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

Chapter 5

Electric Charges and Fields

Coulomb's Law ${\displaystyle {\vec {F}}={\tfrac {1}{4\pi \varepsilon _{0}}}{\tfrac {q_{1}q_{2}}{r_{12}^{2}}}{\hat {r}}_{12}}$ where the vacuum permittivity ${\displaystyle \varepsilon _{0}=}$ 8.85×10−12 F/m.

Elementary charge = e = 1.602×10−19C (electrons have charge q=−e and protons have charge q=+e.)

▭ By superposition, ${\displaystyle {\vec {F}}={\tfrac {1}{4\pi \varepsilon _{0}}}Q\sum _{i=1}^{N}{\tfrac {q_{i}}{r_{Qi}^{2}}}{\hat {r}}_{Qi}}$ where ${\displaystyle {\vec {r}}_{Qi}={\vec {r}}_{Q}-{\vec {r}}_{i}}$
▭ Electric field ${\displaystyle {\vec {F}}=Q{\vec {E}}}$ where ${\displaystyle {\vec {E}}({\vec {r}}_{P})={\tfrac {1}{4\pi \varepsilon _{0}}}\sum _{i=1}^{N}{\tfrac {q_{i}}{r_{Pi}^{2}}}{\hat {r}}_{Pi}}$ is the field at ${\displaystyle {\vec {r}}_{P}}$ due to charges at ${\displaystyle {\vec {r}}_{i}}$
▭ The field above an infinite wire ${\displaystyle {\vec {E}}(z)={\tfrac {1}{4\pi \varepsilon _{0}}}{\tfrac {2\lambda }{z}}{\hat {k}}}$ and above an infinite plane ${\displaystyle {\vec {E}}={\tfrac {\sigma }{2\varepsilon _{0}}}{\hat {k}}}$
▭ An electric dipole ${\displaystyle {\vec {p}}=q{\vec {d}}}$ in a uniform electric field experiences the torque ${\displaystyle \tau ={\vec {p}}\times {\vec {E}}}$

For quiz at QB/d_cp2.5

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