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General Properties:

eec

r c aer as o no possess ree eec

r ccharge and hence do not conduct electricity.

Polar dielectrics: Molecules posses dipolemoment

Non-polar dielectrics: Molecules do not possesdipole moment

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What is dipole and dipole moment?

Electric di ole:

rr

rq=Dipole Moment:

Unit: Coulom -meter Deb

1 Debye = 3.33 x 10-30 C-m

DKR-J IITN-2010-MS

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at appens w en po e s expose to an - e

exerted by an E-field

Er

=

snp=

Potential energy of dipole cos. pEEpV ==r

r

in an E-field

pEV == ,0 pEV == ,

DKR-JIITN-2010-MS

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Dipole moment per unit volume.

If the number of dipoles per unit volume is N, and if each hasmomentp then polarization is given as (assuming that all the

pNP

r

r

=

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 otherthen Polarization

DKR-JIITN-2010-MS

P = 3.34x1028 x 6x10-30 = 0.2004 C/m2

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ELECTRIC FLUX DENSITY AND POLARIZATION0E

According to Gauss law,

'Gaussia

0

.

dE =

'qq

00

qq '=

PEDrrr

+=

A

qE

q '+=

Polarization (P)Electric flux density (D)

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rrrr

, 0

PEErrrr

+= 00

PE rrr

= )1(0

Pr

r

r

)1( ==0

=

Here is known aselectric susceptibility andr is known as

0

relative dielectric constantof the medium.

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POLARIZABILITY

r

r

o arza on o a me um s pro uce y e ere ore, sreasonable to assume that,

=

rr

molecule representing dipole moment perunit applied electric field

, =Thus, PED

rrr

+=0

ENErr

+=0

rr

EDrr

)1(0

0

+=

N rr

, r0=

N

r0

00 +=

0r

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,density , molar mass M of the material and Avogadro'snumber NA as

M

NN A=

Thus dielectric constant can be written as:

)(10M

r +=+=0

1

r

However, experiments show that though above equationshold good in gases but not for liquids and solids i.e. in thecondensed physical systems.

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LOCAL FIELD

3210 EEEEEloc +++=rr

=

E1 = Field due to polarization chargeslying on the surface of the sample.

E2 = Field due to polarization charges

lying on the surface of Lorentz sphere.

0Er

1Er

+ E3 = Field due to other dipoles lying withinthe Lorentz sphere.

Lorentzsphere

2Er

Central dipole

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Calculation of various fields:

0

1

E =Depolarizing field E1:

surface. Above equation is for a simple case of an infinite slab.Field for a standard geometry is given as

0

1

E =

Here N is known as de olarizin factor. The values of N forother regular shapes are given below:

Shape Axis N

Sphere any 1/3Thin slab normal 1

Thin 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 2=

Charge on the surface dA would be)sin2(cos 2 drPdq =

Field due to this charge at the

centre of the sphere would be

204 r

dqdE

=

dE

Field in the direction of applied field would be

cosd 2

0

24

cosr

==

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Field due to charges on the entire cavity thus would be,

=

22 dEE

=

0

204

cosr

dq

=

0

20

22

4

cossin2

r

drP

0

23

PE =

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The field due to other dipoles in the cavity may be calculatedby using the equation

5

2).(34

1r

prrrpE

rrrr

=

The result depends on crystal structure of the solid under

cubic it sum sup to zero. Thus

=E

(In other structure E3 may not vanish and it should be included.

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Thus Eloc

would be

00

03

PPE

rr

r

+=3210 EEEEEloc +++=rr

0

03

2

PE

r

r

=

PEE

r

rr

+=03

Eloc = EL=Lorentz field

.E

is known as Maxwell field.

Now the polarization would be given as

E = Maxwell field.

LENPrr

= )3

(0

P

EN

r

r

+=03

PNEN

r

r

+=

EN

N

P

rr

= )31( 0

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EN

=

r

rN rr

031

N

Now

3 0 1