Gauss's Law (Notational confusion)

[Mandelbroth]

I'm familiar with the differential form of Gauss's Law, which reads that $\nabla\cdot\vec{E}=\frac{\rho}{\epsilon_0}$, where E is the electric field, ρ is the charge density, and $\epsilon_0$ is the permittivity of free space. We can take the volume integral of both sides, and then use the divergence theorem to obtain $\displaystyle \iint\limits_{\partial V}(\vec{E}\cdot\hat{n}) \, dA = \frac{1}{\epsilon_0}\iiint\limits_{V}\rho \, dV$.

A friend of mine says that this is wrong, and that the statement of Gauss's Law in integral form is $\displaystyle \oint\limits_{S}(\vec{E}\cdot\hat{n}) \, dA = \frac{1}{\epsilon_0}\iiint\limits_{V}\rho \, dV$. Is this just a notational issue, where the apparent closed line integral is just a physics shorthand for an integral over a surface, or am I misunderstanding what Gauss's Law is saying?

 

[Doc Al]

Is this just a notational issue, where the apparent closed line integral is just a physics shorthand for an integral over a surface, or am I misunderstanding what Gauss's Law is saying?

I'd say it was a notational issue. The S under the "line integral" means that it's actually an integral over a closed surface.

 

[WannabeNewton]

As long as it is understood in the first expression that $V\subseteq \mathbb{R}^{3}$ is a compact regular surface with boundary (manifold boundary, to be precise), there is no issue. Intuitively, we want to enclose an arbitrary portion of the charge distribution generating the electric field so we wish to use a compact subset (since all compact subsets of $\mathbb{R}^{3}$ are closed and bounded) and the smooth requirement is obvious of course.

 

[jtbell]

We can take the volume integral of both sides, and then use the divergence theorem

Take a close look at the divergence theorem. On one side is the volume integral of the divergence of $\vec E$. On the other side is the surface integral of the flux of $\vec E$ through the surface of that volume.

 

[WannabeNewton]

Take a close look at the divergence theorem. On one side is the volume integral of the divergence of $\vec E$. On the other side is the surface integral of the flux of $\vec E$ through the surface of that volume.

He was talking about taking the volume integral on both sides of $\nabla\cdot E = \frac{\rho}{\epsilon_{0}}$ and then applying Gauss's theorem to the left side i.e. $\int _{S}(\nabla\cdot E )dV = \int _{\partial S}E\cdot dA$ to then say $\int _{\partial S}E\cdot dA = \frac{1}{\epsilon_{0}}\int_{S} \rho dV$ so it isn't at odds with what you said.

 

[Mandelbroth]

Take a close look at the divergence theorem. On one side is the volume integral of the divergence of $\vec E$. On the other side is the surface integral of the flux of $\vec E$ through the surface of that volume.

That's my point. My friend is convinced that the law is stating that the line integral of the electric field (how you'd think of that, I don't know) is given by the net charge divided by the permittivity of free space. I think this is the danger of only memorizing textbook equations without an understanding of where they come from, and I think it's kind of awesome that I can ask questions here to understand what I don't yet fully comprehend.

Thank you for answering, everybody. It's much appreciated.

 

[WannabeNewton]

That's my point. My friend is convinced that the law is stating that the line integral of the electric field (how you'd think of that, I don't know) is given by the net charge divided by the permittivity of free space.

I don't think he is saying that if what you wrote above is what your friend actually said. The appearance of an area element $dA$ in the integral makes it pretty clear that it is a surface integral but what your friend said is notationally not favorable since you really should write the integral over the boundary of something (e.g. $\partial S$) to make it clear that you used Stokes' theorem. I've also never seen the "closed integral" notation (the circle in the middle) show up in physics outside of anything that uses vector calculus so you aren't losing out by not using it.