A1 - electromagnetism

Gauss' law:

\begin{matrix} \int \vec{E} \cdot d \vec{S} = \int_{V} \frac{ρ}{ε_{0}} d^{3} \vec{r} \\ \vec{\nabla} \cdot \vec{E} = \frac{ρ}{ε_{0}} \end{matrix}

in a material, there are free and bound charges, so:

\begin{matrix} (1) & \vec{\nabla} \cdot (ε_{0} \vec{E} + \vec{P}) = \vec{\nabla} \cdot \vec{D} = ρ_{f r e e} \end{matrix}

where, $\vec{P}$ is the polarisation

this module considers only free space, ie. no bound charges, so:

\begin{matrix} \vec{D} = ε_{0} \vec{E} \\ ε_{r} = 1 \end{matrix}

Ampere's law:

\vec{\nabla} \times \vec{H} = {\vec{j}}_{f r e e}

in free space:

\vec{B} = μ_{0} \vec{H}

magnetic monopoles have not been observed, so:

\begin{matrix} (2) & \vec{\nabla} \cdot \vec{B} = 0 \end{matrix}

considering charge conservation, the continuity equation:

\frac{\partial ρ}{\partial t} = - \vec{\nabla} \cdot \vec{j}

if $\vec{j} = \vec{\nabla} \times \vec{H} :$

\frac{\partial ρ}{\partial t} = - \vec{\nabla} \cdot (\vec{\nabla} \times \vec{H}) = 0

note that this is the static condition
instead say:

\vec{\nabla} \times \vec{H} = \vec{j} + \vec{X}

such that:

\begin{matrix} \vec{\nabla} \cdot (\vec{j} + \vec{X}) = 0 \\ ⟹ \vec{\nabla} \cdot \vec{X} = - \vec{\nabla} \cdot \vec{j} = \frac{\partial ρ}{\partial t} \\ - \vec{\nabla} \cdot \vec{j} = \frac{\partial}{\partial t} (\vec{\nabla} \cdot \vec{D}) = \vec{\nabla} \cdot (\frac{\partial \vec{D}}{\partial t}) \\ ∴ \vec{X} = \frac{\partial \vec{D}}{\partial t} \end{matrix}

$\vec{X}$ is therefore the displacement current

\begin{matrix} (3) & \vec{\nabla} \times \vec{H} = \vec{j} + \frac{\partial \vec{D}}{\partial t} \end{matrix}

finally, considering electromagnetic induction, Faraday's law:

\begin{matrix} (4) & \vec{\nabla} \times \vec{E} = - \frac{\partial \vec{B}}{\partial t} \end{matrix}

$(2)$ and $(4)$ have only fields, but $(1)$ and $(3)$ have both fields and source terms
Maxwell's equations in free space:

\begin{matrix} \vec{\nabla} \cdot \vec{B} = 0 \\ \vec{\nabla} \cdot \vec{E} = \frac{ρ}{ε_{0}} \\ \vec{\nabla} \times \vec{E} + \frac{\partial \vec{B}}{\partial t} = 0 \\ \vec{\nabla} \times \vec{B} - μ_{0} ε_{0} \frac{\partial \vec{E}}{\partial t} = μ_{0} \vec{j} \end{matrix}