# Talk:QB/a18ElectricChargeField findE

${\displaystyle \varepsilon _{0}=}$ 8.85×10−12 F/m = vacuum permittivity.

e = 1.602×10−19C: negative (positive) charge for electrons (protons)

${\displaystyle k_{e}={\tfrac {1}{4\pi \varepsilon _{0}}}=}$ = 8.99×109 m/F

${\displaystyle {\vec {F}}=Q{\vec {E}}}$ where ${\displaystyle {\vec {E}}={\tfrac {1}{4\pi \varepsilon _{0}}}\sum _{i=1}^{N}{\tfrac {q_{i}}{r_{Pi}^{2}}}{\hat {r}}_{Pi}}$

${\displaystyle {\vec {E}}=\int {{\tfrac {dq}{{r}^{2}}}{\hat {r}}}}$ where ${\displaystyle dq=\lambda d\ell =\sigma da=\rho dV}$

${\displaystyle E={\tfrac {\sigma }{2\varepsilon _{0}}}}$ = field above an infinite plane of charge.

${\displaystyle {\vec {E}}({\vec {r}})={\frac {1}{4\pi \varepsilon _{0}}}\sum _{i=1}^{N}{\frac {{\widehat {\mathcal {R}}}_{i}Q_{i}}{|{\mathcal {\vec {R}}}_{i}|^{2}}}={\frac {1}{4\pi \varepsilon _{0}}}\sum _{i=1}^{N}{\frac {{\vec {\mathcal {R}}}_{i}Q_{i}}{|{\mathcal {\vec {R}}}_{i}|^{3}}}}$ is the electric field at the field point, ${\displaystyle {\vec {r}}}$, due to point charges at the source points,${\displaystyle {\vec {r}}_{i}}$ , and ${\displaystyle {\vec {\mathcal {R}}}_{i}={\vec {r}}-{\vec {r}}_{i},}$ points from source points to the field point.