PX284 - O4 - maxwell's equations in differential form

gauss' law

\begin{matrix} \subset \supset \iint_{S} \vec{E} \cdot d \vec{S} == ∭_{V} \vec{\nabla} \cdot \vec{E} d V = ∭_{V} \frac{ρ}{ϵ_{0}} d V \\ ∭_{V} (\vec{\nabla} \cdot \vec{E} - \frac{ρ}{ε_{0}}) d V = 0 \\ (1) & ∴ \vec{\nabla} \cdot \vec{E} = \frac{ρ}{ε_{0}} \end{matrix}

solenoidal condition

\begin{matrix} \subset \supset \iint_{S} \vec{B} \cdot d \vec{S} = ∭_{V} (\vec{\nabla} \cdot \vec{B}) d V = 0 \\ (2) & ∴ \vec{\nabla} \cdot \vec{B} = 0 \end{matrix}

faraday-lenz law

this requires stokes' theorem

\begin{matrix} \oint_{C} \vec{E} \cdot d \vec{l} = \iint_{S} (\vec{\nabla} \times \vec{E}) \cdot d \vec{S} = - \frac{d}{d t} \iint_{S} \vec{B} \cdot d \vec{S} \\ \iint_{S} (\vec{\nabla} \times \vec{E} + \frac{\partial \vec{B}}{\partial t}) \cdot d \vec{S} = 0 \\ (3) & ∴ \vec{\nabla} \times \vec{E} = - \frac{\partial \vec{B}}{\partial t} \end{matrix}

ampere's law

\begin{matrix} \oint_{C} \vec{B} \cdot d \vec{l} = \iint_{S} (\vec{\nabla} \times \vec{B}) \cdot d \vec{S} = μ_{0} \iint_{S} \vec{J} \cdot d \vec{S} \\ \iint_{S} (\vec{\nabla} \times \vec{B} - μ_{0} \vec{J}) \cdot d \vec{S} = 0 \\ \vec{\nabla} \times \vec{B} = μ_{0} \vec{J} \end{matrix}

this is almost, but not quite the fourth equation

ampere-maxwell law

\vec{\nabla} \cdot (\vec{\nabla} \times \vec{B}) = μ_{0} \vec{\nabla} \cdot \vec{J}

note: $\vec{\nabla} \cdot (\vec{\nabla} \times \vec{v}) = 0$

⟹ \vec{\nabla} \cdot (\vec{\nabla} \times \vec{B}) = 0

but the $R H S :$

μ_{0} \vec{\nabla} \cdot \vec{J} = - μ_{0} \frac{\partial ρ}{\partial t}

from the continuity equation:

\vec{\nabla} \cdot (\vec{\nabla} \times \vec{B}) = 0 = μ_{0} (\vec{\nabla} \cdot \vec{J} + \frac{\partial ρ}{\partial t})

this was maxwell's solution
so, ampere's law is modified to:

\begin{matrix} (4) & \vec{\nabla} \times \vec{B} = μ_{0} (\vec{J} + ε_{0} \frac{\partial \vec{E}}{\partial t}) \end{matrix}

this is called the ampere-maxwell equation
the added term is historically called the displacement current density
checking:

\begin{aligned} \vec{\nabla} \cdot (\vec{\nabla} \times \vec{B}) & = 0 \\ = μ_{0} (\vec{\nabla} \cdot \vec{J} + ε_{0} \frac{\partial}{\partial t} (\vec{\nabla} \cdot \vec{E})) \\ = μ_{0} (\vec{\nabla} \cdot \vec{J} + ε_{0} \frac{\partial}{\partial t} (ρ / ε_{0})) \\ = μ_{0} (\vec{\nabla} \cdot \vec{J} + \frac{\partial ρ}{\partial t}) \end{aligned}

the reason this was not discovered earlier is because ampere's law alone is a good quasi-static approximation, as the additional term, $| ε_{0} \frac{\partial \vec{E}}{\partial t} | ≪ | \vec{J} |$

parallel-plate capacitor

considering a parallel-plate capacitor
current flow will build up opposite charges on each plate
an electric field is generated between the plates
considering a closed loop around the connecting wire:

\oint_{C} \vec{B} \cdot d \vec{l} = \iint_{S_{1}} (\vec{\nabla} \times \vec{B}) \cdot d \vec{S} = \iint_{S_{2}} (\vec{\nabla} \times \vec{B}) \cdot d \vec{S}

where, $S_{2}$ goes around a plate of the capacitor, but not $S_{1}$

\iint_{S_{1}} μ_{0} (\vec{J} + ε_{0} \frac{\partial \vec{E}}{\partial t}) = \iint_{S_{1}} μ_{0} (\vec{J} + ε_{0} \frac{\partial \vec{E}}{\partial t})

for $L H S$ , $\frac{\partial \vec{E}}{\partial t} = 0$
for $R H S$ , $\vec{J} = 0$
for a parallel-plate capacitor:

\begin{matrix} | \vec{E} | = \frac{σ}{ε_{0}} = \frac{Q}{A ε_{0}} \\ \frac{\partial E}{\partial t} = \frac{1}{A ε_{0}} \frac{\partial Q}{\partial t} = \frac{I}{A ε_{0}} \\ ⟹ \iint_{S_{1}} μ_{0} \vec{J} \cdot d \vec{S} = μ_{0} I \\ ⟹ \iint_{S_{2}} μ_{0} \frac{\partial \vec{E}}{\partial t} \cdot d \vec{S} = μ_{0} I \end{matrix}