Delta Dirac Charge Density question 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 ($\rho = q)$?
 A: 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.
A: 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 $q$.
A: $$ρ(r⃗ )=qδ(r⃗ −r⃗ ′)$$
$$Q=∫ρdV=∫δ(r⃗ −r⃗ ′)q dV=q$$
bingo!!
A: 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.
