PX284 - Q4 - boundary conditions

using faraday-lenz law and stokes' theorem:

\begin{matrix} \iint_{S} \vec{\nabla} \times \vec{E} \cdot d \vec{S} = \oint_{C} \vec{E} \cdot d \vec{l} = \iint_{S} (- \frac{\partial \vec{B}}{\partial t}) \cdot d \vec{S} \\ = (E_{2, ∥} - E_{1, ∥}) \times L + O (d) \end{matrix}

where, $O (d)$ is the order of $d$ , ie. it goes linearly with $d$

in the limit $d \to 0 : S \to 0$

E_{2, ∥} - E_{1, ∥} = 0

this is assuming that $- \partial_{t} \vec{B}$ is finite
this is true for any orientation of loop with $L$ outside the plane of the boundary
writing it as:

\hat{n} \times ({\vec{E}}_{2} - {\vec{E}}_{1}) = 0

where, $\hat{n}$ is the unit vector normal to the boundary

this means that $E_{∥}$ is continuous across the boundary
doing the same with ampere-maxwell law:

\begin{matrix} \iint_{S} \vec{\nabla} \times \vec{H} \cdot d \vec{S} = \oint_{C} \vec{H} \cdot d \vec{l} = \iint_{S} ({\vec{J}}_{f} + \frac{\partial \vec{D}}{\partial t}) \cdot d \vec{S} \\ ⟹ \hat{n} \times ({\vec{H}}_{2} - {\vec{H}}_{1}) = 0 \end{matrix}

ie. $\vec{H}$ is continuous across the boundary
this is assuming that $\iint_{S} {\vec{J}}_{f} \cdot d \vec{S} = 0$
if surface currents are present:

\hat{n} \times ({\vec{H}}_{2} - {\vec{H}}_{1}) = {\vec{j}}_{f}

where, ${\vec{j}}_{f}$ is the surface current density, with units A m $^{- 1}$

when resistivity $\to 0$ , ie. superconductors, ${\vec{j}}_{f}$ acts to expel magnetic fields, ie: $H_{2} = 0$

considering a pill box on the surface

\subset \supset \iint_{S} \vec{B} \cdot d \vec{S} = (B_{1 ⊥} - B_{2 ⊥}) A + O (h) = ∭_{V} \vec{\nabla} \cdot \vec{B} d V = 0

$O (h) \to 0$ as $h \to 0$

∴ B_{1 ⊥} = B_{2 ⊥}

similarly, for $\vec{D} :$

\subset \supset \iint_{S} \vec{D} \cdot d \vec{S} = (D_{1 ⊥} - D_{2 ⊥}) A + O (h) = ∭_{V} \vec{\nabla} \cdot \vec{D} d V = ∭_{V} ρ_{f} d V = Q_{f}

where, $Q_{f}$ is the charge enclosed

$O (h) \to 0$ as $h \to 0$
$Q_{f}$ can contain contribution from surface charges

Q_{f} = ∭_{V} ρ_{f} d V + \iint_{S} σ_{f} d S

where, $ρ_{f}$ is the volume charge density, and $σ_{f}$ is the surface charge density

as $h \to 0$ , the volume charge $\to 0$ , but the surface charge remains $σ_{f} A$

∴ D_{1 ⊥} - D_{2 ⊥} = σ_{f}

additionally, in conductors, the currents ( ${\vec{J}}_{f}$ ) also need to be considered

summarized

\begin{matrix} \hat{n} \times ({\vec{E}}_{2} - {\vec{E}}_{1}) = 0 \\ \hat{n} \cdot ({\vec{D}}_{1} - {\vec{D}}_{2}) = σ_{f} \\ \hat{n} \cdot ({\vec{B}}_{1} - {\vec{B}}_{2}) = 0 \\ \hat{n} \times ({\vec{H}}_{2} - {\vec{H}}_{1}) = \vec{j} f \end{matrix}