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Electric Charge Density

  1. Sep 15, 2005 #1
    (src: Intro to Electrodynamics, Griffith, Problem 1.46a)
    Q: Write an expression for the electric charge density [itex] \rho (r)[/itex] of a point charge [itex]q[/itex] at [itex]r^'[/itex]. Make sure that the volume integral of [itex]\rho[/itex] equals [itex]q[/itex].

    Now, Closest I can seem to come up with is:


    [tex] \rho(r)=\frac{q}{4*Pi*R^2}\delta(r-r^')[/tex]

    But, the problem I see with this, is that while yes, integrating this over any volume [itex]V[/itex] that enclosed the point charge will return q, but that q would have to have units of charge/unit_volume which just dosent make sense. Or am I missing something?

    Any help would be appreciated.
     
  2. jcsd
  3. Sep 16, 2005 #2

    Tom Mattson

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    I think that delta function should be 3D: [itex]\delta^{(3)}(\vec{r}-\vec{r}')[/itex]. Note that the n-D delta function has dimensions of (length)-n.
     
    Last edited: Sep 16, 2005
  4. Sep 16, 2005 #3
    Yeah, sorry missed that. Have the [itex]\delta^3[/itex] on my paper, just forgot to type it in.

    I don't understand how n-D delta functions have a dimension of (length)-n, could you explain that perhaps?
     
  5. Sep 16, 2005 #4

    Tom Mattson

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    Sure, let's look at the 1D case. Consider the following integral:

    [tex]\int_{-\infty}^{\infty}\delta(x)dx=1[/tex]

    The right side of that is 1. Not 1 meter or 1 Joule, just plain old 1. So if the units of dx are meters, then what must the units of the delta function be? Inverse meters.

    Similar results hold for higher dimensional cases.
     
  6. Sep 16, 2005 #5
    Okay, that makes sense.

    Thanks for your help, this was driving me crazy, I couldnt figure out why units were not making sense.
     
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