What is the Application of Dirac Delta in Charge Constellations?

[sunrah]

Homework Statement

We have to give the total charge, dipol and quadrupol moments of a charge constellation, but I seem to be falling at the first hurdle.

Q = \frac{1}{4\pi \epsilon_{0}} \int_{vol} \rho(\vec{r}) d^{3}\vec{r}

whereby the charge density of the group of particles is:

\rho(\vec{r}) =q\delta(\vec{r} - R\vec{e_{x}}) + q\delta(\vec{r} + R\vec{e_{x}}) + q\delta(\vec{r} - R\vec{e_{y}}) + q\delta(\vec{r} + R\vec{e_{y}}) - 2q\delta(\vec{r} - R\vec{e_{z}}) - 2q\delta(\vec{r} + R\vec{e_{z}})

Homework Equations

I'm using the following property of the delta function:

\int_{vol} \delta(\vec{r} - R\vec{e_{x}}) d^{3}\vec{r} = \int_{vol} \delta(x - R) dx \int_{vol} \delta(y)dy \int_{vol} \delta(z)dz = 1

The Attempt at a Solution

ok, so I got zero net charge. Which means I don't have a dipol or quadrupol moment either. help!

 

[vela]

You're right. There's zero net charge. A zero net charge doesn't necessarily mean the dipole and quadrupole moments vanish, however.

 

[sunrah]

That's a relief. I still find the delta function a bit confusing.

ahh, I hadn't realized that about the moments. so I worked out the dipole moment like this:

\vec{P} = \int \rho(\vec{r}) \vec{r} \delta(\vec{r} - R\vec{e_{x}}) d^{3} = \Sigma q_{i}\vec{r}_{i}

which in this case was also zero due to cancelling. I'm guessing the quadrupole moment is non-zero, which will be fun :)