# Regarding charge densities and the amount of charge located on a point for continuous charge distributions

Suppose we have a volume $$V$$ containing a charge distribution defined by $$\rho (\textbf{x})$$.

The amount of charge $$q~(P)$$ located at an arbitrary point $$P(x_{0},y_{0},z_{0})$$ is :

$$q(P)=\int_{x_{0}}^{x_{0}}\int_{y_{0}}^{y_{0}}\int_{z_{0}}^{z_{0}}\rho(\textbf{x}) \,\mathrm dx\,\mathrm dy\mathrm \,dz=0 \,C$$

This means that the charge at each point insde of $$\,V$$ is equal to zero coulomb, yet $$V$$ was defined to be a volume containing charges.

Where does this contradiction come from?

• You're integral isn't covering any volume – Aaron Stevens Aug 10 at 17:39

The charge density of a point charge $$q$$ at $$\mathbf{r}_0$$ is a Dirac delta function:

$$\rho(\mathbf{r})=q\delta^{(3)}(\mathbf{r}-\mathbf{r}_0).$$

Its integral over any volume containing $$\mathbf{r}_0$$ is $$q$$, not 0.

However, you didn’t integrate over any volume at all.

• Technically the OP didn't say the distribution was a single charge. But they accepted so oh well :) – Aaron Stevens Aug 10 at 17:53
• But there is still a small contradiction from the answer posted by tparker in this question. Here's the part of the answer that confuses me "remember that strictly speaking, for a continuous charge distribution with no delta functions in the density, $\rho(\textbf{x})$ is not the amount of charge at point $\textbf{x}$. There is zero charge exactly at point $\textbf{x}$.". And so if the charge at each $\textbf{x}$ equals zero, then how does the total charge not equal zero? – Hilbert Aug 10 at 18:00
• You have to decide if you want to think of charges as discrete point charges or as a continuous, “smeared out,”charge distribution. If the latter, then there is infinitesimal charge $dq=\rho dV$ in infinitesimal volume $dV$. – G. Smith Aug 10 at 18:03
• @G. Smith, my apologies, I think I rushed a little bit before accepting your answer. Indeed as Aaron Stevens pointed, the volume contains infinitesimal charges distributed over a volume,not point charges. – Hilbert Aug 10 at 18:04
• That is like asking, if a point has no volume, how can space have volume. – G. Smith Aug 10 at 18:09

The contradiction is that your integral is not covering any volume. The easy answer is that, by definition, the amount of charge $$\text dq$$ contained in a volume $$\text dV$$ at location $$\mathbf x$$ is just $$\text dq=\rho(\mathbf x)\,\text dV$$

However, if you wanted to relate this to the integral you could just look at a cubic element of length $$2\delta$$ on each side centered on your point, and then look at what happens as $$\delta$$ approaches $$0$$. $$q=\int_{x_0-\delta}^{x_0+\delta}\int_{y_0-\delta}^{y_0+\delta}\int_{z_0-\delta}^{z_0+\delta}\rho(\mathbf x)\,\text dz\,\text dy\,\text dx$$ $$\lim_{\delta\to0}\,q=\text dq=\rho(x_0,y_0,z_0)\,\text dV$$

Something to keep in mind that it seems like is tripping you up: $$\text dV$$ is not a point. It is still a volume. $$\text dV$$ technically still has an "infinite number of points" contained within it.