Next: Energy Density Within Dielectric
Up: Electrostatics in Dielectric Media
Previous: Boundary Conditions for and
Consider a point charge
embedded in a semi-infinite dielectric medium of dielectric constant
, and located a distance
from a plane interface that
separates the first medium from another semi-infinite dielectric medium of
dielectric constant
. Suppose that the interface coincides with the plane
.
We need to solve
![$\displaystyle \epsilon_1 \nabla \cdot {\bf E} = \frac{\rho}{\epsilon_0}$](img1092.png) |
(512) |
in the region
,
![$\displaystyle \epsilon_2\nabla \cdot {\bf E} = 0$](img1094.png) |
(513) |
in the region
, and
![$\displaystyle \nabla\times{\bf E} = {\bf0}$](img1096.png) |
(514) |
everywhere, subject to the following constraints at
:
Figure 1:
The method of images for a plane interface between two dielectric media.
![\begin{figure}
\epsfysize =3in
\centerline{\epsffile{Chapter04/plane.eps}}
\end{figure}](img1103.png) |
In order to solve this problem, we shall employ a slightly modified form of
the well-known method of images. Because
everywhere,
the electric field can be written in terms of a scalar potential: that is,
. 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 dielectric of dielectric constant
, and, in addition to the real charge
at position
,
there is a second charge
at the image position
. (See Figure 1.)
If this is the case then the potential at some point
in the region
is given by
![$\displaystyle \phi(z>0) = \frac{1}{4\pi\,\epsilon_0\, \epsilon_1}\left(\frac{q}{R_1} + \frac{q'}{R_2}\right),$](img1107.png) |
(518) |
where
and
. Here,
,
,
are conventional cylindrical
coordinates.
Note that the potential (519) is clearly a solution of Equation (513) in
the region
: that is, it satisfies
, 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 a dielectric medium of dielectric constant
, and a charge
is located at the point
. If this is the case then the potential in this region is
given by
![$\displaystyle \phi(z<0) = \frac{1}{4\pi\,\epsilon_0\,\epsilon_2} \frac{q''}{R_1}.$](img1112.png) |
(519) |
The above potential is clearly a solution of Equation (514) in the region
: that is, it satisfies
, with
no singularities.
It now remains to choose
and
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,
![$\displaystyle \phi(z=0_+) = \phi(z=0_-).$](img1113.png) |
(520) |
The constraint (516) implies a jump in the normal derivative
of the scalar potential across the interface. In fact,
![$\displaystyle \epsilon_1 \,\frac{\partial\phi(z=0_+)}{\partial z} = \epsilon_2\, \frac{\partial \phi(z=0_-)}{\partial z}.$](img1114.png) |
(521) |
The first matching condition yields
![$\displaystyle \frac{q+q'}{\epsilon_1} = \frac{q''}{\epsilon_2},$](img1115.png) |
(522) |
whereas the second gives
![$\displaystyle q-q' = q''.$](img1116.png) |
(523) |
Here, use has been made of
![$\displaystyle \frac{\partial}{\partial z}\!\left(\frac{1}{R_1}\right)_{z=0} =- ...
...rtial z}\!\left(\frac{1}{R_2}\right)_{z=0} = \frac{d}{(r^{\,2}+d^{\,2})^{3/2}}.$](img1117.png) |
(524) |
Equations (523) and (524) imply that
The polarization charge density is given by
, However,
inside either dielectric, which implies that
![$\displaystyle \nabla \cdot{\bf P} = \epsilon_0\,\chi_e\, \nabla \cdot {\bf E} = 0,$](img1123.png) |
(527) |
except at the point charge
.
Thus, there is zero bound charge density in either dielectric
medium. At the interface,
jumps discontinuously,
![$\displaystyle {\mit\Delta}\chi_e = \epsilon_1-\epsilon_2.$](img1124.png) |
(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
![$\displaystyle \sigma_b = - ({\bf P}_2-{\bf P}_1) \cdot {\bf n}_{21},$](img1125.png) |
(529) |
where
is a unit normal to the interface pointing from
medium 1 to medium 2 (i.e., along the positive
-axis).
Because
![$\displaystyle {\bf P}_i = \epsilon_0\, (\epsilon_i -1)\,{\bf E} = - \epsilon_0\, ({\epsilon}_i - 1)\,\nabla\phi$](img1126.png) |
(530) |
in either medium, it is easily demonstrated that
![$\displaystyle \sigma_b(r) = -\frac{q}{2\pi\,\epsilon_1} \left(\frac{\epsilon_2-\epsilon_1} {\epsilon_2+\epsilon_1}\right) \frac{d}{(r^{\,2}+d^{\,2})^{3/2}}.$](img1127.png) |
(531) |
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 when the plane
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
, lies between the planes
and
. Suppose that this slab is placed in a uniform
-directed
electric field of strength
. Let us calculate the field-strength
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
is the
same in both the dielectric slab and the surrounding vacuum.
In the vacuum region,
, whereas
in the dielectric. It follows that
![$\displaystyle E_1=\frac{E_0}{\epsilon}.$](img1135.png) |
(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
![$\displaystyle \sigma_b(z=b)=-\sigma_b (z=0) = \epsilon_0 \left(\frac{\epsilon-1}{\epsilon}\right)E_0.$](img1136.png) |
(533) |
In the limit
, the slab acts like a conductor, and
.
Let us now generalize this result. Consider a dielectric medium whose
dielectric constant
varies with
. The medium is assumed to
be of finite extent, and to be surrounded by a vacuum. It follows
that
as
. Suppose that this dielectric is
placed in a uniform
-directed electric field of strength
. What is the
field
inside the dielectric?
We know that the electric displacement inside the dielectric is
given by
. We also know from
Equation (501) that, because there are no free charges, and this is
essentially a one-dimensional problem,
![$\displaystyle \frac{d D(z)}{dz} = \epsilon_0\, \frac{d [\epsilon(z)\, E(z)]}{d z} = 0.$](img1143.png) |
(534) |
Furthermore,
as
.
It follows that
![$\displaystyle E(z) = \frac{E_0}{\epsilon(z)}.$](img1145.png) |
(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
![$\displaystyle \rho_b = \epsilon_0 \,\frac{d E(z)}{d z} = \epsilon_0\, E_0\frac{d}{dz} \!\left[\frac{1}{\epsilon(z)}\right].$](img1146.png) |
(536) |
Consider a third, and final, boundary value problem in which a dielectric sphere of radius
, and dielectric constant
, is placed in a
-directed electric field of strength
(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
.
There are no free charges, so Equations (501) and (505) imply that
![$\displaystyle \nabla^{\,2}\phi = 0$](img1147.png) |
(537) |
everywhere. The matching conditions (510) and (512) reduce to
Furthermore,
![$\displaystyle \phi(r,\theta,\varphi) \rightarrow - E_0 \,r\,\cos\theta$](img1152.png) |
(540) |
as
: that is, the electric field asymptotes to uniform
-directed field of strength
far from the
sphere. Here,
,
,
are spherical
coordinates centered on the sphere.
Let us search for an axisymmetric solution,
. 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
![$\displaystyle \frac{1}{r}\frac{\partial^{\,2} (r\,\phi)}{\partial r^{\,2}} +\fr...
...ial\theta}\!\left(\sin\theta \,\frac{\partial\phi}{\partial\theta} \right) = 0.$](img1156.png) |
(541) |
Straightforward separation of the variables yields (see Section 3.7)
![$\displaystyle \phi(r,\theta) = \sum_{l=0,\infty}\left[A_l \,r^{\,l} + B_l \,r^{\,-(l+1)}\right] P_l(\cos\theta),$](img1157.png) |
(542) |
where
is a non-negative
integer, the
and
are arbitrary constants, and the
are Legendre polynomials. (See Section 3.2.)
The Legendre polynomials form a complete set of angular functions, and
it is easily demonstrated
that the
and the
form a complete set of
radial functions. It follows that Equation (543), with the
and
unspecified, represents a completely general (single-valued) axisymmetric solution
to Equation (538). It remains to determine the values of the
and
that
are consistent with the boundary conditions.
Let us divide space into the regions
and
. In the former
region
![$\displaystyle \phi(r,\theta) = \sum_{l=0,\infty} A_l\,r^{\,l}\,P_l(\cos\theta),$](img1163.png) |
(543) |
where we have rejected the
radial solutions because they diverge
unphysically as
.
In the latter
region
![$\displaystyle \phi(r,\theta) = \sum_{l=0,\infty}\left[B_l\,r^{\,l} + C_l \,r^{\,-(l+1)}\right]P_l(\cos \theta).$](img1164.png) |
(544) |
However, it is clear from the boundary condition (541)
that the only non-vanishing
is
. This follows because
. The boundary condition
(540) [which can be integrated to give
for a potential that is single-valued in
] gives
![$\displaystyle A_1 = - E_0 + \frac{C_1}{a^{\,3}},$](img1168.png) |
(545) |
and
![$\displaystyle A_l = \frac{C_l}{a^{\,2\,l+1}}$](img1169.png) |
(546) |
for
. Note that it is appropriate to match the coefficients of
the
because these functions are mutually orthogonal. (See Section 3.2.)
The boundary condition (539) yields
![$\displaystyle \epsilon \,A_1 = - E_0 -2\,\frac{C_1}{a^{\,3}},$](img1172.png) |
(547) |
and
![$\displaystyle \epsilon\, l\, A_l = - (l+1)\,\frac{C_l}{a^{\,2\,l+1}}$](img1173.png) |
(548) |
for
. Equations (547) and (549) give
for
. Equations (546) and (548) reduce to
The solution to the problem is therefore
![$\displaystyle \phi(r,\theta) = - \left(\frac{3}{2+\epsilon}\right)\,E_0\,r\,\cos\theta$](img1179.png) |
(551) |
for
, and
![$\displaystyle \phi(r,\theta) = - E_0\, r\,\cos\theta + \left(\frac{\epsilon-1}{\epsilon+2}\right) \,E_0\,\frac{a^{\,3}}{r^{\,2}}\cos\theta$](img1180.png) |
(552) |
for
.
Equation (552) is the potential of a uniform
-directed electric
field of strength
![$\displaystyle E_1 = \frac{3}{2+\epsilon}\,E_0.$](img1181.png) |
(553) |
Note that
, provided that
. 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
, plus the field of
an electric dipole (see Section 2.7), located at the origin, and directed along the
-axis, whose dipole moment is
![$\displaystyle p = 4\pi\,\epsilon_0 \left(\frac{\epsilon-1}{\epsilon +2} \right) \,a^{\,3}\, E_0.$](img1184.png) |
(554) |
This dipole moment can be interpreted as the volume integral of the
polarization
over the sphere. The polarization is
![$\displaystyle {\bf P} = \epsilon_0 \,(\epsilon -1)\,E_1\,{\bf e}_z = 3\,\epsilon_0 \left(\frac{\epsilon-1}{\epsilon+2}\right)\,E_0\,{\bf e}_z.$](img1185.png) |
(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
[see Equation (530)]. It follows that
![$\displaystyle \sigma_b(\theta) = 3\,\epsilon_0 \left(\frac{\epsilon-1}{\epsilon+2}\right)\,E_0\,\cos\theta.$](img1187.png) |
(556) |
The problem of a dielectric cavity of radius
inside a dielectric medium
of dielectric constant
, and in the presence of an applied electric field
, parallel to the
-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
. Thus, the field
inside the cavity is uniform, parallel to the
-axis, and of magnitude
![$\displaystyle E_1 = \frac{3\,\epsilon}{2\,\epsilon + 1} \,E_0.$](img1189.png) |
(557) |
Note that
, provided that
. The field outside the
cavity is the original field, plus that of a
-directed dipole, located
at the
origin, whose dipole moment is
![$\displaystyle p = - 4\pi\,\epsilon_0 \left(\frac{\epsilon -1}{2\,\epsilon +1}\right)a^{\,3}\, E_0.$](img1191.png) |
(558) |
Here, the negative sign implies that the dipole points in the opposite
direction to
the external field.
Next: Energy Density Within Dielectric
Up: Electrostatics in Dielectric Media
Previous: Boundary Conditions for and
Richard Fitzpatrick
2014-06-27