Is my attempt to derive Gauss' law correct?

[SothSogi]

Hi there. I am trying to derive Gauss's law from the divergence. I would like to know if it is correct:

The divergence is defined as (I saw this on Fuller & Byron "Mathematics of classical and quantum physics")

##
\nabla\cdot\textbf{E}=\lim_{\Delta\tau\rightarrow0}\frac{1}{\Delta\tau}\int_\sigma\textbf{E}\cdot d\boldsymbol\sigma
##

Then I made the following

##
\nabla\cdot\textbf{E}=\lim_{\Delta\tau\rightarrow0}\frac{1}{\Delta\tau}\int_\sigma\textbf{E}\cdot d\boldsymbol\sigma=\frac{1}{d\tau}\int_\sigma\textbf{E}\cdot d\boldsymbol\sigma
##

So then, for a point charge and taking the dot product for a spherical surface element

##
\nabla\cdot\textbf{E}=\frac{1}{d\tau}\int_\sigma\frac{q}{4\pi\varepsilon_0\vert\textbf{r}\vert^2}d\sigma=\frac{1}{d\tau}\frac{q}{4\pi\varepsilon_0\vert\textbf{r}\vert^2}\int_\sigma d\sigma
##

##
\nabla\cdot\textbf{E}=\frac{1}{d\tau}\frac{q}{4\pi\varepsilon_0\vert\textbf{r}\vert^2}\left(4\pi r^2\right)=\frac{1}{d\tau}\frac{q}{\varepsilon_0}
##

Now ##
q=\int_V\rho d\tau
##

So

##
\nabla\cdot\textbf{E}=\frac{1}{d\tau}\frac{1}{\varepsilon_0}\int_V\rho d\tau=\frac{1}{\varepsilon_0}\int_V\rho\frac{d\tau}{d\tau}
##

So

##
\nabla\cdot\textbf{E}=\frac{1}{\varepsilon_0}\rho
##Thanks for taking the time to read it.

 

[Gene Naden]

Hi,
Thank you so much for putting your equations in Tex form!

There are two forms of Gauss's law (see Wikipedia https://en.wikipedia.org/wiki/Gauss's_law), the integral form and the differential form. From your conclusion, it sounds like you are trying to prove the differential form, ##\nabla\cdot\textbf{E}=\frac{1}{\varepsilon_0}\rho##
It looks like you are supposed to use Coulomb's law.

Your solution looks basically OK, following the correct logic. You use of infinitesimals such as ##\Delta \tau## and ##d\tau## seems a little off. For example, you write ##\int_V\rho\frac{d\tau}{d\tau}##. This expression is malformed. ##\frac{d\tau}{d\tau}## is unity and the integral becomes ##\int_V\rho##, which doesn't make sense.

 

[PumpkinCougar95]

Hi,
I am not an expert but if you assume that the E field is that of a point charge then how would this be a general derivation? Personally, i like the derivation using solid angles method but it gives the integral form of Gauss law.

 

[Gene Naden]

Yes, Pumpkin, I see what you mean. Do you think the assignment is to derive the differential form starting with the integral form? I mean, we have to start with something. It seems we could also start with a general form for the electric field, as below. Do you think that is the starting point the instructor intended?

##E(x)=\int_V \frac{\rho (x^\prime) (x-x^\prime)}{|x-x^\prime|^3}d^3 x^\prime##

 

[Gene Naden]

I omitted the factor ##\frac{1}{\epsilon_0}##

 

[PumpkinCougar95]

Yes, i think you should use that Expression instead.

 

[Gene Naden]

Does your course include the use of delta functions? If we take the divergence of the integral we will get an expression involving delta functions.

 

[dRic2]

Personally, i like the derivation using solid angles method but it gives the integral form of Gauss law.

Well, I don't know much about electrodynamics but once you have the integral form it's pretty easy to switch to derivate form:

$$\int_S (\mathbf E⋅\mathbf n) dA = \frac Q {\epsilon}$$
$$ \int_S (\mathbf E⋅\mathbf n) dA = \frac 1 {\epsilon} \int_V \rho dV$$

Using the divergence theorem

$$ \int_S (\mathbf E⋅\mathbf n) dA = \int_V ∇ ⋅ \mathbf E dV $$

so

$$\int_V ∇ ⋅ \mathbf E dV = \frac 1 {\epsilon} \int_V \rho dV $$

Since ##\epsilon## is constant the above equality is true only if

$$ ∇ ⋅ \mathbf E = \frac {\rho} {\epsilon} $$

 

[Gene Naden]

Looks good

 

[SothSogi]

Ok, thank you very much to all! There are some things you mention that I honestly do not know, like for instance the delta functions. But it is now clear where my mistakes were and how to derive it using other methods.

Thank you.