' = x'\hat{i} + y'\hat{j}+z'\hat{k}[/tex] (region of charge)

    From that equation you can use vector calculus to convert to the differential form. I don't want to type down the proof because it involves alot of typing in TEX. That and alot of the details aren't clear in my book. For now just accept that the math works out and you get

    [tex]\nabla \cdot E = \frac{\rho}{\epsilon_0}[/tex]

    Then take the volume integral of both sides.

    [tex]\iiint (\nabla \cdot E) dV = \frac{1}{\epsilon_0} \iiint \rho(r) dV[/tex]

    [tex]\iiint \rho dV = Q_{enc}[/tex] (total charge)

    [tex]\iiint (\nabla \cdot E) dV = \iint E \cdot dA[/tex] (divergence theorm)

    which gives you:

    [tex]\iint E \cdot dA = \frac{Q_{enc}}{\epsilon_0}[/tex]
    Last edited: Jan 21, 2005
  8. Jan 21, 2005 #7
    [tex]\nabla \cdot E = \frac{\rho(r)}{\epsilon_0}[/tex]
    how could you prove the above equation without using gauss law?
  9. Jan 21, 2005 #8


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    That site if nice Al but FulhamFan3's way satisfies me fully.
  10. Jan 21, 2005 #9
    How do you prove it using gauss's law?
  11. Jan 21, 2005 #10
    hey, quasar, [itex]\nabla \cdot E = \frac{\rho(r)}{\epsilon_0}[/itex] is the direct consequence of gauss law... you can't get the above formulas without assuming guass law is right in the first place, therefore, if you prove the guass law by the above formulas, you are doing a circular logics...
  12. Jan 21, 2005 #11


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    No, I'll show you vincent, if FulhamFan3 doesn't want to.
  13. Jan 21, 2005 #12
    [tex]\nabla \cdot \ver{F} = \lim_{\Delta V \rightarrow 0} \frac{\int \vec{F} \cdot d \vec{A}}{\Delta V}[/tex]
    the above is the definition of [itex]\nabla \cdot \vec{F} [/itex]

    replace F by E and the integral by Q/epsilon, see what you get

    actually, people called [itex]\nabla \cdot E = \frac{\rho}{\epsilon_0}[/itex] "the differential form of gauss law", "point form of gauss law" etc... it IS gauss law... what you just did is proving gauss law by gauss law
  14. Jan 21, 2005 #13


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    starting from his equation of the field (which is a direct consequence of Coulom's law)

    [tex]E(r)=\frac{1}{4\pi\epsilon_0} \iiint \frac{r-r'}{|r-r'|^3}\rho(r') d^3r'

    Let's write is vectorially...

    [tex]\vec{E}=\frac{1}{4\pi\epsilon_0} \iiint \frac{\rho(\vec{r'})}{|\vec{r}-\vec{r'}|^2}(\vec{r}-\vec{r'}) dV'[/tex]

    Now, it is done by taking the divergence of this expression. It turns out that

    [tex]\nabla \cdot \left(\frac{\vec{r}-\vec{r'}}{|\vec{r}-\vec{r'}|^2} \right) = 4\pi \delta^3(\vec{r}-\vec{r'})[/tex]

    where delta^3 is the 3-D dirac delta function. Therefor

    [tex]\nabla \cdot \vec{E}=\frac{1}{4\pi\epsilon_0} \iiint 4\pi \delta^3(\vec{r}-\vec{r'}) \rho(\vec{r'}) dV' = \frac{\rho(\vec{r'})}{\epsilon_0}[/tex]

    The last equality is from the property of the delta fucntion.
  15. Jan 21, 2005 #14
    Just because I didn't type in the proof for converting coulomb's law to the gauss differential form doesn't mean it doesn't exist.

    As long as quasar understands i'm happy.
  16. Jan 21, 2005 #15
    quasar at the end that r isn't a vector, it should be a scalar coordinate
  17. Jan 21, 2005 #16
    now you need to prove the inverse square law.......
    actually, my point is... in physics... not everything can be proved... you can assume the inverse square law is true, and prove the gauss law. Or you can do it the other way around... just wanna let you know... if you assume gauss law is true in the first place, the world would look much simpler and make sense...
    what guass law said is all charge emit electric field... (imagine E field is a line start at the charge and goes to infinite) and you wrap the charge by a closed surface.. no matter how your closed surface looks like... number of lines passes through the surface is the same...
    most of the physicsist use gauss law prove the inverse square law, but not the other way around... how come the there is a 1/4pi epsilon constant in the inverse square law but only a 1/epsilon constant in the gauss law...
    and one more thing.. I was wrong... you can prove the gauss law by inverse square law.. but you are making yourself in a mess... try to see it the other way... and understand the definition of div and curl...
  18. Jan 21, 2005 #17


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    Well, Griffiths keeps the vector dependancy of rho... I don't see why that's wrong, as long as rho itself is a scalar.

    Actually Griffiths writes


    (the prime poped out), but that ought to be a typo.. :grumpy:
  19. Jan 21, 2005 #18
    gauss's law came from coulomb's law, not the other way around.

    who? I know you can find it that way but that's to validate that gauss's law works, not to prove the inverse square relation.

    the original formulation of the inverse square law used "k" instead of [itex]\frac{1}{4\pi\epsilon_0}[/itex]. Epsilon is used now because it's more fundamental.
    Last edited: Jan 21, 2005
  20. Jan 21, 2005 #19
    I just don't see how a vector position in a formula makes the function scalar. But that's just arguing semantics. As long as we know it's the same thing it's cool.
  21. Jan 21, 2005 #20
    do you know which 4 formulas (plus F=qE+qvB) summarize the electromagnetics... I really can't find the coulomb law in maxwell equation....
    Last edited: Jan 21, 2005
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