2012-09-26 17:36:01 8 Comments

On a similar note: when using Gauss' Law, do you even begin with Coulomb's law, or does one take it as given that flux is the surface integral of the Electric field in the direction of the normal to the surface at a point?

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## 4 comments

## @Krishna Deshmukh 2018-10-06 06:06:06

vs_292,just missed a small point.The dot product.The surface

dsshould be along the electric field i.e alongr(position vector). Refer the pictureSource:Edupoint Youtube Channel.

## @user191954 2018-10-06 09:12:57

Please don't post formulas as pictures or plain text, but use MathJax instead. MathJax is easy for people on all devices to read, and can show up clearer on different screen sizes and resolutions. Look at this Math SE meta post for a quick tutorial.

## @vs_292 2017-06-14 16:49:41

@Qmechanic's already provided a nice answer. I would like to provide another one.

Consider a charge $q$ be enclosed by any surface (not necessarily a sphere). Something like this -

Now, you write the flux coming out of this weird surface - $$ \phi_E = \displaystyle \oint_S \mathbf{E} \cdot \mathrm{d\mathbf{S}} $$ We know that - $$ \mathbf{E} = E \vec{r} = \dfrac{1}{4\pi \epsilon_0} \dfrac{q}{r^2}\vec{r} $$ So, here in this weird surface. there is no fixed radius, is there? And the surface area that is considered here is not a continuous one. So, I would get - $$ \phi_E = \dfrac{q}{4\pi \epsilon_0} \displaystyle \oint_S \dfrac{\mathrm{d\mathbf{S}}}{r^2} \tag{1} $$ Recall that the term $\dfrac{\mathrm{d\mathbf{S}}}{r^2}$ is the

verydefinition for the steradian - which is equal to $\dfrac{1}{4\pi}$ of a complete sphere. This holds good for any surface. Simply put, this is the 3D analogue of the $2\pi$ rotation in a circle. Here, we have its differential element, i.e, $d\Omega = \dfrac{\mathrm{d\mathbf{S}}}{r^2}$ Integrating it entirely, we have $$ \displaystyle \oint_S \dfrac{\mathrm{d\mathbf{S}}}{r^2} = \displaystyle \oint_S d\Omega = 4\pi$$ Plugging this back into (1), we have - $$ \phi_E = \dfrac{q}{\epsilon_0} $$ Which implies - $$ \displaystyle \oint_S \mathbf{E} \cdot \mathrm{d\mathbf{S}} = \dfrac{q}{\epsilon_0} $$ Ok, since we're done with deriving the integral form of Gauss's law (Which holds true for any closed surface), the following differential form can be obtained by applying the divergence theorem - $$ \nabla \cdot \mathbf{E} = \dfrac{\rho}{\epsilon_0} $$## @gen-z ready to perish 2017-06-27 21:28:22

Excellent choice to provide this as well. This is the proof I always encountered in my first year of electromagnetism.

## @user131786 2016-10-02 07:35:46

@Qmechanic's proof is a nice general way to prove Gauss Law from Coulumb's Law.However I would like to add a simpler proof which I discovered on YouTube.

Source:https://www.youtube.com/watch?v=X_CHPTZfUGo

## @Qmechanic 2012-09-26 19:11:41

Let us for simplicity consider $n$ point charges $q_1$, $\ldots$, $q_n$, at positions $\vec{r}_1$, $\ldots$, $\vec{r}_n$, in the electrostatic limit, with vacuum permittivity $\epsilon_0$.

Now let us sketch one possible strategy to prove Gauss' law from Coulomb's law:

Deduce from Coulomb's law that the electric field at position $\vec{r}$ is $$\tag{1} \vec{E}(\vec{r})~=~ \sum_{i=1}^n\frac{q_i }{4\pi\epsilon_0}\frac{\vec{r}-\vec{r}_i}{|\vec{r}-\vec{r}_i|^3} . $$

Deduce the charge density $$\tag{2} \rho(\vec{r})~=~\sum_{i=1}^n q_i\delta^3(\vec{r}-\vec{r}_i). $$

Recall the following mathematical identity $$\tag{3}\vec{\nabla}\cdot \frac{\vec{r}}{|\vec{r}|^3}~=~4\pi\delta^3(\vec{r}) .$$ (This Phys.SE answer may be useful in proving eq.(3), which may also be written as $\nabla^2\frac{1}{|\vec{r}|}=-4\pi\delta^3(\vec{r})$).

Use eqs. (1)-(3) to prove Gauss' law in differential form $$\tag{4} \vec{\nabla}\cdot \vec{E}~=~\frac{\rho}{\epsilon_0} .$$

Deduce Gauss' law in integral form via the divergence theorem.

## @Meow 2012-09-26 19:17:56

Is 3 Poisson's equation, a generalisation of it or a subdivision of it? And thanks for the answer- not too unfathomable.

## @Qmechanic 2012-09-26 19:35:42

I updated the answer. Eq.(3) is a mathematical identity, while Poisson's equation has physical content.

## @Qmechanic 2012-09-30 16:16:47

Now I realize that the proof is also given near the bottom of this Wikipedia page.

## @Meow 2013-03-12 13:17:59

I intuit equation (2), but why cube $\delta(\mathbf{r}-\mathbf{r_i})$?

## @Qmechanic 2013-03-12 14:27:09

@Alyosha: It's standard notation for the 3-dimensional delta function $\delta^3(\vec{r})~:=~\delta(x)\delta(y)\delta(z)$.