ChpChp--10 Electric fields in Matter10 Electric fields in ...

ChpChp--10 Electric fields in Matter10 Electric fields in Matter

Dielectrics

Matter comes in many forms

Notes mainly based onGriffiths Chp 4

These respond differently to an applied E-field

Conductors & Insulators or DielectricsConductors & Insulators or Dielectrics

We already considered conductors

Conductors have charges that can freely moveConductors have charges that can freely move

In metals it is typically one or twotypically one or twoelectrons per atomThat are not associated with a specific nucleus Electron density

In a metal


Insulators or Dielectrics

All charges are attached to a specific atom or molecule

They are not free to roam

The can move small distances within the confines of anThe can move small distances within the confines of an

atom or molecule

Yet the small displacements of charge in an E field accountYet the small displacements of charge in an E-field account

for the properties of dielectrics

Two types of displacements are possible…stretching and

rotating


Induced Dipoles

What happens when a neutral atom is place in an E-field?

E-Field

+ Positively chargedNucleus

- Negatively chargedElectron cloud

Nucleus is pushed in the direction of the field

Electron cloud is attracted toward the field

If the E-field is strong enough it will ionize the atom (hence

forming ions and it becomes a conductor)


In the Limit of low magnitude E-field

Equilibrium is reached between the displacement induced

by the E-field and Coulombic attraction of nucleus and

electronselectrons

The atom is polarized in the E-field

r dr

+qq

d

+q+q-q

No E-field Applied E-field

+q-q

Get an induced dipole moment p points in the same

direction as the applied E-field


In the Limit of low magnitude E-field

Induced dipole moment is proportional to E

litypolarizabiatomic

Ep

H He Li Be C Ne Na Ar K Cs0.667 0.205 24.3 5.60 1.67 0.396 24.1 1.64 43.4 59.4

Atomic Polarizabilities /4o in units of 10-30 m3

Handbook of Chemistry & Physics 91st Edition CRC Pressy y


Assuming that the electron cloud retains its spherical

symmetry under small displacement in an E-field, calculate

the atomic polarizability for the atom.

dr d

+q-q

No E-field Applied E-field

+q-q

eEE Internal field due to e- pulling nucleus leftIs balanced by the externalIs balanced by the external E-field pushing right

What is the field a distance d from the center of a uniformly charged Sphere?


The charge from outside shell does not contributed4V

3

sphere -q

Applied E-field

rd

d4density arg

3

3

sphere

echApplied E field

eEE 3d4q' so

soqE r4 3

2

334

2

2o

ddq'E'

so r4

E

densityarg3r4Vsphere

ech

32

r4q33

34

o2

o2

o

qddE'

3d4

d4E

3

3r4q so

yg3

What is the field a distance d from the center of a uniformly charged Sphere?

3o

2o r4d4

E


Very crude atomic model but accurate within a factor of ~4

qdpbutqdEE' -q

Applied E-field

rd

3o

3o

e

Epbut E'r4p

qdpbut r4

EE

Applied E fieldeEE

Estimate the atomic polarizability of an H atom

atomo3

o V3r4

241

312123o

Fm10x64.1

)m10x9.52)(F/m10x85.8(4r4

331

o

m10x48.14

330

o

m10x667.04


What about molecules? They are not isotropic in their

polarizability

CO2 /NC1054

m/NC 10x2240

240

If applied field is at an angle then resolve into components

m/NC10x5.4 240||

Applied E-field perpendicular or parallel to molecular axisIf applied field is at an angle then resolve into components

||||EEp

However, the induced dipole moment may not be pointing in

the same direction as the E-field

For an asymmetrical molecule the most general linear

relationship between E and p is via a tensor


Tensor: magnitude and multiple directions

Polarizability tensor (tensor of rank 2 so for 3D space 32=9

elements)

zxzyxyxxxx EEEp

zzzyzyxzxz

zyzyyyxyxy

yy

EEEpEEEp

zzzyzyxzxzp

Tensorility Polarizabelements of ij MatrixThe values of ij depend on orientation of the axes you use

Orient so you have the “principal” axes so that all off

diagonal elements are zero leaving: zzyyxx

Characterize an asymmetric molecule with 3 polarizabilities


Griffiths Prob. 4.1

A H atom (with Bohr radius of 0.5 angstrom) is situated between two metal plates 1 mm apart, which are connected to opposite terminals of a 500 V battery. What fraction of the atomic radius does the separation distance d amount pto, roughly? Estimate the voltage you would need with this apparatus to ionize the atom [Use the value of in Table 4 1]4.1]

Expectation…The displacements due to polarization will be quite small


Griffiths Prob. 4.1

V/10500V5VE 5 V/m10x5m10x

E 53-

330106670

241330123o

330

o

F/m10x42.7)m10x667.0)(F/m10x85.8(4r4

m10x667.04

o

m 10x32.2C106021

)V/m 10x5)(F/m 10x42.7(Ed so edEp 1619

5241

C 10x602.1e 19

x10394m 2.32x10d 6--16

Ratio

%x1039.4Or

x1039.4m 52.9x10r

4-

12-Ratio


Griffiths Prob. 4.1…To ionize let d=radius of atom

VEr

V10x14.1m) 10x1C)( 10.602x1m)( 102.9x5(rexV

exer

8241

31912

soF/m10x42.7 241


Griffiths 4.1.3 Alignment of Polar Molecules

What about molecules with a permanent dipole moment

E.g. Water has a large electronegativity difference between

hydrogen and oxygen & OH bonds are 105o aparthydrogen and oxygen & OH bonds are 105 apart

Electrons are more tightly pulled toward oxygen (although it

is still a covalent compound) pis still a covalent compound) pCm 10x1.6p 30

What happens when anE-field is applied?

Large dipole moment so water is a good polar solvent

E field is applied?


Place water molecule in a uniform E-field

The force on the positive end will be exactly balance the

force on the negative end but there will be a torque

EFEF qandq

dd

FrFrNEFEF

q and q

torque

p EdEdEdN

qq

2q

2p

q FE q

dO

qF


The dipole moment in a uniform E-field experiences a

torquedp q

Note N is in a direction to line up p parallel to E

EpN

Note N is in a direction to line up p parallel to E

A polar molecule that is rotationally free will rotate in an

applied E fieldapplied E-field

A non uniform E-field will mean a net force in addition to Nq F O h l l di iE q

dOOver the molecular dimensionsVariation in E is usually small

qF


A non uniform E-field will mean a net force in addition to N

There will be a difference in E-field from + to - end EEEFFF qq

dzTdyTdxTdT

Recall that

zyxzyx ˆdzˆdyˆdxˆzTˆ

yTˆ

xTdT

zyx

q FE IdTdT

zyx

So if the dipole is very short

dEΔE A d f &q

F

dOE

E+ d xx EΔE And for x & y

EdΔE dp qAnd since

qF

d EpF

E-


Polarization*

What happens when we put slab of a dielectric material in

an E-field?

If the dielectric is made of neutral atoms or nonpolar moleculesIf the dielectric is made of neutral atoms or nonpolar moleculesThe E-field will induce a tiny dipole moment pointing in the field directionIn asymmetric molecules the induced p will not be parallel to the field but if they are Randomly oriented then the components perpendicular to the field will cancel. If it is a crystalline solid then the cancelation does not occur!If th di l t i i d f l l l h l lIf the dielectric is made of polar molecules, each molecular dipole will experience a torque and rotate if freeNote that temperature dependent random thermal motions will always compete withTh li t i th li d fi ld

* Note that this is NOT the same polarization as in light polarization…which refers to the time dependent orientation of the E-field vector during propagation of an EM Wave

The alignment in the applied field


Dielectric material becomes polarized in the E-field

Cartoon representation of dipoles aligning in an E-field

Note that there will still be someStretching of a polar molecule inStretching of a polar molecule inThe fieldBut it is easier to rotate a molecule than stretch it(Energy scale for rotations vs. vibrations?)

We measure this effect as the polarization P

vibrations?)

P Dipole moment per unit volumeThe polarized material will produce its own field…what will this do?


Induced fields in nanoparticles

Shafiei et al. Nature Nanotechnology 8,95–99 (2013)

ChpChp--10 Electric fields in Matter10 Electric fields in MatterInduced fields in nanoparticles

Shafiei et al. Nature Nanotechnology 8,95–99 (2013)


We now have two E-fields

1. External applied E-field

E

Dielectric Material

Induce many dipole moments(each atom or

Materialp

2. ``Internal field`` due to the induced Polarization

(each atom or molecule)

These fields will combine by superposition


The field of a Polarized object

Concept of bound charges

What is the field produced by the induced dipoles?

Strategy…We know the field/potential (far field limit) for aStrategy…We know the field/potential (far field limit) for a

single dipole Field Lines &Potential surfaces f i l di l

Subdivide the dielectric into infinitesimal dipoles & integratefor a single dipole

It is easier to work with the potential


Notational Warning to Physics 242 & 340 students

Purcell’s Notation for the potential (Phys 242) r Purcell s Notation for the potential (Phys 242)

Griffiths Notation for the potential (Phys 340)

r rV

Generalized Multipole Expansion

Rr

dv'P

Or'

R

OrExpansion for the potential for anyLocalized charge distribution( i l id d i l t d(previously we considered one isolated dipole)



rdv'

P'

R

dv'r'1 r

Or'

R4r

2

sourcesallo

Law of cosines

cos

rr'2

rr'1rcosrr'2r'rR

22222

Law of cosines

cos2

rr'

rr' 1rR

i.e. for point P far from r largefor 1

S t bi i l charge distributionSuggests a binomial expansion for1/R (see integral above)

1nn

...x

!21-nnnx1x1 2n



rdv'

Pr'

R 53111111 322

1

r

Or ...1682

1r

1rR

Substitute back in for

cos2r'r'

cos2rr

...cos2r'r'165cos2r'r'

83cos2r'r'

21111 3322

rr16rr8rr2rR

Collect terms in powers of (r’/r)33322 ...

2cos3cos5

rr'

21cos3

rr'cos

rr'1

r1

R1

33322

Pn are the Legendre polynomials…It is a cos

rr'

r1

R1

0n

n

n

P

Pn are the Legendre polynomials…It is a generating function for multipole expansionsNow substitute back into the integral



rdv'

P'

R

dv''1 rr r

Or'

dv''cosPr'11

R4

n rr

r

sourcesallo

...dv''1cos3r'1dv''cos'r1dv''11

dvcosPrr4

2232

n0

1n

rrrr

rr

no

...dv2

cos2

rr

dvcosrr

dvr4 32 rrrr

o

M lti l E i f th t ti l i t f 1/Monopole Dipole Quadrupole

Multipole Expansion for the potential in terms of 1/r(Exact expression if all terms included)

Integral depends on the angle between r & r’If you orient so r is on z’ axis (i.e. P on z’ axis) becomes the polar angle ’


Multipole Expansion as an approximation to the potential dv'

P

R dQhQ1 r

Or' dvQ whererQ

41

o

monopole r

Monopole term dominates at large r

Dipole term

dv''cos'r11 rr 'r'r ˆ'

2

2

dv''ˆ14

1

dv''cos'rr4

rr'rr

rr

dipole

odipole

r'r ˆcosr' so r'r

cos

2

ˆ1

dv''

r4

rr'p

o

dipole

p =dipole moment of the charge distribution

2r41 rpr

o

dipole For a single dipole


Where were we? Bound charges

Subdivide the dielectric into infinitesimal dipoles & integrate

r 2r

ˆ4

1 rpr

odipole

P Dipole moment per unit volume

dv' Ppp

dv'ˆ1

dv

2

rr'Pr

Pp

dvr4 2

volumeo

r

But note ˆ1 ‘ notation indicatesBut note…2rr

1' r

notation indicates differentiation wrt the source coordinates (r’)


Where were we? Bound charges

Subdivide the dielectric into infinitesimal dipoles & integrate

r

dv'r1'P

41

r

r4 volumeo

fff AAAIntegrate by parts using the product rulep

fff Integrate by parts using the product rule

dv''r1dv'

r'

41

ll

PPr

dv''14

1da'ˆ14

1

rr4

volumevolumeo

PnPr

r4r4 volumeoSurfaceo


Resembles potentials of surface & volume charges

dv''14

1da'ˆ14

1 PnPr


nP ˆb Pb

dv'r4

1da'r4

1 l

b

S f

b

r

The potential of a polarized dielectric material is the same as is produced by:


P ˆ1. Surface charge density

2 Volume charge density

nP ˆb

Pb

EThis is true for the E-field too since the gradient is a linear operator

2. Volume charge density b E


Net result we can avoid integrating all the dipoles

We find the bound charges then calculate their fields

E.g. we could apply Gauss’s Law as we have seen

d '1d '1 bb dv'

r4da'

r4 volume

b

oSurface

b

o

r

n̂P ˆb Pb

ChpChp--10 Electric fields in Matter10 Electric fields in MatterFind the electric field produced by a uniformly polarized sphere, radius R

p R

z Set z axis in the direction of polarization

Then determine bound charge densitiesn̂

nP ˆb Pb

p R Then determine bound charge densities

0 Pb Vol. bound charge density zero (since Polarization is uniform) cosPˆ nPb Surface bound charge density integrated over a sphere

d 'cosP1

da'r

cos4

Sphereo

r

RfP Inside

RP

Rrfor cosr3

r,

3

o

Solved in Ex 3.9 inGriffiths

cosrz Inside

Rrfor cosrR

3Pr, 2

o

Griffiths Outside


p R

z n̂ Rrfor cosrPr, Insidep R

zPr

3,

o

cosrz

Potential varies linearly in z inside the sphere

z3

r,o

PotentialRrfor

31 ˆ

3P z

3P

PzE

ooo

ooo

E-field

http://homepage.ntu.edu.tw/~chernrl/


p R

z n̂ Rrfor cosr

3Pr, Insidep R

RrforcosRPr

3,

3

o

Outside

Potential outside the sphere is identical to perfect dipole at the

Rrfor cosr3

r, 2

o

Outside

Potential outside the sphere is identical to perfect dipole at the origin!

rp2 rrforˆ1θ)(r,

Potential E-field

Pp 3

2

R4

rrfor r4

θ)(r,

o

Pp R3

http://homepage.ntu.edu.tw/~chernrl/


Physical Interpretation of bound charges

E-field of a polarized dielectric is produced by the combined

distribution of bound volume and surface chargesnP ˆbPb

Was this just abstract mathematical wizardry/bookkeeping?No the bound charges represent physical accumulations

How do we understand this? Consider a line of dipoles in a

o t e bou d c a ges ep ese t p ys ca accu u at o sOf charge!

polarized material…We get effective cancelation except at

the end points Looks like a large charge displacementbut really many small displacements

but really many small displacements


We call the net charge at the ends a bound charge

It is still attached (bound) to a specific atom or molecule

But it is still a charge!

How do we calculate the amount of bound charge?

Bound Charge

How do we calculate the amount of bound charge?d

A P

n̂

A= q- q

AendAConsider a cylinder

Of dielectric parallel to P qdAdPp p PAq

qdAdPp

Bound charge accumulated on right face


How do we calculate the amount of bound charge?d

A P

n̂

A= q- q

AendAP

Consider a cylinderOf dielectric parallel to P dd Of dielectric parallel to P

PAqqdAdPp

Bound charge accumulated on right facef

PAqb

If the slab is cut perpendicular to the cylinder

If th l b i t t l t th li d h i thIf the slab is cut at angle to the cylinder, charge is the same

nP ˆcosPA

q

end

bnP ˆend

Physically, Polarization “paints” bound charge over the surfacenP b


If the polarization is nonuniform bound charge will also

accumulate inside the dielectric as well as at the surface

d

Uniformly polarized sphere

E+ charges all move up

charges all move down

Without polarizationCh l

With polarizationCh t i t

E- charges all move down(slightly!)

Diverging P Charges cancel(neutral)

Charges separate intoCaps (surface bound charge)

The Integrated bound volume charge will be equal and opposite to the charge pushed out to the surface

d 'dˆd ' PnP ConcludePb

dv' dadv' Surface

PnP VolVol

bConclude


The field inside a dielectric

We started the derivation with formula for the potential for

an ideal or perfect dipole

Atoms or molecules are real or physical dipoles (but tiny!)Atoms or molecules are real or physical dipoles (but tiny!)

Also the Polarization P is a continuous vector function

Does this raise flags of concern?Does this raise flags of concern?

Outside the dielectric we are in the far field (R>>d) so the

dipole term dominates the expansion and we do not notice

the discrete graininess of the real dipole


What about inside the dielectric?

At the microscopic level of atoms/subatomic particles it

would be impossible to integrate

Electric field would vary wildly in space and time at the

But we are interested in the macroscopic field

Electric field would vary wildly in space and time at the microscopic level

But we are interested in the macroscopic field

i.e. the average over regions containing thousands of atoms

But not too large!

So we average (integrate) over regions smaller than object

itself dv' Vol

b Physicists and Mathematicians willDiverge here!


Assume we want to measure the macroscopic field near a

point r 2a1000R oIntegrate over a sphere of radius R radiusatomic ao

o

RrChargesO t id & I idOutside &

sphereInside

Macroscopic Field at r will be the sum of E-fields due to charges Inside & Outside the sphere

inoutr EEE


RCh

Macroscopic Field at r will be the sum of E-fields due to RrCharges

Outside &sphere

Insidecharges Inside & Outside the sphere

inoutr EEE The average E-field over a sphere produced by charges outside, is equal to the field they produce at the center

Eout is the is the field at r due to dipoles outside the sphere

dv'ˆ1

rr'Pr dv

r4 2outsideo

r


RCh


Outside &sphere


inoutr EEE

Ein dipoles are too close…we can’t integrate safely!

But wait…we just need the average fieldFall of Icarus byM ti

The average field within a sphere of radius R due to all the charge contained in the volume is

1 p

Matisse

3Ave R41 pE

o (This is Eqn 3.105 in Griffiths)

(p is the total dipole moment)


RCh


Outside &sphere


EEE inoutr EEE

PppE

R34 where

R41 3

3in

PE

o

31

3R4

in Regardless of the charge distributionInside the sphere

o3 Inside the sphere We implicitly assumed our sphere is small enough that P does not vary much over its volume…Wait what was the field of a uniformly polarized sphere?

1 31 PE

o The term omitted for the outside

integration gets put back in by Ein! See last lecture


Thus the macroscopic E-field can be obtained from the potentialWhich is obtained by integrating over the entire dielectric!

dv'r

ˆ4

12

rr'Pr

r4 Volumeo

We were properly calculating the average macroscopic field for insideaverage macroscopic field for inside the dielectric

All due to the fact that…Th fi ld h d h h i idThe average field over any sphere due to the charge inside Is the same asThe field at the center of a uniformly polarized sphere with y p pthe same total dipole moment

*It would hold for other averaging geometries…but harder to prove


The Electric Displacement

Gauss’s Law in the Presence of Dielectrics

E-field due to Polarization is the field due to bound charges

Bound Bound Surface Charge

nP ˆb Pb

Volume Charge

Let’s put it all together!

The E-field due to all other charges (e.g. electrons in a

conductor, ions in the material, etc.) is called the free charge

Free Charge f Any charge that is not due to Polarization of the dielectric material


In the Dielectric total charge density

Total Charge Densityf b

Gauss’s law for a dielectric material is

ff PE b ff PE bo

E is the total field not just that due to Polarization P

Combine the two divergence terms and re-write defining the Electric Displacement D

PE f

PED

PE

o

o

define D is the Electric Displacement

f D Gauss’s Law for Dielectrics in Differential Form

ChpChp--10 Electric fields in Matter10 Electric fields in MatterGauss’s Law in terms of the electric displacement

f D enclosed fQda ˆ nDSurfaceDifferential Form Integral Form

Griffith’s Eqn 4.22 Griffith’s Eqn 4.23

Note this only refers to the free charge

But we can only control the free chargeTada

“free of charge”But we can only control the free chargeBound charge “comes along for the ride” once we put the free charge in place (creating an E-field…which induces the Polorization P)

free of charge

Polorization P)Initially we start with knowing the free chargeBut we do not initially know the bound charge y gGiven the correct symmetry, we can use Eqn 4.23 to calculate D using the usual Gauss’s Law methods

ChpChp--10 Electric fields in Matter10 Electric fields in MatterEx. 4.4 A long straight wire of charge line density is

surrounded by rubber insulation out to radius a. Find D

aGaussian surface

Flux through the cylinder endss Dn̂

L

Flux through the cylinder endsWill be zero by symmetry

Apply Integral form of Gauss’s Law (Eqn 4.23) Charge enclosed

enclosed fQ daDda ˆ SurfaceSurface

nD

pp y teg a o o Gauss s a ( q 3)

LQ enc f

gBy surface

sL2A

sD ˆ s 2

so LsLD 2 s This equation holds both inside & outside

The insulator…Outside P=0

asfor ˆs2

so sDEPEDoo

o

Outside

(we would need toKnow P to calculateE inside)

ChpChp--10 Electric fields in Matter10 Electric fields in MatterGauss’s Law in terms of the electric displacement



Wait a minute…what about the surface bound charge b ?We cannot apply the differential form of Gauss’s law

f f fprecisely at the surface of a dielectric surface as b blows up along with the DivEThe integral form of Gauss’s law is free of this issuegWe can also treat the edge as having a finite thickness so thePolarization decays to zero…in which case there is no b

* The polarization drops precipitously to zero outside the dielectric material…so its derivative is a delta function…The surface bound charge is this term (see Griffith’s Prob. 1.46)

ChpChp--10 Electric fields in Matter10 Electric fields in MatterA Deception Parallel…beware!



Eq. 4.22 Looks exactly like Gauss’s law except with the free charge density and DCan we treat D and E the same way? No!Can we treat D and E the same way?There is no Coulomb’s law for D

No!

'dv'ˆ1 f r'rD

2'4

1rfrD

Moreover the curl of D is not always zero PPED o

Research Aside Linear & Nonlinear OpticsResearch Aside Linear & Nonlinear Optics

Introduction

Second Harmonic Generation (SHG) Microscopy( ) py

Collagen Imaging

Third Harmonic Generation (THG) Microscopy

Sensitive Imaging of Malaria Infectiong g

4

zNonlinear Optical Effects

-2

0

2

zTightly Focused Femtosecond Laser

-1 -0.5 0 0.5 1-4

Specific Material Properties

Fluorescence Second Harmonic

Nonlinear Harmonic MicroscopyNonlinear Harmonic Microscopy

hh h

Fluorescence Second Harmonic

hSHGhh

hfluor hh

SHG Changes With Membrane PotentialSHG Nonlinear Microscope

Millard et al. BiophysJ 88, L46-48 (2005)SHG Changes With Membrane Potential

Nuriya et al. PNAS 103, 786 (2006)Sacconi et al. PNAS 103, 3124 (2006)

Time Dependent PolarizationTime Dependent Polarization

2

4

Linear optics

Nonlinear optics if E field is very large

-1 -0.5 0 0.5 1-4

-2

0

SHG THG

Linear opticsp

The nonlinear polarization produces an electric field

Second Harmonic Generation (SHG)Second Harmonic Generation (SHG)

Consider the following electric field

The second order nonlinear polarization will give

Virtual State

SHGNo generation of light

Virtual State

SHGNo generation of light

2 photonslaser fundamental

1 photonSHG

Medium Must beNon centro-symmetric

Jablonski energy diagram for SHG

laser fundamentalfrequency

SHG

Third Harmonic Generation (THG)Third Harmonic Generation (THG)

Consider the same electric field

The third order nonlinear polarization will give

Virtual State

Changes

Non resonantNo energy deposition

Virtual State

grefractive index of material

THGdeposition

Sensitive to gradients3 photonslaser fundamental

1 photonTHG

gin n or (3)

Jablonski energy diagram for THG

laser fundamentalfrequency

THG

SHG Imaging of Healthy PericardiumSHG Imaging of Healthy Pericardium

Optical Sections from a z-stack from Human Heart Pericardium (fibrous layer)

SHG (green)

SHG from Collagen: Fine F,Hansen WP Appl Opt 10, 2350 (1971)

SHG (green)Collagen

2PAF (red)

40 m 60

2PAF (red)elastin

40 m 60 m

80 m 100 m

SHG Imaging of Healthy PericardiumSHG Imaging of Healthy PericardiumOptical z-stack (64 m) from Healthy Human Heart Pericardium

SHG (green)C llCollagen

SHG Imaging of Human PericarditisSHG Imaging of Human Pericarditis

Optical Sections from a z-stack from Human Heart Pericarditis Case

SHG (green)Collagen

20 m 40 m

2PAF (red)elastin

Pericarditis is an InflammationInflammationof the pericardium

60 m 80 m

SHG Imaging of Human PericarditisSHG Imaging of Human Pericarditis

Optical z-stack (64 m) from Human Heart Pericarditis Case

SHG (green)Collagen

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