Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ($\rho = q)$?

share|cite|improve this question
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
up vote 6 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.

share|cite|improve this answer
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 $q$.

share|cite|improve this answer

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.

share|cite|improve this answer

$$ρ(r⃗ )=qδ(r⃗ −r⃗ ′)$$

$$Q=∫ρdV=∫δ(r⃗ −r⃗ ′)q dV=q$$


share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.