Lecture Note #4A Chapter 4. Electric Fields in Mattercontents.kocw.net/KOCW/document/2015/inha/kimkyunghon/07.pdfThe force acting on +q : p H O 6.1 10 30C m 2 A dipole moment in an

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Chapter 4. Electric Fields in MatterLecture Note #4A

4.1 Polarization

4.2 The Field of a Polarized Object

4.3 The Electric Displacement

4.4 Linear Dielectrics

00 EP e

p

Polarization P dipole moment per unit volume = p / V

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유전체의분극

[1] 도체와 부도체(유전체)에 외부 전기장이 가해지면 어떻게 될까?

(1A) 유전체 원자의 분극도는 어떻게 기술되는가?

(1B) 유전체의 분극도는 방향성에 따른 특성은 어떠한가?

(1C) 유전체 내에서 일어나는 편광과 외부 전기장과의 관계식은? 또 유전체의 유전 감

수율 (dielectric susceptibility)은 분극과 외부 전기장과 어떤 관계?

[2] 유전체에 외부 전기장이 가해질 때 유전체 내부에 일어나는 분극(polarization)에 의한

전하 분포와 이에 따른 전위 및 전기장 특성은 어떻게 기술되는가?

(2A) 분극에 의한 유전체 내부 전하 밀도와 표면 전하 밀도는?

(2B) 분극된 유전체에 의한 외부에서의 전위와 전기장은?

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전기장속에서의도체와부도체

--------

++++++++

-------

+++++++

I) 도체의경우

II) 부도체의경우

외부전기장

외부전기장

0 E

0 E

No internal charge ( = 0)

No internal field (Eint = 0)

Polarization :

nPbˆ

: surface charge density

Pb

: volume charge density

00 EP e

p


p : electric dipole moment per each atom

v'

0S

0

v'4

1

4

1d

rda'

rrV bb

Potential at a point outside the dielectric medium is


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Electron Configuration in Atoms

H11 He4

2 Li73 Be9

4 B115

C126 N14

7 O168 Cu64

29

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Vibration & Rotational Motions of CO2 Molecules

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Rotational Motions of Linear Molecules1

H H

H2

O C

CO2

O

where

BhcJJ

I

JJE

J

rot1

2

1 2

cIπ

hB

28

2MbI

I : moment of inertial of the molecule

about the axis of rotation

c : speed of light

where I : moment of inertial of motion

: angular velocity of rotation

L : angular momentum = I

I

L

I

ωIωIErot

222

1 22

2

22 1 JJL

where J : angular moment quantum number

J = 0, 1, 2, 3, …

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Rotational Motion of Symmetric-Top Molecules

For symmetric-top molecules,

two quantum numbers J and K specify the rotational energy levels

hcBCKBhcJJErot 21

,,,

.,,,

321

21

K

etcKKKJwhere

(5.7)

B, C : 2 moments of inertia

c

b

cIπ

hC

cIπ

hB

2

2

8

8

where Ic : the moments of inertia about the symmetric axis

Ib : the moments of inertia about the perpendicular axis.

O

O

O3

O

Ic

Ib

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Energy Levels of Atoms, Molecules, Liquids & Solids

Discrete atoms

(Gas, ion in solid)

Energy Levels

Distance between atoms

E

Molecules

(Liquid & gas molecules)Semiconductors

(Atomic distance)close far

Distance between

atoms

E

Valence band

Conduction band

ⓔ

ⓔ ⓔ

ⓔⓔ

ⓔⓔ ⓔ

ⓔⓔ

ⓔⓔ

ⓔ ⓔ

ⓔⓔ

ⓔⓔ

ⓔ

ⓔⓔ

ⓔ

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Energy Bands in Periodic Structures

Conduction band

Valence band

1. When the atoms are far apart, there is no interaction among them, and each of

them has the same electronic energy level structures.

2. As the atoms get close to make covalent bonding, their energy levels are

divided into separate energy bands according to the Pauli’s exclusion

principle.

3. The energy bands are separated into valence band, energy gap, and conduction

band.

6C : 1s2 2s2 2p2

14Si : 1s2 2s22p6 3s2 3p2

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4.1 Polarization (편광)

4.1.1 Dielectrics (유전체誘電體) = Insulator (전기절연체絶緣體)

Filled

Conduction

Band (CB)

Valence

Band (VB) Filled

Eg

Filled

EgPartially

Filled

EmptyEmpty

Metal

(conductor)InsulatorSemiconductor

Fermi levelⓔⓔ

E ED

Polarization

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4.1 Polarization (분극, 편광)

4.1.2 Induced Dipoles (유도된쌍극자)

: atomic polarizability

E

p : electric dipole moment per each atom

+q-q p

ED

i

i

i

i rqpP

Ep

(4.1)

1) For a neutral atom


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p


[Example 4.1]

Using the Gauss’s law, we can obtain the electric field at the center:

v' 1

0

dadE

3

3

03

3

0

2

3

43

4

4a

dq

a

d

dEr

3

04

1

a

qdEr

+q-q p

d

Gaussian surface

of radius d

Charge

density

Eaqdp 3

04

Ep

v34 0

3

0 a

Eq. (4.1)

(4.2)

where v is the volume of the atom.

It is assumed that the charge density is uniform over the entire atomic

volume. Then,

: an approximated model


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4.1.2 Induced Dipoles (유도된쌍극자)

Ex) Carbon dioxide (CO2)

: polarizability tensor

////EEp

NmC 104.5 2-40

//

2) For molecules

E

: polarizability

E

NmC 102 2-40 : polarizability

zzzyzyxzxz

zyzyyyxyxy

zxzyxyxxxx

EEEp

EEEp

EEEp

ij

zz

yy

xx

00

00

00

(4.3)


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4.1.3 Alignment of Polar Molecules (극성분자의정렬)

The force acting on +q :

mCp OH 30101.62

A dipole moment in an uniform electric field:

EqF

Neutral atoms have no dipole moment initially, but the dipole moment is induced by the applied field.

Some molecules have built-in permanent dipole moments “polar molecules”.

Ex.) H2O

The force acting on -q : EqF

The total force : 0 EqEqFFFtotal

The torque acting on the dipole :

EdqEqd

Eqd

FrFr

22

Thus, a dipole p = qd in a uniform field E experiences a torque :Ep

(4.4)

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4.1.3 Alignment of Polar Molecules (극성분자의정렬)

A dipole moment in a nonuniform electric field:

The total force acting on the dipole :

EqEEqFFFtotal

Therefore,

dEE xx

where

ldTdT

- continued (1)

EEE

Since the dipole is very short (i.e., d = small),

From Chapter 1, Eq. (1.35)

dEE yy

dEE zz

EdE

EpEdqEqFtotal

(4.5)

For a “perfect” dipole of infinitesimal length,

The total torque acting on the dipole :

totalFrEp

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4.1.4 Polarization (편광, 분극)

When a dielectric material is placed in an electric field,


The dielectric material is polarized

00 EP e

where e is the dielectric susceptibility of the material.

(유전 감수율)

2

212

0 10858mN

C.

: the permittivity of free space.

(유전율)

(1C) 유전체 내에서 일어나는 편광과 외부 전기장과의 관계

식은? 또 유전체의 유전 감수율 (dielectric susceptibility)

은 분극과 외부 전기장과 어떤 관계?

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Chapter 4. Electric Fields in Matter

4.1 Polarization




A

p

r v'

0S

0

v'4

1

4

1d

rda'

rrV bb

where nPbˆ

: surface charge density

Pb


VE

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유전체의분극

A

p

r




(1C) 유전체 내에서 일어나는 편광과 외부 전기장과의 관계식은? 또 유전체의 유전 감

수율 (dielectric susceptibility)은 분극과 외부 전기장과 어떤 관계?

[2] 유전체에 외부 전기장이 가해질 때 유전체 내부에 일어나는 분극(polarization)에 의한

전하 분포와 이에 따른 전위 및 전기장 특성은 어떻게 기술되는가?



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4.2.1 Bound Charges

For a single dipole p within the polarized dielectric material, the potential at point A becomesA

p

r

2

0

ˆp

4

1

r

r

rVdip

from Eq. (3.151) of lecture note 3C

(4.8)

where r is the vector from the dipole to the point A.

dv' P

p

The total potential becomes

v' 2

0

v'ˆP

4

1d

'rrV

r

r

(4.9)

Since ,ˆ1

'2r

r

r

v'0

v'1

'P4

1drV

r

(4.10)

b

a

b

a

b

afgdx

dx

dfgdx

dx

dgf

from Chapter 1,

the integration by parts relation is used:

v'0

S0

v'v'0

v'P'1

4

1P

1

4

1

v'P'1

v'P

'4

1

d'ad

ddrV

rr

rr

SSV

danaddV ˆvvv

from Chapter 1,

the divergence theorem is used:

where nPbˆ

: surface charge density Pb


(4.11) (4.12)

v'

0S

0

v'4

1

4

1dda'rV bb

rr

(4.13) ndaad ˆ

Appendix 1

P p / V


2

0

cos

4

1

r

qdrV

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(4.14)

,3

4 3PRp

Since P is uniform,

cosˆ PnPb

: the total dipole moment of the sphere.

0 Pb

: identical to that of a perfect dipole

at the origin.

(4.16)

(4.15)

RrPzP

zzP

dz

dVE

for

3

1ˆ

3ˆ

3 000

[Example 4.2]

(Solution)

from Example 3.9 of the lecture note 3B

Rrrk

rV cos3

,0

Rrr

kRrV cos

1

3,

2

0

3

,coscos 10 kPk

Rrr

RP

RrrP

rV

cos3

cos3

,

2

3

0

0

Since r cos = z , the field inside the sphere is uniform :

where

Rrr

rpV

for

ˆ

4

12

0

The potential outside the sphere becomes


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ˆsin

1ˆ1ˆ

V

r

V

rr

r

VVE

[Example 4.2]

, cos3

,2

3

0

Rrr

RPrV

Since

- continued (1)

The electric field outside the sphere becomes

ˆ3

sinˆ

3

cos23

0

3

3

0

3

r

PRr

r

PRE

ErE

E

Rr

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4.2.2 Physical Interpretation of Bound Charges (1)

The dipole moment resulted from the atomic polarization within the tiny volume (v = Ad ) is

Adp Pv' P

Since

where A is the cross-sectional area, and d is the length of the tiny volume.

,qdp Adqd P PAq

The surface charge density at the end section is PA

qb

When the end section is at an oblique angle, the normal component of the surface becomes

cosend AA

n̂P cos Pcos

A

q

A

q

end

bThus, n̂P

b

++++

+-

--

--

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4.2.2 Physical Interpretation of Bound Charges (2)

For nonuniform polarization distribution,

v PPvvv

daddS

b

Pb

PAq

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[Example 4.3] For two charged spheres, a positive sphere and a negative sphere,

P3

4 3

Rdqp

3

04

1

R

dqE

+q - q

+ =2) when the material is uniformly polarized, the

spheres no longer overlap perfectly upon their

superimpose.

All the plus charges moves slightly upward, and

all the minus charges moves slightly downward.

1) without polarization, their charges cancel

out completely when they are superimposed.

+ =The field in the overlap region is

where R is the radius of the sphere.

The polarization is written as

P3

1

0

E

The potential at points outside the spheres is the same as a dipole’s2

0

ˆ

4

1V

r

rp

+q

- q

+q

- q

+++ ++

-- ---

P

Eq. (2.13)

[Appendix 2]

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4.2.3 The Field Inside a Dielectric

- The microscopic electric field inside the dielectric medium is too

complicated to calculate.

inout EEE

The average field over

the sphere due to all

charges outside

[The average field (over a sphere), produced by charges outside] = (the field they produces at the center)

outE

'v

ˆ'

4

12

0

drP

Voutsideout

r

r

Thus,

- The macroscopic electric field inside the dielectric medium is an

average field over a specific region covering many thousands of

atoms. (This is what we are interested in for most of the practical

situations.)

r

R

• The macroscopic electric field at some point r within the dielectric

medium:

The average field

over the sphere due

to all charges inside

= the field at r due to dipoles exterior to the sphere.

(See [Appendix 1] at the end of this lecture note)

Then, we can use Eq. (4.9): (4.17)

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4.2.3 The Field Inside a Dielectric

- The average electric field inside the dielectric medium is

P 3

4 3

Rp

3

04

1

R

pEin

regardless of the details of the charge distribution within the sphere.

r

R

It is related to the total dipole moment:

Thus, (4.18)

(See Eq. (3.105) in [Appendix 1] at the end of this lecture note)

- continued

P3

1

0

inE

For a uniformly polarized sphere,

'vˆ'

4

1

20

drP

rVsphereentireover

r

r

(4.19)

The average field over any sphere (due to the charge inside) is the same as the field at the

center of a uniformly polarized sphere with the same total dipole moment.

No matter what the actual microscopic charge configuration is, we can replace it by a nice

smooth distribution of perfect dipoles.

P p / V

3

00

2

3

4114 rQrEadE enc

r̂

3

1

0

rE

3

3

4Rq

3

0

3

00 44-

3

1

R

p

R

rqrE

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Next Class

Chapter 4. Electric Fields in Matter

4.1 Polarization




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[Appendix 1] - Problems 1.13 & Its Solution

[Problem 1.13] Let s be the separation vector from a fixed point (x’, y’, z’) to the point

(x, y, z), and let s be its length. Show that

(a)

(b)

(c) What is the general formula for ?

zzzyyyxxx ˆ 'ˆ 'ˆ ' r

rr

22

(a)

r

r

2ˆ '2ˆ '2ˆ '2

ˆ '''ˆ '''ˆ '''2222222222

zzzyyyxxx

zzzyyxxz

yzzyyxxy

xzzyyxxx

2/ˆ/1 rrr

nr

222''' zzyyxx r

(b)

23

23-222

23-222

23-22223-222

21-222

21-22221-222

ˆ1ˆ 'ˆ 'ˆ ''''

ˆ '2'''2

1

ˆ '2'''2

1ˆ '2'''

2

1

ˆ '''

ˆ '''ˆ '''1

r

rr

r

r

zzzyyyxxxzzyyxx

zzzzzyyxx

yyyzzyyxxxxxzzyyxx

zzzyyxxz

yzzyyxxy

xzzyyxxx

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[Problem 1.13]

(a)

(b)

(c) What is the general formula for ?

(c) x

nx

nn

rrr

1

rrrr

r

rr

rrrrr

ˆ1

ˆ 'ˆ 'ˆ '1

ˆ ˆ ˆ

11

1

1

nn

n

nn

nn

zzzyyyxxxn

zx

yx

xx

n

r

r ''2'''

2

1'''

21-22221222 xxxxzzyyxxzzyyxx

xx

Back

- continued

zzzyyyxxx ˆ 'ˆ 'ˆ ' r

rr

22

2/ˆ/1 rrr

nr

222''' zzyyxx r

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[Problem 3.47] Show that the average field inside a sphere of radius R, due to all the charge

within the sphere, is

'v ˆ

3

4

1

4

12

30

dq

R

rr

where r is the vector from r to dv’.

304

1

R

pEave

where p is the total dipole moment. There are several ways to prove this delightfully simple

result. Here’s one method:

(a) Show that the average field due to a single charge q at point r inside the sphere is the

same as the field at r due to a uniformly charged sphere with , namely

3

3

4Rq

(b) The latter can be found from Gauss’s law (see Prob. 2.12). Express the answer in terms

of the diploe moment of q.

(c) Use the superposition principle to generalize to an arbitrary charge distribution.

(d) While you’re at it, show that the average field over the sphere due to all the charges

outside is the same as the field they produce at the center.

(3.105)

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r

R

-rdv

E

[Appendix 2] Solution of Problem 3.47 (1)

i) The average field due to the point charge q at r is

,v

3

4

1

3

dE

R

Eave

where r

rˆ

4

12

0

qE

,v ˆ

4

1

3

4

12

03

dq

R

Eave

r

r

Thus,

(Here r is the source point, dv’ is the field point, so s direction: from r to dv’)

ii) The field at r due to uniform charge over the sphere is

r

R

qrdv

,v ˆ

4

12

0

dE r

r

In this case, dv is the source point and r is the field point.

Thus, ,

3

4 3

R

q

EEave

where r is the vector from r to dv’.

[Solution]

(a) Show that the average field due to a single charge q at point r inside the sphere is the

same as the field at r due to a uniformly charged sphere with , namely

3

3

4Rq

Thus, r direction: from dv to r so the vector direction should have the opposite sign.

'v ˆ

3

4

1

4

12

30

dq

R

rr

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The electric field inside a uniformly charged solid sphere (with a charge density )

is obtained from the Gauss’s law:

3

00

2

3

4114 rQrEadE enc

Thus, r̂ 3

1

0

rE

r

R

qrdv

R

r

Gaussian surface

3

3

4Rq r̂

4r̂

343

13

0

3

0 R

rqr

R

qE

Therefore, 3

0

3

00 44-

3

1

R

p

R

rqrE

If there are many charges inside the sphere, Eave is the sum of the individual

averages, and Ptot is the sum of the individual dipole moments. Thus,

3

013

013

0

ave4

-1

4

1-

4-

1

R

pp

nRR

p

nE ave

n

i

i

n

i

i

[Solution]

[Solution]

(b) The latter can be found from Gauss’s law (see Prob. 2.12). Express the answer in terms

of the diploe moment of q.

(c) Use the superposition principle to generalize to an arbitrary charge distribution.

3

3

4Rq From (a)

r

R

q1rdv

q2

q3q4

q5

Inha University 33


(field at r due to uniformly charged sphere)

The same argument, only with q placed at r outside the sphere, gives

rr

R

EE ˆ3

4

4

12

3

0

ave

[Solution]

rr

qEq

ˆ4

12

0

This is precisely the field produced by q (at r) at the center of the sphere.

So the average field (over the sphere) due to a point charge outside the sphere is

the same as the field that same charge produces at the center.

By superposition, this holds for any collection of exterior charges.

Back

(d) While you’re at it, show that the average field over the sphere due to all the charges

outside is the same as the field they produce at the center.

r

R

r

dv

r

R

q

Eq

EE

ave

3

3

4Rq From (a)

Documents

Lecture Note #4A Chapter 4. Electric Fields in Mattercontents.kocw.net/KOCW/document/2015/inha/kimkyunghon/07.pdfThe force acting on +q : p H O 6.1 10 30C m 2 A dipole moment in an