4.4 Linear Dielectrics F stable

image dipole

magnetic dipole

superconductor F

stable θ

r

4.4.1 Susceptibility, Permitivility, Dielectric Constant If E

r is not too strong, the polarization is proportional to the field.

EP e

rr0εχ= (since , PED

rrr+= 0ε D

r is electric displacement from freeρ )

0=×∇=×∇ PDrr

eχ is called the electric susceptibility. E

Pe

0εχ = ,

HM

m =χ

( ) EEPED e

rrrrrεχεε =+=+= 100 , ε is called the permittivity.

er χεεε +== 1

0

is called the relative permittivity, or dielectric constant.

Material Dielectric Constant Material Dielectric ConstantVacuum 1 Benzene 2.28 Helium 1.000065 Diamond 5.7 Air 1.00054 Water 80.1 Nitrogen 1.00055 KTaNbO3 34,000 What is high K material? It may discharge for an electric field high

E - - - - - - - - -

+++++++++ Vacuum E

Example: A metal sphere of radius a casurrounded, out to radius b, by linearpermittivity ε . Find the potential at the

rr

QD ˆ4 2π

=r

, br > rr

QE ˆ4 2

0πε=

r an

2=rε++++++++++++++++++

er than E. - - - - - - - - -- - - - - - - - -

ba

rries a charge Q. It is dielectric material of center.

d b , ar >> rr

QE ˆ4 2πε

=r

+−=

−+−= ∫∫

∞ abbQdr

rdr

rQV

a

b

b

εεεπεεπ111

411

4 022

0

arb >> , rrQEP e

e ˆ4 2

00 πε

εχεχ ==rr

0=⋅−∇= Pb

rρ

ar = , ( ) 20

4ˆ

aQrP e

b πεεχσ −=−⋅=

r and br = , ( ) 2

0

4ˆ

bQrP e

b πεεχσ =⋅=

r

If the space is entirely filled with a homogeneous linear dielectric, then

fD ρ=⋅∇r

and ∇ 0=× Dr

Pvacuum

E

When crossing the boundary: 0≠×∇=×∇ PD

rr

Example: A parallel-plate capacitor is filled with insulating material of dielectric constant rε . What effect does this have on its capacitance?

- - - - - - - - - -

++++++++++

+= σAAD2 2

++ =

σD , ε

σ2,

++ =upE ,

0,, 2ε

σ ++ =outupE ,

0, 2ε

σ ++ −=downE

−= σAAD2 22

+−− −==

σσD , 0

, 2εσ +

− −=upE , ε

σ2,

+− =downE ,

0,, 2ε

σ +− =outdounE

εσ

=inE dAQdVεε

σ==∆ , vacuumrCd

AV

QC εε==

∆=

4.4.2 Boundary Value Problem with Linear Dielectrics

+If we place free charge inside a linear dielectric,

( ) fe

eeeb

DEP ρχ

χε

εχεχρ+

−=

⋅−∇=⋅−∇=⋅−∇=

10

0

rrr

If we place the free charge on the boundary:

fbelowabove DD σ=− ⊥⊥ ,, fbelowbelowaboveabove EE σεε =− ⊥⊥ ,,

fbelow

belowabove

above nV

nV σεε =

∂∂

−−

∂∂

− , continuity: belowabove VV =

Example: A sphere of homogeneous linear dielectric material is

placed in an otherwise uniform electric field 0Er

. Find the

electric field inside the sphere. E0

Method 1:

The free charge is on the boundary 0==

∂∂

−−

∂∂

− fin

inout

out rV

rV σεε

Only bound charge exists, no free charge. Rr >> , θcos0rEVout −→

Rr = , r

Vr

V outin

∂∂

=∂

∂0εε and V outin V=

θθ coscos 20 rArEVout +−= , θcosBrVin =

BRRARE =+− 20 ,

−−= 300

2RAEB εε 0

0/23 EB

εε+−= ,

εεεε+

−=

0

03

2RA

zErEVrr

in 00 23cos

23

εθ

ε +−=

+−= zEz

zVE

r

ˆ2

3ˆ 0ε+=

∂∂

−=r

Method 2: Uniformly polarized sphere with polarization zPP ˆ=

r may produce electric field

zPE ˆ3 0ε

−=r

inside.

Total field inside: zPEE ˆ3 0

0

−=

ε

r inducing the polarization:

zPE ˆ3 0

0

−

εEzP eeˆ 00 == εχεχr

0033 EP

e

e εχ

χ+

= zEzEEre

ˆ2

3ˆ3

300 εχ +

=+

=r

Example: Suppose the entire region below the plane z = 0 is filled with uniform linear dielectric material of susceptibility eχ . Calculate the force on a point charge q situated a distance d above the origin. Considering q without dielectric material:

( ) 22220

,1

4 drd

drqE belowz

++−=

πε

Considering dielectric material without q: zPb ˆ⋅=r

σ 0

, 2εσ b

abovezE = and

0, 2ε

σ bbelowzE −=

Total effect:

−

+−===⋅

03

220

0,,0 24ˆ

εσ

πεεχεχσ b

etotalbelowzebdr

qdEzPr

( ) 2/322221

drqd

e

eb

++−=

χχ

πσ

Total bound charge:

qdaQe

ebb

+

−== ∫ 2χχσ

Use image charge qQe

eb

+

−=2χ

χ at z = -d to solve the problem.

( ) ( )

++++

−++=

22222204

1

dzyxQ

dzyxqV b

πε 0>z

( )

−++

+=

22204

1

dzyxQqV b

πε 0<z

attractive force on q: ( )

zd

qQF b ˆ24

12

0πε=

r

4.4.3 Energy in Dielectric Systems To charge up a capactor, it takes energy of

VdQdW = 22

21

2CV

CQdQ

CQW === ∫

Change of capacitance in dielectric materials: vacuumrdielectric CC ε= increase the energy because of an increase of charge

∫∆=∆ τρ VdW f , Df

r⋅∇=ρ ( ) ( )∫∫ ∆⋅∇=⋅∇∆=∆ ττ VdDVdDW

rr

( ) ( ) VDVDVD ∇⋅∆+∆⋅∇=∆⋅∇rrr

( ) ( ) EDVDVDrrrr

⋅∆+∆⋅∇=∆⋅∇

Choose V=0 at ∞→r .

( )∫ ⋅∆=∆ τdEDWrr

for a linear dielectric ( )EDEDrrrr

⋅∆=⋅∆21

∫ ⋅= τdEDWrr

21

l

4.4.4 Forces on Dielectrics x fringing field

dwl

d

wlVQCvacuum

0

0

ε

εσ

σ===

vacuumrdielectric CC ε= ( ) ( )xldwxl

dwx

dwC err χεεεεε

−=−+= 000

Assume that the total charge on the plate is constant (Q CV= ), CQW

2

21

=

2022

2

221

2V

dw

dxdCV

dxdC

CQ

dxdWF eχε

−===−= attractive force

If you start from constant voltage, the work supplied by a battery must be included.

dxdQVCV

dxdF +

−= 2

21

If the voltage is constant and the charge is varying, you must include the force due to the battery for maintaining a constant voltage.

++++++++ + + + + ++++

Exercise: 4.18, 4.23, 4.28 1. Dielectric material increase the capacitance of a capacitor store much more charges 2.

E

P

03εPE +

P

P

0/εPE + E

E

P

0εP

−

03ε−

3. Ferroelectricity: BaTiO3

ferroelectric antiferroelectric Transition temperature? Curie-Weiss law?