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 ($\rho = q)$?

• 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. Dec 8 '12 at 3:17
• I wrote the question verbatim from my text. I don't understand how a point has volume either 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. Dec 9 '12 at 2:16
• @Cactus BAMF: Is your text published? If yes, please provide reference. Dec 9 '12 at 10:27
• @Qmechanic Introduction to Electrodynamics, by Griffiths, pg. 52. Sep 5 '16 at 21:14

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.

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

$$ρ(r⃗ )=qδ(r⃗ −r⃗ ′)$$

$$Q=∫ρdV=∫δ(r⃗ −r⃗ ′)q dV=q$$

bingo!!

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.

• Why is the expression not $\rho (\overrightarrow r ) = q \delta^3 (\overrightarrow r - \overrightarrow r')$ ??? Aren't we supposed to be using the 3-dimensional dirac delta function here? We are dealing with volume after all. Sep 5 '16 at 21:16
• @whatwhatwhat Yes, I was liberal with notation here, assuming the context would suffice without explicit mention of the dimensionality of the Dirac. This is not uncommon (as you can also see from the other answers), but when the focus is on the dimensionality of the Dirac or when multiple versions are used in the same article, it is indeed usually mentioned explicitly as $\delta^{(n)}$. Sep 6 '16 at 7:26