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A. F. Peterson:
Notes on Electromagnetic Fields & Waves
10/04
Fields & Waves Note #11
Capacitors in Series and Parallel: Field Mapping
Objectives:
Continue the study of capacitance, by considering examples
with multiple layers or regions of different permittivities.
Expand the
concept of series and parallel capacitance to describe the field mapping
approach for general capacitor shapes.
Example:
A parallelplate capacitor with two layers of dielectric
Figure 1 depicts the cross section of a parallelplate capacitor with plates located at
x
= 0
and
x
=
d
, as in Note #10.
However, in this situation the region between the plates is divided
into two materials, at location
x
=
c
, as illustrated.
As in the previous examples, we assume
that the fields are “ideal” in the sense that they are only functions of
x
, and neglect any
fringing of the fields near the edges of the plates.
As in the preceding examples, the problem can be posed mathematically as determining the
voltage between the plates, using Laplace’s equation. However, we must separately treat the
two regions between the plates.
Therefore the differential equation to consider is separated
into two equations, by region:
—=
=
<
<
2
1
2
1
2
00
V
dV
dx
xc
,
(11.1)
=
<
<
2
2
2
2
2
0
V
dx
cxd
(11.2)
The solutions
V
1
and
V
2
are subject to the boundary conditions
Vx
x
1
0
0
()
=
=
(11.3)
V
xd
20
=
=
(11.4)
at the plates.
In addition, the continuity of the voltage may be imposed at
x
=
c
:
12
==
=
(11.5)
A fourth condition is required, and must be obtained from the conditions that are satisfied
by the electric field at the material interface.
Since
EV
=—
, the only component of
E
present between the plates is
E
dV
dx
x
=
(11.6)
Because
E
x
is the normal component at the interface, the appropriate condition (Note #9) is
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Notes on Electromagnetic Fields & Waves
10/04
ee
11
2 2
EE
x
xc
x
==
=
(11.7)
By combining the relations in (11.6) and (11.7), we obtain the fourth boundary condition as
1
1
2
2
dV
dx
dV
dx
=
(11.8)
The general solutions to the differential equations in (11.1) and (11.2) are easily obtained as
Vx K Kx
x c
2
0
()
,
=+
<
<
(11.9)
c x d
23
4
,
<
<
(11.10)
There are four constants to be determined by the boundary conditions.
Imposing (11.3)
yields
Vx
K K
x
1
0
12
00
=
=
(11.11)
or equivalently,
K
1
0
=
(11.12)
Imposing (11.4) results in the equation
K Kd V
xd
4
0
=
=
(11.13)
The condition in (11.5) produces
Kc Vx
K K c
4
=
+
(11.14)
Finally, (11.8) results in
e
1
1
2
2
24
dV
dx
K
dV
dx
K
=
(11.15)
In this situation, the application of the four boundary conditions results in one of the four
constants being given directly (
K
1
= 0), and three equations that must be solved to determine
the other constants:
Kd
KV
cK
K
cK
KK
34
0
4
0
0
+=

=
=
(11.16)
A. F. Peterson:
Notes on Electromagnetic Fields & Waves
10/04
From the last equation, we determine
KK
2
2
1
4
=
e
(11.17)
Substituting this result into the second equation, we obtain
Kc
K
3
2
1
4
1
=
Ê
Ë
Á
ˆ
¯
˜
(11.18)
Finally, by combining (11.18) with the first equation of the three, we obtain the result
K
V
cd
4
01
21
1
=
+
ee
()
(11.19)
By substituting back into (11.17) and (11.18), we obtain
K
V
2
02
1
=
(11.20)
K
Vc
3
02 1
1
=

(11.21)
Therefore, the solution for voltage as a function of
x
is given by
Vx V
x
xc
10
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This note was uploaded on 01/27/2011 for the course ECE 3025 taught by Professor Citrin during the Spring '08 term at Georgia Institute of Technology.
 Spring '08
 CITRIN
 Electromagnet

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