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ENGINEERING PHYSICS II
DIELECTRICS
4.5 VARIOUS POLARIZATION MECHANISMS IN DIELECTRICS
Dielectric polarization is the displacement of charged particles under the action of the electric
field. There are number of devices based on this concept. Those devices are rectifiers, resonators,
amplifiers and transducers which converts electrical energy in to other forms of energy.
In modern computers, memory devices are also based upon this concept.
Several microscopic mechanisms are responsible for electric polarization. Specially, in the case
of d.c. electric field, the macroscopic polarization vector P arises due to following four types of
microscopic polarization mechanisms
i.
Electronic polarization
ii.
Ionic polarization
iii.
Orientation polarization
iv.
Space-charge polarization.
4.5.1 ELECTRONIC POLARIZATION
Electronic polarization occurs due to the displacement of positively charged nucleus and
negatively charged electrons in opposite directions, when an external electric field is applied and
thereby creates a dipole moment in the dielectric.
The induced dipole moment is given by
µ=αe E
Where, αe is the electronic polarizability.
Monoatomic gases exhibit this kind of polarization. Electronic polarizability is proportional to
the volume of the atoms and is independent of temperature.
Calculation of electronic polarizability
Without field
Let us consider a classical model of an atom. Assume the charge of nucleus of that atom is Ze.
The nucleus is surrounded by an electron cloud of charge –Ze, which is distributed in a sphere of
radius R as shown in the fig
The charge density of the charged sphere =
−Ze
4πR3
3
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ENGINEERING PHYSICS II
(Or)
DIELECTRICS
charge density =
−3Ze
…(1)
4πR3
With field
When the dielectric is placed in a d.c. electric field E, two phenomenon occur
i.
Lorentz force due to the electric field tends to separate the nucleus and the electron
cloud from their equilibrium position.
ii.
After separation, an attractive coulomb force arises between the nucleus and the
electron cloud which tries to maintain the original equilibrium position.
Let x be the displacement made by the electron cloud from the positive core, as shown
Since the core is heavy, it will not move when compared to the movement of the electron
cloud. Here x <<R, where R is the radius of the atom.
Since Lorentz and Coulomb forces are equal and opposite in nature, equilibrium is reached.
At Equilibrium,
Lorentz force = Coulomb force
Lorentz force = Charge × Field
… (2)
= -ZeE
The negative sign indicates the repulsive force.
Coulomb force = Charge × Field
= +Ze ×
𝑄
4𝜋∈𝑜 𝑥 2
The positive sign indicates the attractive force.
Coulomb force =
Charge × Total negative charges (Q) enclosed in the sphere of radius x
4𝜋∈𝑜 𝑥 2
….(3)
The total number of negative charges (Q) enclosed in the sphere of radius ‘x’
= Charge density of electrons × volume of the sphere
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ENGINEERING PHYSICS II
DIELECTRICS
Substitute the charge density from eqn (1), we get
The total number of negative charges (Q) enclosed in the sphere of radius x
−3Ze
=
4
× π x3
4πR3
−𝑍𝑒
∴Q=
is
3
x3
….. (4)
R3
Substitute eqn (4) in (3)
Coulomb force =
Ze
4𝜋∈𝑜 𝑥 2
=
.
.
−𝑍𝑒 x3
R3
−Z2 e2 x
..… (5)
4π∈o R3
At equilibrium position, eqn (2)= eqn (5)
∴ −ZeE
=
x =
(Or)
−Z 2 e2 x
4π ∈𝑜 R3
4π∈𝑜 R3 E
.....(6)
𝑍𝑒
Therefore, the displacement of electron cloud (x) is proportional to the applied electric field E.
Dipole moment
Now the two electric charges +Ze and –Ze are displaced by a distance x under the influence of
the field and form an induced dipole moment which is given by
Induced dipole moment (µe) = Magnitude of charge × displacement
= Zex
Substituting the value of x from eqn (6), we get
µe =
Ze4π∈𝑜 R3 E
𝑍𝑒
µe = 4π∈𝑜 R3E
µe α E
µe = αe E
Where, αe = 4π∈𝑜 R3 (Farad-m3) is called electronic polarization which is proportional to the
volume of the atom.
Relation between αe and Dielectric Constants
We know, the induced electronic dipole moment is proportional to the applied field. This dipole
moment per unit volume is called electronic polarization. This is independent of temperature.
Electronic polarization
Pe=N µe
Where, N is the number of atoms
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ENGINEERING PHYSICS II
DIELECTRICS
Pe=N αe E
Since
Pe
E
…….(8)
= ∈𝑜 (∈𝑟 -1), we can write
αe =
∈𝑜 (∈𝑟 −1)
…….(9)
𝑁
4.5.2 IONIC POLARIZATION
Ionic polarization arises due to the displacement of cations (+ve ions) and anions (- ve ions) from
its original position, in opposite directions in the presence of electric field as shown in figure.
Explanation
Let us assume that there are one cation and one anion present in each unit cell of the ionic crystal
NaCl. When the electric field is applied, let x1 and x2 be the distances to which positive and negative
ions move from their equilibrium positions. The resultant dipole moment per unit cell, due to ionic
displacement is given by
Induced dipole moment = magnitude of charge × displacement
……… (1)
µi = e(x1 + x2)
Where, x1 is the shift of +ve ion and x2 is the shift of –ve ion, from their equilibrium positions.
When the field is applied, the restoring force produced is proportional to the displacements.
For +ve ion,
Restoring force F α x1 or F = β1 x1
……….(2)
For –ve ion,
Restoring force F α x2 or F = β2 x2
……….(3)
Here, β1 and β2 are restoring force constants, which depend on the masses of the ions and angular
frequency of the molecule in which the ions are present.
If ‘m’ is the mass of +ve ion and ‘M’ be the mass of the –ve ion and ωo is the angular frequency, then
β1 =m ω2o
….….(4)
β2 =M ω2o
…….(5)
Substituting for β1 in eqn (2), the restoring force for +ve ion can be written as
F = mω2o x1
…….(6)
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ENGINEERING PHYSICS II
We know,
DIELECTRICS
…….(7)
F = eE
Equating eqn (6) and (7), we get
eE = mω2o x1
x1 =
Similarly for –ve ion we can write
𝑒𝐸
……..(8)
mω2o
eE
x2 = Mω2
……..(9)
o
Adding eqn (8) and (9) we get
x1 + x2 =
𝑒𝐸
ω2o
1
(m +
1
M
)
..…..(10)
Substituting eqn (10) in (1), we get
𝑒 2𝐸 1
1
( + )
2
ωo m
M
µi = αi E
µi =
Or
Where, αi is the ionic polarization given by
αi =
𝑒2
ω2o
1
(m +
1
M
)
so, the ionic polarizability αi is inversely proportional to the square of the natural frequency of the
1
ionic molecule and directly proportional to its reduced mass which is given by (m +
79
1
M
)