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Lecture 7

Polarization

In this lecture you will learn:

• Material Polarization

• Mathematics of Polarization

• Conductors Vs

• Appendix

ECE 303 – Fall 2007 – Farhan Rana – Cornell University

Charge and Moments

Consider a charge dipole:

+ q r d − q

Dipole of the charge dipole is a vector p r such that:

r pr = q d

ECE 303 – Fall 2007 – Farhan Rana – Cornell University

1 Conductors in Electric Fields – A Review φ = V φ = 0 • Consider the problem when a conductive + - plate was placed inside an + - + - + - • Conductors have “free charges” that are able + - to move around (mostly these “free charges” + - are that are not attached to any + - particular ) + - • Under the influence of external E-field these +- free charges move to completely screen out V the E-field from within the conducting material φ = V φ = 0 + - + - Conductors + - + - Sea of Free Electrons + - + - + - + - +++ + - + - + +ve nucleus + - + - +++ + - + - -ve sea of free + - + - +++ electrons

+- +++ V σ

ECE 303 – Fall 2007 – Farhan Rana – Cornell University

Dielectrics in Electric Fields - Polarization

Many materials do not have “free electrons” that can move around, but have electrons bound to , as shown below

Dielectric Dielectric in an E-field E + + + + +ve nucleus - + - + - +

+ + + -ve - + - + - + cloud + + + - + - + - +

+ + + - + - + - +

+ + + - + - + - +

+ + + - + - + - +

In the presence of an E-field, the electron cloud in each atom distorts ALMOST INSTANTANEOUSLY so that each atom looks like a charge dipole

ECE 303 – Fall 2007 – Farhan Rana – Cornell University

2 Dielectrics in Electric Fields – Polarization Vector r d r r - + p = Qd = dipole moment of each dipole -Q +Q

r The polarization vector P is a vector such that: E r P = N pr - + - + - + v = NQd - + - + - + Where N is the number of charge dipoles per unit - + - + - + in the material - + - + - + r The units of P are: Coumlombs/m2 - + - + - + r The polarization vector P characterizes the - + - + - + of the medium under the influence of the electric field

ECE 303 – Fall 2007 – Farhan Rana – Cornell University

Dielectrics in Electric Fields – Electrical Susceptibility

Naturally, one would expect the polarization of the material to be proportional to the strength of the electric field: r r P ∝ E r r ⇒ P = εo χe E

The constant of proportionality χe is called the electrical susceptibility of the material E - + - + - +

- + - + - +

- + - + - +

- + - + - +

- + - + - +

- + - + - +

ECE 303 – Fall 2007 – Farhan Rana – Cornell University

3 Material Polarization and Densities r P = P xˆ

- + - + - + - + - + E - + - + - + - + - + σ = −P p σ p = +P - + - + - + - + - +

- + - + - + - + - +

- + - + - + - + - +

- + - + - + - + - + x

• The stuff inside the box in on the average charge neutral (same number of positive and negative charges)

• There is a net negative surface on the left facet of the material as a result of material polarization

• There is a net positive surface charge density on the right facet of the material as a result of material polarization

ECE 303 – Fall 2007 – Farhan Rana – Cornell University

Material Polarization and Surface Charge Densities

d

Area = A E - + - + - + - + - + How much charge is r in this volume?? - + - + - + - + - + P = P xˆ - + - + - + - + - +

- + - + - + - + - +

- + - + - + - + - +

Total interface negative charge due to dipoles in the volume Ad = - Q N A d

If we divide the total interface charge by the A we get the interface charge per unit area which would be the surface charge density σp QNAd ⇒ σ = − = −NQd = −P p A

ECE 303 – Fall 2007 – Farhan Rana – Cornell University

4 Material Polarization and Volume Charge Densities

More generally, one can write a volume polarization volume charge density due to material polarization as:

r ρp = −∇ .P

(A formal proof is given in the Appendix)

There will be a net non-zero volume charge density inside a material if the material polarization is varying in space

In 1D situations: ∂P ρ ()x = − x p ∂x

ECE 303 – Fall 2007 – Farhan Rana – Cornell University

Material Polarization and Charge Densities E σ p = −P σ p = +P r P = Pxˆ x

x=0 x=L

Px P

0 L x r ρp = −∇ .P ∂P ρ = − x p ∂x Pδ (x − L)

0 L x ∂P − Pδ ()x ρ = − x = −P δ ()x + P δ (x − L ) p ∂x

ECE 303 – Fall 2007 – Farhan Rana – Cornell University

5 Mathematics of Polarization – The “D” Field

Gauss’ Law states: r ∇ .εo E = ρ

But charge densities could be of two types:

1) Paired charge density ρp (due to material polarization) 2) Unpaired charge density ρu (due to everything else – the usual stuff)

So: r r r ρ = −∇ .P ∇ .εo E = ρu + ρp = ρu − ∇ .P Using: p r r ⇒ ∇ .()εo E + P = ρu

If one defines the D-field inside materials as: r r r D = εo E + P Then inside materials Gauss’ Law becomes: r ∇ .D = ρu

ECE 303 – Fall 2007 – Farhan Rana – Cornell University

Mathematics of Polarization – Dielectric Permittivity

If one defines the D-field as: r r r r Then: ∇ .D = ρ D = εo E + P u

r r r r r r Note that: D = εo E + P = εo E + εo χe E = εo (1+ χe )E

If one defines the dielectric permittivity of a material as:

ε = εo (1+ χe ) then one can write the D-field inside materials as: r r D = ε E

Inside materials the D-field obeys the Gauss’ Law:

r r ∇ .D = ∇ .ε E = ρu

ECE 303 – Fall 2007 – Farhan Rana – Cornell University

6 Mathematics of Polarization – Polarization

So far we have looked at the current density due to the motion of free unpaired charges: r r E Ju = σ E - + - + - + The motion of paired charges also results in + + + a current density: - - - r r - + - + - + r ∂P ∂ (εo χe E ) Jp = = ∂t ∂t - + - + - + r Jp is called the polarization current density r r r r ∂ (ε E ) ’s Law is correctly given by: ∇ × H = J + J + o u p ∂t

r r r ∂ (ε E) Which can also be written as: ∇ × H = J + u ∂t

ECE 303 – Fall 2007 – Farhan Rana – Cornell University

Mathematics of Polarization – Modified Maxwell’s Equations

Putting it all together …. One can forget all about material polarization, and polarization charge densities, and polarization current densities, as long as one uses the dielectric permittivity ε instead of the free space permittivity εo

Maxwell’s equations in dielectric materials take the form: r r ∇. ε E = ρ ∇. (εo E ) = ρu + ρp ( ) u r r ∇. µo H = 0 Here the charge density ∇. µo H = 0 OR ρu is the unpaired charge density r r r ∂ µ H r ∂ µoH ∇ × E = − o ∇ × E = − ∂t ∂t r r r r r ∂ ()ε E r r ∂ ()ε E ∇ × H = J + J + o ∇ × H = Ju + u p ∂t ∂t

Here the current density is due to the unpaired charges

ECE 303 – Fall 2007 – Farhan Rana – Cornell University

7 Dielectric Permittivity – Boundary Conditions - I σu + σ p

How to relate the electric fields on both sides of E E the dielectric interface ?? 1 2

Gauss’ Law in the presence of dielectric material is: ε1 ε2 r r ∇. (D) = ∇. (ε E ) = ρu r r or ∫∫ D . da =∫∫∫ ρu dV σu + σ p

Draw a at the interface:

D D (D2 − D1)A = σu A 1 2 ⇒ D2 − D1 = σu A ε1 ε2 or ε2 E2 − ε1 E1 = σ u

ECE 303 – Fall 2007 – Farhan Rana – Cornell University

Dielectric Permittivity – Boundary Conditions - II

σu + σ p How else can one relate the electric fields on both sides of the dielectric interface ?? E E Gauss’ Law in the presence of dielectric material is 1 2 also: r A ∇. (εo E ) = ρu + ρp ε1 ε2 Draw a Gaussian surface at the interface:

εo (E2 − E1) A = (σu + σ p ) A

or εo (E2 − E1) = σu + σ p σu + σ p

The parallel component of the E-field at a material interface is always continuous at the interface (no E1 E2 change here)

ε1 ε2 (E2 − E1) = 0

ECE 303 – Fall 2007 – Farhan Rana – Cornell University

8 Dielectrics Vs Conductors φ = 0 Conductors: φ = V + - + - + - Free unpaired charges move to completely + - + - + - + - + screen the E-field on time scales longer than - + - + - + the dielectric relaxation time - + - + - + - + - + - + - + - + - + Dielectrics: - + - + - + - Paired charges originating due to material polarization partially screen the E-field +- almost instantaneously V φ = V φ = 0 When σu = 0: + - + - + - + - ε2 E2 − ε1 E1 = 0 + - + - + - + - also: + - + - + - + - εo (E2 − E1) = σ p + - + - The discontinuity of the normal component of the + - + - E-field is due to the paired charges at the interface +- (even when σu = 0) V

ECE 303 – Fall 2007 – Farhan Rana – Cornell University

Conductors or Dielectrics

Some materials are conductors and some are dielectrics ………….

Dielectrics Conductors Tightly Bound Electrons Sea of Free Electrons

+ + + + +ve nucleus +++ + +ve nucleus

+ + + -ve electron +++ -ve sea of free cloud electrons + + + +++

+ + + +++

+ + + +++

+ + + +++ ε σ

ECE 303 – Fall 2007 – Farhan Rana – Cornell University

9 Conductors and Dielectrics

……. but many important materials are conductors as well as dielectrics

Tightly bound core electrons and a sea of free electrons

• Most conductors like gold, copper, + +ve nucleus and silver, and like + + + Silicon, are both conductors and + + + -ve electron dielectrics cloud + + + • They have a sea of free electrons -ve sea of free electrons that results in a finite value of + + + conductivity and they also have tightly bound core electrons that + + + result in a value for the dielectric permittivity + + + ε σ

ε Dielectric relaxation time for these materials = τ = d σ

ECE 303 – Fall 2007 – Farhan Rana – Cornell University

Appendix: Polarization Charge Density - I

The expression relating the polarization charge density to the divergence of the polarization vector, r ρp = −∇ .P can be proved more formally as shown below:

The potential of an isolated dipole sitting at the origin and pointing in the z-direction is: pr cos()θ r r + q rr p = qd φ()= 2 θ 4πεo r r d − q More generally, the potential of a dipole sitting at position r’ and pointing in an arbitrary direction is:

r r + q 1 pr . (rr − rr') p = qd φ()rr = r r r 3 d 4πεo r − r '

− q

ECE 303 – Fall 2007 – Farhan Rana – Cornell University

10 Appendix: Polarization Charge Density - II

r r If a material has a polarization density vector P(r ') - + - + - + then the potential due to all the dipoles can be found using superposition: - + - + - +

- + - + - + r 1 P()(rr' . rr − rr' ) φ()rr = dV ' ∫∫∫ r r 3 4πεo r − r ' ⎡ ⎛ ⎞⎤ 1 r r r 1 Now integrate by parts in 3D………. = ∫∫∫ P()r ' . ⎢∇'⎜ r r ⎟⎥ dV ' 4πεo ⎣ ⎝ r − r ' ⎠⎦ 1 r r r ⎡ 1 ⎤ = ∫∫∫ []− ∇' . P()r ' ⎢ r r ⎥ dV ' 4πεo ⎣ r − r ' ⎦ r r − ∇' . P()rr' This looks like the superposition = ∫∫∫ dV ' 4πε rr − rr' integral for a volume charge o density given by: r r r ρp (r ') = −∇ .P(r ')

ECE 303 – Fall 2007 – Farhan Rana – Cornell University

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