# Electrodynamics¶

Note for electrodynamics course.

Coulomb Potential Energy for a point charge Q with the appearance of a test charge q at distance r

$V(r) = k \frac{q Q}{r}.$

The ability to keep storage of charge is called capacitance, which is straight forward to have such a definition as

$C = \frac{q}{U},$

where $$U$$ is the electric potential (not the potential energy).

Maxwell’s equations are

$\begin{split}\mathbf{E}\cdot\mathrm{d}\mathbf{S} &= \frac{1}{\varepsilon_0} \iiint_\Omega \rho \,\mathrm{d}V \\ \mathbf{B}\cdot\mathrm{d}\mathbf{S} &= 0 \\ \oint_{\partial \Sigma} \mathbf{E} \cdot \mathrm{d}\boldsymbol{\ell} & = - \frac{d}{dt} \iint_{\Sigma} \mathbf{B} \cdot \mathrm{d}\mathbf{S} \\ \oint_{\partial \Sigma} \mathbf{B} \cdot \mathrm{d}\boldsymbol{\ell} &= \mu_0 \iint_{\Sigma} \mathbf{J} \cdot \mathrm{d}\mathbf{S} + \mu_0 \varepsilon_0 \frac{d}{dt} \iint_{\Sigma} \mathbf{E} \cdot \mathrm{d}\mathbf{S}\end{split}$

or

$\begin{split}\nabla \cdot \mathbf{E} &= \frac {\rho} {\varepsilon_0} \\ \nabla \cdot \mathbf{B} &= 0 \\ \nabla \times \mathbf{E} &= -\frac{\partial \mathbf{B}} {\partial t} \\ \nabla \times \mathbf{B} &= \mu_0\left(\mathbf{J} + \varepsilon_0 \frac{\partial \mathbf{E}} {\partial t} \right)\end{split}$