I'll try to give the simplest solution I can think of to the classical problem of a dielectric sphere in a constant external field, see for instance Landau & Lifshitz, vol. 8, chapter II, §8.

The sphere radius is \(R\) and its dielectric constant is \(\epsilon_i\), while that of the surrounding medium is \(\epsilon_e\). The electric field at infinity is along the \(z\) axis: \(\mathbf{E}_0 = E_0\, \hat{z}\). Considering the symmetry of the object we will work in spherical coordinates \((r,\theta,\phi)\). Before writing down any equations, let us note the following points:

The sphere radius is \(R\) and its dielectric constant is \(\epsilon_i\), while that of the surrounding medium is \(\epsilon_e\). The electric field at infinity is along the \(z\) axis: \(\mathbf{E}_0 = E_0\, \hat{z}\). Considering the symmetry of the object we will work in spherical coordinates \((r,\theta,\phi)\). Before writing down any equations, let us note the following points:

- The system has rotational symmetry about the \(z\) axis. Thus, the field has no component along \(\hat{\phi}\).
- The problem is antisymmetric with respect to \(z\) (changing \(E_0\) to \(-E_0\) reverses the sign of the field everywhere).
- The field scales by \(E_0\) (doubling the applied field doubles the field at any point in the system).
- The only length parameter is the sphere radius \(R\).

The last two points express scaling relations, while the first two relate to the symmetry properties of the "complete" system: sphere + applied field and we can discuss them in terms of the multipole expansion of the potential created by a charge distribution:

\begin{equation}

\Phi(r, \theta, \phi) = \frac{1}{4 \pi \epsilon_0} \sum_{\ell = 0}^{\infty} \frac{1}{r^{\ell + 1}} \sum_{m = - \ell}^{m = \ell} Q_{\ell m} Y_{\ell m}(\theta, \phi)

\label{eq:finalpot}

\end{equation}

Before doing that, we should notice that a constant field along \(z\) is necessary for describing the potential outside the sphere (and is a trivial solution to Laplace's equation). However, such a field is not accounted for in the expansion, since it does not decay at infinity, and we must consider it separately. Aside from that, condition 1 amounts to forbidding any azimuthally modulated contribution, so that only terms with \(m=0\) should be preserved. Condition 2 further limits the sum over \(\Phi_{\ell}\) to odd values of \(\ell\). For instance, a net charge \(q\) at the origin (corresponding to \(\ell = 0\)) is symmetric, so that it cannot be engendered by the source field \(\mathbf{E}_0\); thus, \(q=0\).

\begin{equation}

\Phi(r, \theta, \phi) = \frac{1}{4 \pi \epsilon_0} \sum_{\ell = 0}^{\infty} \frac{1}{r^{\ell + 1}} \sum_{m = - \ell}^{m = \ell} Q_{\ell m} Y_{\ell m}(\theta, \phi)

\label{eq:finalpot}

\end{equation}

Before doing that, we should notice that a constant field along \(z\) is necessary for describing the potential outside the sphere (and is a trivial solution to Laplace's equation). However, such a field is not accounted for in the expansion, since it does not decay at infinity, and we must consider it separately. Aside from that, condition 1 amounts to forbidding any azimuthally modulated contribution, so that only terms with \(m=0\) should be preserved. Condition 2 further limits the sum over \(\Phi_{\ell}\) to odd values of \(\ell\). For instance, a net charge \(q\) at the origin (corresponding to \(\ell = 0\)) is symmetric, so that it cannot be engendered by the source field \(\mathbf{E}_0\); thus, \(q=0\).

Instead of the potential \(\Phi\), we will work directly with the field \(\mathbf{E}\), written as the sum of the constant field along \(z\) and a dipole field (\(\ell = 1\)) along the same direction. We will show that those terms are enough to fulfill the boundary conditions at the surface of the sphere; consequently, they describe the solution completely and terms of higher order than the dipole (\(\ell = 3, 5, \ldots\)) are not necessary.

\begin{equation}

\mathbf{E}_e(r,\theta) = \mathbf{E}_0 + A E_0 \left ( \frac{R}{r} \right )^3 \left [ 3 \cos (\theta) \hat{r} - \hat{z} \right ]

\label{eq:Eext}

\end{equation}

Inside the sphere the second term vanishes (otherwise, it would diverge at the origin), so we are left with a constant field:

\begin{equation}

\mathbf{E}_i(r,\theta) = B \, \mathbf{E}_0 = B E_0\, \hat{z}

\label{eq:Eint}

\end{equation}

The constants \(A\) and \(B\) must be determined from the boundary conditions at the sphere surface:

\begin{equation}

\begin{split}

{E}_i^{\|}(R, \theta) &= {E}_e^{\|}(R, \theta) \qquad \mbox{and}\\

\epsilon_i{E}_i^{\bot}(R, \theta) &= \epsilon_e{E}_e^{\bot}(R, \theta)

\end{split}

\label{eq:BCsphere}

\end{equation}

where subscripts \(\|\) and \(\bot\) denote the field components parallel and normal to the interface, respectively. Since the field has no component along \(\hat{\phi}\) and the normal to the surface of the sphere is always \(\hat{r}\), \(\mathbf{E} = E^{\bot} \hat{r} + E^{\|} \hat{\theta}\). We therefore plug equations \eqref{eq:Eext} and \eqref{eq:Eint} in \eqref{eq:BCsphere}, using the identity \(\hat{z} = \hat{r} \cos (\theta) - \hat{\theta} \sin (\theta)\). We are left with the system of linear equations:

\begin{equation}\begin{split}

(1-A) &= B\\

\epsilon_e (1+2A) &= \epsilon_i B

\end{split}

\label{eq:BCcoeff}

\end{equation}

which yields \(\displaystyle A = \frac{\epsilon_r - 1}{\epsilon_r + 2}\) and \(\displaystyle B = \frac{3}{\epsilon_r + 2}\), with \(\epsilon_r = \epsilon_i / \epsilon_e\) the relative dielectric constant of the sphere with respect to the surrounding medium. Finally:

\begin{equation}

\begin{split}

\mathbf{E}_e(r,\theta) &= \mathbf{E}_0 + E_0 \frac{\epsilon_r - 1}{\epsilon_r + 2} \left ( \frac{R}{r} \right )^3 \left [ 3 \cos (\theta) \hat{r} - \hat{z} \right ]\\

\mathbf{E}_i(r,\theta) &= \frac{3}{\epsilon_r + 2} \mathbf{E}_0

\end{split}

\label{eq:FinalE}

\end{equation}

We see that the field does not depend on \(\epsilon_i \) and \(\epsilon_e\) separately, but only on their ratio \(\epsilon_r\); furthermore, for \(\epsilon_r = 1\) (\(\epsilon_i = \epsilon_e\)) the field \(\mathbf{E}(r,\theta) = \mathbf{E}_0\) everywhere; the sphere is "transparent" and induces no perturbation to the applied field.

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