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PHYS2012
EMP10_03
DIELECTRICS – MACROSCOPIC VIEW
DIELECTRIC MATERIALS
The term dielectric comes from the Greek dia + electric, where dia means through, thus
dielectric materials are those in which a steady electric field can be set up without casuing an
appreciable current.
Matter is usually neutral with an equal numbers of negative and positive charges. In dielectric
materials, these charges are not free to move far under the influence of an applied external
electric field, as are conduction electrons in a metal conductor. However, the forces due to an
external field do cause small relative displacements (on an atomic scale) of the charges of
each sign. The extent of such displacements depends upon the tightness with which the
charges are held fixed. Also, polar molecules rotate in the external electric field to try an
align with the electric field. This displacement of the charges and rotation of molecules
resulting from an applied external electric field is called polarization P of the material. The
dielectric constant is a measure of the extent of the polarization. The parameter that directly
relates the polarization of the material to the applied electric field is called the electric
susceptibility  e where e   r  1
P  e  0 E
We can start with a very crude model to explain the behaviour of dielectric materials. We
assume to a continuum of two uniform charge distributions of opposite signs. In the absence
of an applied electric field, the positive and negative charge distributions are exactly
superimposed. When an external electric field is applied, the positive distribution is displaced
in the direction of the external electric field and the negative distribution is displaced in the
opposite direction. This results in a cancellation of the charges in the interior of the dielectric
material (bound charge density [C.m-3] b  0 and results in induced bound (polarized)
surface charges +qb and -qb at the end surfaces of the dielectric material. The resulting surface
charge density  b [C.m-2] is
dq
b  b
dA
In this case, the bound surface charge density determines the polarization
where n̂ is a unit normal vector pointing away from the dielectric.
 b  P nˆ
On the atomic level, there is another possible point of view. In the dielectric material each
atom or molecule is distorted to produce an electric dipole and a dielectric material can be
though of as consisting of large number of electric dipoles. We can attribute the effect of
polarization of the material to be the sum of all the fields of all the dipoles.
Separation of charge  electric dipole
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A dipole consists of two equal and opposite charges + and –q
separated by a vector distance d
d
pe  qd
dipole moment p = pe = q d
-q
+q
pe
points from negative to positive
Induced dipole moment – helium atom
E
+2e
-e
Zero electric field –
helium atom
symmetric  zero
dipole moment
-e
-e
+2e
A
-e
d
B
effectively charge +2e at A and -2e at B
dipole moment
p = 2ed
p
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Potential and electric field from an electric dipole
V ( P) 
1 q q
q  r2  r1 
  


4   r1 r2  4   r 1r2 
Er
E
r  d
V ( P) 

P
q d cos 


d2
4   r 2  cos2  
4


r2  r + (d/2)cos
q d cos  p cos 
p r


2
2
4  r
4  r
4  r 3
r1  r – (d/2)cos
r
r extends from the centre of the
dipole to the point P
(d/2)cos
The radial and tangential components
of the field at point P are

V 2 p cos 

r
4  r 3
1 V p sin 
E  

r  4  r 3
-q
Er  
along the axis of the dipole
along the right bisector of the dipole
d
+q
 = 0  E = 0
 = /2  Er = 0
Electric field approaches zero much more quickly than a point charge. ??? Why ?
ELECTRIC DIPOLE PLOT– MATLAB
Why is difficult to plot the
potential in a plane passing
through the axis of the dipole?
Potential: Electric Dipole
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
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% electric_dipole.m
% Ian Cooper School of Physics,University of Sydney
close all
clear all
clc
% emconstants ------------------------------------------------------c = 3.00e-8;
% speed of light
e = 1.602e-19;
% elementary charge
eps0 = 8.85e-12;
% permittivity of free space
NA = 6.02e23;
% Avogadro constant
me = 9.11e-31;
% electron rest mass
mp = 1.673e-27;
% proton rest mass
mn = 1.675e-27;
% neutron rest mass
h = 6.626e-34;
% Planck's constant
kB = 1.38e-23;
% Boltzmann's constant
kC = 8.988e9;
% Coulomb constant
mu0 = 4*pi*1e-7;
% permeability of free space
amu = 1.66e-27;
% atomic mass unit
% Setup ------------------------------------------------------------q = e;
% dipole charge
d = 1.6795e-018;
% dipole separation distance
q1 = q; q2 = -q;
% separated charges
kc = 1/(4*pi*eps0);
% constant in Coulomb's Law
x1 = d/2; x2 = -d/2;
% position of dipole
y1 = 0; y2 = 0;
scale = 1.25;
% plotting region
xmax = scale * d;
ymax = xmax; xmin = -xmax; ymin = -ymax;
% plane above diople
num = 100;
x = linspace(xmin,xmax,num);
y = x;
[xx yy] = meshgrid(x,y);
r1 = sqrt((xx-x1).^2 + (yy-y1).^2);
% distance from charges
% to test point to calc. potential
r2 = sqrt((xx-x2).^2 + (yy-y2).^2);
V1 = kc .* q1 ./ (r1);
% potential from each charge
V2 = kc .* q2 ./ (r2);
Vtot = V1 + V2;
Vmax = max(max(Vtot));
sat = 0.5;
% saturate the potential
Vtot(Vtot > sat*Vmax) = sat * Vmax;
% potential near a charge
%
is extremely large
Vtot(Vtot < -0.5*Vmax) = -sat * Vmax;
Vtot = Vtot/(max(max(Vtot)));
figure(2);
% [3D] plot
surf(xx/d,yy/d,Vtot,'FaceColor','interp',...
'EdgeColor','none',...
'FaceLighting','phong')
daspect([1 1 1])
axis tight; view(-45,20)
camlight left; colormap(jet)
grid off; axis off
colorbar
title('Potential: Electric Dipole')
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POLARIZATION
The quantity of real interest is not an individual dipole moment but the electric dipole
moment per unit volume. In a region of uniform polarization, the polarization is then
Pn p
Where p is the induced atomic dipole moment and n is the number of electyric dipoles per
unit volume.
The word polarization has two meanings: a qualitative one referring to any relative
displacements of positive and negative charge and the quantative one, giving the resulting
vector dipole moment per unit volume, P
How is the macroscopic measureable quantity, the dielectric constant related to quantities at
an atomic level?
Conductors
Contain charges that are free to move and in the presence of an electric field,
redistribute themselves on the surface of the conductor so that the electric field is zero
in the interior.
Dielectrics

Induced dipoles (electronic)
– in an external electric field,
positive charge (nucleus) and
negative charges (electron
cloud) pushed in opposite
directions  induced electric
dipole moment.
+
-
+
-
Electric Field

Polar molecules eg H2O, N2O – neutral but a lopsided charge distribution – one side
excess positive and the other excess negative charge  permanent electric dipole
moment – zero electric field, random orientation of molecules in gases and liquids 
net electric dipole moment is zero. In an external electric field – dipoles experience a
torque to orientate them with the electric field – thermal agitation of the molecules
opposes the alignment  align is not perfect.

Ionic contribution to electric dipole moment – in a molecule some of the atoms have an
excess positive or negative charge resulting from the ionic nature of the bond: in an
electric field, the + ions and – ions are shifted in opposite directions.
Net electric dipole moment = (induced dipoles - electronic + permanent dipoles –
orientation + ionic dipoles - ionic)
p  pe  po  pi
Net polarization = (electronic polarization + orientation polarization
+ ionic polarization)
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P  n p  Pe  Po  Pi
The net polarization is related to a surface charge density bound and the density of bound
charges bound
 bound   b  P nˆ
n̂ is the normal pointing out of the volume
 Px Py Pz  three dimensional variation of the


 polarization
 x y z 
bound  b   P   
Simple model
In an electric field the alignment of the dipoles  layer of charge on each surface of the
dielectric  surface charge density bound (induced) – these bound charges are not free to
move as they are bound to the molecules. In the interior of the dielectric, the net dipole
moment is zero because of the cancellation of the charges. This phenomenon is called
polarization and the material is said to be polarized.
This is the reason why a charged object can attracts a neutral object.
If the polarization depends on time, then we may expect that the effect is similar to that of a
current
P
polarization current density J b 
t
and needs to be added to possible currents associated with free charges. Consider a medium
when an applied electric field is turned on. As a consequence the atoms or molecules form
small dipoles where none existed before – the alignment of the molecules constitutes a
current.
RESPONSE OF A MOLECULE TO AN ELECTRIC FIELD
We can assume the electric dipole moment and polarization are proportional to the electric
field experienced by the molecule (local electric field Eloc). Taking into account the three
contributions:
p  (e  o  i ) Eloc   Eloc
Electric dipole moment
P  n ( e   o  i ) Eloc  n  Eloc
Polarization

atomic polarizability
e
electronic or molecular polarizability
o
orientation polarizability
i
ionic polarizabiltiy
Macroscopic view (homogeneous electric field and r independent of E)
P  e  0 E   r 1  0 E
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E is the electric field within the dielectric
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We can now relate the macroscopic quantities - the electric susceptibility e and the dielectric
constant r to a property of the molecules, called the atomic polarizability . The atomic
polarizability relates to the ease in which electric dipoles moments can be formed giving rise
to the polarization of the material and hence to the dielectric constant of the material.
E073
Eloc
– the local (inner) field
– electric field acting upon a molecule within a dielectric
Dielectric – not continuous – composed of molecules
S
What is the local (inner) field Eloc that acts
upon an individual molecule within the
dielectric?
Consider dielectric between the plates of a
parallel plate capacitor.
The local electric field at the point O consists
of 4 parts:
+f
-b
O
+b
r
-f
1 Field at O due only to the charged plates
E1 
f
0
2 Polarization of the charges on the surface of the dielectric

P
E2   b  
0
0
3
Polarization of charges on the surface of S which would be formed if the spherical
section of the dielectric was removed
P
E3 
3 0
4
Polarization from the polar molecules within the spherical section, E4
P
E4  K 4
K4 some constant, usually, E4 can’t be calculated exactly.
0
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Hence the local electric field at O is
Eloc  E1  E2  E3  E4
Eloc 
Eloc 
f P P
P
 
 K4
 0  0 3 0
0
DP
0
Eloc  E 

P
P
 K4
3 0
0
P
P
 K4
3 0
0
Eloc  E  K
P
electric field inside dielectric
0
E
DP
0
K is some positive constant.
This equation gives the electric field Eloc that acts upon a single molecule of the dielectric.
For dielectrics with E4  0, K4  0 and K = 1/3, the total local electric field at O is
Eloc  E 
P
3 0
This is a useful result as this equation is applicable to cubic crystals, dilute solutions and
gases.
Calculation of E3
E3 
P
3 0
Assume a spherical section is removed from the dielectric. E3 is found by summing the
contributions to the field of all ring elements of polarization charge on the surface S.
Width of ring r d 
Radius of ring r sin
+ + +
E
surface S
r 
r sin 
emp10_04.doc
-
Area of the shaded ring
between  and  + d
 2 r sin   rd 
d
-

Pcos
P
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The charge density s on the spherical surface is given by the component of the polarization
normal to S
where  is measured w.r.t. E
 s  P nˆ  P cos
Surface area of the ring element is
dS  2 r sin  r d
The charge on the surface element dS that lies between  and  + d is
dq    P cos  2 r sin  r d 
By symmetry, all components of
the electric field that are not normal
to the capacitor plates cancel, the
electric field due to an element of
charge is in the vertical direction.s
+ + +
E
electric f ield at O
due to charge dq e
Hence, the electric field due to an
element of charge dqe (Coulomb’s
Law) iis
1 dqe
E0 cos  
cos 
4 0 r 2
E0 cos
-

-
E0
-
element of charge dq e
The electric field due to the ring with charge dq is
dE3 
dq
1
4  0 r
2
cos  
1
  P cos   2 r sin   rd 
4  0
r
2
cos  
P
20
cos2  sin  d
The resultant field at the centre of the sphere is obtained by integrating over  = 0  
E3 
E3 
P
20


0
cos 2  sin  d 

P
P
cos3   
 1  1
0
(2)(3)  0
(2)(3)  0
P
3 0
 the local electric field is greater than the electric field within the dielectric
because of the contribution of E3.
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NONPOLAR DIELECTRICS
Molecules without any intrinsic dipole moment will acquire an induced dipole moment in an
electric field, and so such molecules have a dielectric constant. The electrons shift position
slightly inside their molecules and do so very quickly, around 10-15 s, so that temperature
and frequency have little effect.
Monatomic gases (nonpolar)
We will consider the rare gases such as helium and argon because of the simple theoretical
model that can be used, although for most practical purposes it is not very useful.
Simple model of a single atom (gives results that are correct to an order of magnitude)
Positive nucleus +Ze and electrons -Ze
Atomic nucleus: diameter ~ 10-10 m nuclear diameter ~ 10-15 m
Nucleus point charge and electron cloud of charge –Ze distributed homogeneously
throughout a sphere of radius a  10-10 m
When the atom of radius a placed into external electric field E (E =Eloc)  nucleus and
electron cloud move in opposite directions to create an induced electric dipole  equilibrium
established with the nucleus shifted slightly relative to the centre of the electron cloud by a
distance d.
+Ze
+Ze
d
a
a
d << a
E
The nucleus will experience a force in the direction of the electric field FE
FE = Ze E
and an opposing force Fec due to the electric field of the negative charge of the electron cloud
which is assumed to acts as a point charge at the centre of the cloud The electric field Eec
experienced by the nucleus at a distance d from the centre of the negatively charged electron
cloud is determined by the application of Gauss’s Law
 Z e (d 3 / a 3 )
Eec  4 d 2  
0
2
Z e2 d
Fec  Ze Eec 
4  0 a3
emp10_04.doc
where the charge enclosed is (-Ze)(d3/a3)
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Fec  FE
4  0 a3
 d
E
Ze
The displacement distance d is proportional to the external electric field E.
For the single atom Eloc = E, the molecular (electronic) polarizability of a monatomic gas is
  e
pe   Eloc
4  0 a3
  E  ( Ze) d  ( Ze)
E
Ze
 e    4  0 a3
The electronic (molecular) polarizability is proportional to the volume of the electron cloud
(  a3)  the larger the atom, the greater the charge separation and the greater the induced
dipole moment: bigger the atom  the larger   r
10-40 F.m2

He
0.18
Ne
0.35
A
1.43
Kr
2.18
Xe
3.54
Now we consider a rare gas containing n molecules.m-3 and we can neglect any interactions
between the induced dipoles in the atoms (good approximation for a gas).
Microscopic view: the polarization of the gas P is
P  n pe  n  E
Macroscopic view: the polarization of the gas is
P  e  0 E   r  1  0 E
Therefore we can relate the microscopic molecular polarizability  with the macroscopic
dielectric constant r
n
r  1 
 1  4 n a 3
0
We have obtained a relationship between the measurable quantity r and the microscopic
quantities  and a. How good is our simple model?
Helium gas: temperature, T = 300 K and pressure, p = 1 atm = 1.1105 Pa
Dielectric constant r = 1.0000684
1/ 3
  1 
p V  N k T  n= N/V = 2.510 atoms.m  a   r

 4 n 
25
-3
 6  1011 m
Correct order of magnitude !!! our simple model not too bad
We can estimate the relative shift d between the nucleus and the centre of the electron cloud
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E ~ 105 V.m-1
a ~ 10-10 m
Z ~ 2  d ~ 10-17 m
very small – very slight perturbing influence of the applied electric field on the atom
Number density n
The number density for a gas is obtained from the ideal gas equation
N
p
kB T  n kB T
n
V
kB T
o
For a gas at atmospheric pressure and 20 C, the number density n is
pV  N k B T
p
p = 1 atm = 1.013105 Pa T = 20 oC = 293 K
kB = 1.381023 J.K-1
n = 2.51023 molecules.m-3
For a solid or liquid (density , molecular mass m, number of molecules N,mass of sample
msample) the number density n is obtained as follows

msample
n
V

Nm
V
M  NA m
m
M
NA
n
N
V
 n
M
NA
NA
M
Avogadro’s number NA = 6.021023 molecules.mol-1
Molar mass M (in kilograms)
For copper
 = 8.93103 kg.m-3
M = 63.5 g = 63.510-3 kg
n = 8.51028 atoms.m-3
Note: the number density of solids is much greater than that of gases.
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Gases, dilute solutions and simple solids (nonpolar)
Gases, dilute solutions, solids - one kind of atom eg diamond, phosphorus (cubic crystals),
No permanent dipole moments or ions
Polarization due to relative displacement of electron clouds and nuclei
Local electric field same for all atoms
K = 1/3 (no polar molecules E4 = 0 K4 = 0)
 = e
P
P    r  1  0 E
3 0
Combining these three equations gives the Clausius-Mossotti relationships
P  n pe  n  Eloc

Eloc  E 
3  0   r  1
n  r  2 
 r  1  n 
 r  2  3  0
The distance between atoms in a solid is affected only slightly by temperature and therefore,
n, , K and r are in a first approximation independent of the temperature.
For a solid, a typical value for the number density is n ~ 51028 m-3.
The dielectric constant for three solids with a diamond structure are:
r(C) = 5.68
r(Si) = 12
r(Ge) = 16
The dielectric constant for the gases are very close to 1 eg r(H2) = 1.000132.
Why is the dielectric constant for a solid much greater than for a gas?
If r very close to 1  r+2  3   r  1 
emp10_04.doc
n
0
22-Jun-10
the same equation as for monatomic gases
13
POLAR DIELECTRICS
Consider the dielectric material containing n molecules.m-3. Assume that each molecule has a
permanent electric dipole moment p. Water is a typical polar liquid. Its constituent molecules
have a permanent dipole moment.
The polarization is due to the electronic polarization Pe (nucleus shifted slightly relative to
the centre of the electron cloud) and the ionic polarization Pi (ionic nature of bond between
atoms) and the orientation polarization Po (rotation and alignment of the polar molecules in
the external electric field).
Pe and Pi are essentially independent of the temperature but Po is very temperature dependent.
At a temperature T and zero external electric field, the molecules will be randomly oriented
 zero polarization.
When there is an external electric field, the molecules will try to align with the field. Each
polar molecule can be considered to be a simple dipole. The force on the dipole provides the
torque to rotate the molecule so that they will be in the lowest energy state where they are
parallel to the field. If there were no thermal motion, all dipoles would line up along the
external field direction.
F
+Q
p

d
F
-Q
E
The electric force on the dipole produces a couple and the torque acting to rotate the dipole
about its centre is
d
d
  F sin   F sin   Q E d sin   p E sin 
  p E
2
2
Set the potential energy U( ) of the dipole to zero when  = 90o. The potential energy of the
dipole for an arbitrary angle  is then given by
 90o
+
-
emp10_04.doc
U=0
-
U=- p E
Lowest energy state
=0
+
E
p E sin  d   p E cos    p E
-

22-Jun-10
 = 90o
+
U ( )  
U=+ p E
highest energy state
 = 180o
14
The dipole has the lowest potential
energy when the dipole is parallel to
the electric field and the highest
energy when anti-parallel to the field
 small angles are preferred over
larger ones. And if they were no
thermal motion, all dipoles would line
along the direction of the external
electric field. The greater the
temperature, the greater the thermal
motion  reduced alignment of the
dipoles with the field.
+pE
U
0
-p E
0
π/2

π
The orientation polarization Po is given by
the Langevin function (1905)



 pE
1 

Po  n p  coth 



 kT  pE 

k T 

pE
kt
1
Po
T
Po  1
1
Complete alignment – this does
not occur in gases
pE
kt
Po
np
 pE 
Po  n p 

 3k T 
Most practical case:
1
Po  p 2
Po 
T
1
slope = 1/3
0
pE/kT
10
The total polarization of a polyatomic gas is given by where Eloc = E
P  Pe  Pi  Po

p2 
P  n  e  i 
E
3k T 

Macro view P  e  0 E   r 1  0 E
 r is related to the molecular properties by
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
 r  1  0  n  e  i 

p2 

3k T 
How well does this prediction agree with experiment?
If dielectric constant r plotted against 1/T  straight line
 measurement of p
Slope
n p2 / 3 k
Intercept
n (e + i)  measurement of (e + i)
r - 1
intercept 
n
0
slope 
n P2
3 kB T
 e   i 
1/T
Dipole moments of gases in debye units (3.3310-30 C.m)
NO
0.1
CO
0.11
HCl
HBr
0.79
HI
0.38
NO2
CO2
0
CH4
0
H2O
H2
0
A
0
NH3
1.04
0.4
1.84
1.4
Dielectric constant measurements have played an important part in determining molecular
structure: CO2 has zero resultant dipole moment, whereas each CO bond does have a nonzero dipole moment  O=C=O. H2O molecule must have a triangular structure.
At ordinary temperatures and electric fields, the average dipole moment and hence
polarization are greatly reduced by the thermal agitation – the molecules point in every way
with only a slight alignment with the external electric field. In the low electric field
approximation
 pE 
pE
1
Po  n p 

kt
 3k T 
kT is the thermal energy and pE is the energy of the dipole molecule when aligned with the
effective electric field
The polarization is reduced by the factor (pE / 3kT) and the greater the temperature the
greater the reduction in the polarization and the polarization increases with increasing electric
field strength.
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The dielectric constant of water is 80 and the molecule dipole moment of a water molecule is
6.210-30 C.m. Find the electric field required to maximize the polarization of water.
Water:
M = 1810-3 kg  = 103 kg.m-3
p = 6.210-30 C.m
NA = 6.0210-23 mol-1
To maximize the polarization, all the water molecules need to be aligned. The polarization
and electric field are given by
P
P  e  0 E   r  1  0 E  E 
 r  1  0
N
n   A  3.34 1028 molecule.m3
Number density
M
P  n p = 0.20 C.m-2
Polarization
Electric field
E
P
0.20

V.m -1  3.0 108 V.m -1
 r  1  0 80  1 8.854 1012 
This means that the water molecules are not all aligned up until the electric field reaches
300 MV.m-1, which is enough to break down water and produce an electric arc.
E073
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