FI 2201 Electromagnetism Electric Fields in Matter - Multisite ITB

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FI 2201 Electromagnetism

Alexander A. Iskandar, Ph.D.

Physics of Magnetism and Photonics Research Group

Electric Fields in Matter

POLARIZATION AND ELECTRIC DISPLACEMENT

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Dipole in an Electric Field

• Consider a (physical) dipole, i.e. there is a finite distance of separation between the positive and negative charges.

• When placed in a region ofWhen placed in a region of constantconstantelectric field, the totalelectric field, the total force on the dipole is zero.

+q

–q

Er Fr qEr

+

+ = E

q Fr r

−

− =

d x

z

Electromagnetism

• However, the electric field produces a torque on the dipole.

Alexander A. Iskandar 3

( ) ( ) [ ( ) ( ) ] [ ( ) ( ) ]

E p E d q

E q d

E q d F r F r

N r r r r

r r

r r r r

r

×

=

×

=

−

×

− +

×

=

× +

×

= ₊ ₊ ₋ ₋ 2 2

Dipole in an Electric Field

• Consider a dipole initially perpendicular to the field (position a). The field tends to pull it into alignment

(position b). To push it to position c, we have to do a work:

Er

x

z

d d

d r r r r

r ′

∫

′

∫

′ ′

∫

^θ ^θ ^θ ^θ ^θ ^θ ^θ

θ

c a b

θ

• The work that we do to oppose the force of the electric field is the potential energy

E p cos pE d

sin pE d

E p d

N

W =

∫

_π ^θ′=

∫

_π r× ^θ′=

∫

_π ^θ′^θ′=− ^θ =−r⋅

2 2

2

E p U =−r⋅ r

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Dipole in an Electric Field

• If the electric field is notconstant, then there is a net force on the dipole.

(

^E ^E

)

^q ^E

q

Fr r r r

∆

=

−

=

From the definition of a gradient

(

^E ^E

)

^q ^E

q

F ₊ ₋ ∆

+q

–q

Er ⁺ ⁺

+

= qE Fr r

−

− =−qE Fr r

d x

α z

From the definition of a gradient

• If we also include the xand ycomponents of the electric field then

(

_z

)

_x _z

(

_z

)

_x

z E E dcos E

cos d

E = ∇ → ∆ = ∇

∆ α

α

( )

^d ^E ^F ^q

( )

^d ^E

(

^p

)

^E

Er= r⋅∇ r → r= r⋅∇ r= r⋅∇ r

∆

Dipole in an Electric Field

• And since there is a net force, then the torque on the dipole is

F r E p

Nr r r r r

× +

×

=

x

z

F r E p

N × + ×

rr

Er Fr pr Nr

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Induced Dipoles

• We have learned about conductors, substances that contains “unlimited” supply of charges (electrons) that are

“free” to move around.

• Other type of substances, the dielectrics, the electrons are attached to specific atoms or molecules. These electrons can move a bit in the atom or molecule.

• Hence there is no dramatic change of charge distribution can be caused by external electric field. However, the cumulative effectsresulted in a characteristic behavior of

the dielectric materials.

• There are two principal mechanism by which electric fields can distort the charge distribution : stretchingand rotating.

Induced Dipoles

• Molecules or atoms, such as the Hydrogen atom, known as non-polarmolecules or atoms, although they are neutral consist of a positively charged small hard nucleus and negatively charged electron cloud.

• When placed in an external electric field, the nucleus is pushed in the direction of the field and the electrons in the opposite direction. Until the force on the proton by the displaced electron balances the force on the proton by the external field.

Er

proton electron

Frp

Fre

pr

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Induced Dipoles

• The atom becomes polarizedand there occurs an

induced dipole moment, and typically this dipole moment is proportional to the external electric field

αis called atomic polarizability.

• Suppose (crudely) that the electron has uniform charge density and stays spherical, with radius a, through this process. If the equilibrium position of the proton is a distanced from the center of the electron then the field

E pr=αr

distance d from the center of the electron, then the field of the electron (considered as a uniformly charged sphere) on the proton is in equilibrium with the external field, hence

(

^a

)

^E ^E

p qd a E

E_e qd πε α

πε ⁼ ^→ ⁼ ⁼ ⁼

= ₃ ₀ ³

0

4 4 1

Induced Dipoles

• For molecules, the situation is not quite simple, because they polarize more readily in some direction than others.

• The general relation between the external field and theThe general relation between the external field and the dipole moment becomes

• The set of nine constants α_ijconstitute the polarizability

t f th l l

z zz y zy x zx z

z yz y yy x yx y

z xz y xy x xx x

E E

E p

E E

E p

E E

E p

α α

α

α α

α

α α

α

+ +

=

+ +

=

+ +

=

tensorfor the molecule.

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Alignment of Polar Molecules

• Some naturally-occurring charge distributions, such as many simple molecules (for example water molecule), have permanent, built-in dipole moments.

• Some of this materials are easily polarizable.

• On the large scales, there is zero dipole moment, since the orientation of permanent dipoles is generally random.

• Immersion in an electric field polarizes atoms and molecules and tends to align their permanent dipole

molecules, and tends to align their permanent dipole moments.

Alignment of Polar Molecules

• This polarizationis characterizedby a dipole moment per unit volume, in the same direction as the applied field.

volume unit

per moment dipole

Pr=

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Polarization and Bound Charges

• What is the electric field produced by a polarized object with polarization (dipole per unit volume) ?

• As usual it’s easier to work with the electric potential, and Pr

As usual it s easier to work with the electric potential, and after that to obtain the field we simply take the gradient of this potential.

• From previous lecture, the potential of a single dipole momentis

( )

₂

4 1

r r p r ˆ

V

r = ⋅r

πε ^r^r

since a dipole moment , then

4πε0 r

( ) ( )

∫

^⋅ ^′ ^′

=

V r

r τ

πε ^d

r P r ˆ

V ₂

4 0

1 r r

r

dτ P p r

r= dτ rr

rr′ Pr Er

Polarization and Bound Charges

• Observe that

r t r w iation

diffferent ′

←

⎟=

⎠

⎜ ⎞

⎝

∇′⎛1 ˆ . .

r2

r r then

• Integrating by parts yields

⎠

⎝r r

( )

⁼

∫

^⋅^∇′^⎜_⎝^⎛ ^⎟_⎠^⎞ ^′

V r τ

πε ^P ^d

r

V 1

4 1

0

r r

( ) ( )

⎥⎤

⎢⎡

′

⎞ ′

⎜⎛

′

∫

^P^r ^r

r 1 1

( ) ( )

∫

^⋅ ^′ ^′

=

V r

r τ

πε ^d

r P r ˆ

V ₂

4 0

1 r r

r

• Using divergence theorem

( ) ( )

⎥⎥

⎢ ⎦

⎢⎣ ⎟⎠ ′− ∇′⋅ ′

⎜⎜ ⎞

⎝

⋅⎛

∇′

=

∫ ∫

V

V r τ r τ

πε ^d ^P ^d

r P

V r 1

4 1

0

( )

⁼

∫

^⋅ ^′⁻

∫ ( )

^∇′^⋅ ^′

V

S r r τ

πε

πε ^d^a ^P ^d

r P

V r r r

r 1

4 1 4

1

0 0

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Polarization and Bound Charges

• The first term can be considered as potential resulted from a surface charge density and the second term is a potential caused by a volume charge density,

with

( )

⁼

∫

^′⁺

∫

^′

V

S r ρr τ

πε σ

πε ^d^a ^d

r

V ^b ^b

0

0 4

1 4

r 1

vector unit normal :

nˆ , nˆ

b=Pr⋅ σ

b P

⋅ r

−∇

ρ =

• These charge density are called bound charges.

• Example 4.2

• Problem 4.10

Polarization and Bound Charges

• Another way of analyzing the uniformly polarized sphere of the previous example is by considering it as a

superimpose two spheres of positive and negative charges.

• With out polarization, the two cancels completely.

• With uniform polarization, all the positive charges move slightly upwards and vice-versa, hence the two spheres no longer overlaps completely.

• The electric field of the overlap

d

The electric field of the overlap region is (prob. 2.18)

with . R P d

E qr r

r

0 3

1 4

1

ε πε ⁼⁻

−

=

(

^R

)

^P

p d

qr= r= ₄π ₃ r Pr

=

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Polarization and Bound Charges

• If the polarization is uniform, then the volume bound charge density is zero,

=0

⋅

−∇

= P

b

ρ r

leaving only surface bound charges.

• This surface bound charges resulted from the alignment of many electric dipoles. Since head and tail of

neighboring dipoles effectively cancels, the end results is only end point of the starting and ending dipoles

contributed as a surface charge density ρb

contributed as a surface charge density

• These end charges are boundto the molecules, they are not free to move around.

• In the previous example (Ex. 4.2), the field inside the dielectric is calculated from a puredipole. However, in reality the dipole is resulted from alignment of the

Field Inside a Dielectrics

molecules, hence it is a physical dipole.

• Further, this alignment of physical dipole is not static.

• Hence, the field inside a dielectric is actually fantastically complicated on a microscopic level. So what is that field we calculated in the last example ?

• It is themacroscopic volume-averaged field: an average

It is the macroscopic, volume averaged, field : an average taken over a size large compared to intermolecular

distances in the medium.

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Electric Displacement

• Now let us consider the total fields contributed by the bound chargesand everything else (the free charges).

• Within the dielectrics, the total charge densities can thenWithin the dielectrics, the total charge densities can then be written as

• And Gauss’s law now reads, or,

f

b ρ

ρ ρ= +

f f

b P

E ρ ρ ρ ρ

ε ∇⋅ r= = + =−∇⋅ r+

0

(

^r ^r

)

• The term in brackets is known as Electric Displacement vector field

(

ε E+P

)

=ρf

⋅

∇ r r

0

fenc

f or D da Q

D P

E

D= + → ∇⋅ =

∫

⋅ =

S

r r r

r r

r ε₀ ρ

Electric Displacement

• Using the Electric Displacement field, we can calculate the electric field (when we know the polarization vector).

• Example 4.4

• Problem 4.15

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Electric Displacement

• Note the similarity of Gauss’s Law for and Er Dr

fenc

f or D da Q

D= ⋅ =

⋅

∇ ^r ^ρ

∫

^r ^r

ε ε

ρ Q_enc

a d E or

E= ⋅ =

⋅

∇ ^r

∫

^r ^r

• However, this does not mean that other relations are similar, in particular there is no Coulombs law for

• Further recall that in determining a vector field we need

( )

^≠

∫ ( )

^′ ^′

outside

f r d

r ˆ

D ρ τ

π ^r

r r 4 2

1 r

r

Dr

0 S

0 ε

ε

∫

S

• Further, recall that in determining a vector field, we need information bothof its divergence and its curl, which for electric field it is zero.

• But for the displacement field it is not zero

Er

Dr

P P

E

Dr r r r

×

∇

=

×

∇ +

×

∇

=

×

∇ ε₀

When to use Electric Displacement

• In situations in which Gauss’ Law helps, one can use this new relation to calculate , from the free charges alone, and then to determine

Dr Er

• In other words, is the same, whether or not there is polarizable material present.

Dr

fenc

f or D da Q

D P

E

D= + → ∇⋅ =

∫

⋅ =

S

r r r

r r

r ε₀ ρ

• But be cautioned !!This is not as useful as it sounds, it doesn’t really allow one to ignore the presence of

polarizable media.

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When to use Electric Displacement

• Example 4.4 and Problem 4.15 which we have done can be solved using the Gauss’ Law for Electric Displacement field, because the symmetry of the problem.

• Furthermore, this symmetry will dictates the answer, and we don’t need to worry about the curl of .

• However, for problem that has no symmetry, bewarewhen calculating electric field from the displacement field , for which you need to know the effect of polarization

Dr

Dr Er

for which you need to know the effect of polarization.

• Like Ex. 4.2, although in this problem , however, because the problem does not have a symmetry, we cannot solve this problem using Gauss’ Law.

• We cannot ignore the presence of the polarizable media.

=0

×

∇

=

×

∇ Dr Pr

When to use Electric Displacement

• The use of turns out to be most helpful in cases in

which the polarization is not built in, but instead is induced by an external applied electric field.

Dr

• This will be the topic of the next section (lecture).

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Boundary Conditions for Electric Displacement

• The electrostatic boundary conditions can also be recast in terms of , discontinuity of the normal component is caused by the surface charge density and discontinuity in

Dr

the parallel direction is resulted from discontinuity of the polarization

f below

above D

D^⊥ − ^⊥ =σ

fenc

f Dd D da Q

D= → ∇⋅ = ⋅ =

⋅

∇

∫ ∫

S V

r r r

r ρ τ

( ) ∫

∫

^∇^× ^⋅ ⁼ ^⋅

→

×

∇

=

×

∇ r r r r r rl

d D a d D P

D

• Compare this with

||

below

||

above

||

below

||

above D P P

Dr r r r

−

=

−

ε0

=σ

− ^⊥

⊥

below

above E

E Er^||_above−Er_below^|| =0

( ) ∫

∫

^∇^× ^⋅ ⁼ ^⋅

→

×

∇

=

×

∇

C S

l d D a d D P

D