Past exam of the physics 2 paper of the University of Oxford / 2014 / 10

Maxwell's equations in matter, with free charge and current densities, areFor vacuum, $\mathbf{D}={ϵ}_0\mathbf{E}$ and $\mathbf{H}=\mathbf{B}/{\mu }_0$, so the last equation is $\nabla \times \mathbf{B}={\mu }_0\mathbf{J}+{\mu }_0{ϵ}_0\partial \mathbf{E}/\partial t$.

For a charging parallel-plate capacitor of area $A$, neglect fringing. The field in the gap is $E=Q/({ϵ}_{0}A)$, so the displacement current through the gap isAn Amperian loop around the feed wire can bound a surface cutting the wire, which has conduction current $I$, or a bulging surface passing through the dielectric gap, which has no conduction current. Including displacement current makes both surfaces give the same magnetic circulation. It is also required by charge conservation: the divergence of the Ampère-Maxwell equation and Gauss's law yields $\nabla \cdot \mathbf{J}+\partial \rho /\partial t=0$, the continuity equation.

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 givesSince $\nabla \times (\nabla \times \mathbf{E})=\nabla (\nabla \cdot \mathbf{E})-{\nabla }^2\mathbf{E}$, it follows thatSubstituting a plane wave with phase $\mathbf{k}\cdot \mathbf{r}-\omega t$ gives $k^2={\mu }_0{ϵ}_0{\omega }^2$ and thereforeThe divergence equations make the waves transverse, and the curl equations fix $\mathbf{B}=\hat{\mathbf{k}}\times \mathbf{E}/c$.

The phase $kz-\omega t$ describes propagation along $+z$. With $\psi =kz-\omega t$, the vacuum plane electromagnetic wave hasThus $B_x=-3A\sin \psi /c$, $B_y=2A\sin \psi /c$, and $B_z=0$. The field amplitudes are $E_0=\sqrt{13}|A|$ and $B_0=\sqrt{13}|A|/c$, givingThe instantaneous magnitude ratio also equals $c$ wherever the fields are nonzero; at a node both vanish together. Their Poynting vector points along $+z$, checking the relative component signs.