$$ \\ \\ $$ $$ \text{Take the curl of Faraday's Law and substitute in }\text{Amp}\grave{\mathrm{e}}\text{re-Maxwell Law:} $$ \[ \begin{aligned} \nabla \times (\nabla \times \mathbf{E}) &= \nabla \times \left( -\frac{\partial \mathbf{B}}{\partial t} \right) \\ &= -\frac{\partial}{\partial t} (\nabla \times \mathbf{B}) \\ &= - \mu_0 \epsilon_0 \frac{{\partial}^2 \mathbf{E}}{{\partial t}^2} \\ \end{aligned} \] $$ \\ \\ $$ $$ \text{Use the vector calculus identity: } \nabla \times (\nabla \times \mathbf{F}) = \nabla (\nabla \cdot \mathbf{F}) - {\nabla}^2 \mathbf{F} $$ $$ \text{and substitute in Gauss's Law for Electric Fields:} $$ \[ \begin{aligned} \nabla \times (\nabla \times \mathbf{E}) &= \nabla (\nabla \cdot \mathbf{E}) - {\nabla}^2 \mathbf{E} \\ &= - {\nabla}^2 \mathbf{E} \\ \end{aligned} \] $$ \text{Now we have the ...} $$ $$ \\ \\ $$ $$ {\large \text{Electromagnetic Wave Equation}} $$ \[ \begin{aligned} {\nabla}^2 \mathbf{E} &= \mu_0 \epsilon_0 \frac{{\partial}^2 \mathbf{E}}{{\partial t}^2} \\ {\nabla}^2 \mathbf{B} &= \mu_0 \epsilon_0 \frac{{\partial}^2 \mathbf{B}}{{\partial t}^2} \\ \end{aligned} \\ \] $$ {\small \text{(Derivation of magnetic field wave equation parallels the steps above.)}} $$ $$ \\ \\ $$ \[ \begin{array}{ll} \text{General Wave Equation Form: }\ c^2 {\nabla}^2 u = \frac{{\partial}^2 u}{{\partial t}^2} \\ \text{where c in the electromagnetic wave equation is:} \\ \end{array} \] \[ \begin{aligned} c^2 &= \frac{1}{\mu_0 \epsilon_0} \\ c &= \frac{1}{\sqrt{\mu_0 \epsilon_0}} \\ c &= 2.99792 \times 10^8\ \text{m/s} \\ \end{aligned} \]