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I can understand the Differential form of Gauss's Law ∇⋅𝐄= $\frac{ρ}{ɛ_0}$

as saying that the source of electric field vectors or flow disperse(The divergence of the electric field) is equal to the charge density divided by the permittivity of free space.

However I do not understand how this is equivalent to the integral form of Gauss's law: $$\oint \textbf{E} \cdot d\textbf{A} = \frac{q}{ɛ_0}$$ Which states that the total electric flux out of an arbitrary closed surface is equal to the electric charge inside that surface divided by the permittivity of free space.

I understand that there might be a mathematical way to show that both equations are equivalent but is there a physically intuitive way to understand why both statements are saying the same thing?

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2 Answers 2

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An easy way to understand this equivalence is by using the intuitive definition of the divergence as the surface integral per volume (in the limit of small volume):

$$\nabla\cdot\textbf{E} = \lim_{V\to 0} \frac{\oint\textbf{E}\cdot d\textbf{A}}{V} \tag{1}$$

Begin with the integral form of Gauss's law: $$\oint\textbf{E}\cdot d\textbf{A}=\frac{1}{\epsilon_0}q$$

Divide both sides of this equation by the enclosed volume $V$ and do the limit $V\to 0$: $$\lim_{V\to 0}\frac{\oint\textbf{E}\cdot d\textbf{A}}{V} =\frac{1}{\epsilon_0}\lim_{V\to 0}\frac{q}{V}$$

Here you recognize the divergence $\nabla\cdot\textbf{E}$ from (1) on the left side and the definition of the charge density $\rho$ on the right side, and immediately get the differential form of Gauss's law: $$\nabla\cdot\textbf{E}=\frac{1}{\epsilon_0}\rho$$

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  • $\begingroup$ In fact, this shows that the integral form is more general because the divergence assumes differentiability while the integral form of Mawell's equation also works for discontinuous medium. In other words, when the vector field is discontinuous you have to include explicitly the transition formulas of the normal and tangential fields at the interface of discontinuity along with the the divergence formula valid in the continuous medium. $\endgroup$
    – hyportnex
    Commented Mar 23 at 9:31
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Basically you can think of the divergence of a vector field as its tendency to converge (or diverge) to a point in space. For example, think of a fluid: the velocity of every element of fluid is a field in space; while we define sinks as points in space where the fluid converge; once it goes in, it can't exit anymore. Now imagine a opaque surface surrounding the sink: even if we can't see the sink inside it, we can see the fluid going in without exiting; obviously the volume inside the surface is finite, therefore we can logically assume that the fluid is getting sucked by a sink even if we can't directly see it (assuming the fluid is incompressible). I hope it makes kind of sense, but this means that we can know where the sinks are located by simply looking to how the fluid is flowing. Analogously, in electrostatics, charges are sinks (negative) or sources (positive) and the electric field is the fluid flowing. The differential form connects the divergence of the field to the local charges (the charge density is a function of space). Meanwhile the integral form connects the flow of the field through a surface to the charges inside it. You can notice that Gauss's Law in integral form doesn't give us any information on how many charges there are and where exactly, we only know they are enclosed by surface and how strong their total effect is, because you're not effectively looking inside it.

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  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Commented Mar 23 at 2:01

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