# Dielectric Sphere in Electric Field

### Dielectric sphere in uniform field

A classic problem that illustrates the mechanics of matching boundary conditions across an interface is the dielectric sphere of relative permittivity εr = ε / ε0 > 1.0 immersed in an otherwise constant electric field (Seen in Figure 2).

Illustration of a dielectric sphere immersed in an otherwise constant electric field. The electric field is directed along the z-axis.

Given the symmetry of the sphere about the z-axis, variations in the azimuthal angle φ can be neglected. The Laplace Equation in the remaining spherical coordinates is

$\frac{1}{r^2}\frac{\partial^2}{\partial r^2} (r^2\Phi)+\frac{1}{r\sin \theta} \frac{\partial}{\partial \theta}\left (\sin \theta\frac{\partial \Phi}{\partial \theta}\right ) = 0$

The potential can be expressed as a product

$\Phi = R\Theta \,$

where $R\,$ is a function of radial distance r alone and $\Theta\,$ is a function of elevation angle $\theta\,$ alone. Taking the derivatives and rearranging, we get

$\frac{r^2 R^{\prime\prime}}{R} + \frac{2rR^\prime}{R}= -\frac{\Theta^{\prime\prime}}{\Theta}-\frac{\Theta^\prime}{\Theta\tan\theta}$.

This expression can hold only if both sides are equal to the same constant, which for convenience, we can call $m(m+1)\,$. Hence,

$r^2\frac{d^2R}{dr^2}+2r\frac{dR}{dr}-m(m+1)R = 0$
$\frac{d^2\Theta}{d\theta^2} + \frac{1}{\tan \theta}\frac{d\Theta}{d\theta} +m(m+1)\Theta = 0$

The equation for $R\,$ has the solution

$R = C_1 r^m + C_2 r^{-m-1} \,$.

The equation for $\Theta\,$ is not so obvious. However, by using a series solution for $\Theta\,$, one can verify that the solutions are given by the Legendre polynomials

$P_m(\psi) = \frac{1}{2^m}\left ( \frac{d}{d\psi}\right )^m(\psi^2-1)^m,$

where $\psi=\cos\theta\,$. The first few are tabulated below:

$P_0 = 1\,$
$P_1 = \psi\,$
$P_2 = \frac{1}{2}\left (3\psi^2 - 1\right )\,$
$P_3 = \frac{1}{2}\left (5\psi^3 - 3\psi\right )\,$
$P_4 = \frac{1}{8}\left (35\psi^4 - 30\psi^2 + 3\right )\,$

Going back to the problem at hand, we recgnise that the constant E-field directed along the z-axis has a potential (that must represent the behaviour at infinity)

$\Phi_c = -E_0 z = -E_0 r\cos\theta\,$.

The contribution to the field due to the presence of the dielectric sphere must be computed in two parts: one for the region outside the sphere and another for the potential inside the sphere. Outside the sphere, we select the solution such that the perturbation in field vanishes as distance tends to infinity:

$\Phi = \sum_{m=0}^\infty b_m \left (\frac{a}{r}\right )^{m+1} P_m(\cos\theta) -E_0r P_1(\cos\theta)$.

Note that this solution has the correct behaviour at infinity.

Within the sphere, if the solution is to "behave" as $r\rightarrow 0$, we have

$\Phi_s = \sum_{m=0}^\infty c_m \left (\frac{r}{a}\right )^m P_m(\cos\theta)$

The key to the solution is to "match" the potential and the normal electric flux density across the air-dielectric boundary, such that

$\Phi_{\rm inside} = \Phi_{\rm outside}\,$ at $r=a\,$

and, since there are no surface charges on the sphere,

$\epsilon_r \left . \frac{\partial \Phi}{\partial r}\right \vert_{\rm inside} = \left . \frac{\partial\Phi}{\partial r}\right \vert_{\rm outside}$ at $r = a\,$.

Equating terms with the same $\theta\,$ dependence gives

$\begin{matrix} b_0 = c_0 \\ b_1 - E_0a = c_1 \\ b_2 = c_2 \\ \vdots \end{matrix}$

and

$\begin{matrix} b_0 = 0 \\ -E_0 - 2\frac{b_1}{a} = \epsilon_r \frac{c_1}{a} \\ 3b_2 = 2c_2 \\ \vdots \end{matrix}$

The zeroth coefficient is clearly zero. Coefficients with indices greater than 1 are also zero. We are left with a contribution only from the first coefficients

$b_1 = E_0 a\frac{\epsilon_r - 1}{\epsilon_r + 2}$
$c_1 = -E_0 a\frac{3}{\epsilon_r + 2}$

The total potential outside the sphere is

$\Phi = E_0 r\left (\frac{\epsilon_r - 1}{\epsilon_r + 2} \frac{a^3}{r^3} - 1\right )\cos\theta$.

Notice how the the presence of the dielectric sphere introduces a "dipole field" contribution (the term that varies as 1 / r2) to the overall field solution outside the sphere. Indeed, the dielectric sphere will look like an electric dipole as it is becomes polarised as a result of the external field.

The potential inside the sphere is found to be

$\Phi = -E_0 r\frac{3}{\epsilon_r + 2}\cos\theta$.

The figures show the equipotentials and the field lines for spheres of dielectric constant 2.2 and 10.

Equipotential contours and flux lines for a sphere of relative dielectric constant 10.
Equipotential contours and flux lines for a sphere of relative dielectric constant of 2.2. In comparison to the sphere of dielectric constant of 10, the confinement of the flux lines is significantly less.

Observe how the higher dielectric constant causes a strong concentration of flux lines within the sphere. In fact, it is easy to see the the flux achieves a maximum of

$\textbf{D} = 3\epsilon_0E_0 \hat{\textbf{z}}\,$

inside the sphere as the relative dielectric constant $\epsilon_r \rightarrow\infty\,$.