Net charge of a distribution that includes delta function

[coaxmetal]

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.

 

[nickjer]

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.

 

[coaxmetal]

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?

 

[nickjer]

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]

 

[paranormal]

I would approach this problem by first clarifying the meaning of Z in the given charge distribution. Z typically represents the atomic number or number of protons in an atom, so it is likely that this distribution is describing the charge distribution of an atom with a nucleus at the origin. This makes sense given the use of spherical polar coordinates and the presence of the exponential term representing the electron cloud.

Moving on to the actual problem, it is clear that the first term in the charge distribution, Z\delta^3(\vec{r}), represents the charge of the nucleus concentrated at the origin. The second term, -\frac{Ze^{-r/R}}{4\pi R^2r}, represents the charge of the electron cloud surrounding the nucleus.

To show that the net charge of this distribution is zero, we can use Gauss's law, which states that the flux of the electric field through a closed surface is equal to the net charge enclosed by that surface. In this case, we can choose a spherical surface centered at the origin and enclosing the entire charge distribution.

By symmetry, the electric field at any point on this surface will be directed radially outward. We can then use Gauss's law to relate the electric field to the charge enclosed by the surface:

\oint_S \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0}

Since the electric field is constant over the spherical surface, we can pull it out of the integral and solve for the enclosed charge:

\vec{E}\cdot 4\pi r^2 = \frac{Q_{enc}}{\epsilon_0}

Q_{enc} = 4\pi r^2 \epsilon_0 \vec{E}

Now, using the given charge distribution, we can calculate the enclosed charge:

Q_{enc} = \displaystyle\int_V \rho(\vec{r})d\tau = \displaystyle\int_V Z\delta^3(\vec{r})-\frac{Ze^{-r/R}}{4\pi R^2r} d\tau

The first term in this integral, Z\delta^3(\vec{r}), will only contribute to the enclosed charge at r=0, since the delta function is zero everywhere else. However, at r=0, the exponential term in the second term is also zero, so the entire integral becomes zero.

Therefore,