Application of Dirac delta

  • Thread starter sunrah
  • Start date
  • #1
198
22

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.

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

whereby the charge density of the group of particles is:

[itex]\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}})[/itex]


Homework Equations


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

[itex] \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 [/itex]

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!
 

Answers and Replies

  • #2
vela
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You're right. There's zero net charge. A zero net charge doesn't necessarily mean the dipole and quadrupole moments vanish, however.
 
  • #3
198
22
That's a relief. I still find the delta function a bit confusing.

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

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

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

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