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

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## Chapter 6

#### Gauss's Law

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

▭ Closed surface integral $\Phi =\oint {\vec {E}}\cdot d{\vec {A}}=\oint {\vec {E}}\cdot {\hat {n}}\,dA$ ▭ Gauss's Law $=q_{enc}=\varepsilon _{0}\oint {\vec {E}}\cdot d{\vec {A}}$ . In simple cases: $E\int dA=EA^{*}={\tfrac {q_{enc}}{\varepsilon _{0}}}$ ▭ Electric field just outside the surface of a conductor ${\vec {E}}={\tfrac {\sigma }{\varepsilon _{0}}}$ #### For quiz at QB/d_cp2.6

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

$q_{enclosed}=\varepsilon _{0}\oint {\vec {E}}\cdot d{\vec {A}}$ $d\,{\text{Vol}}=dxdydz=r^{2}drdA$ where $dA=r^{2}d\phi d\theta$ $A_{\text{sphere}}=r^{2}\int _{0}^{\pi }\sin \theta d\theta \int _{0}^{2\pi }d\phi =4\pi r^{2}$ 