Prof. David R. Jackson ECE Dept. - University of Houstoncourses.egr.uh.edu/ECE/ECE3318/Class...
Transcript of Prof. David R. Jackson ECE Dept. - University of Houstoncourses.egr.uh.edu/ECE/ECE3318/Class...
Prof. David R. Jackson Dept. Of ECE
Spring 2018
Notes 20 Dielectrics
ECE 3318 Applied Electricity and Magnetism
1
Dielectrics
Water
0 rε ε ε=
Single H2O molecule:
2
+
-
H+
O-
H+
+
Dielectrics (cont.)
Water
0 rε ε ε=
“Dipole” model: p p
p qd
=
=
“Vector dipole moment” of molecule p:
3
+
-
d
q+
q−
+
-
H+
O-
H+
+
Dielectrics (cont.)
4
The dipoles representing the water molecules are normally pointing in random directions.
+
+ +
-
+
-
+ -
- -
- +
Note: The molecules are not floating in water – they make up the water.
Dielectrics (cont.)
5
The dipoles partially align under an applied electric field.
+ -
+
+
+
+
-
-
-
-
+ -
+ -
+ -
+ -
x
0V
xE
Define: P = total dipole moment/volume
1i
VP p
V ∆
≡∆ ∑
0D E Pε≡ +
Define the electric flux density vector D:
V∆
Dielectrics (cont.)
We can also write this as
( )( )1 avei v
V
NP p N pV N ∆
= = ∆ ∑
pave = average vector dipole moment
N molecules (dipoles) inside ∆V
6
/vN N V= ∆
Nv = number of molecules per unit volume
Linear material: 0 eP Eε χ=
Define:
0 rε ε ε=
0 rD Eε ε=Then we have
Note: χe > 0 for most materials
Dielectrics (cont.)
The term χe is called the “electric susceptibility.”
( )0 0
0 1e
e
D E EE
ε ε χε χ
= +
= +
1r eε χ≡ +
7
D Eε=or where
Note: A negative value of χe
would mean that the dipoles align against the field.
Typical Linear Materials
Teflon Water Styrofoam Quartz
2.2811.035
r
r
r
r
εεεε
====
(a very polar molecule, fairly free to rotate)
8
Note: εr > 1 for most materials: 1 , 0r e eε χ χ≡ + >
Dielectrics: Bound Charge
0vbρ <
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.”
ρvb = “bound charge density”: it is bound to the molecules.
ρv = “free charge density”: this is charge that 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.
9
+ -
+ -
+ -
+ -
+ -
+ - + - + - + -
Dielectrics: Bound Charge (cont.)
totalv v vbρ ρ ρ= +
Total charge density inside the material:
0vbρ <
This total charge density may be viewed as being in free space (since there is no material left after the molecules are removed).
0sbρ <0sbρ >
10
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
11
Dielectrics: Bound Charge (cont.)
The derivation of these formulas is included in the Appendix.
vb Pρ = −∇ ⋅
ˆsb P nρ = ⋅
( ), ,r x y zε
- - - - - - - - - - -
+ +
+
-
- -
+
+ n̂
The unit normal points outward from the dielectric.
( )0 0 1e rP E Eε χ ε ε= = −
E = electric field inside material
Dielectrics: Gauss’s Law
( )0totalv v vbEε ρ ρ ρ∇⋅ = = +
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.
12
Inside a material:
The free-space form of Gauss’s law is
( )0totalv v vb vE Pε ρ ρ ρ ρ∇⋅ = = + = −∇⋅
Hence
vD ρ∇ ⋅ =
Dielectrics: Gauss’s Law (cont.)
( )0 vE Pε ρ∇ ⋅ + =
or
This is the usual Gauss’s law: This is why the definition of D is so convenient! 13
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 ρ∇ ⋅ =
ˆ enclS
D n dS Q⋅ =∫
14
0 rε ε ε=
D Eε=
Example Point charge inside dielectric shell
Dielectric spherical shell with εr
Gauss’ law: ˆ enclS
D n dS Q q⋅ = =∫
(The point charge q is the only free charge in the problem.)
15
Find D, E
qr
ba
r
S
Example (cont.)
22ˆ C/m
4qD rrπ
=
[ ]
[ ]
20
20
ˆ V/m ,4
ˆ V/m4 r
qE r r a r br
qE r a r br
πε
πε ε
= < >
= < <
16
( )24rD r qπ =
Hence
We then have
qr
ba
r
S
Example (cont.)
17
qrε
Flux Plot
Note that there are less flux lines inside the dielectric region (assuming that the flux lines represent the electric field).
Homogeneous Dielectrics
( )0
0
00
00
1
1
vb
e
e
er
er
rv
r
PE
E
D
D
ρε χ
ε χ
ε χε ε
ε χε ε
ε ρε
= −∇ ⋅
= −∇ ⋅
= − ∇ ⋅
= − ∇ ⋅
= − ∇ ⋅
−= −
( ), ,r x y zε = constant
Homogeneous dielectric No free charge density inside.
Note: All of the interior bound charges cancel from
one row of molecules to the next. 0vbρ =
18
0vρ =
rε
Practical case:
Hence
Homogeneous Dielectrics (cont.)
( ), ,r x y zε = constant
19
0vρ =
rε
+ -
+ -
+ -
+ -
+ -
+ -
+ -
+ - + -
0vbρ =
Blown-up view of molecules
Practical case (homogeneous)
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 ε.
When using the D vector, the problem is then solved using only the (known) free-charge density on the right-hand side of Gauss’ law.
Important points:
20
Example A point charge surrounded by a homogeneous dielectric shell.
21
20
ˆ4
qE rrπε
=
20
ˆ4 r
qE rrπε ε
=
Find the bound surface charges densities and verify that we get the correct answer for the electric field by using them in free space.
From Gauss’s law:
Dielectric shell
b
aq
rε
Example (cont.)
The alignment of the dipoles causes two layers of bound surface charge density.
22
sbρ
0 0v vbρ ρ= ⇒ =
Dielectric shell
+ -
+ -
+ -
+ -
+ - + - + - + -
rε
q
Free-space model
23
( )( )( )
( )
0
0
00
ˆˆ
ˆ1
ˆ1
1 ˆ
sb
e
r
rr
r
r
P nE n
E n
D n
D n
ρε χ
ε ε
ε εε ε
εε
= ⋅
= ⋅
= − ⋅
= − ⋅
−
= ⋅
2ˆ4
qD rrπ
=
2
14
a rsb
r
qa
ερε π
− = −
2
14
b rsb
r
qb
ερε π
− = +
Hence
Example (cont.)
Note: We use the electric field
inside the material, at the boundaries.
+
+ +
+
+ +
+ +
- -
-
-
-
-
- -
n̂
n̂ q
sbρ
Free-space model
24
1a rb
r
q qεε
−= −
1b rb
r
q qεε
−= +
20
ˆ:4
qr a E rrπε
< =
: a
encl ba r b Q q q< < = +
20
ˆ:4 r
qa r b E rrπε ε
< < =
From Gauss’s law:
Example (cont.)
1a rb
r r
qq q q q εε ε
−+ = − =
b ab bq q= −Note :
20
ˆ4
enclQE rrπε
=
24a ab sbq aπ ρ=
24b bb sbq bπ ρ=
where
+
+ +
+
+ +
+ +
- -
-
-
-
-
- -
abq
bbq
qS
a r b< <
Free-space model
25
1a rb
r
q qεε
−= −
1b rb
r
q qεε
−= +
: a bencl b br b Q q q q> = + +
20
ˆ:4
qr b E rrπε
> =
From Gauss’s law:
Example (cont.)
0a bb bq q+ =
b ab bq q= −Note :
where
+
+ +
+
+ +
+ +
- -
-
-
-
-
- -
abq
bbq
q
S
r b>
26
Observations
We get the correct answer by accounting for the bound charges, and putting them in free space.
However, it is much easier to not worry about bound charges,
and simply use the concept of εr and the free charge.
ˆ enclS
D n dS Q⋅ =∫
0 rD E Eε ε ε= =
From the previous example, we have the following observations:
Free charge enclosed
Exotic Materials
27
Plasmas have a relative permittivity that is less than one (and can even be negative)
Plasmas
2
0 1 pωε ε
ω
= −
( )
2
0 1 p
jω
ε εω ω υ
= − −
Lossless plasma:
Lossy plasma:
ωp = plasma resonance frequency
υ = plasma collision frequency (loss term)
(derived in ECE 6340)
Exotic Materials (cont.)
28
Low frequencies (f < 30 MHz) will reflect off the ionosphere.
Plasmas
Shortwave radio signals propagate around the earth by
skipping off the ionosphere.
The ionosphere has a relative permittivity that is less than one, so the waves will bend (refract) “away from the normal” and
travel back down to the earth, bouncing off of the earth.
https://en.wikipedia.org/wiki/Skywave
Exotic Materials (cont.) Artificial “metamaterials” that have been designed that have
exotic permittivity and/or permeability performance.
29
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
00
r
r
εµ<<
(over a certain bandwidth of operation)
Exotic Materials (cont.)
30
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.
Appendix
In this appendix we derive the formulas for the bound change densities:
31
ˆsb P nρ = ⋅
vb Pρ = −∇ ⋅
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
32
Appendix (cont.)
1vb bQ
Vρ =
∆
+ - + - + -
+ - + -
Observation point
d
x/ 2x d+/ 2x d−
bQ
V∆
2 2
1vb b
d db v vx x
QV
Q q N V q N V
ρ
+ −
=∆
= − ∆ + ∆
Hence
Nv = # of aligned dipoles per unit volume
2 2d dvb v vx x
v
q N N
q N
ρ+ −
= − − = − ∆
33
Appendix (cont.)
+ - + - + -
+ - + -
d
x/ 2x d+/ 2x d−
bQV∆
vb vq Nρ = − ∆
Hence
Nv = # of aligned dipoles per unit volume
x v
x v
P pNP p N=
⇒ ∆ = ∆
1
1
1
vb x
x
x
x
q Pp
q Pqd
Pd
Px
ρ
= − ∆
= − ∆
= − ∆
∆= −
∆or
xvb
dPdx
ρ = −34
Appendix (cont.)
+ - + - + -
+ - + -
d
x/ 2x d+/ 2x d−
bQV∆
In general,
yx zvb
dPdP dPdx dy dz
ρ = − − −
or
vb Pρ = −∇⋅35
xvb
dPdx
ρ = −
For dipole aligned in the x direction,
Appendix (cont.)