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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

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

1 Answer 1

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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