Talk:QB/d cp2.5

$\varepsilon _{0}=$ 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.

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