OpenStax University Physics/E&M/Electromagnetic Waves

For quiz at QB/d_cp2.16
Displacement current $$I_d=\varepsilon_0\tfrac{d\Phi_E}{dt}$$ where $$\Phi_E=\int\vec E\cdot d\vec A$$ is the electric flux.

Maxwell's equations: $$\epsilon_0\mu_0=1/c^2$$ $$\oint_S \vec{E} \cdot \mathrm{d}\vec{A} = \frac 1{\epsilon_0}Q_{in} \qquad     $$ $$\oint_S \vec{B} \cdot \mathrm{d}\vec{A} = 0$$ $$\oint_C \vec{E} \cdot \mathrm{d}\vec{\ell} = -  \int_S \frac{\partial\vec{B}}{\partial t} \cdot \mathrm{d} \vec{A}$$ $$\oint_C \vec{B} \cdot \mathrm{d}\vec{\ell} = \mu_0I + \epsilon_0\mu_0\frac{\mathrm{d}\Phi_E}{\mathrm{d}t}$$

$$\frac{\partial^2E_y}{\partial x^2}=\varepsilon_0\mu_0\frac{\partial^2E_y}{\partial t^2}$$ and $$\tfrac{E_0}{B_0}=c$$

Poynting vector $$\vec S = \tfrac 1{\mu_0} \vec E\times\vec B$$=energy flux

Average intensity $$I=S_{ave}=\tfrac{c\varepsilon_0}{2}E_0^2=\tfrac{c}{2\mu_0}B_0^2=\tfrac{1}{2\mu_0}E_0B_0$$

Radiation pressure $$p=I/c$$ (perfect absorber) and $$p=2I/c$$ (perfect reflector).