Divergence and Curl of Electrostatic Field

Section 2.1: Field Lines, Flux and Gauss' Law

\( \require{enclose} \enclose{box}[mathcolor="black"]{\frac{1}{4\pi \epsilon_0} \int_S \frac{\rho (\boldsymbol{r'})}{r^2}\hat{\boldsymbol{r}} \partial{\tau'}} \rightarrow\)

1. Tells all we need.
2. Can be formidable.
3. Need tricks to avoid such integrals.

Magnitude of E field is indicated by density \(\frac{\text{# of lines}}{\text{surface area}} = \frac{n}{4 \pi r^2}\)
Field lines:

1. Emanate from point charges symmetrically in all directions.
2. Begin on positive charges, end on negative charges.
3. Don't terminate midair -> may extend to infinity.
4. Never cross.

Flux of \(\overrightarrow{\boldsymbol{E}}\) through a surface \(\boldsymbol{S}\): \(\Phi_{E} = \int_S \overrightarrow{E} \cdot \overrightarrow{\partial{a}}\)
For a given sampling rate the flux is proportional to the # of lines drawn.
Suggestion: Flux through any closed surface is a measure of total charge inside.
Outside charge will contribute nothing to the total flux - field lines pass through (in and out)

Flux of point charge q through sphere (radius r) =>
\(\oint \overrightarrow{E} \cdot \overrightarrow{\partial{a}} = \int \frac{1}{4\pi \epsilon_0} \cdot (\frac{q}{r^2}\hat{\boldsymbol{r}}) \cdot (r^2 \sin{\theta} \partial{\theta} \partial{\phi}\hat{\boldsymbol{r}}) = \frac{q}{4\pi \epsilon_0} \int \sin{\theta} \partial{\theta} \partial{\phi} = \frac{q 4\pi}{4\pi \epsilon_0} = \frac{q}{\epsilon_0}\)

As \(E\ \alpha\ \frac{1}{r^2}\) goes down, Area \(\alpha\ r^2\) goes up -> product is constant
Thus flux through any surface enclosing the charge is \(\frac{q}{\epsilon_0}\)

\( \left\{ \begin{array}{c} \text{Now, bunch of charges => invoke }\boldsymbol{superposition}\ \rightarrow \overrightarrow{E}= \sum_{i=1}^{n}\overrightarrow{E_i} \\ \text{Flux:}\ \oint \overrightarrow{E} \cdot \overrightarrow{\partial{a}} = \sum_{i=1}^{n} (\oint \overrightarrow{E_i} \cdot \overrightarrow{\partial{a}}) = \sum_{i=1}^{n} \frac{q_i}{\epsilon_0} \end{array} \right. \Longrightarrow \text{For any closed surface: } \enclose{box}[mathcolor="black"]{\oint_S \overrightarrow{E} \cdot \overrightarrow{\partial{a}} = \frac{1}{\epsilon_0}Q_{enc}} \)

Gauss Law hinges on \(\frac{1}{r^2}\) character of Coulomb's Law. Without cancelation flux would depend on the surface chosen.

Apply divergence theorem: \( \left\{ \begin{array}{c} \oint_S \overrightarrow{E} \cdot \overrightarrow{\partial{a}} = \int_V (\nabla \cdot \overrightarrow{E}) \partial{\tau} \\ Q_{enc} = \int_V \rho \partial{\tau} \end{array} \right. \Longrightarrow \int_V (\nabla \cdot \overrightarrow{E}) \partial{\tau} = \int_V (\frac{\rho}{\epsilon_0}) \partial{\tau} \)

Holds for any volume, thus Gauss Law in differential form \(\enclose{box}[mathcolor="black"]{\nabla \cdot \overrightarrow{E} = \frac{\rho}{\epsilon_0}}\)