Talk:Quizbank/University Physics Semester 2

transclustion problem with cp2.5 and 2.6

$\varepsilon _{0}=$ 8.85×10−12 F/m = vacuum permittivity. e = 1.602×10−19C: negative (positive) charge for electrons (protons) $k_{e}={\tfrac {1}{4\pi \varepsilon _{0}}}=$ = 8.99×109 m/F ${\vec {F}}=Q{\vec {E}}$ where ${\vec {E}}={\tfrac {1}{4\pi \varepsilon _{0}}}\sum _{i=1}^{N}{\tfrac {q_{i}}{r_{Pi}^{2}}}{\hat {r}}_{Pi}$ ${\vec {E}}=\int {{\tfrac {dq}{{r}^{2}}}{\hat {r}}}$ where $dq=\lambda d\ell =\sigma da=\rho dV$ $E={\tfrac {\sigma }{2\varepsilon _{0}}}$ = field above an infinite plane of charge. $\Phi ={\vec {E}}\cdot {\vec {A}}$ $\to \int {\vec {E}}\cdot d{\vec {A}}=\int {\vec {E}}\cdot {\hat {n}}\,dA$ = electric flux $q_{enclosed}=\varepsilon _{0}\oint {\vec {E}}\cdot d{\vec {A}}$ $d\,{\text{Vol}}=dxdydz=r^{2}drdA$ where $dA=r^{2}d\phi d\theta$ $A_{\text{sphere}}=r^{2}\int _{0}^{\pi }\sin \theta d\theta \int _{0}^{2\pi }d\phi =4\pi r^{2}$ OpenStax University Physics/E&M/Electric Charges and Fields $\Phi ={\vec {E}}\cdot {\vec {A}}$ $\to \int {\vec {E}}\cdot d{\vec {A}}=\int {\vec {E}}\cdot {\hat {n}}\,dA$ = electric flux $q_{enclosed}=\varepsilon _{0}\oint {\vec {E}}\cdot d{\vec {A}}$ $d\,{\text{Vol}}=dxdydz=r^{2}drdA$ where $dA=r^{2}d\phi d\theta$ $A_{\text{sphere}}=r^{2}\int _{0}^{\pi }\sin \theta d\theta \int _{0}^{2\pi }d\phi =4\pi r^{2}$ OpenStax University Physics/E&M/Electric Charges and Fields $\Phi ={\vec {E}}\cdot {\vec {A}}$ $\to \int {\vec {E}}\cdot d{\vec {A}}=\int {\vec {E}}\cdot {\hat {n}}\,dA$ = electric flux $q_{enclosed}=\varepsilon _{0}\oint {\vec {E}}\cdot d{\vec {A}}$ $d\,{\text{Vol}}=dxdydz=r^{2}drdA$ where $dA=r^{2}d\phi d\theta$ $A_{\text{sphere}}=r^{2}\int _{0}^{\pi }\sin \theta d\theta \int _{0}^{2\pi }d\phi =4\pi r^{2}$