Student Name :
Maysaa Al – Darabseh
Instructor Name :
Dr. Hassan Al - Ghanem
- Abstract
When a dielectric material is placed in an external
electric field, it will be polarized with time, the
polarization goes through a relaxation process by
making successive collisions inside the material,
suffering energy loss in these processes. These processes
of energy loss extends over a wide range of frequencies,
covering all kinds of polarizations, electronic, ionic and
orientational.
It is the purpose of this project to study the relaxation
processes due to orientation in the dipole moments, to
derive the components of the dielectric constant and
their dependence on frequency.
We will check also the validity of Debye theory for
diluted liquids.
- Introduction
The electric properties of dielectric materials are usually
described in terms of the dielectric constant .For most materials
this quantity is independent of the strength of the electric field
over a wide range of it, but in the case of alternating fields it
depends on the frequency . It also depends on other parameters
such as the temperature.
When an ideal dielectric body is exposed to an electric field
there exist only bound charges that can be displaced from their
equilibrium positions. This phenomenon is called displacement
polarization. In molecular dielectrics, bound charges form
permanent dipoles. The molecular dipoles can only be rotated by
an electric field. Usually, their dipole moments are randomly
oriented. In an external field the permanent dipoles rotate with
the electric field direction and this process is called orientational
polarization.
The most important phenomenon in dielectric materials when
exposed to an external alternating electric field is the relaxation
in these dipoles, which is the subject of our project.
2
- Theory
- When we apply an external static field to a dielectric
®
D
material (  = 0 ) the electric displacement written as:
®
®
Ds = es E
-----[1]
®
And, the electric displacement D equal:
®
®
®
D = e0 E + P
®
®
-----[2]
®
e0 er E = e0 E + P
Then,
Where, er is the relative dielectric constant defined as e\ e0 .
From these two equations the polarization is given by:
®
HL
HL
H
L
®
P = e0 er - 1 E
-----[3]
- In static field case equation[3] becomes:
®
®
P s = e0 es - 1 E
And we can write Ps as:
H
L
Ps = P¥ + Ps
Where,
dip
-----[4]
-----[5]
P¥ :The part due to the polarizability of the particles.
Ps dip : The part of Ps due to permanent dipoles.
HL
From Eq.[4] we can write:
®
®
P ¥ = e0 e¥ - 1 E
3
-----[6]
The time needed by dipoles to reach the equilibrium distribution
-6
- 13
is in the range ( 10 to 10 s) .[due to static external electric
field].
H
L
- During this interval of time :
[1]- Ps dip built up as shown in Fig.[1].
[2]- P increased from zero to Ps .
( The time needed to reach P¥ will be neglect )
Fig.[1]- The polarizability as a function of time.
H
L HL
where, the polarization is given by :
P E = P 0 L E, T
where,
L is Langevin function.
4
Now, after an interval of time ( t ) :
P t = P ¥ + P dip
-----[7]
Then, due to relaxation mechanism, the rate of which dipole
build up in the material is given by:
dP dip
1
=
dt
t
@
HL D
Ps
dip
- P dip
-----[8]
Where,
 : The relaxation time of the mechanism.
- When the constant electric field is suddenly switched off,
then :
H
L
Ps
dip
=
0
Then, Eq.[8] becomes:
dP dip
1
= P
dt
t dip
HLHL
à à
H
L
[ Pdip
- Integrating Eq.[9] we get :
t
0
then,
-----[9]
dP dip
1
= P dip
t
P dip = P s
5
dip
e
- t
t
t =0
= Ps
dip
]
t
ât
0
-----[10]
-----------------------------------------------------------------------
H
L
H
L HLHL
H
L HL
BHL F
- For alternating field at time t :
®
E w = E0 ei w t
From Eq's.[4],[5]and[6] we get :
Ps
dip
= Ps - P¥
®
®
= e0 es - 1 E - e0 e¥ - 1 E
Then,
®
Ps
Then , From Eq.[8]
dip
= e0 es - e¥ E
-----[11]
®
dP dip
1
=
e0 es - e¥ E - P dip
dt
t
then the solution of this equation give:
P dip = C e
- t
t
+
HL
e0 es - e¥ ®
E
1 + iwt
-----[12]
- t
t
The term C e in Eq.[12] will be neglect because it's
decreases to small value after some time.(transient term).
Then, the steady state solution,
P dip =
HL
e0 es - e¥
E0 ei wt
1 + iwt
6
-----[13]
We know that:
HL HL
B
HLHLF
A
HL E
A E
P t = P ¥ + P dip
= e0 e¥ - 1 E0 ei w t +
Then,
e0 es - e¥
1 + iwt
es - e¥
1 + iwt
P t = e0 e¥ - 1 +
E0 ei w t
E0 ei w t
-----[14]
Using Eq.[2] and Eq.[14]:
®
®
®
D = e0 E + P
®
= e0 E + e0 e¥ - 1 +
This means that:
®
D = e0 e¥ +
es - e¥
1 + iwt
es - e¥
1 + iwt
E0 ei wt
E0 ei wt
-----[15]
®
®
*
From the relation of the complex electric displacement D =
e* E and comparing it with Eq.[15] : The complex permitivity is
given by :
e* = e¥ +
es - e¥
1 + iwt
and,
e* = e` - i e``
7
-----[16]
From Eq.[16]
e* = e¥ +
Then,
HLHL
HL
HH
L
L
es - e¥
1 - iwt
1 + wt 2
e` = e¥ +
&
e`` =
es - e¥
1 + wt 2
es - e¥ wt
1 + wt 2
-----[17]
-----[18]
- Eq's[17] and [18] are called Debye equations.
We plotted on figures [1, 2 and 3] equations [17] and[18 ]for
different relaxation times and for different dielectric strengths.
8
- Results and Conclusions
25
ε8 = 2
εs = 20
20
τ = 10-2
15
έ
10
ε?
5
0
2.5
8
13.5
19
24.5
30
lnω
ε8 = 2
25
εs = 20
έ
20
τ = 10-6
15
10
ε?
5
0
4
9
14
19
lnω
24
29
25
ε8 = 2
20
εs = 20
έ
τ = 10-4
15
10
ε?
5
0
4
9
14
19
24
29
lnω
Fig.[2] variation of the real and imaginary parts of dielectric
constant,έ andε˝ with frequency .
( For different values of relaxation time )
9
25
έ
έ
20
τ = 10-6
έ
15
τ = 10-4
τ = 10-2
10
ε?
ε?
ε?
5
0
3
8
13
18
Fig.[3] variation of the real and imaginary parts of dielectric
constant, έ and ε˝ with frequency .
( For different values of relaxation time )
- In the figure we showed that when the relaxation time is large
the variation of the real and imaginary parts of dielectric
constant, έ and ε˝ with frequency is very fast, although the
strength of the dielectric material remains constant.
- The difference between εs and ε∞ gives the polarizibilitystrength- of dielectric material.
10
ε8 = 1
30
έ
25
εs = 25
τ = 10-6
20
15
ε?
10
5
0
4
9
14
19
24
29
lnω
ε8 = 2
25
εs = 10
20
τ = 10-6
15
έ
10
5
ε?
0
4
9
14
19
24
29
lnω
ε8 = 2
25
εs = 20
έ
20
τ = 10-6
15
ε?
10
5
0
4
9
14
19
24
29
lnω
Fig.[4]- variation of the real and imaginary parts of dielectric
constant, έ and ε˝ with frequency .
( For different values of ε∞ and εs with constant relaxation time )
ε˝
ε∞ = 2
εs = 20
έ
Fig [5] – The relation ship between the real and imaginary parts of dielectric
constant, έ and ε˝ for ε∞ = 2 and εs = 20 .
Figure[5] shows the relation between έ and ε˝ which is called
“ Niquist plots” or Cole-Cole plots.
It gives a semi circle with its center on the real axes of έ.
20
ε˝
17.5
15
12.5
ε∞ = 2
εs = 20
10
7.5
5
ε∞ = 2
εs = 10
2.5
2.5
5
7.5
10
12.5
15
17.5 έ
20
Fig[6]- The dependence of Niquist plots on the dielectric strength of the
material for an ideal diluted polar system.
Generally, we can apply Debye theory in polarization in a
general way by considering a series of relaxation processes.
then, we can write Equ.[16] as :
e* = e¥ +
HL
es - e¥
1 + iwt
n
-----[19]
where, the variable n equals one for ideal case which satisfies
Debye’s formula. but, if it deviates from the ideal case ,
Equ.[19] can be written as:
HL
es - e¥
1 + wt n in
e* = e¥ +
ip
but, i is a phaser (i = e 2 )
then,
in = e
this gives:
ip n
2
np
np
+ i sin
2
2
= cos
HLH L
L
HLHH
L HL
H
L
HLH
L HL
e* = e¥ +
1 + wt
n
es - e¥
cos n2p + i sin
np
2
-----[20]
Separation of the real and imaginary parts leads to :
`
e - e¥ = es - e¥
and
``
1 + 2 wt
wt
e = es - e¥
1 + 2 wt
`
n
1 + wt
``
n
n
n
cos n
cos n
sin n
sin n
p
2
p
2
p
2
+ wt
2n
-----[21]
p
2
+ wt
2n
-----[22]
The distribution of the e and e which is described in Equ’s
[21] and [22] are showed in Fig.[7]for the case of n = 0.8.
ε˝
ε∞ = 2
εs = 20
ideal
έ
Fig [7] – The relation between έ and ε˝ displaying the difference
between the ideal Debye model and the deviation from simple
theory.
From Fig.[7] we can see clearly that the center of a semi
circle of the Cole-Cole plots when we consider a series of
relaxation processes is not on the real axes of έ .but, it deviates
by a small angle θ from the έ axes which is called the dip
angle.
The dip angle θ equal some parameter h multiplied with π/2.
where, the parameter h equal zero when the dielectric has only
one relaxation time .but, for the case of dielectric which has a
series of relaxation times the parameter h is between zero and
one. Then ,the dip angle can be written as:
q=
HL
p
p
h=
1- n
2
2
-----[23]
From Fig.[7], the center of the deviation semi circle is at the
point ( 11 , -2.9 ).then, we can get the dip angle from
e``
2.9
=
= 0.322
e`
9
q = tan- 1 0.322 = 17.86 0
q = 0.0992 p rad
tan q =
then, from Equ.[23] we can find that the value of n is 0.8. where
this value agrees with our assumption.
- References :
- C.J.F.Bottchetr,” Theory of electric polarization”.
- H.Frohlich,”Theory of dielectrics”. second edition.1958.
- J.R.Hook,”Solid state physics”. Second edition.1991.