# Why is the tangential component of electric field is equal on both sides of interface?

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 $$E_{1\parallel}= E_{2\parallel}$$ ?

• Using another Maxwell equations (Gauss-Faraday law in integral form). Commented Apr 6, 2022 at 13:45
• edited to include gauss-faraday's law. Commented Apr 6, 2022 at 13:54
• This is not correct - in the integral form you have derivative of the flux (rather than the magnetic field) in the right-hand-side. And for a contour with vanishingly small surface it vanishes. Commented Apr 6, 2022 at 13:59
• Since I hadn't included ds in $\hat{n} \times (\vec{E_2}-\vec{E_1})= -\frac{\partial B}{\partial t}$ , I thought it was okay ..I mean with ds and integral it would indeed take the form $\int ds \space \hat{n} \times (\vec{E_2}-\vec{E_1})= -\frac{\partial \int B \cdot \hat{n} \space ds}{\partial t}$ , however why would integral containing B would vanish ? (if we assume an infinite interface spread out in yz plane) I am thinking of deriving snell's law and assuming light entering from one medium to other via this interface. Commented Apr 6, 2022 at 14:05
• Without integral the dimensionality of your equation would be incorrect, since your replaced $\nabla\times$ by a unit vector (see the correct for of the equations). If you use a rectangular contour that runs along one side of the border and then back on the other side, you can evaluate the integrals exactly. Commented Apr 6, 2022 at 14:09

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.