2010-2011 Dielectric Materials

8/6/2019 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


2/66

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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33/66

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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35/66

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


40/66

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.


43/66

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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47/66

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


48/66

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


49/66

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


51/66

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?


54/66

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.


55/66

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()()(


57/66

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(


58/66

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


59/66

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


60/66

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


63/66

)]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:

Date post:	07-Apr-2018
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