OpenStax University Physics/E&M/Electric Charges and Fields

For quiz at 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.