- #1
Living_Dog
- 100
- 0
Hi,
After reading DJGriffiths sections on the DDF I have a question about evaluating it in regards Prob. 1.46 (a), to wit:
"Write an expression for the electric charge density [tex]\rho(\vec{r})[/tex] of a point charge q at [tex]\vec{r}[/tex]'. Make sure that the volume integral of [tex]\rho[/tex] equals q."
This is easily done in rectangular coordinates, namely, let [tex]\rho(\vec{r}) = q\delta(\vec{r}-\vec{r}')[/tex].
Checking that this resultsin a volume integral of q I did:
[tex] \int_V \rho(\vec{r}) d\tau = \int_V q\delta^3(\vec{r}-\vec{r}') d\tau = q \int_V \delta^3(\vec{r}-\vec{r}') d\tau = q \cdot 1 = q.[/tex]
Here is my question: doing the integral as is I get the correct answer. However, when I wrote it out in spherical coord's. It seems that I should rewrite [tex]\rho(\vec{r})[/tex] as [tex]\frac{q}{4\pi}[/tex]?
Why? Because:
[tex] \int_V \rho(\vec{r}) d\tau = \frac{q}{4\pi} \int_V \delta^3(\vec{r}-\vec{r}') dr d\Omega = \frac{q}{4\pi} \cdot \int_0^R \delta^3(\vec{r}-\vec{r}') dr \cdot \int_0^{4\pi} d\Omega = \frac{q}{4\pi} \cdot 4\pi \cdot \int_0^R \delta^3(\vec{r}-\vec{r}') dr = q \int_0^R \delta^3(\vec{r}-\vec{r}') dr = q[/tex].
So my question is: is this second form of evaluation correct? And if it is, then why does [tex]\rho[/tex] have two different expressions? Also, what about other curvilinear coordinates? Shouldn't his expression be independent of the geometry being used? So I suspect there is something wrong with my second evaluation. Can you not integrate over [tex]\delta^3[/tex] over one dimension?
tia!
-LD
EDIT: I see it now. I was missing the fact that [tex]\delta^3(\vec{r}-\vec{r}') = \delta(r-r')\delta(\theta-\theta')\delta(\phi-\phi')[/tex] and so the integral over [tex]\Omega[/tex] is still 1 and not [tex]4\pi[/tex].
PS: I wasn't sure if I should have deleted this post having figured out the correct answer, or leave it and post what I found out. Let me know and I will do whatever is the correct procedure here.
thx!
-LD
After reading DJGriffiths sections on the DDF I have a question about evaluating it in regards Prob. 1.46 (a), to wit:
"Write an expression for the electric charge density [tex]\rho(\vec{r})[/tex] of a point charge q at [tex]\vec{r}[/tex]'. Make sure that the volume integral of [tex]\rho[/tex] equals q."
This is easily done in rectangular coordinates, namely, let [tex]\rho(\vec{r}) = q\delta(\vec{r}-\vec{r}')[/tex].
Checking that this resultsin a volume integral of q I did:
[tex] \int_V \rho(\vec{r}) d\tau = \int_V q\delta^3(\vec{r}-\vec{r}') d\tau = q \int_V \delta^3(\vec{r}-\vec{r}') d\tau = q \cdot 1 = q.[/tex]
Here is my question: doing the integral as is I get the correct answer. However, when I wrote it out in spherical coord's. It seems that I should rewrite [tex]\rho(\vec{r})[/tex] as [tex]\frac{q}{4\pi}[/tex]?
Why? Because:
[tex] \int_V \rho(\vec{r}) d\tau = \frac{q}{4\pi} \int_V \delta^3(\vec{r}-\vec{r}') dr d\Omega = \frac{q}{4\pi} \cdot \int_0^R \delta^3(\vec{r}-\vec{r}') dr \cdot \int_0^{4\pi} d\Omega = \frac{q}{4\pi} \cdot 4\pi \cdot \int_0^R \delta^3(\vec{r}-\vec{r}') dr = q \int_0^R \delta^3(\vec{r}-\vec{r}') dr = q[/tex].
So my question is: is this second form of evaluation correct? And if it is, then why does [tex]\rho[/tex] have two different expressions? Also, what about other curvilinear coordinates? Shouldn't his expression be independent of the geometry being used? So I suspect there is something wrong with my second evaluation. Can you not integrate over [tex]\delta^3[/tex] over one dimension?
tia!
-LD
EDIT: I see it now. I was missing the fact that [tex]\delta^3(\vec{r}-\vec{r}') = \delta(r-r')\delta(\theta-\theta')\delta(\phi-\phi')[/tex] and so the integral over [tex]\Omega[/tex] is still 1 and not [tex]4\pi[/tex].
PS: I wasn't sure if I should have deleted this post having figured out the correct answer, or leave it and post what I found out. Let me know and I will do whatever is the correct procedure here.
thx!
-LD
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