Divergence Theorem and Gauss Law

[Caglar Yildiz]

Divergence theorem states that
$\int \int\vec{E}\cdot\vec{ds}=\int\int\int div(\vec{E})dV$
And Gauss law states that

$\int \int\vec{E}\cdot\vec{ds}=\int\int\int \rho(x,y,z)dV$
If $\vec{E}$ to be electric field vector then i could say that
$div(\vec{E})=\rho(x,y,z)$
However i can't see any reason for that since $\rho(x,y,z)$ to be unit charge

 

[BvU]

Divergence theorem states that$$
\int \int\vec{E}\cdot\vec{ds}=\int\int\int div(\vec{E})dV
$$ And Gauss law states that $$\int \int\vec{E}\cdot\vec{ds}=\int\int\int \rho(x,y,z)dV
$$ If $\vec{E}$ to be electric field vector then i could say that $div(\vec{E})=\rho(x,y,z)$
However i can't see any reason for that since $\rho(x,y,z)$ to be unit charge

Hi. (Use double $ for displayed, double # for in-line $LaTeX$ ).

Nevertheless, $\operatorname {div} \vec E = { \rho\over \epsilon_0 }$ is one of the Maxwell equations. So $\left | \vec E \right | \propto \rho$ as you would expect.

 

[vanhees71]

The point is that your integral equations are valid for any volume $V$ and its boundary $\partial V$ (which of course is a closed surface):
$$\int_{\partial V} \mathrm{d}^2 \vec{F} \cdot \vec{E}=\int_V \mathrm{d}^3 x \rho(\vec{x}).$$
This is not very useful in practice, because integrals are not as easy to use as derivatives. The idea thus is to make the volume very small around some specific point $\vec{x}$. On the right-hand side of Gauss's Law you can make $V$ so small that $\rho$ doesn't vary too much over its extension, i.e., you can write
$$\int_V \mathrm{d}^3 x' \rho(\vec{x}') \simeq \rho(\vec{x}) V.$$
Now to the left-hand side. Think of the volume as a little cube with edges parallel to a Cartesian coordinate system. It's best to draw the meaning of the integral to see that the surface integral can be approximated by
$$V (\partial_x E_x(\vec{x})+\partial_y E_y(\vec{x}) + \partial_z E_z(\vec{x}))=V \mathrm{div} \vec{E}(\vec{x}).$$
Now you can cancel the little volume $V$ from both sides of the equation to get the local form of Gauss's Law, which is one of Maxwell's equations (here written in Heaviside-Lorentz units, where you don't have the confusing conversion factors as in the SI units):
$$\mathrm{div} \vec{E}(\vec{x})=\rho(\vec{x}).$$

 

[Caglar Yildiz]

The point is that your integral equations are valid for any volume $V$ and its boundary $\partial V$ (which of course is a closed surface):
$$\int_{\partial V} \mathrm{d}^2 \vec{F} \cdot \vec{E}=\int_V \mathrm{d}^3 x \rho(\vec{x}).$$
This is not very useful in practice, because integrals are not as easy to use as derivatives. The idea thus is to make the volume very small around some specific point $\vec{x}$. On the right-hand side of Gauss's Law you can make $V$ so small that $\rho$ doesn't vary too much over its extension, i.e., you can write
$$\int_V \mathrm{d}^3 x' \rho(\vec{x}') \simeq \rho(\vec{x}) V.$$
Now to the left-hand side. Think of the volume as a little cube with edges parallel to a Cartesian coordinate system. It's best to draw the meaning of the integral to see that the surface integral can be approximated by
$$V (\partial_x E_x(\vec{x})+\partial_y E_y(\vec{x}) + \partial_z E_z(\vec{x}))=V \mathrm{div} \vec{E}(\vec{x}).$$
Now you can cancel the little volume $V$ from both sides of the equation to get the local form of Gauss's Law, which is one of Maxwell's equations (here written in Heaviside-Lorentz units, where you don't have the confusing conversion factors as in the SI units):
$$\mathrm{div} \vec{E}(\vec{x})=\rho(\vec{x}).$$

Hi
Why did you put $$d^2$$ and $$ \vec{F}$$? I saw it on wikipedia too but did not get the idea.

 

[Caglar Yildiz]

Hi. (Use double $ for displayed, double # for in-line $LaTeX$ ).

Nevertheless, $\operatorname {div} \vec E = { \rho\over \epsilon_0 }$ is one of the Maxwell equations. So $\left | \vec E \right | \propto \rho$ as you would expect.

Well after your reply, i went to wiki and i saw the derivations of those equetion. Also thanks for $$ thing :) i did not know.

 

[vanhees71]

Hi
Why did you put $$d^2$$ and $$ \vec{F}$$? I saw it on wikipedia too but did not get the idea.

$\mathbb{d}^2 \vec{F}$ is my convention for the surface-element vector. If your surface is parametrized with generalized coordinates $q^k$ ($k \in \{1,2 \}$) then
$$\mathbb{d}^2 \vec{F}=\mathrm{d} q^1 \, \mathrm{d} q^2 \; \frac{\partial \vec{x}}{\partial q^1} \times \frac{\partial \vec{x}}{\partial q^2}.$$