Boundary value problems with dielectrics - III

Suppose that a dielectric sphere of radius $a$ and dielectric constant $\epsilon$ is placed in a $z$-directed electric field of strength $E_0$ (in the absence of the sphere). What is the electric field inside and around the sphere?

Since this is a static problem we can write ${\bfm E} =-\nabla \phi$. There are no free charges, so Eqs. (3.5) and (3.9) imply that

$$\nabla^2\phi = 0$$ (456)

everywhere. The boundary conditions (3.14) and (3.15) reduce to

$$\epsilon\left. \frac{\partial\phi}{\partial r}\right\vert _{r=a^{-}} = \left. \frac{\partial\phi}{\partial r}\right\vert _{r=a^{+}},$$ (457)

$$\left.\frac{\partial\phi}{\partial \theta}\right\vert _{r=a^{-}} = \left. \frac{\partial\phi}{\partial \theta}\right\vert _{r=a^{+}}.$$ (458)

Furthermore,

$$\phi(r,\theta,\varphi) \rightarrow - E_0 \,r\,\cos\theta$$ (459)

as $r\rightarrow 0$. Here, $(r,\theta, \varphi)$ are spherical polar coordinates centred on the sphere.

Let us search for an axisymmetric solution, $\phi =\phi(r,\theta)$. Since the solutions to Poisson's equation are unique, we know that if we can find such a solution which satisfies all of the boundary conditions then we can be sure that this is the correct solution. Equation (3.37) reduces to

$$\frac{1}{r}\frac{\partial^2 (r\phi)}{\partial r^2} +\frac{1}... ...(\sin\theta \,\frac{\partial\phi}{\partial\theta} \right) = 0.$$ (460)

Straightforward separation of the variables yields

$$\phi(r,\theta) = \sum_{l=0}^\infty (A_l \,r^l + B_l \,r^{-(l+1)}) P_l(\cos\theta),$$ (461)

where $l$ is a non-negative integer, the $A_l$ and $B_l$ are arbitrary constants, and $P_l(x)$ is a solution to Legendre's equation,

$$\frac{d}{dx}\!\left[ (1-x^2)\,\frac{dP_l}{dx}\right] + l(l+1)\,P_l=0,$$ (462)

which is single-valued, finite, and continuous in the interval $-1\leq x \leq +1$. It can be demonstrated that Eq. (3.42) only possesses such solutions when $l$ takes an integer value. The $P_l(x)$ are known as Legendre polynomials (since they are polynomials of order $l$ in $x$), and are specified by Rodrigues' formula

$$P_l(x) = \frac{1}{2^l l!}\frac{d^l}{dx^l}(x^2-1)^l.$$ (463)

Since Eq. (3.42) is a Sturm-Liouville type equation, and the Legendre polynomials satisfy Sturm-Liouville type boundary conditions at $x=\pm 1$, it immediately follows that the $P_l(\cos\theta)$ are orthogonal functions which form a complete set in $\theta$-space. The orthogonality relation can be written

$$\int_{-1}^1 P_{l'}(x) P_l(x)\,dx = \frac{2}{2l+1} \,\delta_{l l'}.$$ (464)

The Legendre polynomials form a complete set of angular functions, and it is easily demonstrated that the $r^l$ and the $r^{-(l+1)}$ form a complete set of radial functions. It follows that Eq. (3.41), with the $A_l$ and $B_l$ unspecified, represents a completely general axisymmetric solution of Eq. (3.37) which is well behaved in $\theta$-space. We now need to find values of the $A_l$ and $B_l$ which are consistent with the boundary conditions.

Let us divide space into the regions $r\leq a$ and $r > a$. In the former region

$$\phi(r,\theta) = \sum_{l=0}^\infty A_l\,r^l\,P_l(\cos\theta),$$ (465)

where we have rejected the $r^{-(l+1)}$ radial solutions because they diverge unphysically as $r\rightarrow 0$. In the latter region

$$\phi(r,\theta) = \sum_{l=0}^\infty (B_l\,r^l + C_l \,r^{-(l+1)}\,) P_l(\cos \theta).$$ (466)

However, it is clear from the boundary condition (3.39), and Eq. (3.43), that the only non-vanishing $B_l$ is $B_1=-E_0$. This follows since $P_1(\cos\theta) = \cos\theta$. The boundary condition (3.38b) (which integrates to give $\phi(r=a^-)= \phi(r=a^+)$ for a potential which is well behaved in $\theta$-space) gives

$$A_1 = - E_0 + \frac{C_1}{a^3},$$ (467)

and

$$A_l = \frac{C_l}{a^{2l+1}}$$ (468)

for $l\neq 1$. Note that it is appropriate to match the coefficients of the $P_l(\cos\theta)$ since these functions are orthogonal. The boundary condition (3.38a) yields

$$\epsilon A_1 = - E_0 -2\,\frac{C_1}{a^3},$$ (469)

and

$$\epsilon\, l\, A_l = - (l+1)\,\frac{C_l}{a^{2l+1}}$$ (470)

for $l\neq 1$. Equations (3.48) and (3.50) give $A_l = C_l = 0$ for $l\neq 1$. Equations (3.47) and (3.49) reduce to

$$A_1 = -\left(\frac{3}{2+\epsilon}\right) \,E_0,$$ (471)

$$C_1 = \left(\frac{\epsilon-1}{\epsilon+2}\right)\,a^3 E_0.$$ (472)

The solution is therefore

$$\phi = - \left(\frac{3}{2+\epsilon}\right)\,E_0\,r\cos\theta$$ (473)

for $r\leq a$, and

$$\phi = - E_0\, r\cos\theta + \left(\frac{\epsilon-1}{\epsilon+2}\right) \,E_0\,\frac{a^3}{r^2}\cos\theta$$ (474)

for $r > a$.

Equation (3.52) is the potential of a uniform $z$-directed electric field of strength

$$E_1 = \frac{3}{2+\epsilon}\,E_0.$$ (475)

Note that $E_1<E_0$ provided that $\epsilon >1$. Thus, the electric field strength is reduced inside the dielectric sphere due to partial shielding by polarization charges. Outside the sphere the potential is equivalent to that of the applied field $E_0$ plus the field of a point electric dipole, located at the origin and pointing in the $z$-direction, whose dipole moment is

$$p = 4\pi\epsilon_0 \left(\frac{\epsilon-1}{\epsilon +2} \right) \,a^3 E_0.$$ (476)

This dipole moment can be interpreted as the volume integral of the polarization ${\bfm P}$ over the sphere. The polarization is

$${\bfm P} = \epsilon_0 (\epsilon -1)\,E_1\,\hat{\bfm z} = 3\e... ...\left(\frac{\epsilon-1}{\epsilon+2}\right)\,E_0\,\hat{\bfm z}.$$ (477)

Since the polarization is uniform there is zero polarization charge density inside the sphere. However, there is a polarization charge sheet on the surface of the sphere whose density is given by $\sigma_{\rm pol} = {\bfm P} \!\cdot\!\hat{\bfm r}$ (see Eq. (3.29)). It follows that

$$\sigma_{\rm pol} = 3\epsilon_0 \left(\frac{\epsilon-1}{\epsilon+2}\right)\,E_0\,\cos\theta.$$ (478)

The problem of a dielectric cavity of radius $a$ in a dielectric medium with dielectric constant $\epsilon$ and with an applied electric field $E_0$ parallel to the $z$-axis can be treated in much the same manner as that of a dielectric sphere. In fact, it is easily demonstrated that the results for the cavity can be obtained from those for the sphere by making the transformation $\epsilon\rightarrow 1/\epsilon$. Thus, the field inside the cavity is uniform, parallel to the $z$-axis, and of magnitude

$$E_1 = \frac{3\epsilon}{2\epsilon + 1} \,E_0.$$ (479)

Note that $E_1 > E_0$ provided that $\epsilon >1$. The field outside the cavity is the original field plus that of a $z$-directed dipole, located at the origin, whose dipole moment is

$$p = - 4\pi\epsilon_0 \left(\frac{\epsilon -1}{2\epsilon +1}\right)a^3 E_0.$$ (480)

Here, the negative sign implies that the dipole points in the opposite direction to the external field.