Boundary value problems with dielectrics

Consider a point charge $q$ embedded in a semi-infinite dielectric $\epsilon_1$ a distance $d$ away from a plane interface which separates the first medium from another semi-infinite dielectric $\epsilon_2$. The interface is assumed to coincide with the plane $z=0$. We need to find solutions to the equations

$$\epsilon_1 \nabla\!\cdot\!{\bf E} = \frac{\rho}{\epsilon_0}$$ (818)

for $z>0$,

$$\epsilon_2 \nabla\!\cdot\!{\bf E} = 0$$ (819)

for $z<0$, and

$$\nabla\times {\bf E} = {\bf0}$$ (820)

everywhere, subject to the boundary conditions at $z=0$ that

$$\epsilon_1 E_z(z=0^+) = \epsilon_2 E_z (z=0^-),$$ (821)

$$E_x (z=0^+) = E_x (z=0^-),$$ (822)

$$E_y (z=0^+) = E_y(z=0^-).$$ (823)

Figure 47:

In order to solve this problem, we shall employ a slightly modified form of the well-known method of images. Since $\nabla\times{\bf E} = {\bf0}$ everywhere, the electric field can be written in terms of a scalar potential. So, ${\bf E} = - \nabla\phi$. 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 the dielectric $\epsilon_1$, and, in addition to the real charge $q$ at position $A$, there is a second charge $q'$ at the image position $A'$ (see Fig. 47). If this is the case, then the potential at some point $P$ in the region $z>0$ is given by

$$\phi(z>0) = \frac{1}{4\pi \epsilon_0 \epsilon_1}\left(\frac{q}{R_1} + \frac{q'}{R_2}\right),$$ (824)

where $R_1= \sqrt{r^2+(d-z)^2}$ and $R_2= \sqrt{r^2+(d+z)^2}$, when written in terms of cylindrical polar coordinates, $(r, \theta, z)$, aligned along the $z$-axis. Note that the potential (824) is clearly a solution of Eq. (818) in the region $z>0$. It gives $\nabla\cdot {\bf E}=0$, 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 the dielectric $\epsilon_2$, and a charge $q''$ is located at the point $A$. If this is the case, then the potential in this region is given by

$$\phi(z<0) = \frac{1}{4\pi \epsilon_0 \epsilon_2} \frac{q''}{R_1}.$$ (825)

The above potential is clearly a solution of Eq. (819) in the region $z<0$. It gives $\nabla\!\cdot\!{\bf E} = 0$, with no singularities.

It now remains to choose $q'$ and $q''$ in such a manner that the boundary conditions (821)-(823) are satisfied. The boundary conditions (822) and (823) are obviously satisfied if the scalar potential is continuous at the interface between the two dielectric media:

$$\phi(z=0^+) = \phi(z=0^-).$$ (826)

The boundary condition (821) implies a jump in the normal derivative of the scalar potential across the interface:

$$\epsilon_1 \frac{\partial\phi(z=0^+)}{\partial z} = \epsilon_2 \frac{\partial \phi(z=0^-)}{\partial z}.$$ (827)

The first matching condition yields

$$\frac{q+q'}{\epsilon_1} = \frac{q''}{\epsilon_2},$$ (828)

whereas the second gives

$$q-q' = q''.$$ (829)

Here, use has been made of

$$\frac{\partial}{\partial z}\!\left(\frac{1}{R_1}\right)_{z=0... ...!\left(\frac{1}{R_2}\right)_{z=0} = \frac{d}{(r^2+d^2)^{3/2}}.$$ (830)

Equations (828) and (829) imply that

$$q' = -\left(\frac{\epsilon_2-\epsilon_1}{\epsilon_2 + \epsilon_1} \right) q,$$ (831)

$$q''&= \left(\frac{2 \epsilon_2}{\epsilon_2+\epsilon_1}\right) q.$$ (832)

The bound charge density is given by $\rho_b = - \nabla\!\cdot\! {\bf P}$, However, inside either dielectric, ${\bf P}=\epsilon_0 \chi_e {\bf E}$, so $\nabla \cdot {\bf P} = \epsilon_0 \chi_e \nabla {\cdot} {\bf E} = 0$, except at the point charge $q$. Thus, there is zero bound charge density in either dielectric medium. However, there is a bound charge sheet on the interface between the two dielectric media. In fact, the density of this sheet is given by

$$\sigma_b = \epsilon_0 (E_{z 1}-E_{z 2})_{z=0}.$$ (833)

Hence,

$$\sigma_b = \epsilon_0 \frac{\partial\phi(z=0-)}{\partial ... ...{\epsilon_1(\epsilon_2+\epsilon_1)} \frac{d}{(r^2+d^2)^{3/2}}.$$ (834)

In the limit $\epsilon_2\gg \epsilon_1$, the dielectric $\epsilon_2$ behaves like a conducting medium (i.e., ${\bf E}\rightarrow 0$ in the region $z<0$), and the bound surface charge density on the interface approaches that obtained in the case where the plane $z=0$ coincides with a conducting surface (see Sect. 5.10).

As a second example, consider a dielectric sphere of radius $a$, and uniform dielectric constant $\epsilon$, placed in a uniform $z$-directed electric field of magnitude $E_0$. Suppose that the sphere is centered on the origin. Now, for an electrostatic problem, we can always write ${\bf E} = - \nabla\phi$. In the present problem, $\nabla\cdot{\bf E}={\bf0}$ both inside and outside the sphere, since there are no free charges, and the bound volume charge density is zero in a uniform dielectric medium (or a vacuum). Hence, the scalar potential satisfies Laplace's equation, $\nabla^2 \phi =0$, throughout space. Adopting spherical polar coordinates, $(r, \theta, \varphi)$, aligned along the $z$-axis, the boundary conditions are that $\phi\rightarrow -E_0 r \cos\theta$ at $r\rightarrow \infty$, and that $\phi$ is well-behaved at $r=0$. At the surface of the sphere, $r=a$, the continuity of $E_\parallel$ implies that $\phi$ is continuous. Furthermore, the continuity of $D_\perp=\epsilon_0 \epsilon E_\perp$ leads to the matching condition

$$\left.\frac{\partial \phi}{\partial r}\right\vert _{r=a+} = ... ...\epsilon \frac{\partial\phi}{\partial r}\right\vert _{r=a-}.$$ (835)

Let us try separable solutions of the form $r^m \cos\theta$. It is easily demonstrated that such solutions satisfy Laplace's equation provided that $m=1$ or $m=-2$. Hence, the most general solution to Laplace's equation outside the sphere, which satisfies the boundary condition at $r\rightarrow \infty$, is

$$\phi(r,\theta) = - E_0 r \cos\theta + E_0 \alpha \frac{a^3 \cos\theta}{r^2}.$$ (836)

Likewise, the most general solution inside the sphere, which satisfies the boundary condition at $r=0$, is

$$\phi(r,\theta) = - E_1 r \cos\theta.$$ (837)

The continuity of $\phi$ at $r=a$ yields

$$E_0 - E_0 \alpha = E_1.$$ (838)

Likewise, the matching condition (835) gives

$$E_0 + 2 E_0 \alpha = \epsilon E_1.$$ (839)

Hence,

$$\alpha = \frac{\epsilon-1}{\epsilon+2},$$ (840)

$$E_1 = \frac{3 E_0}{\epsilon+2}.$$ (841)

Note that the electric field inside the sphere is uniform, parallel to the external electric field outside the sphere, and of magnitude $E_1$. Moreover, $E_1 < E_0$ , provided that $\epsilon>1$. Finally, the density of the bound charge sheet on the surface of the sphere is

$$\sigma_b = -\epsilon_0\left(\left.\frac{\partial\phi}{\parti... ...ht) = 3 \epsilon_0 \frac{\epsilon-1}{\epsilon+2}\cos\theta.$$ (842)

As a final example, consider a spherical cavity, of radius $a$, in a uniform dielectric medium, of dielectric constant $\epsilon$, in the presence of a $z$-directed electric field of magnitude $E_0$. This problem is analogous to the previous problem, except that the matching condition (835) becomes

$$\left.\epsilon \frac{\partial \phi}{\partial r}\right\vert ... ...} = \left. \frac{\partial\phi}{\partial r}\right\vert _{r=a-}.$$ (843)

Hence,

$$\alpha = \frac{1-\epsilon}{1+2 \epsilon},$$ (844)

$$E_1 = \frac{3 \epsilon E_0}{1+2 \epsilon}.$$ (845)

Note that the field inside the cavity is uniform, parallel to the external electric field outside the sphere, and of magnitude $E_1$. Moreover, $E_1 > E_0$, provided that $\epsilon>1$. The density of the bound charge sheet on the surface of the cavity is

$$\sigma_b = -\epsilon_0\left(\left.\frac{\partial\phi}{\parti... ... = 3 \epsilon_0 \frac{1-\epsilon}{1+2 \epsilon}\cos\theta.$$ (846)