Assuming there is a perfect conductor at $x=0$ in $\mathbb{R}^3$ and a plane EM wave $\vec{E}_i(\vec{x},t)=\vec{E_i^0}e^{i(kx-\omega t)}$ is coming from $x=- \infty$. We know, that the wave $\vec{E_r}(\vec{x},t)=-\vec{E_i^0}e^{-i(kx+\omega t)}$ will be reflected, by assuming that \begin{align} \vec{E}(\vec{x},t) = \vec{E}_i(\vec{x},t) + \vec{E}_r(\vec{x},t) \end{align} at $x=0$.
My question now is the following: In my lecture notes I find that the charge density on a surface is given by $\epsilon_0\vec{n}\vec{E}|_{\partial V}$. In this case is $\vec{E_i^0} \perp \vec{n}$ which means that there is no charge density on the surface. Is this really the case? I always thought, that if an electric field hits an perfect conductor, the charges on the surface of the conductor will move to compensate the electric field on the incoming wave, and that is how the reflected wafe is "produced". Since in this example there is no charge density, how is the reflected wave created?
