The boundary conditions in a waveguide

Question

Suppose a cubic waveguide, made of perfect conductor, has only two open parallel sides. And the boundary conditions in this case are that the electric field at the surface must satisfy:

$$\vec{B} \cdot \vec{n}=0,$$

and magnetic field:

$$\vec{E} \times \vec{n}=0,$$

where the $\vec{n}$ is the normal vector pointing outwards from the conductor. These two relations come from the equations:

$$\nabla \cdot \vec{B}=0,$$ $$\nabla \times \vec{E}=0.$$

The question is how to derive the other boundary condition that at the surface the electric field must satisfy:

$$\frac{\partial{E_n}}{\partial n}=0.$$

$E_n$ means the electric field along normal direction.

Answer 1

This simply follows from the Gauss' law:

$$\nabla\cdot\vec E=\frac\rho{\varepsilon_0}.\tag1$$

Since we know that $\vec E\parallel\vec n$ at the boundary inside the waveguide, the divergence of $\vec E$ reduces$^\dagger$ to $$\nabla\cdot\vec E=\frac{\partial E_n}{\partial n}.\tag2$$

Since the waveguide doesn't have any charges inside, $\rho=0$ in the whole internal region including the vicinity of the conductor. Inserting this into $(1)$ simplified by $(2)$, we get our boundary condition.

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^{$^\dagger$ This is true because $\vec E$ is analytic at the interface (with $n=0^+$), in which case $\lim\limits_{n\to0}E_m\to0$, where $m$ is the coordinate along any tangent to the interface and $n$ – along the normal, implies $\frac{\partial E_m}{\partial m}\to0$.}