Source: Cambridge Electromagnetism, Chapter 8 exercises, Question 1; accompanies David Tong’s Electromagnetism .
A dielectric sphere of radius $a$ and permittivity $\varepsilon$ is
placed at rest at the origin in a constant electric field which takes a uniform
value $\mathbf{E}_0$ far from the origin. Show that Maxwell's equations with
the appropriate boundary conditions are solved by an electric field which is
uniform inside the sphere, with value:
\[\mathbf{E}_{\mathrm{int}} = \frac{3}{2+\varepsilon_r}\mathbf{E}_0,\tag{1.1}\]
where $\varepsilon_r = \varepsilon/\varepsilon_0$, and a field outside the
sphere which is the superposition of $\mathbf{E}_0$ and the field
\[\mathbf{E}(\mathbf{x}) = \frac{1}{4\pi\varepsilon_0}
\left(\frac{3(\mathbf{p}\cdot\hat{\mathbf{x}})\hat{\mathbf{x}}-\mathbf{p}}{|\mathbf{x}|^3}\right)\tag{1.2}\]
of an electric dipole at the origin, with dipole moment
\[\mathbf{p} = 4\pi\varepsilon_0\!\left(\frac{\varepsilon_r-1}{\varepsilon_r+2}\right)a^3\mathbf{E}_0.\tag{1.3}\]
By considering the macroscopic polarisation $\mathbf{P}$ inside the sphere,
calculate the bound surface charge density $\sigma$. Verify explicitly that
the electric dipole moment of $\sigma$ is $\mathbf{p}$.
Inside and outside the dielectric, the potential satisfies the Laplace equation,
\[\nabla^2\phi = 0.\tag{1.4}\]
Since the problem is invariant under rotation about $\mathbf{E}_0$, which we will take to be in the $z$-direction, the solution should have no $\phi$ dependence. Then, the general solution for the Laplace equation is
\[\phi(r,\theta) = \sum_{l=0}^\infty (A_lr^{l} + B_lr^{-(l+1)})P_l(\cos\theta).\tag{1.5}\]
Inside the dielectric, we require regularity at the origin; we set $B_l =0$. Outside the dielectric, since there is a constant electric field at $\mathbf{E}_0 = E_0\cos\theta\hat{\mathbf{z}}$ $r\to\infty $, we take $A_l = -E_0\delta_{l1}$. So the general solution we are looking at is
\[\phi(r,\theta) = \begin{cases}
\displaystyle\sum_{l=0}^\infty A_lr^lP_l(\cos\theta),&r<a,\\
\\\displaystyle\sum_{l=0}^\infty D_lr^{-(l+1)}P_l(\cos\theta) -E_0r\cos\theta,&r>a.
\end{cases}\tag{1.6}\]
The boundary condition at $r=a$ is given by
\[\hat{\mathbf{r}}\times(\mathbf{E}(a\hat{\mathbf{r}}^+)-\mathbf{E}(a\hat{\mathbf{r}}^-)) = 0,\quad \text{and} \quad \hat{\mathbf{r}}\cdot(\varepsilon_0\mathbf{E}(a\hat{\mathbf{r}}^+)-\varepsilon\mathbf{E}(a\hat{\mathbf{r}}^-)) = 0,\tag{1.7}\]
which implies
\[\partial_\theta\phi(a^+,\theta) = \partial_\theta\phi(a^-,\theta), \quad \text{and} \quad \varepsilon_0\partial_r\phi(a^+,\theta) = \varepsilon\partial_r\phi(a^-,\theta).\tag{1.8}\]
The first BC gives
\[A_la^l = D_la^{-(l+1)}-E_0a\delta_{l1},\tag{1.9}\]
and the second BC gives
\[\varepsilon_0lA_la^{l-1} =-\varepsilon(l+1)D_la^{-(l+2)}-\varepsilon E_0\delta_{l1}.\tag{1.10}\]
Solving these two, we find
\[A_l = -\frac{3}{2+\varepsilon_r}E_0 \delta_{l1}\quad\text{and} \quad D_1 = -\frac{1-\varepsilon_r}{2+\varepsilon_r}a^3E_0\delta_{l1}.\tag{1.11}\]
Hence, the full potential is
\[\phi(r,\theta) = \begin{cases}
-\dfrac{3}{2+\varepsilon_r}E_0r\cos\theta, & r<a,\\
-\dfrac{1-\varepsilon_r}{2+\varepsilon_r}\dfrac{a^3}{r^2}E_0\cos\theta-E_0r\cos\theta, & r>a.
\end{cases}\tag{1.12}\]
Therefore, the electric field inside the dielectric is
\[\mathbf{E}_{\text{int}} = \frac{3}{2+\varepsilon_r}E_0(\cos\theta\,\hat{\mathbf{r}} - \sin\theta\,\hat{\mathbf{\theta}}) = \frac{3}{2+\varepsilon_r}E_0\hat{\mathbf{x}}.\tag{1.13}\]
The electric field outside is
\[\mathbf{E}(\mathbf{x}) +\mathbf{E}_0 =\frac{1-\varepsilon_r}{2+\varepsilon_r}a^3\frac{(\mathbf{E}_0\cdot\hat{\mathbf{r}})\mathbf{r}-\mathbf{E}_0}{r^3} + \mathbf{E}_0 =\frac{1}{4\pi\varepsilon_0}
\left(\frac{3(\mathbf{p}\cdot\hat{\mathbf{x}})\hat{\mathbf{x}}-\mathbf{p}}{|\mathbf{x}|^3}\right) + \mathbf{E}_0\tag{1.14}\]
with
\[\mathbf{p} = 4\pi\varepsilon_0\!\left(\frac{\varepsilon_r-1}{\varepsilon_r+2}\right)a^3\mathbf{E}_0.\tag{1.15}\]
A sketch of the electric field is shown in Fig. 1.
Figure 1. Electric field lines for a dielectric sphere ($\varepsilon_r=3$, $a=1$) in a uniform applied field $\mathbf{E}_0$. Inside the field is uniform and weaker, $\mathbf{E}_{\mathrm{int}}=\frac{3}{5}\mathbf{E}_0$; outside it is the superposition of $\mathbf{E}_0$ and an electric dipole. Lines refract at the boundary satisfying continuity of $E_\parallel$ and $D_\perp$.