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I have to write an expression for the charge density $\rho(\vec{r})$ of a point charge $q$ at $\vec{r}^{\prime}$, ensuring that the volume integral equals $q$.

The only place any charge exists is at $\vec{r}^{\prime}$. The charge density $\rho$ is uniform:

$$\rho(\vec{r}) = \delta(\vec{r} - \vec{r}^{\prime})\rho$$

But if I evaluate the total charge, I get

$$ q = \int dq = \int^{\infty}_{-{\infty}}\delta(\vec{r} - \vec{r}^{\prime})\rho ~dV $$$$= \rho\int^{\infty}_{-{\infty}}\delta({x} -{x}')dx\int^{\infty}_{-{\infty}}\delta({y} -{y}')dy\int^{\infty}_{-{\infty}}\delta({z} -{z}')dz$$

The Dirac delta functions integrate to one each, but what becomes of the charge density $\rho$? For that matter, how does one integrate a zero dimensional point over 3 dimensinal space? Any help greatly appreciated.

EDIT: So it seems that the charge density is just the charge itself ($\rho = q)$?

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Why didn't you take the $\rho$ out of the integrals since its a constant? If you do so, then you should get something like $q=\rho$. I also think instead of changing one integral into a product of three, you meant to put them as a triple integral. Anyways I didn't quite understand many parts of the question, like saying the charge density is uniform and giving us a clearly non-uniform expression, or askiing howto integrate a zero dimensional point over 3 dimensional space. Really, that part was quite mysterious. –  namehere Dec 8 '12 at 3:17
    
I wrote the question verbatim from my text. I don't understand how a point has volume either –  Cactus BAMF Dec 8 '12 at 3:21
    
Your assignment should raise some dimensional warning flags. Since $\int\delta(\vec{r})dV=1$ a volume delta function has units of density (1/volume). That times a charge will get you a charge density. –  Emilio Pisanty Dec 9 '12 at 2:16
    
@Cactus BAMF: Is your text published? If yes, please provide reference. –  Qmechanic Dec 9 '12 at 10:27

4 Answers 4

up vote 5 down vote accepted

First equation is wrong, it should say $\rho(\vec{r}) = \delta(\vec{r} - \vec{r}')q$. (Note that you had two errors).
You treat it like a normal charge density $\rho(\vec{r})$, if you integrate the density over any volume you get the total charge within that volume.

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No i had the prime in there, somebody edited it out for some reason –  Cactus BAMF Dec 8 '12 at 3:29
    
rho is the charge density, its a function of position, q is the total charge –  Hobo2 Dec 8 '12 at 3:44

The nature (and glory) of the dirac delta function is that the volume integral

$$ \int_{\Delta V} dV' \delta ( \boldsymbol{r-r'} ) = \left\{ \begin{array}{cc} 1 & \text{if } \Delta V \text{ contains } \boldsymbol{r}\\ 0 & \text{if } \Delta V \text{ does not contain } \boldsymbol{r} \end{array} \right. $$

Using this function, you can write the charge density of a point charge so that its integral over a volume containing its location gives $q$.

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Your expression for $\rho$ is off, it should be

$\rho(\vec{r}) = q\delta(\vec{r}-\vec{r}')$

if there is only a single point charge in $\vec{r}'$.

Now, demanding that the volume integral equals $q$, you would get:

$q = \int_{V}{q\delta(\vec{r}-\vec{r}')d\vec{r}} = q\int_{V}{\delta(\vec{r}-\vec{r}')d\vec{r}} = q$

which is quite trivial (using the nature of the Dirac delta, described by Art Brown) and simply shows that your expression for $\rho$ is a good one.

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$$ρ(r⃗ )=qδ(r⃗ −r⃗ ′)$$

$$Q=∫ρdV=∫δ(r⃗ −r⃗ ′)q dV=q$$

bingo!!

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