From Maxwell's equations (gauss's law) :
$ \hat{n} \cdot (\vec{D_2}-\vec{D_1})= \sigma \implies D_{2\perp}-D_{1\perp}=\sigma : $ surface charge density.
How does one go on proving further that $E_{1\parallel}= E_{2\parallel}$ where the parallel component is wrt to the interface that has surface charge density $\sigma$ ?
okay , from gauss faraday's law we have : $ \hat{n} \times (\vec{E_2}-\vec{E_1})= -\frac{\partial B}{\partial t} $ , unless $\vec{B}=0$ we can't write $ \hat{n} \times (\vec{E_2}-\vec{E_1})=0$ and $E_{1\parallel}= E_{2\parallel}$ ?
Best Answer
The argument is: $$\oint \vec{E}\cdot d\vec{l} = - \int \frac{\partial \vec{B}}{\partial t}\cdot d\vec A$$
You construct a rectangle straddling the interface plane at $y=0$, with sides of length $dx$ parallel to the x-axis and $dy$ along the y-axis.
You then consider that over this small rectangle, the components of the E-field tangential ($E_x$) and perpendicular ($E_y$) to the interface can be assumed constant on either side of the interface, but may change as you go across the interface.
The left hand side of the integral Maxwell-Faraday law becomes (evaluating the closed line integral anti-clockwise around the rectangle) $$\oint \vec{E}\cdot d\vec{l} \simeq E_{x,1}dx + E_{y,1}dy/2 + E_{y,2}dy/2 - E_{x,2}dx - E_{y,2}dy/2 - E_{y,1}dy/2 = (E_{x,1}-E_{x,2})dx\ , $$ where the suffixes 1 and 2 refer to the fields on either side of the interface.
On the right hand side we have $$- \int \frac{\partial \vec{B}}{\partial t}\cdot d\vec A \simeq -\frac{\partial}{\partial t}\left( \vec{B_1} + \vec{B_2}\right)\cdot \frac{dx dy}{2}\ \hat{z}\ .$$
The argument then is that we can allow $dy$ to become arbitrarily small and indeed zero. This would not affect the left hand side at all, but on the right hand side, the magnetic flux through the rectangle would be zero.
Hence $$(E_{x,1}-E_{x,2})dx = 0\ ,$$ where $E_{x,1}$ and $E_{x,2}$ are the tangential fields immediately either size of the interface.