- #1
realcomfy
- 12
- 0
I am trying to integrate a charge density over a volume in order to obtain a total charge, but there is a delta function involved and I am not entirely sure how to treat it.
[tex] \rho = q* \delta (\textbf{r})- \frac {q\mu^{2} Exp(- \mu r)} {4 \pi r} [/tex]
[tex]Q = \int \rho (\textbf{r})d^{3}r[/tex]
[tex] d^{3}r = r^{2} dr d(cos \theta )d \phi [/tex]
Plugging the density above in the the Q equation should give me the integral over a delta function times the differential volume element minus the integral of the uglier function times the same volume element.
All this is good and fine except that I get that [tex] \int q \delta (\textbf{r}) r^{2} drd(cos (\theta)d \phi [/tex] and I'm not totally sure how to evaluate it. I am pretty sure it should be something like 4 [tex]\pi[/tex] times the function r evaluated at zero, but I can't find the right answer with this approach. Any help would be appreciated.
[tex] \rho = q* \delta (\textbf{r})- \frac {q\mu^{2} Exp(- \mu r)} {4 \pi r} [/tex]
[tex]Q = \int \rho (\textbf{r})d^{3}r[/tex]
[tex] d^{3}r = r^{2} dr d(cos \theta )d \phi [/tex]
Plugging the density above in the the Q equation should give me the integral over a delta function times the differential volume element minus the integral of the uglier function times the same volume element.
All this is good and fine except that I get that [tex] \int q \delta (\textbf{r}) r^{2} drd(cos (\theta)d \phi [/tex] and I'm not totally sure how to evaluate it. I am pretty sure it should be something like 4 [tex]\pi[/tex] times the function r evaluated at zero, but I can't find the right answer with this approach. Any help would be appreciated.