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Electromagnetism: Charge Density of a Shell

  1. Sep 6, 2009 #1
    1. The problem statement, all variables and given/known data
    We learned in E&M about the Dirac Delta function and its applications to representing the densities of point charges in space. For example, the charge density of space with a point charge q at the origin is given by q*delta(r). How does this density representation work with continuous distributions, such as a charged spherical shell?


    2. Relevant equations
    None really


    3. The attempt at a solution
    Haven't got a clue, just need a hint to point me in the right direction. New to all this dirac delta nonsense
     
  2. jcsd
  3. Sep 7, 2009 #2
    do you havea copy of Riley, Hobson, Bense? it was good when i did this.

    basically the Dirac delta function "is" the continuous analogue of the Kronecker delta.

    one definition is:

    [itex]\int_{-\infty}^{\infty} f(x) \delta(x-x_0) dx=f(x_0)[/itex] for 1D or
    [itex]\int \int \int f(r) \delta(r-r_0) dV=f(r_0)[/itex] for 3D

    now let's say you have a sphere containing a uniformly distributed charge, the total charge of said sphere is [itex]Q=\int \int \int \rho(r) dV[/itex] by definition where [itex]\rho(r)[/itex] is the charge density inside the sphere.

    you're told that instead of having a sphere of uniformly distributed charge, you have instead just a point charge at the centre of the sphere (i.e. located at [itex]r=0[/itex] in that particular coordinate system). In this situation we can describe the charge density as [itex]\rho(r)=q \delta(r)[/itex] where q is the charge of the point charge.

    now we want to find the total charge:
    [itex]Q=\int \int \int q \delta(r) dV=\int \int \int q \delta(r-0) dV = q[/itex] using the definition of the Dirac delta function.

    hopefully that helps as to why you can write [itex]\rho(r)=q \delta(r)[/itex]
     
  4. Sep 7, 2009 #3

    kuruman

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    Think of it this way: For a charged spherical shell of radius R, the volume charge distribution is zero everywhere in space except at r = R where the volume charge distribution is infinite (because the thickness of the shell is zero.) Therefore, we write the volume charge density as

    [tex]\rho(r)=A\delta(r - R)[/tex]

    If you know that the total charge on the shell is Q, you find constant A by integrating the volume charge distribution over all space (triple integral) and setting it equal to Q.
     
  5. Sep 7, 2009 #4
    Ok, thanks. That makes sense now; so now I should say, for a volume containing a spherical shell of radius R centered at the origin:

    [tex]\int_V \rho\cdot\delta (r-R)\cdot r^2\,dr\,sin\,\theta\, d\theta\, d\varphi=Q[/tex]

    [tex]\rho\int_0^\infty r^2 \delta (r-R)\, dr \int_0^\pi sin\,\theta\, d\theta \int_0^{2\pi} \, d\varphi[/tex]

    [tex]\rho\cdot R^2\cdot 4\pi = Q[/tex]

    Therefore the volume charge density should be [tex]\rho = {Q\over 4\pi R^2}\cdot\delta(r-R)[/tex] for the total charge to come out to Q, right? (Sorry, I'm not very experienced with TeX)
     
  6. Sep 7, 2009 #5
    One more thing, does this mean that the one-dimensional delta function is endowed with the units of 1/length?
     
  7. Sep 7, 2009 #6

    kuruman

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    Yes, if the independent variable has dimensions of length.
     
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