Dielectric properties -final-converted-1

Module II: Dielectric Properties
Dr. L. N. Patro
GITAM University
Bangalore
1
Dr. L. N. Patro, GITAM University

Introduction
Fundamental definitions
Local field
Clausius-Mossotti relation
Different types of electric polarizations
Electronic polarization
Ionic polarization
Dipolar polarization
Dielectric loss
Dielectric breakdown
Piezoelectricity and Ferro-electricity
Spontaneous polarization in BaTiO3
Application of dielectrics and Ferro-electrics
Syllabus of
module II
2

3
Introduction
Dielectrics are the materials with no or very small amount of charge carriers. Thus
they are mostly insulators.
In dielectrics, the electrons are tightly bound the nucleus. So there are no free
electrons.
They are characterized by high specific resistance (1010 to 1020 ohm-m), negative
temperature coefficient of resistance and large insulation resistance. The forbidden
gap is very large.
Ex: glass, plastic, rubber, wood, wax etc.
If the dielectric is only used for insulation purpose then it is called insulating
material. If it is used for the charge storage then it is referred as dielectrics
All dielectrics are insulators but not all insulators are dielectrics.

Dr. L. N. Patro, GITAM University 4
The capacitance of a parallel plate capacitor increases by introducing a slab of
dielectric medium between the parallel plate capacitor.
In an external electric field, these materials get electrically polarized and polarization
occurs either due to field induced dipoles or field oriented dipoles.
They find extensive use in the electrical industry for insulation purposes. They also
find many engineering applications involving sensors.
0
0
, ,
, ' r
r
A
We know that without dielectric C
d
A
with dielectric C C
d

 

=
= =

Fundamental definitions in Dielectrics
❖ Electric dipole:
Two equal and opposite charges separated by a small distance constitute an electric
dipole.
❖Electric dipole moment (μ)
✓The product of the magnitude of any one of the charge of the electric dipole and the
distance between two charges is called electric dipole moment.
✓It is denoted by ; where q is the charge and d is the distance between the
charges of the dipole.
✓It is vector quantity and its direction is from negative charge to positive charge.
✓Unit: coulomb-metre.
qd
 =
+q -q
d
qd
 =

❖Polarization (P)
✓The electric dipole moment per unit volume is called polarization.
✓It is denoted by P.
qd q
P
V Ad A

= = =
✓It is numerically equal to surface charge density.
✓It is a vector quantity and its direction is along the direction of the dipole moment.
✓Unit: coulomb/metre2
❖Polarizability (α)
The dipole moment , μ induced in an atom is proportional to the electric field, E i.e.
E
E

 

=
✓α is the constant of proportionality known as polarizability.
✓If there are N atoms per unit volume, the polarization of the solid
is P = Nμ = NαE

❖Electric field
✓The region surrounding an electric charge or group of charges in which another
charge experiences a force is called electric field.
✓The strength or the intensity of electric field is denoted by E;
✓It is a vector quantity.
✓Unit of E is volt/metre
2
0
4
q
E
r

=
❖Electric flux (φ) and electric flux density (D)
✓The number of electric lines of force from a charge is called electric flux.
✓It is denoted by φ
✓Unit: weber

where A is the area of the dielectric. Thus the surface charge per unit area is called
electric flux density.
✓It is a vector quantity and its direction is in the direction of the electric field intensity
✓Unit: coulomb/metre2.
where ε is the absolute permittivity of the medium. ε = ε0 εr, ε0 is the permittivity of the
free space and is given by ε0 = 8.854×10-12 farad/metre and εr is the relative permittivity
of the medium.
2
2 2
4
,
4 4
q
E
r
q q q
therefore D
r r A


 
=
= = =
✓Electric flux density, D is proportional to the electric field intensity, E
0 r
D E
D E E
  

= =
✓The number of electric lines of force passing normally through a surface per unit
cross sectional area is called electric flux density or electric displacement, D.
✓We know

❖Dielectric constant
✓We know the electric field intensity, E and electric flux density, D are related by
0 r
D E E
  
= =
where ε is the absolute permittivity of the medium. ε = ε0 εr ; ε0 is the permittivity of the
free space and is given by ε0 = 8.854×10-12 farad/metre. εr is the relative permittivity of
the medium and is also called dielectric constant.
✓The dielectric constant is thus defined as the ratio of the permittivity of the medium to
the permittivity of the free space.
✓We know that the capacitance of a parallel plate capacitor increases by introducing a
slab of dielectric medium between the parallel plate capacitor.
0
0
,
, ' r
r
A
Without dielectric C
d
A
with dielectric C C
d

 

=
= =

✓Thus the dielectric constant is defined as the ratio of the capacitance of a capacitor
containing dielectric medium to the capacitance of the same capacitor with air as
medium
✓The value of dielectric constant is different for different materials.
✓It is a dimensionless quantity and has no units.
✓It describes the ability of the medium to store the electrical charge.
❖Dielectric Susceptibility, χ
When a dielectric is placed in an electric field, then the dielectric is said to be
polarized. The polarization P is proportional to the applied field E
That is
0
P E
P E
 

=
✓ χ is a constant and is called susceptibility of dielectric material.
✓It characterizes the ease with which the material gets polarized.
✓It has no units.

11
Relation between three electric vectors, D, E and P
When a dielectric material is placed between the plates of a charged capacitor, positive
charges are displaced toward the field and negative charges shift in the opposite
direction. Thus charges are induced on the surface of the dielectric. Let qʹ be the
induced charge on the dielectric and the electric field (Eʹ) due to these charges in the
opposite direction to the electric field (E0) due to the charges on the capacitor plates.
The resultant electric field between the plates is
'
0
'
'
0
0 0
'
0 0
'
0
'
0
0
-
,
, -
-
.
E E E
q q
where E and E
A A
q q
Therefore E
A A
q q
E
A A
q q
E
A A
D E P
This is the relation between three electric vectors
 
 



=
= =
=
=
= +
= +

Relation between dielectric constant (εr) and the dielectric
susceptibility (χ)
,
,
,
, ( -1) .....................(1)
, ..................(2)
o
o r
o r o
o r
o
We know the electric displacement D E P
and D can also defined as D E
therefore E E P
or P E
We know P E
So from equatio

 
  
 
 
= +
=
= +
=
=
1 2, ( -1)
1
tan .
sin 2,
r
r
n and
or
This is the relation between dielecric cons t and dielectric susceptibility
U g the above relation in equation one can arrive the relation between
polarisation an
 
 
=
= +
tan
( -1)
o o r
d dielectric cons t
P E E
   
= =

Problem 1: A parallel plate capacitor has an area of 100 cm2, a plate separation of
1cm and is charged to a potential of 100volts. Calculate the capacitance of the
capacitor and the charge on the plates.
2 -4 2 -2 2
-2
-12
0
-12 -2
-12
0
0 -2
100 100 10 10
1 10
, 100
, 8.85 10 /
8.85 10 10
8.85 10 8.85
10
Area A cm m m
Separation d cm m
Potential applied to the plates V volt
we know F m
A
C farad pF
d
The ch


= = =  =
= = =
=
= 
 
= = =  =
-12 -10
0
arg , 8.85 10 100 8.85 10
e on the capacitor Q C V C
= =   = 

Problem 2: Two parallel plates having equal and opposite charges are separated
by 2cm thickness of the dielectric slab that has dielectric constant of 1.52. if the
electric field inside is 3×106 V/m. Calculate the polarization and displacement
vectors.
-2
6
-12
0
-12 6 -5 2
0
2 2 10
tan , 1.52
, 3 10 /
8.85 10 /
, 8.85 10 1.52 3 10 4.0356 10 /
,
r
r
d cm m
dielectric cons t
Electric field applied E V m
F m
Displacement current D E C m
The relation between D E a


 
= = 
=
= 
= 
= =     = 
0
-5 -12 6 -5 2
0
;
, - 4.0356 10 - 8.85 10 3 10 1.3806 10 /
nd P is D E P
or P D E C m


= +
= =     = 

Problem 3: Determine the dielectric susceptibility for a gas whose dielectric
constant is 1.000057.
1
1.000057
1.000057-1 0.000057
r
r
 


= +
=
= =
Problem 4: When NaCl crystal is subjected to an electric field of 50 V/Cm , the
resulting polarization is 2.215 ×10-7 C/m2. Calculate the relative permittivity of the
NaCl.
o r
o r o
o o r
-7 2
o
r -12
o o
P = ε (ε -1)E
P = ε ε E - ε E
P + ε E = ε ε E
P+ε E P 2.215×10 C/m
ε = = 1+ = 1+ = 6.006
ε E ε E 8.85×10 F/m×50V/cm×100cm/m

16
Dielectric polarization
•Consider a electrically neutral slab of an isotropic dielectric inserted between the plates
of a charged parallel plate capacitor.
•Dielectric materials have no free charge carriers. Hence current does not flow in the
dielectric material.
•However the electric field can act on the bound charges in the dielectric. These bound
charges are not free to migrate through the dielectric. The action of the field Eo on the
bound charges consist in displacing the bound charges relative to one other.
•The negative charges (electrons) are displaced in a direction opposite to the field, while
The positive charges are displaced in the same direction as that of the applied field.
Unpolarised Polarized by an electric field
Eo
Eʹ

•Negative charge is induced adjacent to the positive capacitor plate while positive
charge is induced adjacent to the negative capacitor plate. Thus the action of the electric
field on the dielectric is to induce charges on its surface
•When the charges of opposite polarity are induced on the surface of a dielectric. The
dielectric is said to be polarized. The effect is known as polarization.
•The induced surface charges on the dielectric give rise to an induced electric field Eʹ
which opposes the external field Eo. Therefore the net electric field in the dielectric has
a magnitude of E = E0 – Eʹ.
•Thus each atom or molecules then acts as an elementary dipole and acquires an electric
dipole moment in the direction of the field.

Induced dipoles
•It can be understood by considering the action of the electric field on an atom.
•In an atom the nucleus is about 10-15 m in diameter and can be regarded a point.
similarly the electron cloud (negative charge) may be assumed to be concentrated at
its centre.
• If the centre of gravity of positive and negative charges in atom coincide. Such an
atom does not produce any electric field of its own.
+ -
d
E
+ -

•If the atom is placed in an electric field, the electron cloud will be displaced in the
direction opposite to the electric field by a distance d with respect to nucleus. Thus, the
centre of gravity of positive charge and negative charge no more coincide. The atom is
now equivalent to a system of two charges of equal magnitude separated by a distance .
Such a system is called electric dipole. Thus a dipole is induced under the action of an
electric field.
•The induced dipoles setup an electric field opposite to that of external field.
•The larger the field, the greater is the displacement, hence larger the dipole moment.
•Induced dipole moment vanishes as soon as electric field is switched off
Permanent dipoles
In some molecules known as polar molecules, the centers of gravity of the charges of
opposite sign are separated even in the absence of an external field. Such molecules are
said to have intrinsic dipole moment and carry permanent dipoles.

•When a molecule having permanent dipoles is placed in a uniform electric field, the
field exerts a force +qE on charge +q and –qE on charge -q. So the net force on the
dipole is zero. So there is no translational force acting on the dipole in an uniform
electric field.
•However the force tend to rotate the dipole. The torque acting on the dipole is
τ = qEd Sinθ = μ E Sinθ = μ×E
•The total dipole moment of the molecule, μT = μind+ μper
•In case of polar molecules, μind << μper
•Thus μT = μper

Nonploar and polar dielectrics
•All positive charges of a molecule may be replaced by one equivalent positive charge
located at the center of gravity of positive charge. Similarly, all negative charges of a
molecule may be replaced by one equivalent negative charge located at the center of
gravity of a negative charge. Two resultant charges are of equal magnitude. Their
point of action may or may not coincide.
•When the points of resultant charges of a molecule coincide, the molecule will not
posses a permanent dipole moment. Such molecules are called nonpolar molecules
and the material is called nonpolar dielectric.
•When the points of resultant charges of a molecule do not coincide, the molecule
posses a permanent dipole moment. Such molecules are called polar molecules and the
material is known as polar dielectric.
•In nonpolar dielectrics, the permittivity values are low and range from 1 to 2.2.
however in polar dielectrics, the permittivity values are high and range from 3 to 8

•A molecule is polar or nonpolar could be judged by its structure.
•Symmetric molecules are nonpolar since the centers of gravity of positive and
negative charge coincide with each other Ex: H2, N2, Cl2
•Asymmetric molecules are polar. Ex: KI
•It is always important to consider the actual charge distribution rather than chemical
formula in order to determine the dipole moment of a molecule.
•For ex, CO2 and H2O look identical in form. But CO2 is nonpolar and H2O is polar.
•All hydrocarbons are nonpolar, But hydrocarbon becomes polar when hydrogen atoms
are replaced by other atoms. Ex: Methane , CH4 is nonpolar however methyl chloride
CH3Cl is polar.

•Why CO2 is nonpolar but H2O is polar?
• The dipole moment of two C = O bonds in CO2 molecule are oppositely directed
and cancel each other. Thus the resultant dipole moment is zero. However, the water
molecule has the form of an isosceles triangle with a bond angle of 104.5˙.
Consequently the resulting dipole moment of water molecule is nonzero (6.1×10-30
coulomb. meter) and the molecule is polar.

Types of polarization
Dielectric polarization is classified into four basic types
1. Electronic polarization
2. Ionic polarization
3. Orientation polarization
4. Space charge polarization
Electronic polarization
✓It occurs due to the displacement of the electron clouds of atoms or molecules with
respect to the heavy fixed nuclei to a distance that is less than the dimension of the
atoms or molecules or ions.
✓It sets over a very short period of time of the order of 10-14 to 10-15 s.
✓It is independent of temperature

Expression for electronic polarization
•Consider a single atom with atomic number Z. The charge on the nucleus is +Ze and
there are Z electrons move around the nucleus. Assume that the nucleus is a point
charge and the total negative charge –Ze is homogeneously distributed through out the
sphere of radius R.
•When this atom is subjected to an electric field E, the nucleus and the electron cloud
will move in opposite direction. However the coulomb attractive force opposes the
movement.
•Let us say equilibrium condition will be reached in which the nucleus is displaced
relative to the center of the electron cloud by the amount x .
•The force on the nucleus along the field direction is F = ZeE………………….(1)
•To determine the coulomb attraction of the nucleus, we divide the electron cloud into
two regions. one region is that is inside the sphere of radius x and the other is the
annular region lying between the two spherical surfaces of radii x and R.

•By applying Gauss theorem, the force experience by the nucleus arises due to the
negative charges lying within the spherical region of radius x. The charge inside the
region is given by
The force exerted by this charge on the nucleus is
given by
At equilibrium condition, above two forces balance each other. From equations 1 and 2,
Therefore the displacement of the nucleus is
3
3
Zex
R
−
( )
3
3
2
0
1
............(2)
4
Zex
Ze
R
F
x

 
 
 
= −
( )
3
3
2
0
1
4
Zex
Ze
R
ZeE
x

 
 
 
=
3
0
4 R
x E
Ze

=

3
3
0
0
3
0
4
( ) ( ) 4
,
, 4 ......................(3)
ind
ind e
e
R
Ze x Ze E R E
Ze
Weknow E
therefore R electronic polarisabilty

 
 
 
 
= = =
 
 
=
= =
Now the dipole moment induced in the atom due to the displacement is
Equation 3 is the expression for electronic polarizability.
The electronic polarization per unit volume is given by
e e
e e
r
0 0 0
3
r
P = Nα E
Nα E Nα
P
ε = 1+χ = 1+ = 1+ = 1+
ε E ε E ε
using equation 3, we have ε = 1+4πNR

Problem 5: Calculate the electronic polarizability of argon atom, Given
εr = 1.0024 at NTP and N = 2.7×1025atoms/m3
Problem 6: The number of atoms in hydrogen gas is 9.8×1020 atoms/cc.
The radius of hydrogen atom is 0.053nm. Calculate its electronic
polarizability and relative permittivity.
0
-12
-40 2
0
25
1
8.85 10 (1.0024-1)
( -1)
7.9 10 .
2.7 10
e
r
r
e
N
F
m
F m
N



 

= +
 

 
 
= = = 

3 -12 -9 3 -41 2
0
3 26 3 -9 3
4 4 3.14 8.85 10 / (0.053 10 ) 1.657 10 .
1 4 1 (4 3.14 9.8 10 / (0.053 10 ) 1.0018
e
r
R F m m F m
NR m m
 
 
= =      = 
= + = +      =

Ionic Polarization
✓Ionic polarization occurs in ionic crystals.
✓It occurs due to elastic displacement of positive and negative ions.
Expression for ionic polarization
✓Let us consider an ionic crystal sodium chloride (say NaCl) being subjected to an
electric field.
✓Under the action of the electric field, the positive ions (Na+) displace in the direction
of the electric field, say x1 and the negative ions(Cl-) displace in a direction opposite to
the electric field say -x2 ;the minus sign indicate the opposite direction
✓The net displacement of the ions = x = x1+x2
✓The force on the Na+ ion due to electric field = +eE and the force on Cl- ions due to
the electric field = -eE
✓The restoring force acting on the Na+ ion = -k1x1 and the restoring force acting on the
Cl- ion = k2x2

✓where k1 and k2 are the force constants and they are given by k1=Mω0
2 and k2=mωo
2.
Here M and m is the mass of the positive (Na+) ion and the negative ion (Cl-)
respectively. ω0 is the natural angular frequency of the molecule.
✓At equilibrium condition, the electric force and the restoring force are equal and
opposite to each other.
1 1 2 2
1 2
2 2
1 0 2 0
1 2 2 2
0 0
2
0
2
0
1 1
1 1
.
eE k x and eE k x
eE eE eE eE
x and x
k M k m
eE eE
The net displacement of the ions x x x
M m
eE
or x
M m
The induced dipole moment ex
eE
e
M m
 
 




= =
= = = =
= = + = +
 
= +
 
 
= =
 
= +
 
 
2
2
0
2
2
0
1 1
..........................................................(1)
1 1
ionic , ( sin ).................(2)
i
i
e E
M m
e
The polarizability ce E
M m
The ionic polarization is given by P

  

 
= +
 
 
 
= + =
 
 
=
2
2
0
1 1
...................(3)
i
Ne
N E E
M m


 
= +
 
 

31
✓Equation 1, 2, 3 respectively provide expression for induced dipole moment, ionic
polarizability and ionic polarization.
It is seen from the expression for ionic polarizability (equation 2) that
✓It is inversely proportional to square of the natural frequency of the molecule.
✓It is directly proportional to the reduced mass of the molecule,
✓It does not depend on temperature.
✓For most materials, the ionic polarizability is less than the electronic polarizability
typically
✓Ionic polarization takes 10-11 to 10-14s to build up.
1 1
M m
 
+
 
 
1
10
i e
 
=
Orientation Polarization
✓It is a characteristic of polar dielectrics i.e. the material consisting of molecules
having permanent dipole moment.
✓In the absence of the field, the orientation of dipoles is random, so the net dipole
moment is zero. However in the presence of the field, the molecular dipoles rotate and
align in the direction of the field resulting a net dipole moment.

✓Unlike ionic and electronic polarization, restoring force do not exist. In this case, the
dipole alignment is counteracted by thermal agitation. Thus the orientation polarization
is strongly temperature dependent.
✓It occurs in gases, liquids and amorphous viscous substances.
✓In case of solids the molecules are fixed at their positions and is highly restricted by
the lattice forces, which controls the contribution to the orientation polarization.
Because of this reason the dielectric constant of water is about 80 and that of ice is
about 10.

✓As the process of orientation polarization involves rotation of molecules, it takes
longer time (10-10s) than the electronic and ionic polarization.
Expression for orientation polarization
✓Consider a polar gas dielectric. If the electric field is sufficiently large, all the dipoles
in the gas rotate and align in the field direction. The orientation polarization would
reach saturation value, Po = Nμ, where N is the number of molecular dipoles per unit
volume of the gas.
✓At modest electric fields, the dipole alignment is not complete and the polarization
value is less than the saturation value.
✓In the case of orientation polarization, the effect of temperature tends to randomize
the orientation of dipoles. At thermal equilibrium a modest orientation of dipoles is
achieved.

✓In the equilibrium state, the molecular dipoles are distributed over all the directions
making angles varying from 0 to π radians with the field direction.
✓The potential energy of a dipole is given by U = - μ E cosθ
✓According to statistical mechanics, the number of dipoles dN, having orientations
between θ and θ+dθ is proportional to
Here dΩ is the solid angle and is given by dΩ = 2π sinθ dθ
✓A dipole making an angle θ with the field direction contributes a component of
dipole moment μcos θ. Hence the contribution of dN dipoles to the orientation
polarization is
B
U
k T
e d
 
−
 
 

cos
cos
2 cos sin
B
o
E
k T
o
ave
dP d dN
dP e d
Therefore the average contribution of polarization is givenby
Total polarzation due to all dipoles
P
Total number of dipoles
 
 
    
= 
=
=

( )
( )
cos
0
cos
0
-1
1
-1
1
-
-
cos sin
sin
P cos
1 1
- coth - ( )
-
( )
B
B
E
k T
ave E
k T
B
y
ave
y
ave
e d
P
e d
E
utting and y in to the above equation
k T
ye dy
P
e dy
e e
P
L
e e
L is called La
 

 



 
 
   
 

 

 
  

=
= =
=
+
= = =




2
. ,
( )
1
( )
3 3
3
B
ave
B
o a
ngevein function For electric field that are not high and for temperatures not too low
L is givenby
E
L
k T
E
P
k T
The total orientation polarisation of the dielectric is
P NP


 

= =
=
=
2
.................................(1)
3
ve
B
N E
k T

=

0
0
2
0
0
2
0
,
..................................(2)
3
( -1)
( -1)
3
o
B
r
r
B
As P N E
The orienation polarizability is givenby
k T
we also know that
P E
N
thus
k T




 

 
=
=
=
=
From equation 1 it is observed that the orientation polarization is directly proportional
to the square of the permanent dipole moment and inversely proportional to the
temperature

Space charge Polarization
✓It is also known as interfacial polarization or migrational polarization.
✓It occurs whenever there is accumulation of charge at an interface between two
materials or between two regions with in a material (Ex: homogeneous dielectrics
containing impurities, Grain boundaries).
✓It takes longer time to occur. Thus this polarization occurs at low frequencies.
✓It is complicated to calculate the charge on either interface or their contribution to
the total polarization of a dielectric. It is therefore omitted from the discussions on
dielectric properties. However it is very important when one investigates the
application aspects of dielectric materials in devices

Total Polarization
✓The total polarization is given by PTotal = PElectronic + PIonic + POrientation + PMigrational
In general migration polarization is small and negligible.
✓Thus PTotal = PElectronic + PIonic + POrientation
✓The total polarization of a polar dielectric is given by
 
2 2
3
0 2
0
2 2
3
0 2
0
1 1
4
3
1 1
4
3
e i o
B
e i o
B
P N E
e
P N R E
M m k T
The total polarizability is
e
R
M m k T
  




    

= + +
 
 
= + + +
 
 
 
 
 
= + + = + + +
 
 
✓In materials, one or more contribution may be present depending on the structure.
✓Orientation polarization does not exist in nonpolar dielectrics
✓Ionic polarization will not be found in covalently bonded compounds
✓Electronic polarization is negligible compare to orientation polarization in polar
dielectrics.

Temperature dependent Polarization
N(αe+αi)
Nμ2/3kB
1/T
P
or
ε
0
(ε
r
-1
)
 
2 2
3
0 2
0
1 1
4
3
e i o
B
P N E
e
P N R E
M m k T
  



= + +
 
 
= + + +
 
 
 
 
✓Electronic and ionic polarization do not
depend on temperature. However orientation
polarization is inversely proportional to
temperature.
✓The P vs 1/T plot will be a straight line.
✓Intercept of the line with y axis gives the value of N(αe+αi)
✓Slope of the straight line will give the value of Nμ2/3kB
✓So by knowing the slope, value of N i.e the number of molecules per m3, one can
calculate the dipole moment, μ

40
Category
Range of Wavelengths
(nm)
Range of Frequencies
(Hz)
Gamma rays < 1 > 3 × 1019
X-rays 1–10 3 × 1017 – 3 × 1019
Ultraviolet light 10–400 7.5 × 1014 – 3 × 1017
Visible light 400–700 4.3 × 1014 – 7.5 × 1014
Infrared 700 – 105 3 × 1012 – 4.3 × 1014
Microwave 105 – 108 3 × 109 – 3 × 1012
Radio waves > 108 < 3 × 109
Table 1 lists the wavelength and frequency ranges of the divisions of the
electromagnetic spectrum.
❖Audio waves: 20Hz to 20,000Hz
✓Orientation polarization involves rotation of molecules, it takes longer time (10-10s)
than the electronic and ionic polarization.
✓Ionic polarization takes 10-11 to 10-14s to build up.
✓Electronic polarization sets over a very short period of time of the order of 10-14 to 10-15 s

Frequency dependent Polarization
✓In many practical applications, a dielectric is subjected to an ac field. An ac field
changes its sign with time. With each direction reversal, the polarization components
are required to follow the field reversals in order to contribute the total polarization of
the dielectric.

✓In audio frequency region (20Hz to 20,000Hz) all types of polarization are possible
and the polarization is P = PElectronic + PIonic + Porientation
✓At low frequencies, the dipoles will get sufficient time to orient themselves in the
direction of the field. This orientation occurs first in one direction and then in the other
following the changes in the direction of the field. The average time taken by the
dipoles to reorient in the field direction is known as relaxation time and the reciprocal
of the relaxation time is called relaxation frequency.
✓If the frequency of the applied field is much higher than the relaxation frequency of
the dipoles, the dipoles can not reverse fast enough. Orientation polarization is
effective at lower frequencies.
✓Usually in microwave region (3 × 109 – 3 × 1012 Hz), the permanent dipoles fails to
follow the field reversal. So the polarization falls to the value of P = PElectronic + PIonic

✓In the infrared region (3 × 1012 – 4.3 × 1014Hz) , the ionic polarization fails to follow
the field reversal due to inertia of the system. In this region only electronic polarization
contributes to the total polarization. Therefore P = Pelectronic
✓In the optical region, the electron cloud follows the field variations and the material
exhibits an electronic polarizability. The relative permittivity in the optical region will
be equal to the square of the refractive index of the dielectric.
✓In the ultraviolet region the electron cloud too fails to follow the field alternations and
the electronic contribution to polarization ceases. Consequently, the total polarization is
zero.

Lorentz field (Local field)
✓The electric field acting at a molecule in a dielectric is known as polarizing field or
the local field and is different from the applied external field E.
✓The local field in a three dimensional space is determined by the structure of the solid.
✓Let us consider a dielectric slab kept in a uniform electric field E. Let a molecule be
at the point A and imagine that it will be surrounded by a spherical cavity of radius r. r
is arbitrary but sufficiently large compared to molecular dimensions and sufficiently
small compared to the dimension of the dielectric slab.

45
✓ The spherical cavity contains many molecules within it. The molecule at A
experiences three electric fields acting on it. Therefore the total internal field
intensity Ei is given by
Ei = E + E1 + E2…………………………………………………..(1)
Where E = External electric field
E1 = Field E1due to induced surface charges on the surface of the spherical cavity.
E2 = Field E2 due to molecular dipoles present in the spherical cavity.
✓Evaluation of E1
✓E1 is the field at a point A due to the polarized charges on the surface of spherical
cavity. For this consider a small element dS on the surface of the spherical cavity of
angular width dθ and at an angle θ with the direction of field E.
✓Area of the element dS = 2π (PQ) (QR)…………………………………(2)
✓ From the triangle PQR, sinθ = PQ/r or PQ = r sinθ ………………..(3)
✓We know angle = arc/radius, So dθ = QR/r or QR = r dθ……………(4)
✓From equations 2, 3, 4 we have Area = dS = 2πr2sinθ dθ…………….(5)

1 2 2
0 0
1
1
dq P
dE = = cosθ dS...................................................(6)
4πε r 4πε r
The electric field can be resolved into two components. dE cosθ is parallel to the direction
of the field and dE sin
i
θ is perpendicular to the direction E
Only parallel component contribute to the total field intensity E , since the perpendicular
components of the upper and lower half of the sphere can cancel with ea
π π
2
1 1 2
0
0 0
2
π
2
1
0 0
ch other.
P
Thus, E = dE cosθ = cos θ dS
4πε r
we know from eqaution 5, dS = 2πr sinθ dθ
P
So, E = cos θ sinθ dθ..............................................................(7)
2ε
Let us say co
 

-1
-1 3
2
1
0 0 0 0
1 1
1
0
sθ = x, and therefore -sinθ dθ = dx
P P x 2P P
from equatin 3, E = - x dx = - = =
2ε 2ε 3 6ε 3ε
P
or E = .....................................................................................
3ε
 
 
 

...(8)
At each point of the sphere, the surface charge density is given by σ = Pcosθ
The charge on the element dS of the surface of the sphere will be dq = Pcosθ dS
The charge will produce an electric field intensity dE1 at the centre of the sphere

2
With in the cavity there will be symmetrical distribution of molecular dipoles. So, their contribution
cancel each other. Hence E = 0..........................................................(9)
Thus t
i 1
0
he total internal field from eqation 1 is givenby
P
E = E + E = E + .........................................................................(10)
3ε
The field given by the above equation is called Lorentz field or local field.
Clausius-Mossotti relation
✓ Let us consider the case of a elemental solid dielectric, which exhibits only
electronic polarizability.
✓In solids we have to consider the actual effective field acting on a molecule in order
to estimate the dielectric permittivity.
✓If αe is the electronic polarizability per atom, it is related to the polarization P by the
relation

48
..........................................................(1)
,
e
i
i
P
NE
N is the number of atoms per unit volume and E is the local field
We know for electronic and ionic polarization the local field f
 =
0
0
int ............(2)
3
1 2,
.....................
3
i
e
i
or cubic crystals and isotropic liquids
P
can be given by the Lorenz field or the ernal field E E
From equation and we have
P P
NE P
N E



= +
= =
 
+
 
 
...............(3)
0
0
0 0
0
( -1) ...............................(4)
( -1)
sin 3
( -1) 3
3( -1
1 1
2
1 1
( -1) 3 3( -1)
r
r
e
r
e r
r
r r
We also know that
P
P E or E
U g the above equation in equation
P
P P
N
N
 
 

  
 
 
 
= =
=
 
+
 
 
= = =
   
+
+
   
   
( ) 0
)
2
( -1)
..........................................(5)
2 3
- .
r
e
r
r
N
The above equation is known as Clausius Mossotti relation



 
+
=
+

Problem 7: The atomic weight and density of sulphur are 32 and 2.08 gm/cm3
respectively. The electronic polarizability of the atom is 3.28×10-40F.m2. if sulphur
has cubic symmetry, what will be its relative permittivity.
23 3 3 -40 2
-12
0 0
( -1) (6.023 10 ) (2.08 10 / ) (3.28 10 )
2 3 3 3 32 (8.85 10 / )
( -1)
0.483
2
3.8
e A e
r
r
r
r
r
N N kg m Fm
M F m
 

  



    
= = =
+   
=
+
=
✓In this form, it is known as Lorentz-Lorenz equation.
✓This relation can be used to approximate the static dielectric constant εr of non polar
materials from their optical properties.
✓ In the case of dipolar materials we cannot use the simple Lorentz field
approximation and hence the Clausius–Mossotti equation cannot be used in the case of
dipolar materials.
( )
2
2
2
0
, , .
5
( -1)
3
2
r
e
At optical frequencies n where n is the refractive index of the material
Thus Equation can be written as
N
n
n



=
=
+

Dielectric breakdown
✓When a dielectric material permits very large current to flow through it under the
action of an applied electric field is termed as dielectric breakdown.
✓The maximum electric field that the dielectric can withstand without suffering
electrical breakdown is known as dielectric strength. In other words dielectric strength
is the limiting field intensity above which a breakdown occurs.
✓Dielectric strength = Emax = Vmax/Thickness of the dielectric.
✓There are different mechanisms by which the dielectric breakdown takes place.
❖ Intrinsic breakdown
❖ Thermal breakdown
❖ Discharge breakdown
❖ Electrochemical breakdown
❖ Defect breakdown

❖ Intrinsic breakdown
✓When a dielectric is subjected to high electric field then the electrons in the valence
band acquire energy and go to conduction band by crossing the energy gap and hence
become conducting electrons. Therefore large current flows and is known as intrinsic
breakdown.
✓This conduction electrons on further application of field, may collide with other
atoms and molecules and release more electrons. In this process, the number of
electrons increases very rapidly with time. Ultimately breakdown occurs.
✓Localized melting, burning and vaporization of the material take place at this stage
causing failure of the material. This is called Avalanche breakdown.
Characteristics
•It occurs at ordinary temperatures
•It occurs in thin samples
•The break down time is of the order of microseconds.

52
❖ Thermal breakdown
✓In dielectric material, the energy due to dielectric loss appears as heat. This heat must
be dissipated away to the surrounding.
✓In some cases the amount of heat generated will be very high compared the amount of
heat dissipated. Under such condition the temperature inside the dielectric increases and
it may produce breakdown. This type of breakdown is known as thermal breakdown.
✓Characteristics
•It occurs at higher temperatures
•The breakdown time is of the order of milliseconds
•The breakdown strength depends on the size and shape of material sample.
❖ Discharge breakdown
✓ In some dielectrics, gas bubbles may be present. Under the action of an electric
field, the gas present in the material will easily ionize and hence produce large
ionization current. Such breakdown is known as discharge breakdown.

53
✓ Characteristics
✓ It occurs at low electric fields
✓ It depends upon the frequency of the applied field.
❖ Electrochemical breakdown
✓ It is similar to thermal breakdown.
✓ Many materials have ions which cause leakage current in presence of an electric
field. When the temperature is increased, the mobility of ions increases. The
electrochemical reaction takes place, insulation resistance decreases and finally
results in breakdown.
✓ Characteristics
✓ It depends on the concentration of ions and magnitude of leakage current.
✓ It occurs at ordinary temperatures.
❖ Defect breakdown
✓ If the surface of the dielectric material has defects such as cracks and porosity,
impurities such as dust or moisture may deposit at these defects. This may lead to

Dielectric loss
✓When a dielectric is subjected to an electric field, a part of electrical energy dissipates
in the form of heat known as dielectric loss.
✓The origin of dielectric loss can be understood as follows. An ac field changes its
direction with time. With each direction reversal, the molecules are required to follow
the field reversal in order to contribute to the polarization
✓When a capacitor is charged in one half cycle, the molecules of the dielectric medium
are polarized. When the capacitor is discharged in the second half cycle, the molecules
should revert to their initial condition. However in the process of returning to their
initial state , the molecules jostle each other and lose energy due to friction. This energy
loss takes place in the form of heat known as dielectric loss.

Loss angle and Loss tangent
✓Let us consider a parallel plate capacitor C constituted by plates of area A and
separated by a distance d. Let a dielectric having permittivity εr fill the space between
the plates. Let a sinusoidal voltage V of angular frequency ω be applied to the capacitor.
✓The current through the capacitor is given by
r a
r a
2 2
r a
V
I = jωCV +
R
I = jI + I
V
I = ωCV is the displacement current and I = is the conduction current
R
The resultant current I = I +I legs behind the displacement current by an angle δ
In the case of idea a
o
l dieletric, R = , So I = 0.In such a case the resultant current I
would be ahead of voltage V precisely by an angle = 90 .


tan .
A loosy dielectric is represented by a resis ce parallel to the capacitor

a r
o o
a
r
However for lossy dielectric, the total current is I = I + j I
The phase angle betweenVand I is now slighly less than 90 .The angle δ =90 - is
called the loss angle and it is given by
I
tanδ =
I
 
( ) 2
2
0 r
2 0 r
L a r 2
V/R 1
= =
ωCV ωCR
tanδ is known as the loss tangent, also known as dissipation factor;
The real powerloss in the dielectric is given by
2πfε ε Ad V
ωε ε AV
P = VI = VI tanδ = ωCV tanδ = tanδ = tanδ sin
d d
2
L 0 r
-11 2
0 L r
V
ce E=
d
or P = 2πfε ε vE tanδ (since, Ad = Area.distance = v = volume of the dielectric)
Substituting the value of 2π and ε ,we have P = 5.565×10 fε vE tanδ
 
 
 
From the above expression it is seen that the power loss in a dielectric depends on
1. Dissipation factor
2. Dielectric constant
3. Frequency
4. Electric field
5. Volume of the dielectric

Complex relative permittivity
In most of the materials, the dielectric behavior is more complex indicating the presence
of other sources of dielectric loss. To include losses from all sources, we rewrite the
equation as
0
*
* '
r r
1 - (1- tan )
ε = ε (1 -
a r
a r r
r
r
I j AV
I I j I J I j j
I d
The above equation suggest that the lossy dielectric can be described with the aid of a
complex relative permittivity given by
  


 
= + = =
 
 
* ' ''
r r r
''
'
'
'
''
tan
jtanδ) or ε = ε - jε
tan
( tan ) .
r
r
r
r
r
represents the relative permittivity in capaci ce calculation
repr
The product is known as the loss fac
esents the energy loss in
tor
dielectr



 


=
.
tan
tan .
,
.
L
ic medium
Loss gent represents how lossy the materi
A loosy dielec
alis for ac signal
tric is represented by a resis ce parallel to the capacitor
P
Power dissipation factor per unit volume
v
s

= * 2
0 r E


Dielectric loss spectrum
The typical variation of of a polar dielectric with frequency are shown below
' ''
r r
and
 
✓At audio frequencies, the electric field reverses slowly so that the molecular dipoles
can keep shifting their orientation directions in step with the field alternations. There is
no power loss in this region.

✓In the radio frequency region, the rotation of dipoles in order to align with the field
direction is opposed by the by the internal friction of the material and thermal agitation
of the molecules. Consequently, the dipoles lag behind the forces that cause the motion.
A phase difference develops between polarization and electric field resulting a fall of
as the frequency increases. It is accompanied by heating of the dielectric and
therefore by a loss of energy. Energy supplied to maintain the rotation of the dipoles
accounts for the power loss.
At frequencies of the field near the relaxation frequency of dipoles, the rotation
becomes more rapid and energy loss approaches a maximum at the relaxation
frequency. At frequencies above relaxation frequency, the electric field reverses so
rapidly that the dipoles can not follow the field reversals due to inertia. By the time the
dipoles attempts to align along the particular direction, the field direction changes.
Therefore the dipoles fails to respond and maintain random orientation and no more
becomes alligned with the field.
'
r


✓Consequently, the dielectric constant is reduced and the power loss decreases after
going through a maximum. At much higher frequencies, the losses becomes negligible.
As a results vs frequency variation exhibits a bell shaped profile with a maximum at
the molecular dipolar relaxation frequency and reaching zero on either side. It is seen
thus that the greatest loss occurs at frequencies at which the dipoles can almost but not
completely be reoriented. At lower frequencies, losses are low because the dipoles have
time to rotate. At higher frequencies, losses are low because the dipoles do not rotate at
all. The losses in this process are known as relaxation losses.
✓Up to the infrared region, ionic dipoles follow electric field variations. When one
constituent of the dipole is displaced relative to the other, it again experiences a
restoring force proportional to the displacement and executes simple harmonic motion.
The frequencies of this simple harmonic motion lies in the range of 1013 to 1014 Hz.
When the frequencies of the applied field approached this region, resonance of ionic
dipoles occurs and power is absorbed from the field.
''
r


✓When the frequency of the field exceeds their natural frequency, the ionic dipoles
do not respond and can not absorb power. The losses encountered in this region is
known as resonance losses.
✓Till the optical frequencies, the electron clouds in the molecular dipole respond to
the variation of the field. The electron cloud executes simple harmonic motion . The
natural frequency of this region lying in the range of 1017 to 1018 Hz.
✓When the frequency of the driving field reaches this value, resonance absorption
talks place. Because of which electron get excited to higher energy levels.
Subsequently, they remit this energy in a random fashion in the form of optical
photons. For frequencies greater than these, electrons fails to follow the excursions of
the field and can not absorb power any more.

Piezoelectricity
✓The piezoelectric effect refers to a change in electric polarization that is produced in
certain materials when they are subjected to mechanical stress. This stress dependent
change in polarization manifests as a measurable potential difference across the
material. This is known piezoelectric effect.
✓Piezoelectric Material will generate electric potential when subjected to some kind of
mechanical stress.
✓Polarization P is directly proportional the applied mechanical stress σ.
Thus P = d σ, where d is known as piezoelectric constant.

✓This process is used in conversion of mechanical energy into electrical energy and
also electrical energy into mechanical energy.
✓Examples: Quartz crystal , Rochelle salt etc
✓It is a reversible process. If the piezoelectric material is exposed to an electric field
(voltage), it consequently lengthens or shortens proportional to the voltage. This is
known as inverse piezoelectric effect.
✓In the inverse piezoelectric effect, the electric field E is proportional to strain s.
✓Thus s = d E (Note: we know stress is proportional to strain)

Ferroelectricity
✓The name ferroelectricity comes from the similarities between polarizations of
ferroelectric materials with the magnetization of ferromagnetic materials.
✓Materials which posses special structure that permits spontaneous polarization are
called ferroelectrics. In other words, they possess polarization in the absence of an
electric field. The phenomenon of spontaneous polarization is called ferroelectricity.
✓Ex: Barium Titanate, Potasium Phosphate, Potasium Niobate etc.
✓They posses very high values of permittivity εr of the order of 1000 to 10,000.
Capacitors made from these materials can be significantly smaller than capacitors
made out of other dielectric materials.
✓ All Ferroelectric materials exhibit piezoelectric effect because of lack of symmetry.
All ferroelectric materials are piezoelectric, But all piezoelectric materials are not
ferroelectric. Ex: Quartz is a piezoelectric but not ferroelectric.

✓In a ferroelectric material a transition occurs from a centro symmetric to a noncentro
symmetric unit cell at the Curie point Tc. The shift in structural symmetry affects both
the structural and physical properties of the crystal. Ferroelectricity can be maintained
only below the Curie temperature. Above this temperature the material becomes
paraelectric (P = 0).
✓Ferroelectric materials have great application potential in developing smart
electromagnetic materials, structures, and devices, including miniature capacitors,
electrically tunable capacitors, filters and phase shifters in recent years
✓In ferroelectrics, the dielectric polarization depends nonlinearly on the applied field.
So they are also known as Nonlinear dielectrics. The polarization versus electric field
curve is known as ferroelectric hysteresis loop.

Ferroelectric hysteresis
✓The hysteresis loop is caused by the existence of
permanent electric dipoles in the material. When the
external electric field is increased from zero value, the
polarization increases as more of the dipoles are lined
up along the direction of the field. When the field is
strong enough, all dipoles are lined up with the field, so
the material is in a saturation state.
✓If the applied electric field decreases from the saturation point, P also decreases.
However, when the external electric field reaches zero, P does not reach zero. The
polarization at the zero fields is called the remanent polarization.
✓When the direction of the electric field is reversed, the polarization decreases. When
the reverse field reaches a certain value, called the coercive field, the polarization
becomes zero. By further increasing the field in this reverse direction, the reverse
saturation can be reached. When the field is decreased from this saturation point, the
sequence just reverses itself.

Spontaneous polarization in BaTiO3
✓BaTiO3 exhibits tetragonal structure at temperatures below Curie temperature (TC =
120 ̊C)
✓The Ba2+ ions are located at the corner of the unit cell. O2- ions at the centers of the
faces and Ti4+ ion is nearly at the body center of the unit cell.
✓In BaTiO3, the titanium ion does not occupy the exact body center of the unit cell. It
is seen that the titanium ion is displaced upward from the center of the unit cell while
the six oxygen ions are located below the centers of each of the six faces. Consequently,
a permanent ionic dipole moment arises in each unit cell.

✓Strong interaction between the adjacent permanent ionic dipole cause all dipoles to
mutually align in the same direction with in some volume of the solid. Such regions of
spontaneous polarization are known as ferroelectric domains.
✓In an un-polarized ferroelectric solid, the polarization vectors of different domains
orient in different directions so the net dipole moment is zero.
✓The spontaneous polarization vanishes above Curie temperature because the thermal
energy causes transformation of tetragonal unit cell into a cubic unit cell. All the ions
assume symmetric positions within the cubic unit cell so that the center of the action of
negative charge is coincident with of positive charge. Therefore the net dipole moment
becomes zero and ferroelectric behavior ceases.

Application of dielectrics and Ferro-electrics
✓Ferroelectric materials are used in small sized capacitors of large capacitance.
✓They are used in memory devices due to their property of hysteresis.
✓Ferroelectric materials are used to generate ultrasonic waves since all ferroelectric
materials are piezoelectric.
✓Ferroelectric materials exhibit pyroelectricity. They find applications in heat
sensors.
✓They are used in transducers such as microphones .
✓They are used in electromechanical filters.

Assignments
✓Explain the electronic polarizability in atoms and obtain an expression for electronic
polarizability in terms of the radius of the atom?
✓Explain ionic polarization? Obtain an expression for the ionic polarizability?
✓Deduce Clausius - Mossotti relation?
✓What is local field in a dielectric material ? Explain how the local field could be
calculated for a cubic crystal.
✓Briefly explain ferroelectricity and piezoelectricity.

Dielectric properties -final-converted-1

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