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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?

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  • $\begingroup$ You're integral isn't covering any volume $\endgroup$ – Aaron Stevens Aug 10 at 17:39
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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.

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  • $\begingroup$ Technically the OP didn't say the distribution was a single charge. But they accepted so oh well :) $\endgroup$ – Aaron Stevens Aug 10 at 17:53
  • $\begingroup$ 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? $\endgroup$ – Hilbert Aug 10 at 18:00
  • $\begingroup$ 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 $dV$. $\endgroup$ – G. Smith Aug 10 at 18:03
  • $\begingroup$ @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. $\endgroup$ – Hilbert Aug 10 at 18:04
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    $\begingroup$ That is like asking, if a point has no volume, how can space have volume. $\endgroup$ – G. Smith Aug 10 at 18:09
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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.

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