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.
Best Answer
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.
$^\dagger$ 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 $\frac{\partial E_m}{\partial m}\to0$.