How do you express 3D charge distributions using the Dirac Delta function?

AI Thread Summary
The discussion focuses on expressing three-dimensional charge distributions using the Dirac delta function, particularly in spherical coordinates. The user attempts to formulate the charge density for a uniformly distributed charge over a spherical shell of radius R but encounters issues with integration limits and incorrect results. Participants emphasize the importance of correctly defining the Dirac delta function and ensuring consistent notation throughout the calculations. They point out that the limits for the radius should not include negative values, and clarify that the Dirac delta function should not be treated as a charge density. The conversation highlights the need for careful mathematical treatment to achieve accurate results in expressing charge distributions.
ZetaX
Messages
6
Reaction score
0
Hello community, this is my first post and i start with a question about the famous dirac delta function.
I have some question of the use and application of the dirac delta function.
My first question is:
Using Dirac delta functions in the appropriate coordinates, express the following charge distributions as three-dimensional charge densities [PLAIN]http://www.physics.sfsu.edu/%7Edsoto/704/1.3_files/image002.gif.[/I] [PLAIN]http://www.physics.sfsu.edu/%7Edsoto/704/1.3_files/image003.gif[B][I](a)[/I][/B][I] In spherical coordinates, a charge [PLAIN]http://www.physics.sfsu.edu/%7Edsoto/704/1.3_files/image005.gifuniformly distributed over a spherical shell of radius R.So, i know the definition of dirac delta function in spherical coordinates ist \rho(\vec{r})=\sum_{k=0}^{\infty}\frac{1}{r_k^2\sin\theta}\delta(r-r_k')\delta(\theta-\theta')\delta(\phi-\phi').

Moreover i know Q=\int\rho(\vec{r})d^3r. So i substitute the density function into the integral and transform it into spherical coordinates.

Q=\int\int\int\sum_{k=0}^{\infty}\frac{1}{r_k^2\sin\theta}\delta(r-r_k')\delta(\theta-\theta')\delta(\phi-\phi')r^2\sin\theta drd\theta d\phi
Now i change the sum into the integral and the limits become:

\int_0^{2\pi}\int_0^{\pi}\int_{-\infty}^{\infty}\delta(r-R)drd\theta d\phi
So here is the problem, if i integrate i get a incorrect answer...

What is my mistake?

Greetings! :)
 
Last edited by a moderator:
Physics news on Phys.org
Shouldn't be your limits of integration for ##r## would be from ##0## to ##R##?
 
i.e. does it make sense for the radius to have a negative value?
Make sure you understand what each term in the notation is for.

What does the integral need to look like to come out right?
You certainly don't have the right units - you seem to have used ##\rho(\vec r)## in the first two equations to mean different things. In the first, it is the dirac delta function in spherical coordinates and in the second it is a charge density ... then you treat the dirac delta function as a charge density which it is not.
 
Last edited:
I think it's easist first to watch a short vidio clip I find these videos very relaxing to watch .. I got to thinking is this being done in the most efficient way? The sand has to be suspended in the water to move it to the outlet ... The faster the water , the more turbulance and the sand stays suspended, so it seems to me the rule of thumb is the hose be aimed towards the outlet at all times .. Many times the workers hit the sand directly which will greatly reduce the water...
Back
Top