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Difficulty in differential form of gauss law

  1. Sep 16, 2012 #1
    hi all,
    im new at electromagnetics and vector calculus, so facing these trivial problems. The differential form of gauss's law for empty space states ∇E = ρ/ε .The right side refers to charge density at the point of calculation of divergence? If a point charge(or a small uniformly charged sphere) is placed at (0,0) then the divergence at position vector r comes out positive( k/r). But charge density at that point is 0?
     
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  3. Sep 16, 2012 #2

    Simon Bridge

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    A sphere of radius r and total charge Q, uniformly distributed, will have a non-zero charge density everywhere inside the sphere. The total charge at any point is zero - since a point has zero volume.

    Don't forget to keep thinking "divergence of the electric field".
     
  4. Sep 17, 2012 #3
    To model a point charge, one simply sets the charge density to be the dirac delta function, which is like an infinite density point function.
     
  5. Sep 17, 2012 #4
    Thanks for replying... I am still not clear. Suppose to avoid point charge, I take a uniformly charged sphere(Q) of radius R, centerd at (0,0). Electric field distribution will be same as that of point charge Q outside the sphere. Now can we say that charge density at any position r in space is given by ρ = Q/4∏εr[itex]^{2}[/itex] for r<R and 0 for r>R. So for any point outside the sphere, the charge density is 0 right? But divergence of electric field is not. Basically I find difficulty in the divergence theorem which is used to derive this form of gauss law. Lets take a closed volume surface in the above space, away from the charged sphere. EM states that flux of E through the surface = total charged enclosed by surface(scaled) = integral of divergence of electric field over volume of surface. The first 2 terms should be 0(outside charge) but since divergence is positive at all points in surface(k/r), how is the sum 0? Thanks in advance
     
  6. Sep 17, 2012 #5

    Simon Bridge

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    The density is the total charge Q divided by the total volume V=4πR3/3 inside the sphere. To avoid the peicewise function we can write:

    $$\rho(r)=\frac{3Q}{4\pi R^3}h(R-r)$$... where h(r) is the Heaviside Step function.

    You appear to be conflating two distinct things.
    If there is charge outside the volume, then there is non-zero passing through any place inside the volume - and that flux can have a divergence. Since the same amount of flux leaves the volume as enters it, then the net flux through the volume is zero.

    It's like a train station - nobody stays there ... over time the same number of people leave as enter. However, there can still be a crowd in there.
     
  7. Sep 17, 2012 #6

    BruceW

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    Yes, Simon Bridge has the right idea that the total flux through a Gaussian surface which does not contain the sphere, is equal to zero. In other words, the divergence of the electric field outside the sphere is zero. This may seem weird, because the field lines are 'spreading out' in space, but the divergence is still zero. This is because the field lines are spreading out in such a way that the divergence is zero.
     
  8. Sep 17, 2012 #7

    Simon Bridge

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    I slightly mangled that description didn't I? I shouldn't answer these things before I have my coffee!
     
  9. Sep 19, 2012 #8
    Thanx! that helped :)
     
  10. Sep 19, 2012 #9
    The easiest mathematical way to show it is to take a volume element in spherical polar coordinates. The four side-wall normals are perpendicular to the field (so no flux), and the inner wall has a smaller area than the outer wall by a factor of r^2, which is exactly the factor by which the electric field of a point charge is reduced from the inner wall to the outer wall. The flux entering the inner wall must equal the flux exiting the outer wall for every value of r>0, and so the divergence is zero.
     
  11. Sep 19, 2012 #10
    This is not the differential form of gauss's law fror empty space. For empty space, the divergence of the electric field intensity is equal to zero. This is the differential from of gauss's law for a region in which a distributed charge is present with charge density ρ.
     
  12. Sep 19, 2012 #11
    It's not even a divergence as written - there is an all-important dot missing between ∇ and E, something not picked up on.
     
  13. Sep 19, 2012 #12

    Simon Bridge

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    The dot is often implied, isn't it?
     
  14. Sep 19, 2012 #13
    I'm not an expert in the mathematical customs of people working in electrostatics, but I have to agree with Q-reeus. It really struck me unusual that no responder mentioned this (I didn't mention it in my response because I wanted to keep focused on just the one point I was trying to make). Also, no one mentioned that the gradient operator and the electric field intensity were not represented in boldface to signify vectors. When I see the gradient operator acting on a vector without a dot in-between, I interpret this as the gradient of the vector, which is a second order tensor. Of course, I'm not sure what the practical application of this second order tensor might be.
     
  15. Sep 19, 2012 #14
    I'm not sure what you mean by the gradient of a vector, but said second order tensor would marry the divergence and curl into a single entity, and doing so can be useful.
     
  16. Sep 19, 2012 #15
    This is not so. The divergence of the field outside the charged sphere is zero.
    Just calculate the Div(1/r^2) in spherical (easier) or Cartesian coordinates.
     
  17. Sep 20, 2012 #16
    Ya sorry about the loose notations, meant divergence only... silly mistake :P
     
  18. Sep 21, 2012 #17

    vanhees71

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    You cannot calculate the divergence of a scalar. The correct formula for the divergence of the Coulomb field is
    [tex]\vec{\nabla} \cdot \left (\frac{\vec{r}}{4 \pi |\vec{r}|^3} \right )=\delta^{(3)}(\vec{r}).[/tex]
    In terms of the electrostatic potential
    [tex]\Phi(\vec{r})=\frac{1}{4 \pi |\vec{r}|}[/tex]
    this gives the important formula
    [tex]-\Delta \left (\frac{1}{4 \pi |\vec{r}|} \right )=\delta^{(3)}(\vec{r}).[/tex]
    To prove this take an arbitrary test function to show
    [tex]\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{r} \; f(\vec{r}) \vec{\nabla} \cdot \left (\frac{\vec{r}}{4 \pi |\vec{r}|^3} \right )=f(0).[/tex]
     
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