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Polarisation Explained

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Lecture 7
Polarization
In this lecture you will learn:
• Material Polarization
• Mathematics of Polarization
• Dielectric Permittivity
• Conductors Vs Dielectrics
• Appendix
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Charge Dipoles and Dipole Moments
Consider a charge dipole:
r
d
+q
−q
r
Dipole moment of the charge dipole is a vector p such that:
r
r
p=qd
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
1
Conductors in Electric Fields – A Review
φ=0
φ=V
+
+
+
+
+
+
+
+
+
-
• Under the influence of external E-field these
free charges move to completely screen out
the E-field from within the conducting material
φ=0
φ=V
+
+
+
+
+
+
+
+
• Conductors have “free charges” that are able
to move around (mostly these “free charges”
are electrons that are not attached to any
particular atom)
V
+
+
+
+
+
+
+
+
+
• Consider the problem when a conductive
plate was placed inside an electric field
-
Conductors
Sea of Free Electrons
+
+
+
+
+
+
+
+
+
+
+
+
V
σ
+ +ve nucleus
-ve sea of free
electrons
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Dielectrics in Electric Fields - Polarization
Many materials do not have “free electrons” that can move around, but have
electrons bound to atoms, as shown below
Dielectric in an E-field
Dielectric
E
+
+
+
+ +ve nucleus
-
+
-
+
-
+
-ve electron
cloud
-
+
-
+
-
+
+
+
+
+
+
+
-
+
-
+
-
+
+
+
+
-
+
-
+
-
+
+
+
+
-
+
-
+
-
+
+
+
+
-
+
-
+
-
+
In the presence of an E-field, the electron cloud in each atom distorts
ALMOST INSTANTANEOUSLY so that each atom looks like a charge dipole
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
2
Dielectrics in Electric Fields – Polarization Vector
r
d
-
r
r
p = Qd = dipole moment of each dipole
+
+Q
-Q
r
P is a vector such that:
r
r
P=N p
v
= NQd
The polarization vector
E
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
Where N is the number of charge dipoles per unit
volume in the material
r
The units of P
are: Coumlombs/m2
r
The polarization vector P characterizes the
polarization density of the medium under the
influence of the electric field
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Dielectrics in Electric Fields – Electrical Susceptibility
Naturally, one would expect the polarization of the material to be proportional to
the strength of the electric field:
r
r
P∝E
r
r
⇒ P = ε o χe E
The constant of proportionality χe is called the electrical susceptibility of the
material
E
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
3
Material Polarization and Surface Charge
Densities
r
P = P xˆ
σ p = −P
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
E
σ p = +P
x
• The stuff inside the box in on the average charge neutral (same number of
positive and negative charges)
• There is a net negative surface charge density on the left facet of the material
as a result of material polarization
• There is a net positive surface charge density on the right facet of the material
as a result of material polarization
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Material Polarization and Surface Charge Densities
d
E
Area = A
How much charge is
in this volume??
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
r
P = P xˆ
Total interface negative charge due to dipoles in the volume Ad = - Q N A d
If we divide the total interface charge by the area A we get the interface
charge per unit area which would be the surface charge density σp
⇒ σp = −
QNAd
= − NQd = − P
A
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
4
Material Polarization and Volume Charge Densities
More generally, one can write a volume polarization volume charge
density due to material polarization as:
r
ρ p = −∇ . P
(A formal proof is given in the Appendix)
There will be a net non-zero volume charge density inside a material if the
material polarization is varying in space
In 1D situations:
ρp (x ) = −
∂Px
∂x
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Material Polarization and Charge Densities
E
σ p = −P
σ p = +P
r
P = Pxˆ
x
x=0
x=L
Px
P
ρp = −
∂Px
∂x
0
− Pδ ( x )
x
L
0
r
ρ p = −∇ . P
Pδ ( x − L )
L
∂Px
ρp = −
= −P δ ( x ) + P δ ( x − L )
∂x
x
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
5
Mathematics of Polarization – The “D” Field
Gauss’ Law states:
r
∇ .εo E = ρ
But charge densities could be of two types:
1) Paired charge density ρp (due to material polarization)
2) Unpaired charge density ρu (due to everything else – the usual stuff)
So:
r
r
∇ . ε o E = ρu + ρ p = ρu − ∇ . P
r r
⇒ ∇ . ε o E + P = ρu
(
r
Using: ρ p = −∇ . P
)
If one defines the D-field inside materials as:
r
r r
D = εo E + P
Then inside materials Gauss’ Law becomes:
r
∇ . D = ρu
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Mathematics of Polarization – Dielectric Permittivity
If one defines the D-field as:
r
r r
D = εo E + P
Note that:
Then:
r
∇ . D = ρu
r
r r
r
r
r
D = ε o E + P = ε o E + ε o χ e E = ε o (1 + χ e ) E
If one defines the dielectric permittivity of a material as:
ε = ε o (1 + χ e )
then one can write the D-field inside materials as:
r
r
D=εE
Inside materials the D-field obeys the Gauss’ Law:
r
r
∇ . D = ∇ . ε E = ρu
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
6
Mathematics of Polarization – Polarization Current Density
So far we have looked at the current density due to the motion of free unpaired
charges:
r
r
Ju = σ E
E
The motion of paired charges also results in
a current density:
(
r
r
r
∂P ∂ ε o χ e E
Jp =
=
∂t
∂t
)
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
r
J p is called the polarization current density
r
r
r
Ampere’s Law is correctly given by: ∇ × H = Ju + J p +
Which can also be written as:
(
r
∂ ε oE
∂t
)
( )
r
r r
∂ εE
∇ × H = Ju +
∂t
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Mathematics of Polarization – Modified Maxwell’s Equations
Putting it all together ….
One can forget all about material polarization, and polarization charge densities,
and polarization current densities, as long as one uses the dielectric permittivity ε
instead of the free space permittivity εo
Maxwell’s equations in dielectric materials take the form:
(
( )
)
r
∇ . ε o E = ρu + ρ p
r
∇. µo H = 0
OR
r
r
∂ µoH
∇×E = −
∂t
r
r r
r
∂ ε oE
∇ × H = Ju + J p +
∂t
(
)
r
∇ . ε E = ρu
r
∇. µo H = 0
r
r
∂ µo H
∇×E = −
∂t
r
r r
∂ εE
∇ × H = Ju +
∂t
Here the charge density
ρu is the unpaired
charge density
( )
Here the current density is
due to the unpaired charges
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
7
σu + σ p
Dielectric Permittivity – Boundary Conditions - I
How to relate the electric fields on both sides of
the dielectric interface ??
E1
E2
Gauss’ Law in the presence of dielectric material is:
ε1
ε2
( )
( )
r
r
∇ . D = ∇ . ε E = ρu
r
r
∫∫ D . da = ∫∫∫ ρu dV
or
σu + σ p
Draw a Gaussian surface at the interface:
(D2 − D1)A = σ u A
⇒
or
D1
D2
D2 − D1 = σ u
ε1
ε 2 E 2 − ε 1 E1 = σ u
A
ε2
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Dielectric Permittivity – Boundary Conditions - II
σu + σ p
How else can one relate the electric fields on both
sides of the dielectric interface ??
Gauss’ Law in the presence of dielectric material is
also:
r
(
E1
)
∇ . ε o E = ρu + ρ p
ε1
E2
ε2
A
Draw a Gaussian surface at the interface:
ε o (E2 − E1 ) A = (σ u + σ p ) A
or
ε o (E2 − E1 ) = σ u + σ p
The parallel component of the E-field at a material
interface is always continuous at the interface (no
change here)
(E2 − E1 ) = 0
σu + σ p
E1
ε1
E2
ε2
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
8
Dielectrics Vs Conductors
φ=0
φ=V
Conductors:
+
+
+
+
+
+
+
+
Free unpaired charges move to completely
screen the E-field on time scales longer than
the dielectric relaxation time
Dielectrics:
Paired charges originating due to material
polarization partially screen the E-field
almost instantaneously
--
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
-
V
φ=0
φ=V
When σu = 0:
+
+
+
+
+
+
+
+
ε 2 E2 − ε 1 E1 = 0
also:
ε o (E2 − E1 ) = σ p
The discontinuity of the normal component of the
E-field is due to the paired charges at the interface
(even when σu = 0)
-
+
+
+
+
+
+
+
+
+
-
V
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Conductors or Dielectrics
Some materials are conductors and some are dielectrics ………….
Dielectrics
Tightly Bound Electrons
Conductors
Sea of Free Electrons
+
+
+
+ +ve nucleus
+
+
+
+
+
+
-ve electron
cloud
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
ε
σ
+ +ve nucleus
-ve sea of free
electrons
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
9
Conductors and Dielectrics
……. but many important materials are conductors as well as dielectrics
Tightly bound core electrons and a
sea of free electrons
• Most conductors like gold, copper,
and silver, and semiconductors like
Silicon, are both conductors and
dielectrics
• They have a sea of free electrons
that results in a finite value of
conductivity and they also have
tightly bound core electrons that
result in a value for the dielectric
permittivity
+
+
+
+ +ve nucleus
+
+
+
-ve electron
cloud
+
+
+
+
+
+
-ve sea of free
electrons
+
+
+
+
+
ε
+
σ
Dielectric relaxation time for these materials = τ d =
ε
σ
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Appendix: Polarization Charge Density - I
The expression relating the polarization charge density to the divergence of the
polarization vector,
r
ρ p = −∇ . P
can be proved more formally as shown below:
The potential of an isolated dipole sitting at the origin and pointing in the
z-direction is:
r
r
r
p = qd
+q
r
φ (r ) =
θ
r
d
−q
p cos(θ )
4πε o r 2
More generally, the potential of a dipole sitting at position r’ and pointing in an
arbitrary direction is:
r
r
p = qd
+q
r
d
r
φ (r ) =
r r r
p . (r − r ' )
4πε o rr − rr ' 3
1
−q
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
10
Appendix: Polarization Charge Density - II
r r
If a material has a polarization density vector P (r ' )
then the potential due to all the dipoles can be found
using superposition:
r
φ (r ) = ∫∫∫
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
-
+
r r
r r
P (r ') . (r − r ')
dV '
r r3
4πε o
r − r'
1
r r ⎡ r ⎛ 1 ⎞⎤
P (r ') . ⎢∇ ' ⎜⎜ r r ⎟⎟ ⎥ dV '
4πε o
⎣ ⎝ r − r ' ⎠⎦
r
r
r ⎡ 1 ⎤
1
= ∫∫∫
− ∇ ' . P (r ') ⎢ r r ⎥ dV '
4πε o
⎣r − r'⎦
r r r
− ∇ ' . P (r ' )
= ∫∫∫
r r dV '
4πε o r − r '
= ∫∫∫
-
1
[
Now integrate by parts in 3D……….
]
This looks like the superposition
integral for a volume charge
density given by:
r
r r
ρ p (r ' ) = −∇ . P (r ')
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
11
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