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Net charge of a distribution that includes delta function

  • Thread starter coaxmetal
  • Start date
  • #1
4
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Homework Statement



show that the charge distribution ([tex]|\vec{r}|\equiv r [/tex])
[tex]\rho(r) = Z\delta^3(\vec{r})-\frac{Ze^{-r/R}}{4\pi R^2r}[/tex]
has zero net charge for any Z and R. Explain the meaning of Z.

Homework Equations



none given, but divergence (gauss) theorem and poisson's equation may be necessary.

The Attempt at a Solution



integrate to get net charge:
[tex]Q = \displaystyle\int_V \rho(\vec{r})d\tau[/tex]

I can split this into 2 integrals and use the volume element for spherical polar coordinates ([tex]d\tau = r^2dr \sin{\theta} d\theta d\varphi[/tex]), so the first integral is

[tex]\displaystyle\int_V Z\delta^3(\vec{r})r^2dr \sin{\theta} d\theta d\varphi[/tex]

This is what I get stuck on. I don't know how to handle the delta function (I assume the Dirac delta, I am given no other information) raised to a power. The fact that the volume integral is a triple integral seems like it may be important here, but I don't recall any tricks or anything about volume integrals.
 

Answers and Replies

  • #2
674
2
An integral of a delta function is very simple. If you get confused in spherical coords, then switch to cartesian coords for that integral. It isn't illegal to switch coordinate systems. Although the 2nd integral you will definitely want to use spherical coords. Hopefully you know the integral of a delta function in one dimension.
 
  • #3
4
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what I am having trouble with is how to integrate the cube of the delta function... Do you think it is supposed to be the third derivative, not the cube? (the notation given is exactly as shown, and to me that notation means cube). That would make more sense to me.

If it is the cube, how should I deal with it?
 
  • #4
674
2
It doesn't mean cubed. Just like [tex]d^3r[/tex] in a volume integral doesn't mean cubed. It just means it is a 3-dimensional delta function. In cartesian coords it looks like:

[tex]\delta^3(\vec{r}) = \delta(x)\delta(y)\delta(z) \neq \delta(\vec{r}) \delta(\vec{r}) \delta(\vec{r})[/tex]
 

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