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

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## Chapter 6

#### Gauss's Law

closed .. open

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

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

#### For quiz at QB/d_cp2.6

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

${\displaystyle q_{enclosed}=\varepsilon _{0}\oint {\vec {E}}\cdot d{\vec {A}}}$

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

${\displaystyle A_{\text{sphere}}=r^{2}\int _{0}^{\pi }\sin \theta d\theta \int _{0}^{2\pi }d\phi =4\pi r^{2}}$