Boundary Value Problems with Dielectrics

Consider a point charge $q$ embedded in a semi-infinite dielectric medium of dielectric constant $\epsilon_1$ , and located a distance $d$ from a plane interface that separates the first medium from another semi-infinite dielectric medium of dielectric constant $\epsilon_2$ . Suppose that the interface coincides with the plane $z=0$ . We need to solve

$$\epsilon_1 \nabla \cdot {\bf E} = \frac{\rho}{\epsilon_0}$$ (512)

in the region $z>0$ ,

$$\epsilon_2\nabla \cdot {\bf E} = 0$$ (513)

in the region $z<0$ , and

$$\nabla\times{\bf E} = {\bf0}$$ (514)

everywhere, subject to the following constraints at $z=0$ :

$$\epsilon_1\, E_z(z=0_+) = \epsilon_2 \,E_z (z=0_-),$$ (515)

$$E_x (z=0_+) = E_x (z=0_-),$$ (516)

$$E_y (z=0_+) = E_y(z=0_-).$$ (517)

Figure 1: The method of images for a plane interface between two dielectric media.

In order to solve this problem, we shall employ a slightly modified form of the well-known method of images. Because $\nabla\times{\bf E} = 0$ everywhere, the electric field can be written in terms of a scalar potential: that is, ${\bf E} =-\nabla \phi$ . Consider the region $z>0$ . Let us assume that the scalar potential in this region is the same as that obtained when the whole of space is filled with dielectric of dielectric constant $\epsilon_1$ , and, in addition to the real charge $q$ at position $A$ , there is a second charge $q'$ at the image position $A'$ . (See Figure 1.) If this is the case then the potential at some point $P$ in the region $z>0$ is given by

$$\phi(z>0) = \frac{1}{4\pi\,\epsilon_0\, \epsilon_1}\left(\frac{q}{R_1} + \frac{q'}{R_2}\right),$$ (518)

where $R_1= \sqrt{r^{\,2}+(d-z)^{\,2}}$ and $R_2= \sqrt{r^{\,2}+(d+z)^{\,2}}$ . Here, $r$ , $\theta$ , $z$ are conventional cylindrical coordinates. Note that the potential (519) is clearly a solution of Equation (513) in the region $z>0$ : that is, it satisfies $\nabla \cdot {\bf E} = 0$ , with the appropriate singularity at the position of the point charge $q$ .

Consider the region $z<0$ . Let us assume that the scalar potential in this region is the same as that obtained when the whole of space is filled with a dielectric medium of dielectric constant $\epsilon_2$ , and a charge $q''$ is located at the point $A$ . If this is the case then the potential in this region is given by

$$\phi(z<0) = \frac{1}{4\pi\,\epsilon_0\,\epsilon_2} \frac{q''}{R_1}.$$ (519)

The above potential is clearly a solution of Equation (514) in the region $z<0$ : that is, it satisfies $\nabla \cdot {\bf E} = 0$ , with no singularities.

It now remains to choose $q'$ and $q''$ in such a manner that the constraints (516)-(518) are satisfied. The constraints (517) and (518) are obviously satisfied if the scalar potential is continuous across the interface between the two media: that is,

$$\phi(z=0_+) = \phi(z=0_-).$$ (520)

The constraint (516) implies a jump in the normal derivative of the scalar potential across the interface. In fact,

$$\epsilon_1 \,\frac{\partial\phi(z=0_+)}{\partial z} = \epsilon_2\, \frac{\partial \phi(z=0_-)}{\partial z}.$$ (521)

The first matching condition yields

$$\frac{q+q'}{\epsilon_1} = \frac{q''}{\epsilon_2},$$ (522)

whereas the second gives

$$q-q' = q''.$$ (523)

Here, use has been made of

$$\frac{\partial}{\partial z}\!\left(\frac{1}{R_1}\right)_{z=0} =- ... ...rtial z}\!\left(\frac{1}{R_2}\right)_{z=0} = \frac{d}{(r^{\,2}+d^{\,2})^{3/2}}.$$ (524)

Equations (523) and (524) imply that

$$q' = -\left(\frac{\epsilon_2-\epsilon_1}{\epsilon_2 + \epsilon_1} \right) q,$$ (525)

$$q''&= \left(\frac{2\,\epsilon_2}{\epsilon_2+\epsilon_1}\right) q.$$ (526)

The polarization charge density is given by $\rho_b = - \nabla \cdot {\bf P}$ , However, ${\bf P}=\epsilon_0\,\chi_e \,{\bf E}$ inside either dielectric, which implies that

$$\nabla \cdot{\bf P} = \epsilon_0\,\chi_e\, \nabla \cdot {\bf E} = 0,$$ (527)

except at the point charge $q$ . Thus, there is zero bound charge density in either dielectric medium. At the interface, $\chi_e$ jumps discontinuously,

$${\mit\Delta}\chi_e = \epsilon_1-\epsilon_2.$$ (528)

This implies that there is a bound charge sheet on the interface between the two dielectric media. In fact, it follows from Equation (498) that

$$\sigma_b = - ({\bf P}_2-{\bf P}_1) \cdot {\bf n}_{21},$$ (529)

where ${\bf n}_{21}$ is a unit normal to the interface pointing from medium 1 to medium 2 (i.e., along the positive $z$ -axis). Because

$${\bf P}_i = \epsilon_0\, (\epsilon_i -1)\,{\bf E} = - \epsilon_0\, ({\epsilon}_i - 1)\,\nabla\phi$$ (530)

in either medium, it is easily demonstrated that

$$\sigma_b(r) = -\frac{q}{2\pi\,\epsilon_1} \left(\frac{\epsilon_2-\epsilon_1} {\epsilon_2+\epsilon_1}\right) \frac{d}{(r^{\,2}+d^{\,2})^{3/2}}.$$ (531)

In the limit $\epsilon_2\gg \epsilon_1$ , the dielectric $\epsilon_2$ behaves like a conducting medium (i.e., ${\bf E}\rightarrow {\bf0}$ in the region $z<0$ ), and the bound surface charge density on the interface approaches that obtained in the case when the plane $z=0$ coincides with a conducting surface.

The above argument can easily be generalized to deal with problems involving multiple point charges in the presence of multiple dielectric media whose interfaces form parallel planes.

Consider a second boundary value problem in which a slab of dielectric, of dielectric constant $\epsilon$ , lies between the planes $z=0$ and $z=b$ . Suppose that this slab is placed in a uniform $z$ -directed electric field of strength $E_0$ . Let us calculate the field-strength $E_1$ inside the slab.

Because there are no free charges, and this is essentially a one-dimensional problem, it is clear from Equation (501) that the electric displacement $D$ is the same in both the dielectric slab and the surrounding vacuum. In the vacuum region, $D= \epsilon_0\, E_0$ , whereas $D= \epsilon_0\,\epsilon\, E_1$ in the dielectric. It follows that

$$E_1=\frac{E_0}{\epsilon}.$$ (532)

In other words, the electric field inside the slab is reduced by polarization charges. As before, there is zero polarization charge density inside the dielectric. However, there is a uniform bound charge sheet on both surfaces of the slab. It is easily demonstrated that

$$\sigma_b(z=b)=-\sigma_b (z=0) = \epsilon_0 \left(\frac{\epsilon-1}{\epsilon}\right)E_0.$$ (533)

In the limit $\epsilon\gg 1$ , the slab acts like a conductor, and $E_1\rightarrow 0$ .

Let us now generalize this result. Consider a dielectric medium whose dielectric constant $\epsilon$ varies with $z$ . The medium is assumed to be of finite extent, and to be surrounded by a vacuum. It follows that $\epsilon(z)\rightarrow 1$ as $\vert z\vert\rightarrow \infty$ . Suppose that this dielectric is placed in a uniform $z$ -directed electric field of strength $E_0$ . What is the field $E(z)$ inside the dielectric?

We know that the electric displacement inside the dielectric is given by $D(z) = \epsilon_0\, \epsilon(z)\,E(z)$ . We also know from Equation (501) that, because there are no free charges, and this is essentially a one-dimensional problem,

$$\frac{d D(z)}{dz} = \epsilon_0\, \frac{d [\epsilon(z)\, E(z)]}{d z} = 0.$$ (534)

Furthermore, $E(z)\rightarrow E_0$ as $\vert z\vert\rightarrow \infty$ . It follows that

$$E(z) = \frac{E_0}{\epsilon(z)}.$$ (535)

Thus, the electric field is inversely proportional to the dielectric constant of the medium. The bound charge density within the medium is given by

$$\rho_b = \epsilon_0 \,\frac{d E(z)}{d z} = \epsilon_0\, E_0\frac{d}{dz} \!\left[\frac{1}{\epsilon(z)}\right].$$ (536)

Consider a third, and final, boundary value problem in which 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). Let us calculate the electric field inside and around the sphere.

Because this is a static problem, we can write ${\bf E} =-\nabla \phi$ . There are no free charges, so Equations (501) and (505) imply that

$$\nabla^{\,2}\phi = 0$$ (537)

everywhere. The matching conditions (510) and (512) reduce to

$$\epsilon\left. \frac{\partial\phi}{\partial r}\right\vert _{r=a_{-}} = \left. \frac{\partial\phi}{\partial r}\right\vert _{r=a_{+}},$$ (538)

$$\left.\frac{\partial\phi}{\partial \theta}\right\vert _{r=a_{-}} = \left. \frac{\partial\phi}{\partial \theta}\right\vert _{r=a_{+}}.$$ (539)

Furthermore,

$$\phi(r,\theta,\varphi) \rightarrow - E_0 \,r\,\cos\theta$$ (540)

as $r\rightarrow 0$ : that is, the electric field asymptotes to uniform $z$ -directed field of strength $E_0$ far from the sphere. Here, $r$ , $\theta$ , $\varphi)$ are spherical coordinates centered on the sphere.

Let us search for an axisymmetric solution, $\phi =\phi(r,\theta)$ . Because the solutions to Poisson's equation are unique, we know that if we can find such a solution that satisfies all of the boundary conditions then we can be sure that this is the correct solution. Equation (538) reduces to

$$\frac{1}{r}\frac{\partial^{\,2} (r\,\phi)}{\partial r^{\,2}} +\fr... ...ial\theta}\!\left(\sin\theta \,\frac{\partial\phi}{\partial\theta} \right) = 0.$$ (541)

Straightforward separation of the variables yields (see Section 3.7)

$$\phi(r,\theta) = \sum_{l=0,\infty}\left[A_l \,r^{\,l} + B_l \,r^{\,-(l+1)}\right] P_l(\cos\theta),$$ (542)

where $l$ is a non-negative integer, the $A_l$ and $B_l$ are arbitrary constants, and the $P_l(x)$ are Legendre polynomials. (See Section 3.2.)

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 Equation (543), with the $A_l$ and $B_l$ unspecified, represents a completely general (single-valued) axisymmetric solution to Equation (538). It remains to determine the values of the $A_l$ and $B_l$ that are consistent with the boundary conditions.

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

$$\phi(r,\theta) = \sum_{l=0,\infty} A_l\,r^{\,l}\,P_l(\cos\theta),$$ (543)

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

$$\phi(r,\theta) = \sum_{l=0,\infty}\left[B_l\,r^{\,l} + C_l \,r^{\,-(l+1)}\right]P_l(\cos \theta).$$ (544)

However, it is clear from the boundary condition (541) that the only non-vanishing $B_l$ is $B_1=-E_0$ . This follows because $P_1(\cos\theta) = \cos\theta$ . The boundary condition (540) [which can be integrated to give $\phi(r=a_-)= \phi(r=a_+)$ for a potential that is single-valued in $\theta$ ] gives

$$A_1 = - E_0 + \frac{C_1}{a^{\,3}},$$ (545)

and

$$A_l = \frac{C_l}{a^{\,2\,l+1}}$$ (546)

for $l\neq 1$ . Note that it is appropriate to match the coefficients of the $P_l(\cos\theta)$ because these functions are mutually orthogonal. (See Section 3.2.) The boundary condition (539) yields

$$\epsilon \,A_1 = - E_0 -2\,\frac{C_1}{a^{\,3}},$$ (547)

and

$$\epsilon\, l\, A_l = - (l+1)\,\frac{C_l}{a^{\,2\,l+1}}$$ (548)

for $l\neq 1$ . Equations (547) and (549) give $A_l = C_l = 0$ for $l\neq 1$ . Equations (546) and (548) reduce to

$$A_1 = -\left(\frac{3}{2+\epsilon}\right) \,E_0,$$ (549)

$$C_1 = \left(\frac{\epsilon-1}{\epsilon+2}\right)\,a^{\,3}\, E_0.$$ (550)

The solution to the problem is therefore

$$\phi(r,\theta) = - \left(\frac{3}{2+\epsilon}\right)\,E_0\,r\,\cos\theta$$ (551)

for $r\leq a$ , and

$$\phi(r,\theta) = - E_0\, r\,\cos\theta + \left(\frac{\epsilon-1}{\epsilon+2}\right) \,E_0\,\frac{a^{\,3}}{r^{\,2}}\cos\theta$$ (552)

for $r>a$ .

Equation (552) is the potential of a uniform $z$ -directed electric field of strength

$$E_1 = \frac{3}{2+\epsilon}\,E_0.$$ (553)

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 an electric dipole (see Section 2.7), located at the origin, and directed along the $z$ -axis, whose dipole moment is

$$p = 4\pi\,\epsilon_0 \left(\frac{\epsilon-1}{\epsilon +2} \right) \,a^{\,3}\, E_0.$$ (554)

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

$${\bf P} = \epsilon_0 \,(\epsilon -1)\,E_1\,{\bf e}_z = 3\,\epsilon_0 \left(\frac{\epsilon-1}{\epsilon+2}\right)\,E_0\,{\bf e}_z.$$ (555)

Because the polarization is uniform there is zero bound charge density inside the sphere. However, there is a bound charge sheet on the surface of the sphere, whose density is given by $\sigma_b={\bf P} \cdot {\bf e}_r$ [see Equation (530)]. It follows that

$$\sigma_b(\theta) = 3\,\epsilon_0 \left(\frac{\epsilon-1}{\epsilon+2}\right)\,E_0\,\cos\theta.$$ (556)

The problem of a dielectric cavity of radius $a$ inside a dielectric medium of dielectric constant $\epsilon$ , and in the presence of 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

$$E_1 = \frac{3\,\epsilon}{2\,\epsilon + 1} \,E_0.$$ (557)

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

$$p = - 4\pi\,\epsilon_0 \left(\frac{\epsilon -1}{2\,\epsilon +1}\right)a^{\,3}\, E_0.$$ (558)

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