Maxwell’s Equations in Vacuum - Trinity College Dublin · 2012. 4. 30. · Maxwell’s Equations...

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Maxwell’s Equations in Vacuum (1) .E = ρ /ε o Poisson’s Equation (2) .B = 0 No magnetic monopoles (3) x E = -B/t Faraday’s Law (4) x B = µ o j + µ o ε o E/t Maxwell’s Displacement Electric Field E Vm -1 Magnetic Induction B Tesla Charge density ρ Cm -3 Current Density Cm -2 s -1 Ohmic Conduction j = σ E Electric Conductivity Siemens (Mho)

Transcript of Maxwell’s Equations in Vacuum - Trinity College Dublin · 2012. 4. 30. · Maxwell’s Equations...

  • Maxwell’s Equations in Vacuum (1) ∇.E = ρ /εo Poisson’s Equation (2) ∇.B = 0 No magnetic monopoles (3) ∇ x E = -∂B/∂t Faraday’s Law (4) ∇ x B = µoj + µoεo∂E/∂t Maxwell’s Displacement Electric Field E Vm-1 Magnetic Induction B Tesla Charge density ρ Cm-3 Current Density Cm-2s-1 Ohmic Conduction j = σ E Electric Conductivity Siemens (Mho)

  • Constitutive Relations (1) D = εoE + P Electric Displacement D Cm-2 and Polarization P Cm-2 (2) P = εo χ E Electric Susceptibility χ (3) ε =1+ χ Relative Permittivity (dielectric function) ε (4) D = εo ε E (5) H = B / µo – M Magnetic Field H Am-1 and Magnetization M Am-1 (6) M = χB B / µo Magnetic Susceptibility χB (7) µ =1 / (1- χB) Relative Permeability µ (8) H = B / µ µo µ ~ 1 (non-magnetic materials), ε ~ 1 - 50

  • Electric Polarisation • Apply Gauss’ Law to right and left ends of polarised dielectric

    • EDep = ‘Depolarising field’

    • Macroscopic electric field EMac= E + EDep = E - P/εo

    σ± surface charge density Cm-2 σ± = P.n n outward normal E+2dA = σ+dA/εo Gauss’ Law E+ = σ+/2εo E- = σ-/2εo

    EDep = E+ + E- = (σ++ σ-)/2εo EDep = -P/εo P = σ+ = σ-

    σ-

    E

    P σ+

    E+ E-

  • Electric Polarisation Define dimensionless dielectric susceptibility χ through

    P = εo χ EMac EMac = E – P/εo εo E = εo EMac + P εo E = εo EMac + εo χ EMac = εo (1 + χ)EMac = εoεEMac Define dielectric constant (relative permittivity) ε = 1 + χ

    EMac = E /ε E = ε EMac

    Typical values for ε: silicon 11.8, diamond 5.6, vacuum 1 Metal: ε →∞ Insulator: ε∞ (electronic part) small, ~5, lattice part up to 20

  • Electric Polarisation Rewrite EMac = E – P/εo as

    εoEMac + P = εoE

    LHS contains only fields inside matter, RHS fields outside

    Displacement field, D

    D = εoEMac + P = εoε EMac = εoE

    Displacement field defined in terms of EMac (inside matter, relative permittivity ε) and E (in vacuum, relative permittivity 1). Define

    D = εoε E

    where ε is the relative permittivity and E is the electric field

  • Gauss’ Law in Matter • Uniform polarisation → induced surface charges only

    • Non-uniform polarisation → induced bulk charges also Displacements of positive charges Accumulated charges

    + + - -

    P - + E

  • Gauss’ Law in Matter Charge entering xz face at y = 0: Px=0∆y∆z Charge leaving xz face at y = ∆y: Px=∆x∆y∆z = (Px=0 + ∂Px/∂x ∆x) ∆y∆z Net charge entering cube: (Px=0 - Px=∆x ) ∆y∆z = -∂Px/∂x ∆x∆y∆z

    ∆x

    ∆z

    ∆y

    z

    y

    x

    Charge entering cube via all faces:

    -(∂Px/∂x + ∂Py/∂y + ∂Pz/∂z) ∆x∆y∆z = Qpol ρpol = lim (∆x∆y∆z)→0 Qpol /(∆x∆y∆z)

    -∇.P = ρpol

    Px=∆x Px=0

  • Gauss’ Law in Matter Differentiate -∇.P = ρpol wrt time ∇.∂P/∂t + ∂ρpol/∂t = 0 Compare to continuity equation ∇.j + ∂ρ/∂t = 0

    ∂P/∂t = jpol

    Rate of change of polarisation is the polarisation-current density Suppose that charges in matter can be divided into ‘bound’ or polarisation and ‘free’ or conduction charges

    ρtot = ρpol + ρfree

  • Gauss’ Law in Matter Inside matter ∇.E = ∇.Emac = ρtot/εo = (ρpol + ρfree)/εo Total (averaged) electric field is the macroscopic field -∇.P = ρpol ∇.(εoE + P) = ρfree

    ∇.D = ρfree

    Introduction of the displacement field, D, allows us to eliminate polarisation charges from any calculation. This is a form of Gauss’ Law suitable for application in matter.

  • Ampère’s Law in Matter

    Ampère’s Law

    ( )

    currentssteady -non for t

    1

    1

    0.

    0..

    ≠−=∇⇒

    =×∇∇=∇⇒

    ×∇=⇒=×∇

    ∂∂ρ

    µ

    µµ

    j

    Bj

    BjjB

    o

    oo

    Problem!

    A steady current implies constant charge density, so Ampère’s law is consistent with the continuity equation for steady currents Ampère’s law is inconsistent with the continuity equation (conservation of charge) when the charge density is time dependent

    Continuity equation

  • Ampère’s Law in Matter Add term to LHS such that taking Div makes LHS also identically equal to zero:

    The extra term is in the bracket extended Ampère’s Law

    Displacement current (vacuum)

    ( )

    ( )0..

    0..=∇−∇

    =∇+∇

    ×∇=+

    jj?j

    B?j

    or

    1oµ

    ( ) jEE

    EE

    ...

    ..

    −∇=

    ∇⇒=∇∴

    =∇⇒=∇

    ttt oo

    oo

    ∂∂ε

    ∂∂ρε

    ∂∂

    ρεερ

    +=×∇

    ×∇=

    +

    t

    t

    oo

    oo

    ∂∂εµ

    µ∂∂ε

    EjB

    BEj

    1

  • Ampère’s Law in Matter

    • Polarisation current density from oscillation of charges in electric dipoles • Magnetisation current density variation in magnitude of magnetic dipoles

    PMf jjjj ++=

    t∂∂PjP=

    tooo ∂∂µεµ EjB +=×∇

    M = sin(ay) k

    k

    i j

    jM = curl M = a cos(ay) i

    Total current

    MjM x ∇=

  • Ampère’s Law in Matter

    ∂D/∂t is the displacement current postulated by Maxwell (1862) In vacuum D = εoE ∂D/∂t = εo ∂E/∂t In matter D = εo ε E ∂D/∂t = εo ε ∂E/∂t Displacement current exists throughout space in a changing electric field

    ( )tt

    tt

    t1

    t

    ff

    f

    PMf

    ∂∂ε

    ∂∂

    µ

    ∂∂ε

    ∂∂

    ∂∂ε

    µ

    ∂∂µεµ

    DjHPEjMB

    EPMj

    EjjjB

    EjB

    +=×∇⇒++=

    −×∇

    ++×∇+=

    +++=×∇⇒

    +=×∇

    oo

    o

    oo

    ooo

  • Maxwell’s Equations

    in vacuum in matter ∇.E = ρ /εo ∇.D = ρfree Poisson’s Equation ∇.B = 0 ∇.B = 0 No magnetic monopoles ∇ x E = -∂B/∂t ∇ x E = -∂B/∂t Faraday’s Law ∇ x B = µoj + µoεo∂E/∂t ∇ x H = jfree + ∂D/∂t Maxwell’s Displacement D = εoε E = εo(1+ χ)E Constitutive relation for D H = B/(µoµ) = (1- χB)B/µo Constitutive relation for H

  • Divergence Theorem 2-D 3-D • From Green’s Theorem

    • In words - Integral of A.n dA over surface contour equals integral of div A over surface area

    • In 3-D • Integral of A.n dA over bounding surface S equals integral of div V dV

    within volume enclosed by surface S • The area element n dA is conveniently written as dS

    dV .dA .V

    S∫∫∫ ∇= AnA

    A.n dA ∇.A dV

  • Differential form of Gauss’ Law • Integral form

    • Divergence theorem applied to field V, volume v bounded by surface S

    • Divergence theorem applied to electric field E

    ∫∫∫ ∇== VSS

    dv .d .dA . VSV nV V.dS ∇.V dv

    o

    V

    d )(d .

    ε

    ρ∫∫ =

    rrS E

    S

    ∫∫

    ∫∫

    =∇

    ∇=

    VV

    V

    )dv(1dv .

    dv ..d

    rE

    ES E

    ρε o

    Soε

    ρ )( )( . rrE =∇

    Differential form of Gauss’ Law (Poisson’s Equation)

  • Stokes’ Theorem 3-D • In words - Integral of (∇ x A).n dA over surface S equals integral of A.dr

    over bounding contour C • It doesn’t matter which surface (blue or hatched). Direction of dr

    determined by right hand rule.

    (∇ x a) .ndA

    n outward normal

    dA

    local value of ∇ x A

    local value of A

    dr A. dr ( ) dA . x .d

    SC

    ∫∫∫ ∇= nArA

    A

    C

  • Faraday’s Law

    ( )

    ( )

    form aldifferenti inLaw sFaraday' t

    x

    .d x t

    d . x d .dtd

    Theorem Stokes' d . x d .

    .ddtd.d

    fluxmagnetic of change of rate minus equals emf Induced:law of form Integral

    ∂∂

    −=∇

    =

    ∇+

    ∂∂

    ∇=−

    ∇=

    −=

    ∫∫

    ∫∫

    ∫∫

    BE

    SEB

    SES B

    SE E

    SBE

    0S

    SS

    SC

    S

  • Integral form of law: enclosed current is integral dS of current density j Apply Stokes’ theorem Integration surface is arbitrary Must be true point wise

    Differential form of Ampère’s Law

    ∫∫ ==S

    S j B d .d . oenclo µµ I

    S

    j

    B

    dℓ

    S j d .d =I( )

    ( ) 0.

    .

    =×∇

    =×∇=

    ∫∫∫

    S

    SS

    SjB

    S jSB B

    d -

    d .d d .

    o

    o

    µ

    µ

    jB oµ=×∇

  • Maxwell’s Equations in Vacuum Take curl of both sides of 3’ (3) ∇ x (∇ x E) = -∂ (∇ x B)/∂t = -∂ (µoεo∂E/∂t)/∂t = -µoεo∂2E/∂t2 ∇ x (∇ x E) = ∇(∇.E) - ∇2E vector identity - ∇2E = -µoεo∂2E/∂t2 (∇.E = 0) ∇2E -µoεo∂2E/∂t2 = 0 Vector wave equation

  • Maxwell’s Equations in Vacuum Plane wave solution to wave equation E(r, t) = Re {Eo ei(ωt - k.r)} Eo constant vector ∇2E =(∂2/∂x2 + ∂2/∂y2 + ∂2/∂z2)E = -k2E ∇.E = ∂Ex/∂x + ∂Ey/∂y + ∂Ez/∂z = -ik.E = -ik.Eo ei(ωt - k.r) If Eo || k then ∇.E ≠ 0 and ∇ x E = 0 If Eo ┴ k then ∇.E = 0 and ∇ x E ≠ 0 For light Eo ┴ k and E(r, t) is a transverse wave

  • λπ

    λπ

    2k i.e.

    r 2r kr k .

    0 . . .e wavesD-3e wavesD1- kz

    =

    =⇒=

    =

    +=+=→

    ||||

    ||||

    ||||

    r kr k)r(r kr k

    k.rii

    r r||

    r� k

    Consecutive wave fronts

    λ

    Plane waves travel parallel to wave vector k Plane waves have wavelength 2π /k

    Maxwell’s Equations in Vacuum

    Eo

  • Maxwell’s Equations in Vacuum Plane wave solution to wave equation E(r, t) = Eo ei(ωt - k.r) Eo constant vector µoεo∂2E/∂t2 = - µoεoω2E µoεoω2 = k2

    ω =±k/(µoεo)1/2 = ±ck ω/k = c = (µoεo)-1/2 phase velocity

    ω = ±ck Linear dispersion relationship

    ω(k)

    k

  • Maxwell’s Equations in Vacuum Magnetic component of the electromagnetic wave in vacuum From Faraday’s law ∇ x (∇ x B) = µoεo ∂(∇ x E)/∂t = µoεo ∂(-∂B/∂t)/∂t = -µoεo∂2B/∂t2 ∇ x (∇ x B) = ∇(∇.B) - ∇2B - ∇2B = -µoεo∂2B/∂t2 (∇.B = 0) ∇2B -µoεo∂2B/∂t2 = 0 Same vector wave equation as for E

  • Maxwell’s Equations in Vacuum If E(r, t) = Eo ex e

    i(ωt - k.r) and k || ez and E || ex (ex, ey, ez unit vectors) ∇ x E = -ik Eo ey e

    i(ωt - k.r) = -∂B/∂t From Faraday’s Law

    ∂B/∂t = ik Eo ey ei(ωt - k.r)

    B = (k/ω) Eo ey ei(ωt - k.r) = (1/c) Eo ey ei(ωt - k.r)

    For this wave E || ex, B || ey, k || ez, cBo = Eo

    More generally

    -∂B/∂t = -iω B = ∇ x E

    ∇ x E = -i k x E

    -iω B = -i k x E

    B = ek x E / c

  • Maxwell’s Equations in Matter Solution of Maxwell’s equations in matter for µ = 1, ρfree = 0, jfree = 0

    Maxwell’s equations become

    ∇ x E = -∂B/∂t

    ∇ x H = ∂D/∂t H = B /µo D = εoε E

    ∇ x B = µoεoε ∂E/∂t

    ∇ x ∂B/∂t = µoεoε ∂2E/∂t2

    ∇ x (-∇ x E) = ∇ x ∂B/∂t = µoεoε ∂2E/∂t2

    -∇(∇.E) + ∇2E = µoεoε ∂2E/∂t2 ∇. ε E = ε ∇. E = 0 since ρfree = 0

    ∇2E - µoεoε ∂2E/∂t2 = 0

  • Maxwell’s Equations in Matter ∇2E - µoεoε ∂2E/∂t2 = 0 E(r, t) = Eo ex Re{ei(ωt - k.r)}

    ∇2E = -k2E µoεoε ∂2E/∂t2 = - µoεoε ω2E

    (-k2 +µoεoε ω2)E = 0

    ω2 = k2/(µoεoε) µoεoε ω2 = k2 k = ± ω√(µoεoε) k = ± √ε ω/c Let ε = ε1 - iε2 be the real and imaginary parts of ε and ε = (n - iκ)2

    We need √ε = n - iκ

    ε = (n - iκ)2 = n2 - κ2 - i 2nκ ε1 = n2 - κ2 ε2 = 2nκ

    E(r, t) = Eo ex Re{ ei(ωt - k.r) } = Eo ex Re{ei(ωt - kz)} k || ez

    = Eo ex Re{e

    i(ωt - (n - iκ)ωz/c)} = Eo ex Re{ei(ωt - nωz/c)e- κωz/c)}

    Attenuated wave with phase velocity vp = c/n

  • Maxwell’s Equations in Matter Solution of Maxwell’s equations in matter for µ = 1, ρfree = 0, jfree = σ(ω)E

    Maxwell’s equations become

    ∇ x E = -∂B/∂t

    ∇ x H = jfree + ∂D/∂t H = B /µo D = εoε E

    ∇ x B = µo jfree + µoεoε ∂E/∂t

    ∇ x ∂B/∂t = µoσ ∂E/∂t + µoεoε ∂2E/∂t2

    ∇ x (-∇ x E) = ∇ x ∂B/∂t = µoσ ∂E/∂t + µoεoε ∂2E/∂t2

    -∇(∇.E) + ∇2E = µoσ ∂E/∂t + µoεoε ∂2E/∂t2 ∇. ε E = ε ∇. E = 0 since ρfree = 0

    ∇2E - µoσ ∂E/∂t - µoεoε ∂2E/∂t2 = 0

  • Maxwell’s Equations in Matter ∇2E - µoσ ∂E/∂t - µoεoε ∂2E/∂t2 = 0 E(r, t) = Eo ex Re{e

    i(ωt - k.r)} k || ez ∇2E = -k2E µoσ ∂E/∂t = µoσ iω E µoεoε ∂2E/∂t2 = - µoεoε ω2E (-k2 -µoσ ω +µoεoε ω2 )E = 0 σ >> εoε ω for a good conductor

    E(r, t) = Eo ex Re{ e

    i(ωt - √(ωσµo/2)z)e-√(ωσµo/2)z} NB wave travels in +z direction and is attenuated The skin depth δ = √(2/ωσµo) is the thickness over which incident radiation is attenuated. For example, Cu metal DC conductivity is 5.7 x 107 (Ωm)-1 At 50 Hz δ = 9 mm and at 10 kHz δ = 0.7 mm

    ωσµωσµωσµ )(121k k2 iio −=−=−=

  • Bound and Free Charges

    Bound charges All valence electrons in insulators (materials with a ‘band gap’) Bound valence electrons in metals or semiconductors (band gap absent/small ) Free charges Conduction electrons in metals or semiconductors

    Mion k melectron k Mion Si ion Bound electron pair

    Resonance frequency ωo ~ (k/M)1/2 or ~ (k/m)1/2 Ions: heavy, resonance in infra-red ~1013Hz Bound electrons: light, resonance in visible ~1015Hz Free electrons: no restoring force, no resonance

  • Bound and Free Charges

    Bound charges Resonance model for uncoupled electron pairs

    Mion k melectron k Mion

    tt

    t

    t

    t

    t

    e Emqe )A(

    mk

    e Emqx(t)

    mk

    hereafter) assumed (Re{} x(t)(t)x x(t)(t)xsolution trial }e )Re{A(x(t)

    }Re{e Emqx

    mkxx

    }Re{e qEkxxmxm

    o

    o

    o

    o

    ωω

    ω

    ω

    ω

    ω

    ωωω

    ωω

    ωω

    ω

    ii

    i

    i

    i

    i

    i

    i

    i

    ++

    +

    +

    +

    +

    =

    +Γ+−

    =

    +Γ+−

    −=+=

    =

    =+Γ+

    =+Γ+

    2

    2

    2

  • Bound and Free Charges

    Bound charges In and out of phase components of x(t) relative to Eo cos(ωt)

    Mion k melectron k Mion ( )

    ( )

    ( ) ( )( ) ( ) ( )( )

    ( )( ) ( )( ) ( ) ( )( )

    Γ+−

    Γ+

    Γ+−

    −=

    −==

    Γ+−

    Γ−=

    Γ+−

    −=

    Γ+−=

    =+Γ+−

    =

    +++

    22222222

    22

    22222222

    22

    22

    222

    ωωωωω

    ωωωωωω

    ωωω

    ωωωωω

    ωωωωωω

    ωωωω

    ωωωω

    ω

    ωωω

    oo

    o

    oo

    o

    o

    oo

    iii

    i

    i

    t)sin(t)cos(m

    qE

    })}Im{eIm{A(})}Re{eRe{A(})eRe{A( x(t)

    mqE )}Im{A(

    mqE )}Re{A(

    1m

    qE )A(

    mk1

    mqE )A(

    o

    oo

    o

    o

    ttt

    in phase out of phase

  • Bound and Free Charges Bound charges Connection to χ and ε

    ( )( ) ( )( ) ( ) ( )( )

    ( ) ( )( ) ( ) ( )( )

    ( ) ( )( )function dielectric model

    Vmq1)( 1 )(

    Vmq )}(Im{

    Vmq )}(Re{

    (t)eERemVq(t)

    qx(t)/V volume unit per moment dipole onPolarisati

    2

    22

    o

    2t

    2222

    22

    22222222

    22

    22222222

    22

    ωωωωωω

    εωχωε

    ωωωω

    εωχ

    ωωωωω

    εωχ

    χεωωω

    ωωωω

    ωω ω

    Γ+−

    Γ−−+=+=

    Γ+−

    Γ=

    Γ+−

    −=

    =

    Γ+−

    Γ−

    Γ+−

    −=

    =≡

    o

    o

    o

    ooo

    o

    o

    o

    oo

    o

    i

    i -i EP

    1 2 3 4

    4

    2

    2

    4

    6 ε(ω)

    ω/ωο

    ω = ωο

    Im{ε(ω)}

    Re{ε(ω)}

  • Bound and Free Charges Free charges Let ωο → 0 in χ and ε jpol = ∂P/∂t

    ( )

    ( )

    tyconductivi Drude

    1 V1N qe

    mNe

    mVq)(

    mVq

    mVq

    mVq)(

    LeteVm

    qt(t)(t)

    eVm

    qt(t)(t)

    e1Vm

    q(t)

    tyconductivi (t)(t)density Current

    2222

    free

    222

    free

    o

    2

    free

    o

    2

    pol

    o

    2

    t

    t

    t

    Γ≡≡≡≡

    Γ=

    Γ+Γ+−

    =Γ+Γ+−

    =Γ+−

    =

    →Γ+−

    +=

    ∂∂

    =

    Γ+−+

    =∂

    ∂=

    Γ+−=

    ≡=

    +

    +

    +

    ττσ

    ωω

    ωωωω

    ωωωωσ

    ωωω

    ωε

    ε

    ωωωω

    εε

    ωωωεε

    σσ

    ω

    ω

    ω

    0

    0

    2224

    23

    2

    2

    22

    22

    iii

    ii

    ii

    ii

    oo

    o

    ooo

    ooo

    i

    i

    i

    EPj

    EPj

    EP

    Ej

    1 2 3 4

    4

    2

    2

    4

    6

    ω

    ωο = 0

    Re{σ(ω)}

    σ(ω)

    Im{ε(ω)}

    Drude ‘tail’

  • Energy in Electromagnetic Waves Rate of doing work on a moving charge W = d/dt(F.dr) F = qE + qv x B = d/dt{(q E + q v x B) . dr} = d/dt(q E . dr) = qv . E = ∫ j . E d3r j(r) = q v δ(r - r’) W = ∫ j . E d3r fields do work on currents in integration volume Eliminate j using modified Ampère’s law ∇ x H = jfree + ∂D/∂t W = ∫ ∇ x H − ∂D/∂t . E d3r

  • Energy in Electromagnetic Waves Vector identity ∇.(A x B) = B . (∇ x A) - A . (∇ x B) W = ∫ ∇ x H − ∂D/∂t . E d3r becomes W = ∫ ∇.(H x E) + H . ∇ x E − E . ∂D/∂t d3r = ∫ ∇.(H x E) − H . ∂B/∂t − E . ∂D/∂t d3r ∇ x E = - ∂B/∂t

    W = ∫ j . E d3r = -∫∇.(E x H) d3r − ddt∫12 H . B + D . E d

    3r U = 12 H . B + D . E Local energy density

    N = E x H Poynting vector

    ∂U∂t

    + ∇. N = − j . E Energy conservation

  • Energy in Electromagnetic Waves Energy density in plane electromagnetic waves in vacuum

    ( )

    ( )

    HEN

    H HE E

    ekr keH r keE

    E D H B

    x Poynting c.f.flux energy mean 21

    c abcU

    densityenergy mean 21kz) - t(cosc ab

    21

    cU

    c kz) - t(cos c

    U

    cHcBE kz) - t(cos c

    E HcHE21U

    kz) - t(cos . .21U

    || . - t(e Re H . - t(e Re E

    . .21U

    2

    2-2

    2

    2

    zyoxo))

    ==

    ==

    ==

    ==

    +=

    +=

    =

    =

    +=

    H E

    H E

    H E

    τ

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    Maxwell’s Equations in VacuumConstitutive RelationsElectric PolarisationElectric PolarisationElectric PolarisationGauss’ Law in MatterGauss’ Law in MatterGauss’ Law in MatterGauss’ Law in MatterAmpère’s Law in MatterAmpère’s Law in MatterAmpère’s Law in MatterAmpère’s Law in MatterMaxwell’s EquationsDivergence Theorem 2-D 3-DDifferential form of Gauss’ LawStokes’ Theorem 3-DFaraday’s LawDifferential form of Ampère’s LawMaxwell’s Equations in VacuumMaxwell’s Equations in VacuumMaxwell’s Equations in VacuumMaxwell’s Equations in VacuumMaxwell’s Equations in VacuumMaxwell’s Equations in VacuumMaxwell’s Equations in MatterMaxwell’s Equations in MatterMaxwell’s Equations in MatterMaxwell’s Equations in MatterBound and Free ChargesBound and Free ChargesBound and Free ChargesBound and Free ChargesBound and Free ChargesEnergy in Electromagnetic WavesEnergy in Electromagnetic WavesEnergy in Electromagnetic Waves