# Talk: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: $\epsilon _{0}\mu _{0}=1/c^{2}$ $\oint _{S}{\vec {E}}\cdot \mathrm {d} {\vec {A}}={\frac {1}{\epsilon _{0}}}Q_{in}\qquad$ $\oint _{S}{\vec {B}}\cdot \mathrm {d} {\vec {A}}=0$ $\oint _{C}{\vec {E}}\cdot \mathrm {d} {\vec {\ell }}=-\int _{S}{\frac {\partial {\vec {B}}}{\partial t}}\cdot \mathrm {d} {\vec {A}}$ $\oint _{C}{\vec {B}}\cdot \mathrm {d} {\vec {\ell }}=\mu _{0}I+\epsilon _{0}\mu _{0}{\frac {\mathrm {d} \Phi _{E}}{\mathrm {d} t}}$ ${\frac {\partial ^{2}E_{y}}{\partial x^{2}}}=\varepsilon _{0}\mu _{0}{\frac {\partial ^{2}E_{y}}{\partial t^{2}}}$ and ${\tfrac {E_{0}}{B_{0}}}=c$ Poynting vector ${\vec {S}}={\tfrac {1}{\mu _{0}}}{\vec {E}}\times {\vec {B}}$ =energy flux

Average intensity $I=S_{ave}={\tfrac {c\varepsilon _{0}}{2}}E_{0}^{2}={\tfrac {c}{2\mu _{0}}}B_{0}^{2}={\tfrac {1}{2\mu _{0}}}E_{0}B_{0}$ Radiation pressure $p=I/c$ (perfect absorber) and $p=2I/c$ (perfect reflector).

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