# How is Gauss's divergence theorem related to Gauss's law?

I have been reading a bit about vector fields and how it relates to the EM field. However, I am confused about the nature of Gauss's law. As far as I understand, it is supposed to relate flux of the lines over a surface to the net charge inside the closed surface. But how is that relationship achieved? Gauss's divergence law allows one to find the divergence through a closed surface, but how is that connected to the charge of the surface?

• Divergence is the microscopic version of Gauss's Law where the volume becomes a differential volume (like $V \rightarrow dV$). if no one else does, i will write an answer that will be more conceptual and won't include Green's theorem. – robert bristow-johnson Dec 19 '16 at 19:34
• Honors, it's gonna take awhile before i can organize it. maybe tonight. in the meantime, look up inverse-square law, Gauss's law, and the concept of flux and flux density. the divergence equation is another way of stating the behavior of "flux" of some physical quantity, and if that quantity is conserved in some manner, the inverse-square law comes out of that notion. And Gauss's law says the same conservation of flux more directly. – robert bristow-johnson Dec 19 '16 at 20:38

# Mathematical Explanation

Gauss's divergence law states that

$$\mathbf{\nabla} \cdot \mathbf{E} = \frac{\rho}{\epsilon_0}$$

So, let's integrate this on a closed volume $V$ whose surface is $S$, it becomes

$$\iiint\limits_{V(S)}\nabla \cdot \mathbf{E}\,\textrm{d}V = \frac{Q}{\epsilon_0}$$

where $Q$ is the total charge in $V$. However, Green-Ostrogradski theorem states that

$$\iiint\limits_{V(S)} \nabla \cdot \mathbf{F}\,\textrm{d}V = \iint\limits_{S}\mathbf{F}\cdot\mathbf{\textrm{d}S}$$

for any field $\mathbf{F}$, so in particular

$$\iiint\limits_{V(S)} \nabla\cdot\mathbf{E}\,\textrm{d}V = \iint\limits_{S}\mathbf{E}\cdot\mathbf{\textrm{d}S} = \Phi_S(\mathbf{E})$$

where $\Phi_S(\mathbf{E})$ is the flux of $\mathbf{E}$ through $S$. Finally, we got

$$\Phi_S(\mathbf{E}) = \frac{Q}{\epsilon_0}$$

which is Gauss's theorem.

# Intuitive Explanation

The previous explanation demonstrates the link between Gauss's divergence law and its theorem, yet we don't really understand why it works. However, once you've understood what the divergence of a field is, it will appear easy to understand.

Let's start from a possible definition of the divergence of a field $\mathbf{F}$:

$$\nabla \cdot \mathbf{F} = \lim\limits_{\delta\tau\rightarrow 0} \frac{\iint\limits_{\delta S}\mathbf{F}\cdot\mathbf{\mathrm{d}S}}{\delta \tau}$$

where $\delta \tau$ is the volume delimited by $\delta S$. Well... this definition isn't really simple. Let's rewrite it as a physicist would do:

$$\nabla \cdot \mathbf{F}\, \delta \tau = \Phi_{\delta S}(\mathbf{F})$$

where $\delta \tau$ is small enough.... So here, we can see that the divergence of $\mathbf{F}$ is how much of $\mathbf{F}$ goes out of a small volume per unit time. So, if we integrate this over a big volume, we will get what comes out of this big volume per unit time (since when we cut this big volume into many small volumes, everything that goes out of a small volume goes in one of its neighbours, except for the extreme small volumes). But what goes out of this big volume is $\Phi_S(\mathbf{F})$, so we can say that

$$\Phi_S(\mathbf{F}) = \sum\limits_{\delta \tau \in V} \Phi_{\delta S}(\mathbf{F}) = \sum\limits_{\delta\tau \in V} \nabla\cdot\mathbf{F}\,\delta \tau = \iiint\limits_{V}\nabla\cdot\mathbf{F}\,\mathbf{\textrm{d}V}$$

# To a possible understanding of Gauss's law

Everyone starts electrostatic with Coulomb's law, that is two charges $q_A$ and $q_B$ with a distance $r$ between them undergo a force

$$F_e = \frac{q_A q_B}{4\pi \epsilon_0 r^2}$$

Now, let's consider a charge $q$ in $O$. It creates a field $$\mathbf{E} = \frac{q}{4\pi\epsilon_0 \| \mathbf{OM} \|^3} \mathbf{OM}$$

Consider a small surface $\textrm{d}^2 S = r^2\textrm{d}\theta\textrm{d}\phi$ in $M$, where $\|OM\| = r$. The flux of $\mathbf{E}$ through $\textrm{d}^2 S$ is $\textrm{d}^2\Phi = \mathbf{E}\cdot\textrm{d}^2\mathbf{S}$ where $\textrm{d}^2\mathbf{S} = \frac{\textrm{d}^2S}{r}\mathbf{OM}$. This finally gives

$$\textrm{d}^2\Phi = \frac{q}{4\pi\epsilon_0 r^3}\mathbf{OM}\cdot\frac{r^2\textrm{d}\theta\textrm{d}\phi}{r}\mathbf{OM} = \frac{q}{4\pi\epsilon_0}\textrm{d}\theta\textrm{d}\phi$$

Now, let's consider a closed surface $S$. If $O$ is in this surface, then

$$\Phi = \iint\limits_S \frac{q}{4\pi\epsilon_0}\textrm{d}\theta\textrm{d}\phi = \frac{q}{4\pi\epsilon_0} \int\limits_{\theta = 0}^{2\pi}\int\limits_{\phi = 0}^{\pi} \textrm{d}\theta\textrm{d}\phi = \frac{q}{\epsilon_0}$$

Now, focus on the case were $O$ is outside $S$. Let $S_{+}$ (resp. $S_{-}$) be the part of $S$ where $\textrm{d}^2\Phi \geq 0$ (resp. $\textrm{d}^2\Phi < 0$), then

$$\iint\limits_{S} \textrm{d}^2\Phi = \frac{q}{4\pi\epsilon_0}(\iint\limits_{S_{+}}\textrm{d}\theta\textrm{d}\phi - \iint\limits_{S_{-}}\textrm{d}\theta\textrm{d}\phi)$$

However, since $S$ is a closed surface, a drawing can convince you that the preceding sum is null (the solid angle of $S_{-}$ is the same as the one of $S_{+}$...). Then, when $O$ is outside of $S$, $\Phi$ = 0.

Since the electrostatic field is linear, we can generalize this idea with the following equation:

$$\Phi_S(\mathbf{E}) = \frac{Q}{\epsilon_0}$$

where $Q$ is the net charge inside $S$.

• Thank you for your answer, but I was hoping for more of a conceptual explanation because I'm not too hot on the math. I think what I'm really confused on is how the connection between charge and divergence is established. – Honors Level Maker Dec 19 '16 at 19:28
• Actually when you calculate the divergence of E due to point charge that equals 0 and volume integral of it then vanishes . I know we explain it with Dirac Delta function , but could you explain it without that , because Lagrange or Gause would have done it in 1800s before Delta function came into existence. – Shashaank Dec 19 '16 at 19:35
• I understand the nature of the divergence over the area now, but how is one able to see that the amount of charge must be determined by this divergence? Is it because the flux represents the charge in a field, and since Gauss's divergence theorem is flux over an area, the net charge must be dependant on that? – Honors Level Maker Dec 19 '16 at 22:43