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Consider a point charge
embedded in a semi-infinite dielectric
a distance
away from a plane interface which
separates the first medium from another semi-infinite dielectric
. The interface is assumed to coincide with the plane
.
We need to find solutions to the equations
![\begin{displaymath}
\epsilon_1 \nabla\!\cdot\!{\bf E} = \frac{\rho}{\epsilon_0}
\end{displaymath}](img1696.png) |
(818) |
for
,
![\begin{displaymath}
\epsilon_2 \nabla\!\cdot\!{\bf E} = 0
\end{displaymath}](img1697.png) |
(819) |
for
, and
![\begin{displaymath}
\nabla\times {\bf E} = {\bf0}
\end{displaymath}](img569.png) |
(820) |
everywhere, subject to the boundary conditions at
that
Figure 47:
![\begin{figure}
\epsfysize =3in
\centerline{\epsffile{plane.eps}}
\end{figure}](img1704.png) |
In order to solve this problem, we shall employ a slightly modified form of
the well-known method of images. Since
everywhere,
the electric field can be written in terms of a scalar potential.
So,
. Consider the region
.
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
, and, in addition to the real charge
at position
,
there is a second charge
at the image position
(see Fig. 47).
If this is the case, then the potential at some point
in the region
is given by
![\begin{displaymath}
\phi(z>0) = \frac{1}{4\pi \epsilon_0 \epsilon_1}\left(\frac{q}{R_1}
+ \frac{q'}{R_2}\right),
\end{displaymath}](img1706.png) |
(824) |
where
and
, when
written in terms of cylindrical polar coordinates,
, aligned along the
-axis.
Note that the potential (824) is clearly a solution of Eq. (818) in
the region
. It gives
, with the
appropriate singularity at the position of the point charge
.
Consider the region
. 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
, and a charge
is located at the point
. If this is the case, then the potential in this region is
given by
![\begin{displaymath}
\phi(z<0) = \frac{1}{4\pi \epsilon_0 \epsilon_2} \frac{q''}{R_1}.
\end{displaymath}](img1711.png) |
(825) |
The above potential is clearly a solution of Eq. (819) in the region
. It gives
, with
no singularities.
It now remains to choose
and
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:
![\begin{displaymath}
\phi(z=0^+) = \phi(z=0^-).
\end{displaymath}](img1713.png) |
(826) |
The boundary condition (821) implies a jump in the normal derivative
of the scalar potential across the interface:
![\begin{displaymath}
\epsilon_1 \frac{\partial\phi(z=0^+)}{\partial z} = \epsilon_2
\frac{\partial \phi(z=0^-)}{\partial z}.
\end{displaymath}](img1714.png) |
(827) |
The first matching condition yields
![\begin{displaymath}
\frac{q+q'}{\epsilon_1} = \frac{q''}{\epsilon_2},
\end{displaymath}](img1715.png) |
(828) |
whereas the second gives
![\begin{displaymath}
q-q' = q''.
\end{displaymath}](img1716.png) |
(829) |
Here, use has been made of
![\begin{displaymath}
\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}}.
\end{displaymath}](img1717.png) |
(830) |
Equations (828) and (829) imply that
The bound charge density is given by
, However, inside either dielectric,
, so
, except at the point charge
.
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
![\begin{displaymath}
\sigma_b = \epsilon_0 (E_{z 1}-E_{z 2})_{z=0}.
\end{displaymath}](img1725.png) |
(833) |
Hence,
![\begin{displaymath}
\sigma_b = \epsilon_0 \frac{\partial\phi(z=0-)}{\partial ...
...{\epsilon_1(\epsilon_2+\epsilon_1)} \frac{d}{(r^2+d^2)^{3/2}}.
\end{displaymath}](img1726.png) |
(834) |
In the limit
, the dielectric
behaves like a conducting medium (i.e.,
in the region
), and the bound surface charge density
on the interface approaches that obtained in the case where the plane
coincides with a conducting surface (see Sect. 5.10).
As a second example, consider a dielectric sphere of radius
, and
uniform dielectric constant
, placed in a uniform
-directed electric field of magnitude
. Suppose that the
sphere is centered on the origin. Now, for an electrostatic problem, we
can always write
. In the present problem,
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,
, throughout space. Adopting spherical polar coordinates,
, aligned along the
-axis, the boundary conditions are that
at
, and that
is well-behaved at
. At the surface of the sphere,
, the continuity of
implies that
is continuous. Furthermore, the
continuity of
leads to the matching condition
![\begin{displaymath}
\left.\frac{\partial \phi}{\partial r}\right\vert _{r=a+} = ...
...\epsilon
\frac{\partial\phi}{\partial r}\right\vert _{r=a-}.
\end{displaymath}](img1733.png) |
(835) |
Let us try separable solutions of the form
. It is
easily demonstrated that such solutions satisfy Laplace's equation
provided that
or
. Hence, the most general solution to Laplace's equation outside
the sphere, which satisfies the boundary condition at
, is
![\begin{displaymath}
\phi(r,\theta) = - E_0 r \cos\theta + E_0 \alpha \frac{a^3 \cos\theta}{r^2}.
\end{displaymath}](img1736.png) |
(836) |
Likewise, the most general solution inside the sphere, which satisfies
the boundary condition at
, is
![\begin{displaymath}
\phi(r,\theta) = - E_1 r \cos\theta.
\end{displaymath}](img1737.png) |
(837) |
The continuity of
at
yields
![\begin{displaymath}
E_0 - E_0 \alpha = E_1.
\end{displaymath}](img1738.png) |
(838) |
Likewise, the matching condition (835) gives
![\begin{displaymath}
E_0 + 2 E_0 \alpha = \epsilon E_1.
\end{displaymath}](img1739.png) |
(839) |
Hence,
Note that the electric field inside the sphere is uniform, parallel
to the external electric field outside the sphere, and of magnitude
. Moreover,
, provided that
. Finally,
the density of the bound charge sheet on the surface of the sphere
is
![\begin{displaymath}
\sigma_b = -\epsilon_0\left(\left.\frac{\partial\phi}{\parti...
...ht) =
3 \epsilon_0 \frac{\epsilon-1}{\epsilon+2}\cos\theta.
\end{displaymath}](img1747.png) |
(842) |
As a final example, consider a spherical cavity, of radius
, in a
uniform dielectric medium, of dielectric constant
, in the presence of a
-directed electric field of magnitude
. This problem is analogous
to the previous problem, except that the matching condition
(835) becomes
![\begin{displaymath}
\left.\epsilon \frac{\partial \phi}{\partial r}\right\vert ...
...} = \left.
\frac{\partial\phi}{\partial r}\right\vert _{r=a-}.
\end{displaymath}](img1748.png) |
(843) |
Hence,
Note that the field inside the cavity is uniform, parallel
to the external electric field outside the sphere, and of magnitude
. Moreover,
, provided that
. The density
of the bound charge sheet on the surface of the cavity
is
![\begin{displaymath}
\sigma_b = -\epsilon_0\left(\left.\frac{\partial\phi}{\parti...
... =
3 \epsilon_0 \frac{1-\epsilon}{1+2 \epsilon}\cos\theta.
\end{displaymath}](img1752.png) |
(846) |
Next: Energy density within a
Up: Dielectric and magnetic media
Previous: Boundary conditions for and
Richard Fitzpatrick
2006-02-02