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What does finite volume charge density mean?

  1. Mar 30, 2013 #1
    1. The problem statement, all variables and given/known data
    "as long as the volume charge density is finite (which is not true of surface charge distributions or point charges), the electric field is continuous.


    2. Relevant equations



    3. The attempt at a solution
    I know that for surface charges distributions and point charges, the electric field isn't continuous, because the electric field changes direction...but what does the sentence mean when it says "finite volume charge density?"
    thanks to anyone who replies.
     
  2. jcsd
  3. Mar 30, 2013 #2

    rude man

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    charge density = charge/volume. The idea is the charges are each infinitesmally small so the density is continuous. Think of water: it has finite density everywhere yet it comprises billions of individual H2O molecules which fact we neglect in doing any kind of macro compuations. Likewise we ignore the fact that charges are finite in size and charge, and we think of continuous density as no. of charges of q each (e.g. electrons), divided by the volume they occupy, as ρ = Ʃq/V.
     
  4. Mar 30, 2013 #3
    so, what would be another example of "infinite" density?
    why would a surface charge distribution not have "finite" density? is it because that a surface has no volume?
     
  5. Mar 30, 2013 #4
    This can be explained by understanding Gauss's Law->

    https://en.wikipedia.org/wiki/Gauss's_law

    Let us consider a point charge.... the point charge creates an electrical field which acts away from it in all directions (or towards it - cant remember the direction of the arrow). This depends on whether the point charge is positive or negative. Some people like to use positive "test charges".

    Now define the electric flux as the rate of flow per unit area of this electric field... essentially how packed together those lines which start/end at that point charge are.

    Consider that all of the electrical field acts perpendicular to every point on the surface of a sphere which surrounds this point charge.... as long as the point charge which creates the field is finite (ie constant and not moving/affected by other electrical and magnetic fields)... this is an example where the electrical field shape is symmetrical.

    Symmetrical electric fields shapes are awesome because they make the maths for these kinds of problems much simpler. If you were to measure the electric field density (or electric flux) through any section or square area on the surface of the sheet enclosing this point charge... you would get the same value irrespective of whether you put your square at the top or the bottom or left or right hand side or in fact anywhere on the skin of the sphere (as long as you use a square of the same side).

    A possible short answer is that a finite volume charge density considered as a point charge creates an electrical field which is symmetrical. This symmetrical electric field has the same flux through it over the same area at any pont when you consider the effect of the field through a square on the skin of a sphere enclosing the charge. Hence this field flux density is continuous. It is actually an example of an equipotential

    In planetary physics non-symmetrical shapes (like the planets) are often assumed to be point charges to make rough calculations about inverse-square gravitational fields. So I guess this is a reasonable assumption. Hope that helps.

    This youtube video explains a little about charges/fluxes/densities/gauss's law


    Good Luck!
     
    Last edited by a moderator: Sep 25, 2014
  6. Mar 30, 2013 #5

    rude man

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    That's a good answer. Surface charges have no depth, therefore no volume. So the charge density around them is zero except where they're located, at which point the density is essentially infinite.

    Some more advanced folks use the Dirac delta to expess this. They say the charge density is ρ = q δ(x)δ(y)δ(z), meaning the charge q = ∫∫∫ρ dx dy dz integrated over all space but δ is zero everywhere except at the origin itself. δ(x) has dimensions of L-1 so the triple product has dimensions of L-3 = 1/volume. So we still have good old q = ρ x volume, though volume → 0 and ρ → ∞ at the origin (0,0,0) and = 0 everywhere else.
    .
     
  7. Apr 4, 2013 #6
    Thank you very much, AugustCrawl and rude man.
    your detailed explanations helped me a lot, thx for taking the time to reply!
     
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