## 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:

\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}

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):

\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}

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

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

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

\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}

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{split}
(1-A) &= B\\
\epsilon_e (1+2A) &= \epsilon_i B
\end{split}
\label{eq:BCcoeff}

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{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}

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.