20 January 2015

Dielectric sphere in a static field

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:
  1. The system has rotational symmetry about the \(z\) axis. Thus, the field has no component along \(\hat{\phi}\).
  2. The problem is antisymmetric with respect to \(z\) (changing \(E_0\) to \(-E_0\) reverses the sign of the field everywhere).
  3. The field scales by \(E_0\) (doubling the applied field doubles the field at any point in the system).
  4. 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, \(q=0\).

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.

Outside the sphere, the constant field is simply \(\mathbf{E}_0\), to fulfill the boundary conditions at infinity. As for the dipole field, from the considerations above (1) it is parallel to \(\hat{z}\) and proportional to \(E_0\) (3); furthermore, we will rescale the distance \(r\) by the sphere radius \(R\) (4):
\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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