= Solution
In vacuum, <Maxwell's equations> in SI units are
$$
\nabla\cdot\mathbf E=\rho_q/\epsilon_0,\qquad
\nabla\cdot\mathbf B=0,\qquad
\nabla\times\mathbf E=-\partial_t\mathbf B,\qquad
\nabla\times\mathbf B=\mu_0\mathbf J+\mu_0\epsilon_0\partial_t\mathbf E.
$$
The <displacement current> term makes the last equation consistent with the <charge continuity equation>. In a source-free region, take the curl of <Faraday's law> and use $\nabla\times(\nabla\times\mathbf E)=\nabla(\nabla\cdot\mathbf E)-\nabla^2\mathbf E$. This gives
$$
\nabla^2\mathbf E=\mu_0\epsilon_0\partial_t^2\mathbf E,
\qquad \nabla^2\mathbf B=\mu_0\epsilon_0\partial_t^2\mathbf B.
$$
A <plane electromagnetic wave> therefore propagates at \b[$c=1/\sqrt{\mu_0\epsilon_0}$]. The divergence equations make both fields transverse; the curl equations give
$$
\mathbf B=\frac1c\widehat{\mathbf k}\times\mathbf E.
$$
The <electric fields> and <magnetic fields> are in <phase> and perpendicular to one another and to the propagation direction. Their <energy densities> are equal:
$$
u=\frac12\epsilon_0E^2+\frac{B^2}{2\mu_0}=\epsilon_0E^2,\qquad
\mathbf S=\frac1{\mu_0}\mathbf E\times\mathbf B=uc\widehat{\mathbf k}.
$$
For electric-field peak <amplitude> $E_a$, the time-averaged <intensity> is $\langle S\rangle=\epsilon_0cE_a^2/2$. \b[Mutually coupled changing <electric fields> and <magnetic fields> transport <energy> through vacuum], even where no charges or conduction currents are present.
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