# Physical interpretation of $\iiint (∇\cdot\vec E)\mbox{d} V$ [duplicate]

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

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

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.

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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