Electric Fields in Matter

Polarization

Electric displacement

Field of a polarized object

Linear dielectrics

Matter

Insulators/Dielectrics

Conductors

All charges are attached to specific atoms/molecules and can only have a

restricted motion WITHIN the atom/molecule.

When a neutral atom is placed in an external electric field (E):

… positively charged core (nucleus) is pushed along E;

• If E is large enough

► the atom gets pulled apart completely

=> the atom gets IONIZED

… centre of the negatively charged cloud is pushed in the opposite direction of E;

• For less extreme fields

► an equilibrium is established

=> the atom gets POLARIZED

……. the attraction between the nucleus and the electrons

AND

……. the repulsion between them caused by E

Induced Dipole Moment:

Ep

α

Atomic Polarizability

(pointing along E)

Ep

To calculate : (in a simplified model)The model: an atom consists of a point

charge (+q) surrounded by a uniformly charged spherical cloud of charge (-q).

At equilibrium, eEE ( produced by the negative charge cloud)

-q

E

d

+q+q

a

-q

304

1

a

qdEe πε

At distance d from centre,

304

1

a

qdE

πε

Eaqdp 304πε

va 03

0 34 επεα

(where v is the volume of the atom)

Prob. 4.4:

A point charge q is situated a large distance r from a neutral atom of polarizability .

Find the force of attraction between them.

Force on q :

Attractiverr

qF

1

42

2

20

πεα

Alignment of Polar Molecules:

when put in a uniform external field:

0netF

Ep

τ

Polar molecules: molecules having permanent dipole moment

Alignment of Polar Molecules: when put in a non-uniform external field:

FFFnet

d

F+

F- -q

+q

F-

d

F+

-q

+qE+

E-

EEqFnet

EpFnet

For perfect dipole of infinitesimal length,

Ep

τ

the torque about the centre :

the torque about any other point:

FrEp

τ

Prob. 4.9:A dipole p is a distance r from a point

charge q, and oriented so that p makes an angle with the vector r from q to p.

(i) What is the force on p?

(ii) What is the force on q?

rrppr

qF pon

ˆˆ.34

13

0

πε

prrpr

qF qon

ˆˆ3

4

13

0πε

Polarization:When a dielectric material is

put in an external field:

A lot of tiny dipoles pointing along the direction of the field

Induced dipoles (for non-polar constituents)

Aligned dipoles (for polar constituents)

A measure of this effect is POLARIZATION

defined as:

P dipole moment

per unit volume

Material becomes POLARIZED

The Field of a Polarized Object

= sum of the fields produced by infinitesimal dipoles

2

0

ˆ

4

1

s

s

r

rprV

πε

p

rs

Dividing the whole object into small elements, the dipole moment in each volume element d’ :

τ dPp

Total potential :

τ

πε τ

dr

rrPrV

s

s2

0

ˆ

4

1

2

ˆ1

s

s

s r

r

r

Prove it !

τπε τ

dr

PVs

1

4

1

0

FffFFf

Use a product rule :

τπε

τπε

τ

τ

dPr

dr

PV

s

s

1

4

1

4

1

0

0

Prr

P

rP

sss

11

Using Divergence theorem;

τπε

πε

τ

dPr

adPr

V

s

S s

1

4

1

1

4

1

0

0

Defining:

nPbˆ

σ

Volume Bound Charge

Pb

ρ

Surface Bound Charge

τρ

πε

σ

πε

τ

dr

adr

V

s

b

S s

b

0

0

4

1

4

1

surface charge density b

volume charge density b

Field/Potential of a polarized object

Field/Potential produced by a

surface bound charge b

Field/Potential produced by a

volume bound charge b

+

=

Physical Interpretation of Bound Charges

…… are not only mathematical entities devised for calculation;

perfectly genuine accumulations of charge !

but represent

-q +q

d

A

Surface Bound Charge

A dielectric tube

Dipole moment of the small piece:

AdP

PAq

=

Surface charge density:

PAq

b σ

P

AP

n̂

A’

θσ cosPAq

b

In general:

nPbˆ

σ

If the cut is not to P :

+

+

+

+

+

+

+

+

____

_

_

___

Volume Bound Charge

A non-uniform polarization

accumulation of bound charge within the volume

diverging P

pile-up of negative charge

Net accumulated charge with a volume

Opposite to the amount of charge pushed out of the volume through the surface

=

S

b adPd

τρτ

ττ

dP

Pb

ρ

Field of a uniformly polarized sphere

Choose: z-axis || P

P is uniform

0 Pb

ρ

θσ cosˆ PnPb

z

P R

n̂

Potential of a uniformly polarized sphere: (Prob. 4.12)

Potential of a polarized sphere at a field point ( r ):

τ

πε τ

dr

rrPV

s

s2

0

ˆ

4

1

P is uniform

P is constant in each volume element

τ

τρ

περ ss

rr

dPV ˆ

4

112

0

Electric field of a uniformly charged

sphere

rEPrV

ρ

θ1

,

rrE

03ε

ρ

At a point inside the sphere ( r < R )

rPrV

03

1,

εθ

z

PE

03ε

kP

E ˆ3 0ε

PE

03

1

ε Inside the sphere

the field is uniform

rr

RrE ˆ

3 2

3

0ε

ρ

rPr

RrV ˆ

3

1, 2

3

0

εθ

At a point outside the sphere ( r > R )

20

ˆ

4

1

r

rpV

πε

(potential due to a dipole at the origin)

prrpr

rE

ˆˆ31

4

13

0πε

Total dipole moment of the sphere: PRp 3

3

4π

Uniformly polarized Uniformly polarized sphere – A physical sphere – A physical

analysisanalysis Without polarization:

Two spheres of opposite charge, superimposed and canceling each other

With polarization:The centers get separated, with the positive

sphere moving slightly upward and the negative sphere slightly downward

At the top a cap of LEFTOVER positive charge and at the bottom a cap of negative charge

Bound Surface

Charge b

+ ++ + + + + +

+ +

+

-d

+ +

- - - - - - - -

Recall: Pr. 2.18

Two spheres , each of radius R, overlap partially.

dE

03ε

ρ+

-

_

+d

_

+

r r

d

dE

03ε

ρ

Electric field in the region of overlap between the two spheres+ +

+ + + + + + + +

+

-d

+ +

- - - - - - - - PE

03

1

ε

For an outside point:

20

ˆ

4

1

r

rpV

πε

Prob. 4.10:A sphere of radius R carries a polarization

rkrP

where k is a constant and r is the vector from the center.

(i) Calculate the bound charges b and b.

(ii) Find the field inside and outside the sphere.

kRb σ kb 3ρ

rkE inside

0ε 0outsideE

The Electric Displacement

Polarization

Accumulation of Bound charges

Total field = Field due to bound charges + field due to free charges

Gauss’ Law in the presence of dielectricsWithin the dielectric the total charge density:

fb ρρρ

bound charge free charge

caused by polarization

NOT a result of polarization

Gauss’ Law :fbE ρρρε

0

fPE ρε

0

Electric Displacement ( D ) :

PED

0ε

Gauss’ Law

fD ρ

enclfQadD

D & E :

τρ drr

rKrE

s

s

2

ˆ

τρ drr

rKrD f

s

s

2

ˆ

PD

0 E

Boundary Conditions:

fbelowabove DD σ

||||||||belowabovebelowabove PPDD

On normal components:

On tangential components:

For some material (if E is not TOO strong)

EP e

χε0

Electric susceptibility of the medium

Linear DielectricsRecall: Cause of polarization is an Electric field

Total field due to (bound + free) charges

Location ► Homogeneous

Magnitude of E

► Linear

Direction of E ► Isotropic

In a dielectric material, if e is independent of :

In linear (& isotropic) dielectrics; EED e

χεε 00

)1(0 ewithED χεεε

Permittivity of the material

The dimensionless quantity:

0

1ε

εχε er

Relative permittivity or Dielectric constant of the material

EP e

χε0 ED

εand / or

Electric Constitutive Relations

Represent the behavior of materials

But in a homogeneous linear dielectric :

0 DandD f

ρ

00 DP

Generally, even in linear dielectrics :

vacED 0ε

DE

ε

1 vacEE

ε

ε0

vacr

EE

ε

1

When the medium is filled with a homogeneous linear dielectric, the field is

reduced by a factor of 1/r .

Capacitor filled with insulating material of dielectric constant r :

vacr

EE

ε

1

vacr

VVε

1

vacrCC ε

Energy in Dielectric Systems

Recall: The energy stored in any electrostatic system:

τε

dEWspaceall 20

2

The energy stored in a linear dielectric system:

τdEDWspaceall

2

1