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 ($\rho = q)$?
Answer
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 comments:
Post a Comment