= Solution
In charge- and current-free vacuum, $\nabla\cdot\mathbf E=\nabla\cdot\mathbf B=0$. Taking the curl of <Faraday's law> and using the <Ampère-Maxwell equation> gives
$$
\nabla\times(\nabla\times\mathbf E)=-\mu_0\epsilon_0\frac{\partial^2\mathbf E}{\partial t^2}.
$$
Since $\nabla\times(\nabla\times\mathbf E)=\nabla(\nabla\cdot\mathbf E)-\nabla^2\mathbf E$, it follows that
$$
\boxed{\nabla^2\mathbf E=\mu_0\epsilon_0\frac{\partial^2\mathbf E}{\partial t^2},\qquad\nabla^2\mathbf B=\mu_0\epsilon_0\frac{\partial^2\mathbf B}{\partial t^2}.}
$$
Substituting a plane wave with phase $\mathbf k\cdot\mathbf r-\omega t$ gives $k^2=\mu_0\epsilon_0\omega^2$ and therefore
$$
\boxed{c=\frac\omega k=\frac1{\sqrt{\mu_0\epsilon_0}}\simeq3.00\times10^8\,\mathrm{m\,s^{-1}}.}
$$
The divergence equations make the waves transverse, and the curl equations fix $\mathbf B=\hat{\mathbf k}\times\mathbf E/c$.
Back to article page
