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

LINEAR DIELECTRICS

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Induced Dipoles and Polarization

• Immersion in an electric field polarizes atoms and

molecules, and there occurs an induced dipole moment, and typically this dipole moment is proportional to the external electric field

where αis called atomic polarizability.

• The polarizationof an object is characterizedby a dipole moment per unit volume,

E pr=αr

l d l

r

Electromagnetism

Alexander A. Iskandar 3

volume unit

per moment dipole

P=

Polarization and Bound Charges

• The electric field produced by a polarized object with polarization (dipole per unit volume) , can be calculated from the potential resulted from a surface charge density

Pr and 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

Pr

∇

• These charge density are called bound charges.

b=−∇⋅P ρ

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Electric Displacement

• 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 Susceptibility

• Recall that an external electric field, , inside the dielectric material will induce an alignment of dipole moments, .

external

Er E

pr=αr

• This alignment of dipole moments will in turn produce a polarization .

• This induced polarization will in turn contributes to the the total electric field inside the dielectric.

• And this total electric field will again modifies the polarization

induced

Pr

polarization.

• This continuous indefinitely, and in practice it is difficult to calculate the polarization.

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Electric Susceptibility

• Induced polarization is related to the local value of the electric field by

tot e

induced E

Pr =ε₀χt ⋅ r

where called the electric susceptibility, is a property of the dielectric medium that is related to the atomic

polarizability, molecular permanent dipole moments, etc.

of its constituents.

• In general, is a second-rank tensor, which could also be a function of the electric field

tot e

induced ₀χ

χe

χ^te

( )

^E^r

χt

be a function of the electric field, .

• When the elements of this tensor differ, it means that a material is easier to polarize with the field in some directions than others. One can imagine how this would be true for crystals.

( )

^E

χe

Linear Dielectric

• Special subset of physical materials: substances that polarize the same in all directions – same for given independent of the direction of (provided it is not too

| P

|r

| E

|r Er

strong)

• Such materials are called linear dielectrics, and in their case is a scalar (and > 0), so that

where is calledpermittivity of a material

( )

E E

P E

Dr =ε₀r+ r=ε₀1+χ_e r=εr χe

(

^χ

)

ε ε = 1+

tot e

induced E

Pr =ε₀χ r

where is called permittivity of a material.

• Further, the dimensionless relativepermittivity or dielectric constant of a material is defined to be

(

χe

)

ε ε = ₀1+

(

_e

)

r χ

ε ε = ε = 1+

0

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Linear Dielectrics

• Example 4.5

• Induced polarization gives fields that “try” to do the same

• Induced polarization gives fields that try to do the same thing that free charges in conductors do: to cancel, at least partially, the applied .Thus is smaller inside a dielectric immersed in an electric field, than it is outside.

Er

+ + + + + + + + +

r

+ + + + + + Di l i Er Er

- - - - Conductor, E=0

- - - - - -

Dielectric, E<E₀ Er0

Er0

−

Linear Dielectrics

• Although we see that the polarization for linear

dielectrics, is proportional to , but this does not always means that the curl of , and hence the curl of , is zero.

Er Pr Pr

Dr

• This is because when we calculate the closed loop integration at the boundary between two type of material, since the permitivities are different at the two sides.

=0 Pr

vacuum

0

0 → = ₀ → ∇× =∇× =

=

×

∇ Er Pr _eEr Dr Pr

χ ε

• However, when the space is entirely filled with a

≠0 Pr

dielectric

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Linear Dielectrics

• Thus for homogeneous space, the last relation means that can be found from the free charges just like as if the dielectric is not there,

Dr

where is the field produced by the free charge distribution in vacuum. Then

i.e. the field in a space of homogeneous dielectric is Evac

Dr r ε0

= Ervac

vac r

vac E

E D

Er r r r

ε ε

1

1 = 0 =

=

p g

reduced by a factor of one over the dielectric constant.

• Hence, when a free charge qis embedded in a large dielectric, the force exerts by nearby charges will be reduced, because the polarization of the medium (the bound charges) partially “shielded” the charge.

Linear Dielectrics

• Example of partially filled capacitor

d

+Q

• With Gauss’s Law

• Then, electric field in the vacuum and inside the dielectric Azˆ

zˆ Q D

A DA Q

a d

D _f _f _f

enc → = → =− =−

=

∫

^r⋅ ^r ^σ ^r ^σ

S

d ε t

–Q

, becomes

• With the electric field known, we can calculate the electric potential difference between the capacitor plates

vac r diel

vac D E D E

Er r r r r

ε ε ε

1 1 1

0

=

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Linear Dielectrics

• With the electric field known, we can calculate the electric potential difference between the capacitor plates

( )

^⎜^⎛ ^⎞ ^∆

∫

^r ^r ^t ^V

• Thus,

• Hence the capacitance of a partially filled capacitor

( )

⎟⎟⎠

⎜⎜ ⎞

⎝

⎛ − +

= ∆

⎟ →

⎠

⎜⎜ ⎞

⎝

⎛ − +

= +

−

=

⋅

=

∆

∫

r vac

diel

vac t

t d E V t t

d E t E t d E l d E V

ε ε

⎟⎟⎠

⎜⎜ ⎞

⎝

⎛ − +

= ∆

=

r

vac t

t d

V A

D Q E

ε ε

ε₀ ₀ z 1

A z Q D^r=−σ_fˆ=− ˆ

• Hence, the capacitance of a partially filled capacitor becomes,

⎟⎟⎠

⎜⎜ ⎞

⎝

⎛ − +

∆ =

=

r

t t d

A V

C Q

ε ε₀

Linear Dielectrics

• When the capacitor is fully filled with dielectrics, then the capacitance becomes

ε A A

Q ε

vac r

r C

d C=ε ε⁰A=ε

⎯→

⎯

⎟⎟⎠

⎜⎜ ⎞

⎝

⎛ − +

∆ =

= ^t^=d

r

t t d

A V

C Q

ε ε₀

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Bound charge and Free Charge

• Note that for linear dielectric, the bound charge density is proportional with the free charge density

(

^E

)

^D ^D

Pr r ^e r⎞ ⎜⎛ ^e ⎞∇ r

⎜⎛

∇

∇ χ χ

since and then

(

^E

)

^D ^D

P

e e e

e

b ⎟∇⋅

⎜⎜ ⎠

−⎝ +

⎠=

⎜ ⎞

⎝

⋅⎛

−∇

=

⋅

−∇

=

⋅

−∇

= χ

χ ε

ε χ χ

ε

ρ ⁰ ⁰ 1

f e e

b ρ

χ

ρ χ _⎟⎟

⎠

⎜⎜ ⎞

⎝

⎛

− +

= 1 D=ρf

⋅

∇ r

(

χe

)

ε ε = ₀1+

Boundary Value Problem with Linear Dielectrics

• In a linear dielectrics, the boundary value problems (BVP) of the Laplace’s equation can be conveniently expressed in terms of the free charges.

• Thus, the discontinuity of electric field can be expressed as

or (in terms of potential)

f below below above above below

above D E E

D^⊥ − ^⊥ =ε ^⊥ −ε ^⊥ =σ

f below below above above

V

V ε σ

ε =−

∂

− ∂

∂

where as the potential itself is continuous

• Example 4.7and 4.8

f below

above

n

n ∂

∂

below

above V

V =

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Energy in Dielectric Systems

• We have derived the energy needed to charged a capacitor as

1 2

V C W =

• When the capacitor is filled completely with dielectrics, the capacitance is given as

• Thus, it suggest that the energy stored in a dielectric

t h ld b itt

2C V W = _vac

vac r

r C

d C=ε ε⁰A=ε

system should be written as

(proof : see textbook)

∫

^→ ⁼ ⁼ ^⋅

=ε τ ε τ τ

d E D d

E W

d E

W r r

2 1 2

2

2 0 2