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Charge Densities & Dirac's Delta Function

  1. Oct 22, 2009 #1
    1. The problem statement, all variables and given/known data

    What is the (volume) charge density of a ring of radius [tex]r_0[/tex] and uniform charge density [tex]\lambda[/tex]?


    2. Relevant equations

    The Dirac Delta Function

    3. The attempt at a solution

    I've done a few line charge densities of straight wires along an axis (usually z, but on x as well), but I'm getting stuck at using the delta functions when wrapping the wire around in a circle. I am pretty sure I'll want the ring to be lying in the y-z plane, as the problem continues with an integral with [tex]\cos\theta[/tex] in the integrand and [tex]\theta[/tex], being measured from the z-axis, should give me a delta function there to make the integral easier.

    Any suggestions?
     
  2. jcsd
  3. Oct 22, 2009 #2

    gabbagabbahey

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    Unless you are told otherwise, you are free to choose whatever coordinate system you like. Personally, I'd use cylindrical coordinates [itex]\{r,\theta,z\}[/itex] oriented so that the ring is in the xy-plane and centered on the origin.....when you do this, the ring has zero extent in both the [itex]z[/itex] and radial directions, so you would expect the volume charge density to be of the form [itex]\rho(\textbf{x})\propto\delta(r-r_0)\delta(z)[/itex].....I'll leave it up to you to find the constant of proportionality by means of a suitable integration...
     
  4. Oct 22, 2009 #3
    I don't think that using cylindrical coordinates will work for me in this case as the "problem continues" part involves the ring existing between two grounded spheres of differing radii, which requires spherical coordinates.


    Small note, which really is a bit of nit-picking from a newbie poster to a certified Homework Helper, but you should use [tex](\rho,\phi,z)[/tex] for cylindrical coordinates and not [tex](r,\theta,z)[/tex] due to the similarity to spherical coordinates and the confusion it can bring using your method.

    Also, the constant of proportionality, in cylindrical coordinates, would be, [tex]\frac{\lambda}{2\pi r}[/tex] ;)
     
  5. Oct 22, 2009 #4

    gabbagabbahey

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    If you know the volume charge density in cylindrical coordinates, what's to stop you from transforming it to the appropriate spherical coordinate system?



    There are only so many letters in the Greek and Latin alphabets. Using [itex]\rho[/itex] as the radial coordinate can also be confusing, since it is the same letter typically used for the volume charge density. There is no harm in using [itex]\{r,\theta,z}[/itex] as long as it is made clear that [itex]r[/itex] in this context, is the distance from the z-axis. Different authors use different notations with varying degrees of sloppiness, so a student must always look to the context in which a variable is used, to understand what it represents.

    Are you sure about that?:wink:
     
  6. Oct 22, 2009 #5
    Good point, working on that now

    See, this goes back to the spherical/cylindrical units--there shouldn't be the r in the denominator as lambda has units of charge/meter and each delta function has units of 1/meter making the three combined to be charge/meter^3; integrating this over all space gives charge, as it should. whoops!
     
  7. Oct 22, 2009 #6

    gabbagabbahey

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    The [itex]2\pi[/itex] also isn't necessary... Integrating the linear charge density over the ring should give the total charge, as should integrating the volume charge density over all space....perform the integrations and compare the results.
     
  8. Oct 22, 2009 #7
    You're right, did the integral and got the constant=[tex]\lambda[/tex].

    Back to the original problem, though I didn't convert from cylindrical to spherical because it didn't look quite right, I end up with a charge density of

    [tex]
    \rho(\mathbf{r})=\frac{\lambda\pi}{2r_0}\delta(r-r_0)\left[\delta\left(\phi-\frac{3\pi}{2}\right)+\delta\left(\phi-\frac{\pi}{2}\right)\right]
    [/tex]

    where [tex]\theta[/tex] is the angle sweeping from +z to -z and [tex]\phi[/tex] sweeps from +x towards +y. This still seems off to me, but [tex]\theta[/tex] runs from [tex]0\rightarrow\pi[/tex] so a point sticking out at radius [tex]r_0[/tex] and sweeping down [tex]\theta[/tex] at [tex]\phi=\pi/2[/tex] and [tex]\phi=3\pi/2[/tex] should give a circle on the y-z plane, right?

    The constant of proportionality then comes from

    [tex]
    Q=2\pi r_0\lambda=M\int_0^\infty r^2dr\delta(r-r_0)\int_0^{2\pi}d\phi\left[\delta\left(\phi-\frac{3\pi}{2}\right)+\delta\left(\phi-\frac{\pi}{2}\right)\right]\int_{-1}^1d(\cos\theta)
    [/tex]

    giving

    [tex]
    M=\frac{2\pi r_0 \lambda}{r_0^24}=\frac{\lambda\pi}{2r_0}
    [/tex]

    Does this make sense to you, because it still seems a little funny to me.
     
  9. Oct 24, 2009 #8

    gabbagabbahey

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    You should double check your value of [itex]M[/itex], but the general form of [itex]\rho[/itex] looks fine to me.
     
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