A dielectric material has a real, frequency-independent relative permittivity $\epsilon_{r}$ with $\left|\epsilon_{r}-1\right| \ll 1$. In this case, the macroscopic polarization that develops when the dielectric is placed in an external electric field $\mathbf{E}_{\text {ext }}(t, \mathbf{x})$ is $\mathbf{P}(t, \mathbf{x}) \approx \epsilon_{0}\left(\epsilon_{r}-1\right) \mathbf{E}_{\text {ext }}(t, \mathbf{x})$. Explain briefly why the associated bound current density is

$$\mathbf{J}_{\text {bound }}(t, \mathbf{x}) \approx \epsilon_{0}\left(\epsilon_{r}-1\right) \frac{\partial \mathbf{E}_{\text {ext }}(t, \mathbf{x})}{\partial t}$$

[You should ignore any magnetic response of the dielectric.]

A sphere of such a dielectric, with radius $a$, is centred on $\mathbf{x}=0$. The sphere scatters an incident plane electromagnetic wave with electric field

$$\mathbf{E}(t, \mathbf{x})=\mathbf{E}_{0} e^{i(\mathbf{k} \cdot \mathbf{x}-\omega t)}$$

where $\omega=c|\mathbf{k}|$ and $\mathbf{E}_{0}$ is a constant vector. Working in the Lorenz gauge, show that at large distances $r=|\mathbf{x}|$, for which both $r \gg a$ and $k a^{2} / r \ll 2 \pi$, the magnetic vector potential $\mathbf{A}_{\text {scatt }}(t, \mathbf{x})$ of the scattered radiation is

$$\mathbf{A}_{\mathrm{scatt}}(t, \mathbf{x}) \approx-i \omega \mathbf{E}_{0} \frac{e^{i(k r-\omega t)}}{r} \frac{\left(\epsilon_{r}-1\right)}{4 \pi c^{2}} \int_{\left|\mathbf{x}^{\prime}\right| \leqslant a} e^{i \mathbf{q} \cdot \mathbf{x}^{\prime}} d^{3} \mathbf{x}^{\prime}$$

where $\mathbf{q}=\mathbf{k}-k \hat{\mathbf{x}}$ with $\hat{\mathbf{x}}=\mathbf{x} / r$.

In the far-field, where $k r \gg 1$, the electric and magnetic fields of the scattered radiation are given by

$$\begin{aligned} &\mathbf{E}_{\text {scatt }}(t, \mathbf{x}) \approx-i \omega \hat{\mathbf{x}} \times\left[\hat{\mathbf{x}} \times \mathbf{A}_{\text {scatt }}(t, \mathbf{x})\right] \\ &\mathbf{B}_{\text {scatt }}(t, \mathbf{x}) \approx i k \hat{\mathbf{x}} \times \mathbf{A}_{\text {scatt }}(t, \mathbf{x}) \end{aligned}$$

By calculating the Poynting vector of the scattered and incident radiation, show that the ratio of the time-averaged power scattered per unit solid angle to the time-averaged incident power per unit area (i.e. the differential cross-section) is

$$\frac{d \sigma}{d \Omega}=\left(\epsilon_{r}-1\right)^{2} k^{4}\left(\frac{\sin (q a)-q a \cos (q a)}{q^{3}}\right)^{2}\left|\hat{\mathbf{x}} \times \hat{\mathbf{E}}_{0}\right|^{2}$$

where $\hat{\mathbf{E}}_{0}=\mathbf{E}_{0} /\left|\mathbf{E}_{0}\right|$ and $q=|\mathbf{q}|$.

[You may assume that, in the Lorenz gauge, the retarded potential due to a localised current distribution is

$$\mathbf{A}(t, \mathbf{x})=\frac{\mu_{0}}{4 \pi} \int \frac{\mathbf{J}\left(t_{\mathrm{ret}}, \mathbf{x}^{\prime}\right)}{\left|\mathbf{x}-\mathbf{x}^{\prime}\right|} d^{3} \mathbf{x}^{\prime},$$

where the retarded time $\left.t_{\text {ret }}=t-\left|\mathbf{x}-\mathbf{x}^{\prime}\right| / c .\right]$