2010-2011 dielectric materials
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DIELECTRIC MATERIALS
Dielectric materials do not possess free electriccharges and hence do not conduct electricity.
-Polar dielectrics: Molecules posses dipolemoment
-Non-polar dielectrics: Molecules do not possesdipole moment
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Review of some basic formula
1. Electric dipole:
rqp
2. Dipole Moment:
3. Torque on the dipole
exerted by an E-field
Exp
sinpE
4. Potential energy of dipolein an E-field
cos. pEEpV
pEV ,0 pEV ,
DKR-JIITN-2010-MS
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5. Polarization: Defined as dipole moment per unit volume.
pNP
If the number of dipoles per unit volume is N, and if each hasmoment p then polarization is given as (assuming that all thedipoles lie in the same direction)
DKR-JIITN-2010-MS
Example: Suppose there are 3.34x1028 molecules per unitvolume of water each having dipole moment 6x10-30 C-m.
Solution: If all dipoles are oriented parallel to each other
then Polarization
P = 3.34x1028 x 6x10-30 = 0.2004 C/m2
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6. ELECTRIC FLUX DENSITY AND POLARIZATION
PED
0
According to Gauss
law,0
'.
qqAdE
00
'
qqEA
A
q
A
qE
'0
A
qE
A
q '0
Where,
A
qD and
A
qP
'= Electric flux density = Polarization
Gaussiansurface
0E
0E
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Further, EED
00
Therefore,
0
0
PEE
Thus polarization results in a reduction of the field insidethe dielectric medium.
Further, PEED
0
PEEr
00 PE r
)1(0
E
Pr
0
)1(
Here is known as electric susceptibility and r is known asrelative dielectric constant of the medium.
0
r
PEED
000
Where,
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POLARIZABILITY
Ep
Polarization of a medium is produced by field therefore, it isreasonable to assume that,
The polarization can now be written as, ENP
Thus, PED
0
But, ED r
0
0
1
Nr
Here is known as polarizability of the
molecule representing dipole moment perunit applied electric field
EN
D
)1(0
0
ENE
0
ENEr
)1(0
00
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In all above expressions, N can be expressed in terms ofdensity , molar mass M of the material and Avogadro's
number NA as
M
NN A
However, experiments show that though above equationshold good in gases but not for liquids and solids i.e. in thecondensed physical systems.
Thus dielectric constant can be written as:
)(1
0M
NAr
0
1
Nr
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LOCAL FIELD
0E
1E
2E
Central dipole
Lorentzsphere
3210 EEEEEloc
E0 = External field
E1 = Field due to polarization chargeslying on the surface of the sample.
E2 = Field due to polarization chargeslying on the surface of Lorentz sphere.
E3 = Field due to other dipoles lying withinthe Lorentz sphere.
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Calculation of various fields:
0
1
PE Depolarizing field E1:
This field depends on the geometrical shape of the externalsurface. Above equation is for a simple case of an infinite slab.Field for a standard geometry is given as
01
NP
E
Here N is known as depolarizing factor. The values of N forother regular shapes are given below:
Shape Axis N Sphere any 1/3 Thin slab normal 1Thin slab in plane 0 Cylinder Longitudinal 0
Cylinder Transverse
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Calculation of E2: Surface area dA of the sphere lyingbetween and +d is given as
drdA sin2 2Charge on the surface dA would be
)sin2(cos2 drPdq
Field due to this charge at thecentre of the sphere would be
2
04 r
dqdE
Field in the direction of applied field would be
2
0
24
coscos
r
dqdEdE
dE
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Field due to charges on the entire cavity thus would be,
0
22 dEE
0
2
04
cos
r
dq
0
2
0
22
4cossin2r
drP
0
2
3
PE
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Calculation of E3:
5
2
0
).(3
4
1
r
prrrpE
The result depends on crystal structure of the solid underconsideration. However for highly symmetrical structure likecubic it sum sup to zero. Thus
03 E
(In other structure E3 may not vanish and it should be included
in the equation).
The field due to other dipoles in the cavity may be calculatedby using the equation
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Thus Eloc would be
000 3
PP
E
03
PEEloc
Eloc = EL= Lorentz field. E is known as Maxwell field.
Now the polarization would be given as
LENP
3210 EEEEEloc
00 3
2
P
E
E = Maxwell field.
)3
(0
P
EN
03
PNEN
EN
N
P
)31( 0
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03
1
N
ENP
Again
Now
PEED r
00
0
00
31
N
EN
EEr
)3
1(1
0
0
N
Nr
Simplifying
ENN
P
)
31(
0 1
032
1
N
r
r CLAUSIUS MOSOTTI RELATION
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032
1
N
r
r
Reconsider CLAUSIUS MOSOTTI relation,
MNM
r
r )3
()2
1(
0
Since,AN
NM
Therefore ,0
3)2
1
(
A
r
r NM
MOLAR POLARIZABILITY
Molar mass
Density
M
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EXAMPLE:
An elemental dielectric material has r = 12 and it contains
5x1028
atoms/m3
. Calculate its electronic polarizabilityassuming Lorentz field.
SOLUTION:
032
1
N
r
r
Using CLAUSIUS MOSOTTI relation,
12
28
1085.83
105
212
112
2201017.4 Fm
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SOURCES OF POLARIZABILITY
1. Electronic Polarizability
2. Ionic Polarizability
3. Dipolar or orientationalPolarizability
0E
0E
0E
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1. ELECTRONIC POLARIZATION:
Volume of the atom is,
34
3V R
If z be the atomic number then
charge/ volume of atom would be
3
3
4
ze
R
Where, R = Radius of sphericallysymmetric atom
0E
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In presence of field E
Force on the charges EZeF
1
Coulomb force between separated charge would be
ZeF 2
X Field produced by displaced charges on nucleus
24 d
Ze
X Charge enclosed in the sphere of radius d
323
4
4d
d
Ze3
0
3
2 4
3
3
4
4 R
Zed
d
Ze
3
0
22
4 R
deZ
This leads to the separation ofcharges.
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In the equilibrium position, the two forces , F1 and F2 areequal, thus
3
0
22
4 RdeZZeE
3
04 R
Zed
E
Ze
ERd
3
04
This is equilibrium separation between charges, whichis proportional to the field.
Now the induced electric dipole moment would be
)4
(3
0
Ze
ERZeZedpe
ERpe3
04
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ERpe3
04
But according to the definition of polarizability,
Ep ee
Comparing,3
04 R
e (e = electronic
polarizability)
Thus, Electronic Polarization can be given as
ENEPere
)1(0
Where N is number of atoms/ m3.
0
1
er
N
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2. IONIC POLARIZATION:
Ep ii
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3. DIPOLAR POLARIZATION
with fieldWithout field
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Consider a molecule which carries a permanent dipolemoment p (like water molecule) is placed in an electricfield. The potential energy of the dipole would be:
cos. pEEpU
According to Boltzmann distribution, no. of moleculeswith energy U at equilibrium temperature T would be:
kTU
enn
0
kTpE
en
n cos
0
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Let n() be the number of dipoles per unit solid angle at ,we have
kT
pE
enn
cos
0)(
The number of dipoles in a solid angle dW
Wden kTpE cos
0
den kTpE
sin2
cos
0
Note: Here dW iscalculated as follows:
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Dipole moment of dipoles making angle with the field
(along x-axis) is
cosppx
Therefore, the dipole moment along the field within angle dW
)cos(sin2
cos
0
pden kTpE
Now, average dipole moment (Total dipole moment dividedby total no. of dipoles) can be written as
0
cos
0
0
cos
0
sin2
)cos(sin2
den
dpen
p
kT
pE
kT
pE
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0
cos
0
cos
sin
cossin
de
de
p
p
kTpE
kT
pE
Let x
kT
pE
cosand a
kT
pE
Therefore, xa cos and, dxda sin
Limits aa
Substituting all above, the integral becomes,
a
a
x
a
a
x
dxe
xdxe
ap
p 1
0
cos
0
0
cos
0
sin2
)cos(sin2
den
dpen
p
kTpE
kT
pE
aee
ee
e
exe
ap
paa
aa
a
a
x
a
a
xx1
][
][1
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aee
ee
p
paa
aa 1
Or,a
ap
p 1)coth(
(Langevin Function)
)(apLp
From the above equation,
)()( aLPaNpLP so
Ps = Saturation polarization
)(aL
Thus polarization would be,
kT
pEa
)(aL
a
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CASE 1: When a is very high (atlow temperature) i.e. a >> 1,
CASE 2: When a is very low (athigh temperature) i.e. a
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3
app
Ep okT
po
3
2
T
o
1
Total polarization
3)(
aaL
p
p
kT
Epp
3
2
Thus orientation polarizability is inversely proportional to T.
oiePPPP ENENEN oie
ENENEP oier )()1(0
oier NN )()1(0
kT
NpN
ier
3
)()1(2
0
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In general, therefore, we may write total polarizability as
oie
kT
pie
3
2
or
orkT
pei
3
2
temperature
independent
Substituting into Clausius-
Mosotti relation, we have
)3
(3
)2
1(
2
0kTpNM eiA
r
rM
k
pNslope A
0
2
9
03
eiAN
Non-polar substances
Polar substances
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Clausius-Mosotti relation may alternatively be written as
0
2
2
321
N
nn
Lorentz-Lorentz Relation
If the material consists of different types of molecules then
Clausius-Mosotti relation may be written as
ii
r
r N
03
1
2
1
Where Ni is the no. of molecules per unit volume and i ispolarizability of ith kind of molecule.
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1. ELECTRONIC POLARIZABILITY IN ALTERNATING FIELD
In presence of electric field E(t) = E0eit, electron cloud would
execute simple harmonic oscillation which will be given as
tieeEfx
dt
dxb
dt
xdm
02
2
2
tieeEfx
dt
dxb
dt
xdm
02
2
2
Where,mc
eb
62
2
0
2
0 = damping factor.
and f = force constant
3
0
2
4 Ref
(Remember we have
obtained earlier)
c = velocity of light
0 = permeability of free space, 0 = natural frequency,
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02 0
2
2
m
eeEx
m
f
dt
dx
m
b
dt
xdti
Assume solution of this equation of the form x(t) = Aeit.Then, substituting x we have
02 02 meeEAe
mfeAi
mbeA
ti
tititi
m
bi
m
eE
A
e
2)(22
0
0
2
0eNatural frequency
of vibration
tieeEfx
dt
dxb
dt
xdm
02
2
2
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Substituting A, the solution can be written as,
m
bi
em
eE
txe
ti
2)(
)( 220
0
Therefore, the induced dipole moment can be written as,
)(texpe
and polarizability can be given as,
]2)[(22
00
0
2
m
bieE
em
Ee
E
p
e
ti
ti
ee
m
bi
m
e
e
2)(22
0
2
m
bi
emEe
e
ti
2)( 2
2
0
0
2
])([ tiAetx
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]2
)][(2
)[(
]2
)[(
22
0
22
0
22
0
2
m
bi
m
bi
m
bi
m
e
ee
e
e
Thus electronic polarizability is an imaginary quantity. Now letus find out real and imaginary parts separately.
''' eee i or,
Thus equating the real and imaginary parts, we have
;4
)(
)('
2
22
222
0
22
0
2
m
bm
e
e
e
2
22
222
0
2
2
4)(
2''
m
b
b
m
e
e
e
]4
)[(
]2)[(
2
22222
0
220
2
m
b
mbi
me
e
e
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Now, the dielectric constant in the alternating field canbe written as
]2
)[(
)1(22
0
2
0
m
bim
Ne
e
r
er N )1(0
]2
)[(
122
00
2
m
bim
Ne
e
r
Thus r is an imaginary quantity. The real and imaginaryparts may be separately written as
;
]4
)[(
)(1'
2
22
222
00
22
0
2
m
bm
Ne
e
e
r
]4
)[(
2"
2
22
222
00
2
2
m
bm
bNe
e
r
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Variation of e and e with frequency of the applied field
2
22222
0
22
0
2
4)(
)(
'
m
bm
e
e
e
e
2
22
222
0
2
2
4)(
2''
m
b
b
m
e
e
e
e0
'e
''e
01.
22
0e
2
0e
2
e)(
me' 2
0e
2
me' e
0"e
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0eIf4.
0eIf3.
0e2.
0'e
0
2
e 2b
e"
2
22
222
0
22
0
2
4)(
)('
m
bm
e
e
e
e
2
22
222
0
2
2
4)(
2''
m
b
b
m
e
e
e
Both e and e are +ve.
e is ve and e is +ve.e0
'e
''e
2
2
0
20
2
2
4
2''
m
b
b
m
e
e
ee
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SELF STUDY
1. Find out max. and min. value of e.
2. Find out full width at half maximum for e.
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2. IONIC POLARIZABILITY IN ALTERNATING FIELD
Equation of motion for the ion pairs can be written as:
dt
dxfxqE
dt
xdM 2
2
2
Consider a pair of oppositely charged ions say, Na+ and Cl-.In presence of an applied field E along x-axis, the Na+ and
Cl- ions are displaced from each other by a distance x.
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Re-organizing we have,
tieEMqx
Mf
dtdx
Mdtxd 02
2
2
Reduced mass
Loss coefficient
Applied field)111
(
MMM
Force constant
The form of the solution of this equation may be considered as
tiexx
0
Defining the resonant or natural vibrational frequency of ionsas 0i2 = 2f/M, above equation may be written as
ti
i eEM
q
dt
dx
Mx
dt
xd 02
02
2
3
0
2
4 R
ef
(Here R is nearestneighbor distance
between +ve andve
ions)
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X, being solution, should satisfy the equation of motion
M
eqEexeixMex
titi
i
titi
0
0
2
00
2
0
M
qExix
Mx i
00
2
00
2
0
M
qE
Mix i
022
00 ))[(
])[( 22
0
00
MiM
qEx
i
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or,
])[(
"'22
0
2
MiM
qi
i
iii
]])][()[(
])[(
22
0
22
0
220
2
Mi
Mi
Mi
Mq
ii
i
2
22222
0
22
0
2
)(
])[("'
M
MiM
q
i
i
i
ii
])[(
)('
2
22222
0
220
2
MM
q
i
ii
])[(
"
2
22222
0
2
2
MM
q
i
i
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Ignoring the damping factor, equation for i can be written as
)( 22
0
2
i
iM
q
The ionic polarization may be written as
)(tNqxPi
The dielectric constant can be written as
Variation of i and i will be same as that ofe and e withonly difference that of the natural frequency of vibration.
E
P
r
0
1)(
E
P
E
Pie
r
00
1)(
E
PPie
0
1
00
1
ieN
E
P
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Above equation may alternatively be written as
2
0
22
00
2
0 1
1
1)(
i
i
er
M
Nq
E
P
2
0
2
1)()0()()(
i
rrrr
00
1)(
ier
N
E
P
Substituting I ,
Where, E
Pe
r0
1)( 01
e
N
m
Ne
e2
00
2
1
]11
[)()0(2
00
2
MM
Nq
i
rr
This equation givesstatic ionic dielectricconstant.
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From equation (2.10), it is obvious that
]11
[)()0(2
00
2
MM
Nq
i
rr
(2.11)
Equation (2.11) gives static ionic dielectric constant.
Example: In a NaCl crystal, N = 2.25x1028/m3. Taking oi =
3.2x1013 radian/ sec, calculate ionic contribution to the total
dielectric constant of the solid.
Solution:
The ionic contribution to the dielectric constant is given as
)11
(1
)()0(2
00
2
MM
Ne
irr
)5.35
1
23
1(
1066.1)102.3(
1
1085.8
)106.1(1025.2)()0(
2721312
21928
xxxx
xxxrr
7.2)()0( rr
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3. DIPOLAR POLARIZATION IN ALTERNATING FIELD
Let us first consider the dipolar polarization in static field:
1. When field E switched on at t = 0.
)1()( t
dsd eptp
Here, pd(t) is the instantaneousdipole moment and pds is the
saturation dipole moment.
dt
tdpd )(
t
dsep
Here = relaxation time orcollision time.
t
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Here, pd() = pds
)()()( tpp
dt
tdp ddd
)(tppdds
dt
tdpd )(
t
dsep
)1()( t
dsd eptpWe know that
)(tppep dds
t
ds
dt
tdpd )(
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2. When field E switched off at t = 0 when substance isfully polarized
t
dsd eptp
)( dt
tdpd )(
Here, pd() = 0
t
dsep
)(tpd
)()()( tpp
dt
tdp ddd
Dipoles in the system tend to follow the field, flipping backand forth as the field reverses its direction during each cycle.
Now, what happens in oscillating field?
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The equation describing this motion of dipolar polarization will be
)]()([1)(
tptpdt
tdp
dds
d
Where, pd(t) = actual dipolar
moment at the instant t, and
pds(t) = saturation value of the moment which would be thevalue of Pd (t) if the field were to retain its instantaneous valuefor a long time.
In the case of ac field tieEtE
0)(
)()0()( tEtp dds
Where, d(0) is the static dipolar polarizability, and
ti
d eE 0)0(
pds (t) is the value which pd(t) would reach if the field wereto remain equal to E(t) at all subsequent times.
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Substituting pds(t) in the equation
ti
ddd eEtp
dt
tdp 0)0()()(
Solution of above equation may be of the form
ti
dd eEtp
0)()(
)]()0([1)(
0
tpeEdt
tdp
d
ti
d
d
Where d() is the ac polarizability.
)]()([1)(
tptpdt
tdpdds
d
ti
dds eEtP
0)0()(
)()0( 0 tpeE dti
d
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Substituting pd(t) in the equation,
)()0()()()()(
tEtEtEi ddd
)0()1)(( dd i
)1(
)0()(
i
dd
Thus ac polarizability is a complex quantity, indicating thatthe polarization is no longer in phase with the field. Thisgives rise to energy absorption.
ti
ddd eEtp
dt
tdp 0)0()()(
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E
Pr
0
1)(
Dielectric Constant
0
)(
)()(
d
rr
N
)1(
)0(
)(0
i
N dr
)1(
)()0()()(
i
rrrr
)1(
)0()(
22
i
nn rr
(3.8)
The dielectric constant being frequency dependent showsthat medium exhibits dispersion.
E
N de
0
)(1
E
N
E
N de
00
1
)1(
)0()(
i
dd
Dipolar polarizability
E
N erer
0
1)(
drr )()0(
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The dielectric constant being a complex quantity,
)1(
)0()(")('
22
i
nni rrr
22
22
1
)1]()0([
inn r
22
22
1
)0()('
nn rr
22
2
1
)0()("
nrr
(3.9)
(3.10)
Debyeequations
)1(
)0()(
22
i
nn rr
2)0( n
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Consider
0)("
d
d r 01
))0((22
2
d
dn
r
0)1(
1))0((
222
222
nr
0122
(1/ is also known ascollision frequency)
221
)0()("
nrr
Let us make
1
Now let us find out at0)("
2
2
d
d r
1
222
22
2
2
)1(4))0(()("
n
dd
rr
Thus22
1
2
2
))0(()("
n
d
dr
r
2)("d
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Thus at = 1/, r() is maximum asshown in the figure.
22
12
2
))0(()("
nd
dr
r
-ve
22
2
1
)0()("
nrr
2)0()("
2
nrr
Thus r() achieves maxima equal to half of the static dipolar
dielectric constant (assuming negligible ionic contribution). Itdecreases as the frequency departs from this value in either dir.
The maximum value of r() is given as (substituting =1/ inequation (3.10)
1
11
)0(
2
2
2
nr
TOTAL POLARIZABILITY
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TOTAL POLARIZABILITY
oie ]11
[200
2
MM
e
i2
0e
2
m
e
kT
p
3
2
1. For static field
2. For oscillatory field
)1(
)0(
i
d
])[(
22
0
2
MiM
e
i
m
bi
m
e
e
2)(
22
0
2
DIELECTRIC LOSS
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DIELECTRIC LOSS
Inability of a dipole to remain in phase with the applied ac fielddue the its interaction with other dipoles in the substance leads
to dielectric loss which appears in the form of heat.
)Re(cos 00 tieEEE (2.1)
t
DJ
(2.2)
)]sin(cosRe[00 titiEJ r
)]sincos)("'Re[(00 ttiiEJ rr
Thus current density is
Let us consider a dielectric in parallel plate capacitor subjectedto an ac field given as
)(Re0
E
tr
)(Re 00
ti
reE
t
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)]sincos)("'Re[(00 ttiiEJ rr
)sin'cos"(00 ttEJ rr
Thus rate of energy loss in unit volume of the material will be
T
JEdtT
W0
1
tdtEttET
W
T
rr cos)sin'cos"(1
0
0
00
2
00 "2
1
EW r
Thus energy loss is proportional to r. The energy loss is alsoexpressed in terms of loss tangent (tan).
(2.3)
(2.4)
The energy loss is also expressed in terms of loss tangent (tan).
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'
"tan
r
r
Where, is the angle complimentary to the angle betweenapplied ac field and the resultant field.
(2.5)
Now, if there is no loss, thenfrom equation (2.3),
tEJ r sin'00
)2
cos('00
tEJ r
Thus current would lead the field by 900. In such a case = 0.However, if there are losses, r are non-zero and there is acurrent component in phase with the field. Thus resultant
current no longer leads the applied field by 900 rather by 900-as shown in the figure above.
(2.6)
The energy loss is also expressed in terms of loss tangent (tan).
Let dielectric be
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The current density in the circuit may be given as
tCER
tEJJJ CR sincos 00
Comparing this equation with equation (2.3) we have
(2.7)
"
1
0 r
R
and '0 r
C
Thus tangent loss can be given as
RCr
r
1
'
"tan
(2.8)
(2.9)
Let dielectric beequivalent to a parallel
combination of R and C(with unit cross
sectional area and unitseparation between its
plates).
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As the frequency increases, the displacement D changes frombeing entirely in phase with applied E to having components
both in phase and out of phase as shown below:
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