Fri. 10/23

(C14) 4.4.1 Linear Dielectrics (read rest at your discretion)

Mon. Wed. Thurs. Fri.

(C 17) 12.1.1-.1.2, 12.3.1 E to B; 5.1.1-.1.2 Lorentz Force Law: fields and forces (C 17) 5.1.3 Lorentz Force Law: currents (C 17) 5.2 Biot-Savart Law

HW6

Polarization & Electric Displacement Example

Consider a huge slab of dielectric material initially with uniform field, and corresponding uniform polarization and electric displacement .

DPEo

adDdQ freefree

oE

oooo PED

oE

You cut a small spherical hole out of it. What is the field in its center in terms of and ? oP

oE

For illustrative purposes only, take the polarization to be anti-parallel to the field, and imagine both to be in the z direction.

Quoting Example 4.2 (which in turn builds on 3.9), the field inside a uniformly polarized sphere is

By Superposition Principle, cutting out a sphere is the same as inserting a sphere of opposite polarization.

oP

sphere

o

sphere PE

3

1

So, we ‘add in ‘ a sphere of polarization oP

Adding field o

o

added PE

3

1

031

. PEEoospherein

Polarization & Electric Displacement Example

Consider a huge slab of dielectric material initially with uniform field, and corresponding uniform polarization and electric displacement .

DPEo

adDdQ freefree

oE

oooo PED

oE

You cut a small spherical hole out of it. What is the field in its center in terms of and ? oP

oE

oP

sphere

o

sphere PE

3

1

There is no material in the sphere, so

031

. PEEoospherein

What is the electric displacement in its center in terms of and ?

oD

oP

sphereshereosphere PED

031 PED

ooosphere

where ooo PDEo

1

so

0311 PPDD

oo ooosphere

032 PDo

You cut out a wafer-shaped cavity perpendicular to .

Polarization & Electric Displacement

Exercise: Consider a huge slab of dielectric material initially with uniform field, and corresponding uniform polarization and electric displacement .

DPEo

adDdQ freefree

oE

oooo PED

oE What is the field in its center in terms of and ?

oP

oE

oP

What is the electric displacement in its center in terms of and ?

oD

oP

PaP bb

andˆ

Hint: Think of inserting the appropriate waver-sized capacitor. oP

d

pdP

Boundary Conditions Electric field, across charged surface

o

enclsidessidesbottombottomtoptop

QadEadEadE

Send side height / area to 0

o

bottomtop EE

topad

Bottomad

bottomE

topE

o

enclQadE

o

surfaceda

o

surface

bottombottomtoptop

daadEadE

o

bottomtop

AAEAE

)1(

o

enclE

Boundary Conditions Electric Displacement, across charged surface

enclfreesidessidesbottombottomtoptop QadDadDadD .

Send side height / area to 0

freebottomtop DD

topad

Bottomad

bottomD

topD

free

enclfreeQadD .

surfacefreeda

surfacefreebottombottomtoptop daadDadD

AADAD freebottomtop )1(

enclfreeD .

Boundary Conditions (static) Electric field, along charged surface

0 sidessidesbottombottomtoptop ldEldEldE

Send side height to 0

0|||| bottomtop EE

topld

bottomE

topE

bottomld

0 ldE

0 bottombottomtoptop ldEldE

0)1(|||| LELE bottomtop

0 E

Boundary Conditions (static) Electric displacement, along charged surface

sidessidesbottombottomtoptopsidessidesbottombottomtoptop ldPldPldPldDldDldD

Send side height to 0

topld

bottomD

topD

bottomldPD

)1()1( |||||||| LPLPLDLD bottomtopbottomtop

ldPldD

bottombottomtoptopbottombottomtoptop ldPldPldDldD

bottomtopbottomtop PPDD ||||||||

Boundary Conditions Electric and Displacement fields

(could have guessed as much from .)

0|||| bottomtop EE

topld

bottomE

topE

bottomld

topad

Bottomad

o

bottomtop EE

aAlong

freebottomtop DD

bottomtopbottomtop PPDD ||||||||

PED o

(could have guessed as much from and .) PED o

aPbfree

ˆ

or o

bottomtop aEaE

ˆˆ

freebottomtop aDaD ˆˆ

bottomD

topD

Across

Boundary Conditions Electric and Displacement fields

0|||| bottomtop EEo

bottomtop EE

Along

freebottomtop DD bottomtopbottomtop PPDD ||||||||

PED o

aPbfreeˆ

Across

Exercise Bar Electret (like an electric bar magnet): uniform P along axis

Sketch 𝑃, 𝐸, and 𝐷 as to obey the boundary conditions.

𝑃

𝐸

Only surface charge is the bound surface charge on two faces

PED o

𝐷

There is no free surface charge, so no discontinuity in D.

Atom on a stick

Recall: Atom’s Response to Electric Field

𝑬𝒆𝒙𝒕𝒆𝒓𝒏𝒂𝒍

𝑭𝒆𝒙𝒕𝒆𝒓𝒏𝒂𝒍 = 𝑞𝑬𝒆𝒙𝒕𝒆𝒓𝒏𝒂𝒍

𝑭𝒊𝒏𝒕𝒆𝒓𝒏𝒂𝒍 = −𝑭𝒆𝒙𝒕𝒆𝒓𝒏𝒂𝒍

...0

intint

ss

FF

ss

For small stretch, first term in Taylor Series (Hook’s Law)

...

0

intint

qsqs

FF

s

...0

intint

pp

FF

s

Electric Dipole moment

...0

intint

pp

FFF

s

ext

...0

intint

pp

FFqE

s

ext

So, for small enough stretch, weak enough field

extqEp or extEp

≡polarizability

Linear Dielectrics Chunk of induced dipoles

𝐸

Ep

Everyone’s field but its own

Point along field Linearly proportional

d

pdP

Polarization = Dipole density

so E

d

d

d

EdP

Define “electric susceptibility” to be the proportionality constant (and provide convenient factor of o.)

P 0eE

d

d

o

e

1

PED o

Always linear dielectric

EE eoo

Eeo

1

ED

0 1 e Permittivity (of not-so-free space)

Dielectric Constant

er o

1

ED ro

Or, in terms of polarization

PDe

11

or PD

e

e

1

for individual induced dipole

For chunk of induced dipoles

Linear Dielectrics Chunk of induced dipoles

𝐸

Example: consider a simplified version of problem 4.18 . Say we have only one dielectric material, of constant r between two capacitor plates distance a apart. a. Electric Displacement, D. enclfreeQadD .

Gaussian box

free

free

free

bottom

enclfree

inside

enclfreebottominside

enclfreebottominsidetopoutside

A

QD

QAD

QADAD

.

.

.

0

Expect only perpendicular to surface and only inside capacitor

outside 0

inside zD

free

b. Electric Field, E.

zor

freeˆ

𝐷

DE

z

freeˆ

c. Polarization, P.

zfreerˆ1 1

EDP o

zz

r

free

freeˆˆ

𝑃

d. Potential Difference across plates, DV.

aor

free

D

top

bottom

ldEV

top

bottom or

freezdz

ˆ

a DV

e. Bound charge, b and b.

bottombottom

toptop

b

nP

nPnP

ˆ

ˆˆ

0 Pb

zntopˆˆ

znbottomˆˆ

zzfreerˆˆ1 1

freer

11

zzfreerˆˆ1 1

freer

11

f. E from charge distribution.

b

or

f

o

enclQadE

o

enclbenclf

bottominside

QQAE

..

o

bf

insideE

o

ff r

11

Linear Dielectrics

𝐸o

Example: Alternate / iterative perspective on field in dielectric. Consider again a simple capacitor with dielectric. We’ll find the electric field in terms of what it would have been without the dielectric. We’ll do this iteratively and build a series solutions.

free

free

𝑃o

b.o

0. Say we start with no dielectric. Initially there’s the field simply due to the free charge; Eo.

oEE

and the associated surface charges contribute a field of their own, where so in the opposite direction.

E1 P0 0 eE0

oeoo EP

zEo

ob ˆ.1

nPobˆ

0.

oe E

𝐸1

1. This field induces a little counter polarization,

P1 0eE1 0e2E0

E2 P1 0 e 2E0

oe E

2

2. See a pattern?

...

𝑃1 𝐸2

o

n

n

e EE

0

As long as e< 1, this converges to

Einside 1

1 e

E0

1

rE0

Same result as we got previously

We insert the dielectric and that field induces a polarization,

Which is means a surface charge and resulting field contribution of its own

Linear Dielectrics

𝐸o

Exercise: Try it for your self. A sphere made of linear dielectric

material is placed in an otherwise uniform electric field 𝐸o. Find the electric field inside the sphere in terms of the material’s dielectric constant, r.

free

free oEE

You can take it as a given that a sphere of uniform polarization contributes field E P 30

Example: A coaxial cable consists of a copper wire of radius a surrounded by a concentric copper tube of inner radius c. The space between is partially filled (from b to c) with material of dielectric constant r as shown below. Find the capacitance per length of the cable.

a b

c

V

QC

D

L

For the sake of reasoning this out, say there’s charge Q uniformly distributed along the surface of the central wire.

r

enclfQadD .

D

c

a

ldEV

QsLD 2

Gaussian cylinder of some radius a<s<c.

sL

QD

2

csbssL

Q

bsassL

Q

E

ro

o

ˆ2

ˆ2

D

c

a

ldEV

D

c

b ro

b

a o

dssL

Qds

sL

QV

22

bc

ab

orL

QV lnln

21

D

csbD

bsaDE

ro

o

1

1

c

b

b

a

ldEldE

bc

ab

o

rL

C

lnln

21

Exercise: There are two metal spherical shells with radii R and 3R. There is material with a dielectric constant r = 3/2 between radii R and 2R. What is the capacitance? R

2R

3R

r

Fri. 10/23

(C14) 4.4.1 Linear Dielectrics (read rest at your discretion)

Mon. Wed. Thurs. Fri.

(C 17) 12.1.1-.1.2, 12.3.1 E to B; 5.1.1-.1.2 Lorentz Force Law: fields and forces (C 17) 5.1.3 Lorentz Force Law: currents (C 17) 5.2 Biot-Savart Law

HW6