# Talk:QB/d cp2.5

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${\displaystyle \varepsilon _{0}=}$ 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.