Gauss' law

Gauss' law

Now recall that flux is the scalar product of a vector field and a bit of surface $$Flux=\vec{F}\cdot \vec{a}$$ where $\vec{F}$ is some vector field and $\vec{a}$ is a surface with the direction defined by the normal to the surface. For a series of connected surfaces ${\vec{a}}_j$ the total flux through the combined surface would be the sum of the individual elements. For a vector field $\vec{E}$ passing through the surface this leads to $$\Phi =\sum {\vec{E}}_j\cdot {\vec{a}}_j$$ or when we go to infinitesimal areas $$\Phi ={\int }_{surface}\vec{E}\cdot d\vec{a}$$ Now lets consider a charge $q$ in the middle of a sphere

$${\displaystyle \begin{array}{c}\Phi =\oint \vec{E}\cdot d\vec{a}=\oint E\cdot da\\ =E\oint da\\ =E(4\pi r^2)\end{array}}$$ but $$E=\frac{kq}{r^2}$$ then $$\Phi =4\pi kq$$ $${\epsilon }_0\equiv \frac{1}{4\pi k}$$ So for this case we get $$\oint \vec{E}\cdot d\vec{a}=\frac{q}{{\epsilon }_0}$$ We can generalize this to any closed surface. It is clear that for an arbitrary closed source, we can draw a sphere around the source within thearbitrary surface.. Think of bullets being fired from a gun, it is clear that the bullets originating in the inner sphere all pass through the outer surfaceand so one would expect that the flux would be the same. For example consider $\vec{a}$ to be a patch on the inner sphere and $\vec{A}$ to be its projection onto the outer arbitrary surface (with its normal making an angle $\theta$ with respect to the normal to $\vec{a}\text{.}$

On the inner patch $${\Phi }_r={\vec{E}}_r\cdot \vec{a}=E_{r}a$$ and at the outer patch $${\displaystyle \begin{array}{c}{\Phi }_R={\vec{E}}_R\cdot \vec{A}\\ =E_{R}A\cos \theta \\ =\left[E_r{\left(\frac{r}{R}\right)}^2\right]\left[a{\left(\frac{R}{r}\right)}^2\frac{1}{\cos \theta }\right]\cos \theta \\ =E_{r}a\\ ={\Phi }_r\end{array}}$$ So the two have equivalent fluxes.

Any electric field is the sum of fields of its individual sources so we can write $${\displaystyle \begin{array}{c}\Phi =\oint \vec{E}\cdot d\vec{a}=\oint \sum _i{\vec{E}}_{i}d\vec{a}\\ =\frac{1}{{\epsilon }_0}\sum _i{q}_i\end{array}}$$ or for charge distributed throughout the volume $$\oint \vec{E}\cdot d\vec{a}=\frac{1}{{\epsilon }_0}\int \rho dV$$

Now we can apply Gauss' Theorem $$\oint \vec{E}\cdot d\vec{a}=\int \vec{\nabla }\cdot \vec{E}dV=\frac{1}{{\epsilon }_0}\int \rho dV$$

The equation $$\int \vec{\nabla }\cdot \vec{E}dV=\frac{1}{{\epsilon }_0}\int \rho dV$$ must be true for any volume of any size, shape or location. The only way that can be true is if:

$$\vec{\nabla }\cdot \vec{E}=\frac{\rho }{{\epsilon }_0}$$ Initially one may think that this is a much less clear way of posing Gauss' Law. In practice it is much more useful than the integral form. Given anarbitrary distribution of charge we can calculate the electric field anywhere in space.

Gauss' law for magnetism

We can consider the same arguments for magnetic fields however there is one major difference! There are no isolated source of magnetism. That is there areno magnetic monopoles. This is an experimental fact. In fact people continue to search for them but they have never been found. (Finding one would almostcertainly be a discovery worthy of a Nobel Prize). So we have $$\oint \vec{B}\cdot d\vec{a}=0$$ or $$\vec{\nabla }\cdot \vec{B}=0.$$