prof. david r. jackson ece dept. fall 2014 notes 20 ece 2317 applied electricity and magnetism 1
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Prof. David R. JacksonECE Dept.
Fall 2014
Notes 20
ECE 2317 Applied Electricity and Magnetism
1
Dielectrics
Water
0 r
Single H2O molecule:
+
-
+
H+
O-
H+
2
Dielectrics (cont.)
Water
0 r
“Dipole” model: p p
p qd
“Vector dipole moment” of molecule p:
d
+q
-q
+
-
+
-
+
H+
O-
H+
3
Dielectrics (cont.)
4
The dipoles representing the water molecules are normally pointing in random directions.
+
+
+
-
+
-
+-
--
-+
Dielectrics (cont.)
5
The dipoles partially align under an applied electric field.
+ -V
x
+
+
+
+
-
-
-
-
Ex
+- +-
+-
+-
+- +-
Define: P = total dipole moment/volume
1i
V
P pV
0D E P
Define the electric flux density vector D:
V+
-
+-
Dielectrics (cont.)
We can also write this as
1 avei v
V
NP p N p
V N
pave = average vector dipole moment
Nv = number of molecules per unit volume
N molecules (dipoles) inside V
6
/vN N V
Linear material: 0 eP E
Define:
0 r
0 rD E Then we have
Note: e > 0 for most materials
Dielectrics (cont.)
The term e is called the
“electric susceptibility.”
0 0
0 1e
e
D E E
E
1r e
7
D Eor where
Typical Linear Materials
Teflon
Water
Styrofoam
Quartz
2.2
81
1.03
5
r
r
r
r
(a very polar molecule, fairly free to rotate)
8
Note: r > 1 for most materials.
1 , 0r e e
Note: A negative value of e would
mean that the dipoles align against
the field.
Exotic MaterialsArtificial “metamaterials” that have been designed that have
exotic permittivity and/or permeability performance.
9
Negative index metamaterial array configuration, which was constructed of copper split-ring resonators and wires
mounted on interlocking sheets of fiberglass circuit board. The total array consists of 3 by 20×20 unit cells with overall dimensions of
10×100×100 mm.
http://en.wikipedia.org/wiki/Mhttp://en.wikipedia.org/wiki/Metamaterialetamaterial
0
0r
r
(over a certain bandwidth of operation)
Exotic Materials (cont.)
10
Cloaking of objects is one area of research in metamaterials.
The Duke cloaking device masks an object from one wavelength at
microwaves.
Image from Dr. David R. Smith.
Dielectrics: Bound Charge
0vb
Assume (hypothetically) that Px increases as
x increases inside the material:
A net volume charge density is created inside the material, called the “bound charge density” or “polarization charge density.”
vb = “bound charge density”: it is bound to the molecules.
v = “free charge density”: this is charge you place inside the
material. You can freely place it wherever you want.
Note: For simplicity, the dipoles are shown perfectly aligned, with the number of aligned dipoles increasing with x.
11
+-
+-
+-
+-
+-
+-
+-
+-
+-
Dielectrics: Bound Charge (cont.)
totalv v vb
Total charge density inside the material:
0vb
This total charge density may be viewed as being in free space (since there is no material left after the molecules are removed).
0sb 0sb
12
Bound surface charge density
Bound charge densities
Note: The total charge is zero since the object is assumed to be neutral.
+-
+-
+-
+-
+-
+-
+-
+-
+-
Formulas for the two types of bound charge
13
Dielectrics: Bound Charge (cont.)
The derivation of these formulas is included in the Appendix.
vb P
ˆsb P n
, ,r x y z
- - - - - -
-- - -
-
++
+
-
-
-
+
+ n̂
The unit normal points outward from the dielectric.
0 0 1e rP E E
E = electric field inside material
Dielectrics: Gauss’s Law
0totalv v vbE
This equation is valid, but we prefer to have only the free-charge density in the equation, since this is what is known.
The goal is to calculate (and hopefully eliminate) the bound-charge density term on the right-hand side.
14
Inside a material:
The free-space form of Gauss’s law is
0totalv v vb vE P
Hence
vD
Dielectrics: Gauss’s Law (cont.)
0 vE P
or
This is why the definition of D is so convenient!15
Dielectrics: Gauss’s Law (Summary)
Important conclusion:
Gauss’ law works the same way inside a dielectric as it does in vacuum, with only the free charge density (i.e., the charge that is
actually placed inside the material) being used on the right-hand side.
vD
ˆ encl
S
D n dS Q
16
0 r
D E
ExamplePoint charge inside dielectric shell
Dielectric spherical shell with r
S
r
q
r
ab
Gauss’ law: ˆ encl
S
D n dS Q q
(The point charge q is the only free charge in the problem.)
17
Find D, E
Example (cont.)
22
ˆ C/m4
qD r
r
20
20
ˆ V/m ,4
ˆ V/m4 r
qE r r a r b
r
qE r a r b
r
18
S
r
q
r
ab
24rD r q
Hence
We then have
Example (cont.)
19
qr
Flux Plot
Note that there are less flux lines inside the dielectric region.
Homogeneous Dielectrics
0
0
00
00
00
1
1
vb
e
e
er
er
e vr
P
E
E
D
D
, ,r x y z constant
Assume a homogeneous dielectric with no free charge density inside.
Note: All of the interior bound charges cancel from one row of molecules to the next.
0v
0vb 20
r
Practical case:
Hence
Summary of Dielectric Principles
A dielectric can always be modeled with a bound volume charge density and a bound surface charge density.
A homogeneous dielectric with no free charge inside can always be modeled with only a bound surface charge density.
All of the bound charge is automatically accounted for when using the D vector and the total permittivity .
When using the D vector, the problem is solved using only the (known) free-charge density on the right-hand side of Gauss’ law.
Important points:
21
ExampleA point charge surrounded by a homogeneous dielectric shell.
22
Dielectric shell
qr
20
ˆ4
qE r
r
20
ˆ4 r
qE r
r
a
b
Find the bound surface charges densities and verify that we get the correct answer for the electric field by using them.
From Gauss’s law:
Example (cont.)
The alignment of the dipoles causes two layers of bound surface charge density.
23
Dielectric shell
q
r+-
+-
+-
+-
+- +-+ -+ -
+-
+-
+-
+-
+-
+-
+-
+-
sb
0 0v vb
Free-space model
24
0
0
ˆ
ˆ
ˆ1
1ˆ
sb
e
r
r
r
P n
E n
E n
D n
2ˆ
4
qD r
r
2
1
4a rsb
r
q
a
q
+
++
+
++
++
--
-
-
-
-
--
sb
n̂
n̂
2
1
4b rsb
r
q
b
Hence
Example (cont.)
Note: We use the electric field inside the material, at
the boundaries.
20
ˆ4
enclQE r
r
(free-space Gauss’s law)
Free-space model
25
1a rb
r
q q
1b rb
r
q q
20
ˆ:4
qr a E r
r
20
ˆ:4
abq q
a r b E rr
2
0
ˆ:4 r
qa r b E r
r
From Gauss’s law:
Example (cont.)
1a rb
r r
qq q q q
b ab bq qq
+
++
+
++
++
--
-
-
-
-
--
abq
bbq
20
ˆ4
enclQE r
r
24a ab sbq a
24b bb sbq b
Free-space model
26
1a rb
r
q q
1b rb
r
q q
20
ˆ:4
a bb bq q q
r b E rr
2
0
ˆ:4
qr b E r
r
From Gauss’s law:
Example (cont.)
0a bb bq q
b ab bq q
q
+
++
+
++
++
--
-
-
-
-
--
abq
bbq
27
Observations
We get the correct answer by accounting for the bound charges, and putting them in free space.
It is much easier to not worry about bound charges, and simply use the concept of r.
ˆ encl
S
D n dS Q
0 rD E E
Appendix
In this appendix we derive the formulas for the bound change densities:
28
ˆsb P n
vb P
ˆ ˆbn n outward normal to the dielectric boundary
where
Model:
Dipoles are aligned in the x direction end-to-end. The number of dipoles that are aligned (per unit volume) changes with x.
Nv = # of aligned dipoles
per unit volume
29
1vb bQ
V
V
x+d/2x-d/2
x
+-+-+-
+-
+-
d
Qb
Observation point
Appendix (cont.)
2 2
1vb b
d db v vx x
QV
Q q N V q N V
Hence
Nv = # of aligned dipoles
per unit volume
2 2
d dvb v vx x
v
q N N
q N
V
x+d/2x-d/2 x
+-+-+-
+-
+-
VV
Qb
30
Appendix (cont.)
vb vq N
Hence
Nv = # of aligned dipoles
per unit volume
V
x+d/2x-d/2x
+-+-+-
+-
+-
x v
x v
P pN
P p N
1
1
1
vb x
x
x
x
q Pp
q Pqd
Pd
P
x
or
xvb
dP
dx
31
Appendix (cont.)
In general,
yx zvb
dPdP dP
dx dy dz
or
vb P 32
xvb
dP
dx
For dipole aligned in the x direction,
Appendix (cont.)
33
After applying the divergence theorem, we have the integral form
ˆ bencl
S
P n Q
Applying this to a surface at a dielectric boundary, we have
ˆdiel air bb encl sbP P n S Q S
ˆb sbP n
DenotingdielP P
we have
S
Sˆbn
r
Appendix (cont.)
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