Boundary value problems with dielectrics - I

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

$$\epsilon_1 \nabla\!\cdot\!{\bfm E} = \frac{\rho}{\epsilon_0}$$ (432)

for $z>0$,

$$\epsilon_2\nabla\!\cdot\!{\bfm E} = 0$$ (433)

for $z<0$, and

$$\nabla\wedge{\bfm E} = 0$$ (434)

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

$$\epsilon_1 E_z(z=0^+) = \epsilon_2 E_z (z=0^-),$$ (435)

$$E_x (z=0^+) = E_x (z=0^-),$$ (436)

$$E_y (z=0^+) = E_y(z=0^-).$$ (437)

In order to solve this problem we will employ a slightly modified form of the well known method of images. Since $\nabla\wedge{\bfm E} = 0$ everywhere, the electric field can be written in terms of a scalar potential. So, ${\bfm E} =-\nabla \phi$. 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 diagram). If this is the case then the potential at some point $P$ in the region $z>0$ is given by

$$\phi(z>0) = \frac{1}{4\pi\epsilon_0 \epsilon_1}\left(\frac{q}{R_1} + \frac{q'}{R_2}\right),$$ (438)

where $R_1= \sqrt{\rho^2+(d-z)^2}$ and $R_2= \sqrt{\rho^2+(d+z)^2}$, when written in terms of cylindrical polar coordinates $(\rho,\varphi, z)$. Note that the potential (3.20) clearly is a solution of Eq. (3.16) in the region $z>0$. It gives $\nabla\!\cdot\!{\bfm E} = 0$, 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

$$\phi(z<0) = \frac{1}{4\pi\epsilon_0\epsilon_2} \frac{q''}{R_1}.$$ (439)

The above potential is clearly a solution of Eq. (3.17) in the region $z<0$. It gives $\nabla\!\cdot\!{\bfm E} = 0$, with no singularities.

It now remains to choose $q'$ and $q''$ in such a manner that the boundary conditions (3.19) are satisfied. The boundary conditions (3.19b) and (3.19c) are obviously satisfied if the scalar potential is continuous at the interface between the two dielectric media:

$$\phi(z=0^+) = \phi(z=0^-).$$ (440)

The boundary condition (3.19a) implies a jump in the normal derivative of the scalar potential across the interface:

$$\epsilon_1 \,\frac{\partial\phi(z=0^+)}{\partial z} = \epsilon_2\, \frac{\partial \phi(z=0^-)}{\partial z}.$$ (441)

The first matching condition yields

$$\frac{q+q'}{\epsilon_1} = \frac{q''}{\epsilon_2},$$ (442)

whereas the second yields

$$q-q' = q''.$$ (443)

Here, use has been made of

$$\frac{\partial}{\partial z}\!\left(\frac{1}{R_1}\right)_{z=0... ...eft(\frac{1}{R_2}\right)_{z=0} = \frac{d}{(\rho^2+d^2)^{3/2}}.$$ (444)

Equations (3.24) and (3.25) imply that

$$q' = -\left(\frac{\epsilon_2-\epsilon_1}{\epsilon_2 + \epsilon_1} \right) q,$$ (445)

$$q''&= \left(\frac{2\epsilon_2}{\epsilon_2+\epsilon_1}\right) q.$$ (446)

The polarization charge density is given by $\rho_b = - \nabla\!\cdot\! {\bfm P}$, However, inside either dielectric ${\bfm P}=\epsilon_0 \chi_e {\bfm E}$, so $\nabla\!\cdot\!{\bfm P} = \epsilon_0 \chi_e \nabla\!{\cdot}\!{\bfm E} = 0$, except at the point charge $q$. Thus, there is zero polarization charge density in either dielectric medium. At the interface $\chi_e$ takes a discontinuous jump,

$${\mit\Delta}\chi_e = \epsilon_1-\epsilon_2.$$ (447)

This implies that there is a polarization charge sheet on the interface between the two dielectric media. In fact,

$$\sigma_{\rm pol} = - ({\bfm P}_2-{\bfm P}_1)\!\cdot\!{\bfm n}_{21},$$ (448)

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

$${\bfm P}_i = \epsilon_0 (\epsilon_i -1){\bfm E} = - \epsilon_0 ({\epsilon}_i - 1)\nabla\phi$$ (449)

in either medium, it is easy to demonstrate that

$$\sigma_{\rm pol} = -\frac{q}{2\pi} \frac{\epsilon_2-\epsilon... ...psilon_1(\epsilon_2+\epsilon_1)} \frac{d}{(\rho^2+d^2)^{3/2}}.$$ (450)

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

The above method can clearly be generalized to deal with problems involving many point charges in the presence of many different dielectric media whose interfaces form parallel planes.