How does one write the boundary condition for the electric field in this tidy package?

The usual process for deriving the boundary condition for the tangential electric field across a surface uses Faraday's Law to prove that (when there are no dipole layers etc.) $$\hat{\textbf{t}} \cdot(\textbf{E}_1 - \textbf{E}_2) = 0 \tag{A}$$ where $$\hat{\textbf{t}}$$ is any vector in the tangent plane at the given point on the surface, and where 1 and 2 denote the two sides of the given surface. I have denoted this statement by $$A$$. If this is true, then

$$\hat{\textbf{n}} \times (\textbf{E}_1 - \textbf{E}_2) = 0 \tag{B}$$

follows immediately, for otherwise $$\textbf{E}_1 - \textbf{E}_2$$ is not completely along $$\hat{\textbf{n}}$$ (which is to say there is a component of $$\textbf{E}_1 - \textbf{E}_2$$ which is tangent to the plane, contradicting $$A$$).

Indeed, it is always written that $$A \iff B$$ and yet I cannot prove the converse, $$B \implies A$$. I'm hoping someone can supply that converse in full generality, without taking a particularly convenient surface etc.

$$\hat{\textbf{t}} \cdot(\textbf{E}_1 - \textbf{E}_2) = 0 \tag{A}$$ where $$\hat{\textbf{t}}$$ is any vector in the tangent plane at the given point on the surface, and where 1 and 2 denote the two sides of the given surface. I have denoted this statement by $$A$$. If this is true, then $$\hat{\textbf{n}} \times (\textbf{E}_1 - \textbf{E}_2) = 0 \tag{B}$$ follows immediately, for otherwise $$\textbf{E}_1 - \textbf{E}_2$$ is not completely along $$\hat{\textbf{n}}$$ (which is to say there is a component of $$\textbf{E}_1 - \textbf{E}_2$$ which is tangent to the plane, contradicting $$A$$).

Indeed, it is always written that $$A \iff B$$ and yet I cannot prove the converse, $$B \implies A$$.

Let $$\vec v = \vec E_1 - \vec E_2$$.

Your Eq (B) means that the angle between $$\hat n$$ and $$\vec v$$ is either zero degrees or 180 degrees. This means that $$\vec v$$ is either parallel to or anti-parallel to $$\hat n$$.

$$\vec v = \pm |v|\hat n$$

Because, by definition, $$\hat n$$ is perpendicular to any $$\hat t$$, so too is $$-\hat n$$, and so too is $$\vec v$$.

$$\vec v \cdot \hat t = \pm |v| \hat n \cdot \hat t = 0$$

• Very simple, I'm mortified I missed this. Thanks!
– EE18
Commented Oct 19, 2022 at 17:44