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  1. Can anybody explain the physical interpretation of Gauss's law $$\iiint (\nabla\cdot \vec E)~\mbox{d}V~=~\frac{Q}{\epsilon_0}? $$

  2. Also, how is the differential form of Gauss's law obtained from the integral form?

  3. How can volume multiplied by a gradient equals flux through a closed surface?

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marked as duplicate by Emilio Pisanty, Qmechanic Sep 14 '13 at 14:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Possible duplicate: physics.stackexchange.com/q/74788/2451 –  Qmechanic Sep 14 '13 at 9:31

2 Answers 2

The Physical interpretation of Gauss's law is covered in various textbooks, e.g. Jewett & Serway.

It essentially means that the flux of the electric field lines is proportional to the electric charge of the charged particle in question.

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This is of course, stated in differential form as:

$$\nabla\cdot\vec E =\frac{q}{\epsilon_0}$$

This can be derived trivially from Gauss's law by using Gauss's theorem (Divergence theorem) applied to Gauss's law.

$$\frac{Q}{\epsilon_0} = \iint\vec E \cdot\hat n\mbox{ d}S=\iiint{\nabla\cdot\vec E}\mbox{d}V $$

The conclusion follows that:

$$\nabla\cdot\vec E =\frac{q}{\epsilon_0}$$

Where $Q$ is the charge, and $q$ is the charge density.

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For a basic and easy to understand answer, read Griffiths' standard reference for undergraduate electromagnetism: 'Introduction to Electrodynamics'. My answer will be based on his discussion in the book.

Gauss' law reads in integral form reads $\oint\limits_{\mathcal{S}}\vec{E}\cdot d\vec{a}=\frac{1}{\epsilon_0}\int\limits_{\mathcal{V}}\rho\ dV=\frac{Q_{enclosed}}{\epsilon_0}$

You can convince yourself by drawing that a charge that is not located in the volume you're integrating over will produce a field which can be visualized by field lines that, in the absence of charge inside the volume, will always either not go through the volume, or will pass through it (that is, they enter on one side, but exit as well). Thus, it is clear that a charge that is not located in the volume will not give rise to any net flux across the closed surface that encloses the volume of interest. Hopefully, this will serve as enough motivation.

Once you have Gauss' law in this form, it is really just a matter of applying the general Stokes' theorem, which in this case takes on the form of the divergence theorem, and the differential form of Gauss' law, $\vec{\nabla}\cdot\vec{E}=\rho/\epsilon_0$, follows immediately.

If you find the pure mathematical logic not convincing enough (?), you can try to form some intuition on why only enclosed charges contribute to the integral of the divergence over the volume. This is essentially because a charge provides beginning/end point for electric field lines. This way of thinking about matters also show immediately why $\vec{\nabla}\cdot\vec{B}=0$, is always true. There are no beginning or end points for magnetic field lines: they are always closed!

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No, that is not Ampere's law. Ampere's law is different. The law that you stated is Gauss's law for magnetism. –  Dimensio1n0 Sep 14 '13 at 8:23
Thank you for your comment, I edited the mistake out. Incidentally, I would like to point out that your post contains an incorrect statement of the Divergence theorem. –  Danu Sep 14 '13 at 8:42
Are you talking about the fact that I said that $\nabla\cdot\vec E =\frac{q}{\epsilon_0}$? If so, by $q$, I meant the charge density, and not the charge itself. –  Dimensio1n0 Sep 14 '13 at 8:46
No, the integrals. It should read: $\oint\limits_{\mathcal{S}}\vec{E}\cdot\hat{n}\ \mbox{d}A=\int\limits_{\mathcal{V}}\vec{\nabla}\cdot\vec{E}\ \mbox{d}V$ –  Danu Sep 14 '13 at 8:47
Oh, yes there should be 2 integral signs (or I could do as you say, only use 1 and state the manifold of integration.). –  Dimensio1n0 Sep 14 '13 at 8:51

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