OpenStax University Physics/E&M/Gauss's Law

$ℰ\mathrm{\&}ℳ$ (Vol. 2):    5:Electric Charges and Fields    6:Gauss's Law    7:Electric Potential    8:Capacitance    9:Current and Resistance    10:Direct-Current Circuits    11:Magnetic Forces and Fields    12:Sources of Magnetic Fields    13:Electromagnetic Induction    14:Inductance    15:Alternating-Current Circuits    16:Electromagnetic Waves

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Chapter 6

Gauss's Law

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Flux for a uniform electric field $\mathrm{\Phi }=\overrightarrow{E}\cdot \overrightarrow{A}$ $\to \mathrm{\Phi }=\int \overrightarrow{E}\cdot d\overrightarrow{A}=\int \overrightarrow{E}\cdot \hat{n}dA$ in general.

▭ Closed surface integral $\mathrm{\Phi }=\oint \overrightarrow{E}\cdot d\overrightarrow{A}=\oint \overrightarrow{E}\cdot \hat{n}dA$
▭ Gauss's Law $=q_{enc}={\epsilon }_0\oint \overrightarrow{E}\cdot d\overrightarrow{A}$. In simple cases: $E\int dA=E{A}^{\ast }=\frac{q_{enc}}{{\epsilon }_0}$
▭ Electric field just outside the surface of a conductor $\overrightarrow{E}=\frac{\sigma }{{\epsilon }_0}$

For quiz at QB/d_cp2.6

$\mathrm{\Phi }=\overrightarrow{E}\cdot \overrightarrow{A}$ $\to \int \overrightarrow{E}\cdot d\overrightarrow{A}=\int \overrightarrow{E}\cdot \hat{n}dA$ = electric flux

$q_{enclosed}={\epsilon }_0\oint \overrightarrow{E}\cdot d\overrightarrow{A}$

$d\text{Vol}=dxdydz=r^{2}drdA$ where $dA=r^{2}d\varphi d\theta$

$A_{\text{sphere}}=r^2{\int }_0^{\pi }\sin \theta d\theta {\int }_0^{2\pi }d\varphi =4\pi r^2$