Electromagnetism - Universität Luxemburg · Grant, Phillips “Electromagnetism”, Wiley...

1 Electromagnetism WS 07/[email protected] I-1

Physique expérimentale 1b

Electromagnetism

WS 07/08

Susanne Siebentritt

BS 1.13 phone: 466644-6304

[email protected]

travaux dirigées by Katja Hönes

(together with Physique expérimentale 1a)

http://www.uni.lu/recherche/fstc/research_unit_physics/

photovoltaics_lpv/teaching


Outline

Part I The static fields

I.1 Introduction to the electrostatic field

I.2 The electrostatic field in isolators

I.3 Introduction to the (static) magnetic field

I.4 The magnetic field in matter

Part II Electrodynamics

II.1 Electromagnetic induction

II.2 Maxwell’s equations

II.3 The basics of electromagnetic waves

II.4 Alternating currents

II.5 Network analysis


Requirements

• Curiosity: lightning, transformer, field concept

• Math: basic calculus, vectors

∫ xdy

x

Ex

∂

∂ ϑcosab=⋅ba


Literature

Grant, Phillips “Electromagnetism”, Wiley

Demtröder “Experimentalphysik 2: Elektizität und Optik”, Springer

Feynman, Leighton, Sands “Lectures on Physics II”, Adison-Wesley

Raith “Bergmann-Schäfer 2: Elektromagnetismus”, de Gruyter

Dransfeld, Kienle “Physik II: Elektrodynamik und spez. Relativitätstheorie”, Oldenbourg

Benson “Physique 2: Electricité et Magnétisme”, De Boeck Université

Tipler, Mosca “Physik”, Elsevier

Haliday, Resnick, Walker “Fundamentals of Physics”, Wiley



I.1 Introduction to the electrostatic field


The electric charge and Coulomb’s law

Experiment →→→→• two different sorts of charges

• like charges repel, unlike charges attract

• the force acts along the line between the two charges

• the force is proportional to the amount of each charge

• the force is inversely proportional to square of the distance

21221

2121 rF

r

qq∝→→→→

constant of proportionality depends on system of units!

SI:

21221

21

021 ˆ

4

1rF

r

qq

πε= Coulomb’s law [ ]

3

42

2

2

0 mkg

sA

Vm

C

Nm

C===ε

where 21

21

21

2121ˆ

r

rrr ==

r

(in cgs: → defines: ) 21221

2121 rF

r

qq= [ ] dyncmq =

Vm

C120 108542.8 −⋅=ε


Many charges

31231

31

0212

21

21

031211 ˆ

4

1ˆ

4

1rrFFF

r

qq

r

qq

πεπε+=+=

in general: ( )∑∑∑≠

∧

≠≠

−−

===ji

ij

ij

ij

jiij

ij

ij

jiijj

qq

r

qqrr

rr

rFF2

02

0 4

1ˆ

4

1

πεπε

Rutherford experiment: scattering of He nuclei at a Au foil

qHe = +2e, qAu = +79e e = 1.602.10-19 C

r = 2.10-14 m

→→→→ Nm

F 90)102(

e 792

4

1214

2

0

≈⋅

⋅=

−πε

for any charge q @ r: ( )∑∧

−−

=i

i

i

iqqrr

rrF

204

1

πε( )∑

∧

−−

=→i

i

i

iq

qrr

rr

F2

04

1

πε

This is the electric field E of the charge distribution:

( ) ( )∑∑ −−

=−−

=∧

ii

i

i

ii

i

i qqrr

rrrr

rrrE

30

20 4

1

4

1)(

πεπε[ ]

m

V

mVm

C

CE ==

2


The electric field

( )030

0

04

1)( rr

rrrE −

−=

q

πεThe electric field of a single charge at r0:

The superposition principle: ∑=i

i )()( rErE

The direction of the electric field:

field lines start at positive charges and end at negative charges

The electric field is a vector field.

The electric field is an intensive quantity!


Field line pictures


The atomic electric field

The charge of the electron in a hydrogen atom: e)()( −==∫ ∫ ∫ ∫∞

∞−

∞

∞−

∞

∞−atomV elel dxdydzdV rr ρρ

The atomic (detailed) electric field in matter:

nuclelatomic ρρρ +=

( )( )

∫∑−

−=−

−=

Vatomic

ii

i

ilocal dV

q'

'

')'(

4

1

4

1)(

30

30 rr

rrrrr

rrrE

ρ

πεπε

“outside” a neutral atom: E ≈ 0

BUT: consider e.g. an HCl molecule: H+Cl-


The macroscopic electric field

in “large” systems: ρatomic too complicated!

average charge density: ')'(1

)( dVV V atomic∫∆∆

= rr ρρ

consider conductor in external electric field:

→ macroscopic electric field:( )

∫−

−=

VdV '

'

')'(

4

1)(

30 rr

rrrrE

ρ

πε

surface charge density:

')'(1

)( dVS V atomic∫∆∆

= rr ρσ

'')'(')'(2

1

2

1

dxdydSqx

x

y

ySS ∫ ∫∫ == rr σσ

superposition principle → macroscopic electric field:

( ) ( )∫∫

−

−+

−

−=

surfaces all 30

space all 30 '

')'(

4

1

'

')'(

4

1)( dSdV

rr

rrr

rr

rrrrE

σ

πε

ρ

πε

limited: E < 108 V/m, in air: < 2…3.106 V/m

electrostatic induction


The flux of the electric field

Flux in hydrodynamics:

can be defined for

any vector field v

“rate of flow”: Sv∆=ϕ

The flux of the electric field through a surface: ∫∑ →∆=S

Sel dSESEϕ

point charge q at origin: rrE ˆ4

1)(

20 r

q

πε=

flux through sphere with radius R

EdSd =→ SE

43421Ω

==

d

ddrdrrddS ϕθθϕθθ sin sin 2

0

4

2

0 00

2

0 0 0

sin4

sin4 ε

ϕθθπε

ϕθθπε

ϕ

π

π ππ π qdd

qdd

qel ===→ ∫ ∫∫ ∫

44 344 21

∆S

∆S

Sv ∆⋅=∆= )cos(ψϕ Sv


Flux through arbitrary surfaces

more complicated surfaces:

• what if dS not r ?^

Ω=

===

dEr

SEdd

2

cosαϕ SE

as before!

• what if r cuts several dS?

Ω±== dq

dd04πε

ϕ SEfor each:

out

in

0

2

0 0 0

sin4 ε

ϕθθπε

ϕπ π q

ddq

dS

el === ∫ ∫∫ SE

correct for any surface inclosing q

0== ∫S

el dSEϕand

→→→→

for q outside S

due to !21

rE ∝


Gauss’ law

superposition principle: QdVqdSV

ii

Sel =→== ∫∑∫ )(

11

00

rSE ρεε

ϕ

∫∫ =SV

S

dVd )(1

0

rSE ρε

→→→→ Gauss’ law

flux of E out of S ∝ total charge within S

independent of the charge distribution or shape of the surface !

consequence of 1/r2 behaviour of E →→→→ Coulomb’s law ↔ Gauss’ law

adapt S to the symmetry of the problem:

alkali ion, e.g. Na+ - noble gas configuration

→ spherical symmetry (@ r > large enough)

∫ ∫ ∫∫ ∫ =π ππ π

ϕθϕθρε

ϕθϕθ2

0 0 0

2

0

2

0 0

2 ' sin'),,'(1

sin),,(R

r ddθdrrrddθRRE

spherical symmetry → )(),,( REREr =ϕθ

0

2 e4)(

επ =⋅ RRE

204

e)(

RRE

πε=→

as if all the charge were at the origin


Macroscopic fields and Gauss’ law

remember:

→ Gauss’ law applies equally

( ) ( )∫∫

−

−+

−

−=

surfaces all 30

space all 30 '

')'(

4

1

'

')'(

4

1)( dSdV

rr

rrr

rr

rrrrE

σ

πε

ρ

πε

the field of a surface charge:

- the field is always dS

- the field is 0 inside

S’∫∫ =

SVS

dVd ')(1

'0

rSE ρε

'1

'''0

0'0

Sddd

sideouterinner S

SE

SS

∆⋅=++ ∫∫∫

∆⋅

σε

3214342143421

SESESE E0εσ =→

max. macroscopic fields: 108 V/m → σmax ≈ 10-3 C/m2 ≈ 1016 e/m2

vs. 1019 atoms/m2


The divergence of a vector field

Definition:V

dVS

V ∆=∇=

∫∆

→∆

SFFF

0lim:div

divF > 0: source

divF < 0: drain

Cartesian coordinates:

∂

∂

∂

∂

∂

∂=∇

zyx:F

FF ∇=

∂

∂+

∂

∂+

∂

∂==

∫→ z

F

y

F

x

F

zyx

dSzyxS

zyx δδδδδδ 0limdiv

zyxx

FzyFzyx

x

FF x

xx

xxF δδδδ

δδδδδδ

δ

δϕ =−+= )(,

zyxz

F

y

F

x

Fd zyx

Sδδδ

δ

δ

δ

δ

δ

δ

++=→ ∫ SF

→ Gauss’ theorem, divergence theorem, Gauss’ law of vector fields:

∫∫ =VSV

ddV SFFdiv


Differential form of Gauss’ law

0

00 )()(

1

lim

)(1

limlim)(div000 ε

ρρ

ερ

ε rrr

SErE =

∆

∆

=∆

=∆

=→∆

∆

→∆→∆

∫∫

∆

V

V

V

dV

V

d

V

V

V

S

V

V

Gauss’ law

0

)()()(div

ε

ρ rrErE =∇= differential form of Gauss’ law

The charges are the sources (and drains) of the electric field

consider a point charge @ origin 0 ,0)(div ≠= rrE

( ) 2/32220

30 44 zyx

qx

r

xqEx

++==

πεπε

( ) ( )( ) 5

22

03222

2/122222/3222

0

3

4

3

4 r

xrq

zyx

zyxxzyxq

x

Ex −=

++

++−++=

∂

∂

πεπε

03333

4div

5

2222

0

=

−−−=

∂

∂+

∂

∂+

∂

∂=

r

zyxrq

z

E

y

E

x

E zyx

πεE


A conservative field

consider the energy of a point charge in an electric field

The electrostatic field is conservative

check path dependence for the field of a single charge:

rr

qqrEqlEqqW t

ttt δπε

δθδδδ2

04cos ==== lE

−===∆ ∫∫

BA

tr

r

tB

At

rr

qqdr

r

qqdqW

B

A

11

44 02

0πεπε

lE

independent of the path!

superposition principle →

∫ = 0 lEd → There are no closed electric field lines. Electric field lines start and end at the charges

consider conductor with cavity (without charges):

within conductor E = 0 → Gauss:net charge in cavity = 0

for any path l → E = 0 within cavity

→ Faraday’s cage

∫ = 0 lEd


The rotation of a vector field

Definition: S

dsl

SS ∆

=∫

∆

→∆∆⊥

lFF

0limrot (rot F = curl F)

Cartesian coordinates:

yzyxxFxzyyxF

yzyxxFxzyyxFd

yx

yxl S

δδδδ

δδδδ

),,(),,(

),,(),,(

21

21

21

21

−−+−

++−≈∫∆

lF

( )y

F

x

F

yx

dxyl

Sz

s

∂

∂−

∂

∂==→

∫∆

→∆ δδ

lFF

0limrot

and ( ) ( )x

F

z

F

z

F

y

F zxy

yzx ∂

∂−

∂

∂=

∂

∂−

∂

∂= FF rot ,rot FF ×∇=→ rot

→ Stoke’s theorem: ∫∫ =Sl

ddS

SFlF rot

electric field:

∫∫ ==S

dd SElE rot0 for any surface S: 0rot =E


The electrostatic potential

work performed → difference in potential energy

∫−=−=∆B

AtAPBPP dqWWW lE,,

electrostatic potential is a function of position only t

P

q

W=Φ )(r

∫ ⋅−=Φ−Φ=B

AAB dU lErr )()(

consider a point charge @ origin

bring qt from ∞

r

qdr

r

qr

02

044

)()(πεπε

∫∞

=−=∞Φ−Φ r

usually: Φ(∞) = 0 →→→→r

q

04)(

πε=Φ r

voltage = potential difference: [ ] [ ] VU =Φ=

qUdqWB

A

=−=→ ∫ lE


The gradient of the potential

rElErrrrr

r

δδδδ

−≈⋅−=+Φ−Φ=Φ ∫+

d)()(

Cartesian coordinates , ,z

Ey

Ex

E zyx∂

Φ∂−=

∂

Φ∂−=

∂

Φ∂−=→

Φ−∇=Φ−= gradE

(For any field with rot F = 0: F = grad ΦF )

superposition principle: ( )212121 gradgradgrad Φ+Φ−=Φ−Φ−=+ EE

''

)'(

4

1)(

0

dVatomicatomic ∫

−=Φ

rr

rr

ρ

πε→→→→

and ''

)'(

4

1'

'

)'(

4

1)(

00

dSdV ∫∫−

+−

=Φrr

r

rr

rr

σ

πε

ρ

πε


Equipotential surfaces

the gradient is always equipotential surfaces


A charged rod

consider an evenly charged rod

with radius R and ∞ long

charge per unit length: πρλ 2R=

cylinder symmetry → E radial outward and coordinates: r,z,ϕ→ E depends only on r

flow through cylinder with radius r and length L:

r > R: Q = λL

r < R: Q = ρπr2L

boundary condition: Φ(R) = 0

rrE ˆ2

)(0επ

λ

r=→

rrrE

02

0 2ˆ

2)(

πε

λ

ε

ρ

R

r==→

∫ ⋅−=Φr

R

drEr)(

R

rrRr ln

2)(:

0πε

λ−=Φ>

−=Φ<

2

2

0

14

)(:R

rrRr

πε

λ

0

2)(ε

πϕQ

rLrEdcylinderel === ∫ SE

Gauss

application: coaxial cable


The dipole potential

−=Φ

−+ rr

q 11

4)(

0πεr

θcos2

4122

ararr m+=±

θcos4

1

111

2

2

r

a

r

arrm+

=±

→ Taylor expansion for r » a

remember: ...)(2

)(''f))(('f)f()f(

20

0000

+−+−+= xx

xxxxxx

...cos2

112

+±=→±

θr

a

rrθ

πεcos

4)(

20 r

aq≈Φ→ r

Dipole potential

p

With the dipole moment p: from –q to +q and ap q=

204

ˆ)(

rdipole

πε

rpr

⋅=Φ

(exact for a → 0)

[ ] AsmCmp ==


The electric dipole

The electric dipole field:

⋅−∇=Φ−∇=

304 r

q

πε

raE

( )prararE −=

∇+∇

−= ˆcos3

4

1)(

11

4

30

330

θπεπε

prrr

q

aararr

=∇=−

=∇ )( cos ,3153

andθarrr

dipole in a homogeneous field:

EFEF qq =−= 21 and → no net force on C

but there is a torque: TEpEa

Ea

T =×=

×+

×

−−=

22qq

T = 0 if p E

potential energy:

→ minimum in parallel configuration (not antiparallel)

WdqqqW =−=−=Φ+Φ−= ∫ pElE2

121

(in an inhomogeneous field the dipole is pulled towards increasing field)


Multipole expansion

' '

)'(

4

1)(

0

dV∫−

=Φrr

rr

ρ

πεfor any distribution of charges:

for r » r’ expansion of : '

1

rr −

...'

8

3''

2

3)'(

2

3

2

''1

1

11

)'(

1

'

15

4

5

2

5

2

3

2

3''22

2

2

2

++−+−+=+−

=−

=− rrrrrrr

r

r

r

rrrrrrrrr

rrrr rr

[] )...' )''3''3''3(2

)''3()''3()''3(2

)'(1

'')'(1

')'(1

4

1)(

222222222

5

30

++++

+−+−+−+

++=Φ→

∫

∫∫

dVyzzyxzzxxyyx

zrzyryxrxr

dVr

dVr

r

rrrrr

ρ

ρρπε

monopoleCoulomb’s law dipole potential

quadrupole potential

2

2

2''2

1

1 ,)(r

r

rxxxf +==

−rr


The quadrupole

introducing abbreviations:

''')'(3

''')'(3

''')'(3

')''3()'(

')''3()'(

')''3()'(

22

22

22

dVzyQMQM

dVzxQMQM

dVyxQMQM

dVrzQM

dVryQM

dVrxQM

zyyz

zxxz

yxxy

zz

yy

xx

∫

∫

∫

∫

∫

∫

==

==

==

−=

−=

−=

r

r

r

r

r

r

ρ

ρ

ρ

ρ

ρ

ρ

and define quadrupole tensor:

=

zzzyzx

yzyyyx

xzxyxx

QMQMQM

QMQMQM

QMQMQM

QM

→ quadrupole potential (

( ))

rQMrrQMr ˆˆ 8

1

8

1

2

8

1

30

50

222

50

⋅⋅=⋅⋅=

=+++

+++=Φ

rr

yzQMxzQMxyQM

zQMyQMxQMr

yzxzxy

zzyyxxq

πεπε

πε


The energy within the atomic field

consider the fission of a U nucleus: n3KrBaU 92141

..

236 ++→ge

eVJm

As

rr

qqqqW

VmAs

811

1512

219

BaKr0

KrBaKrBaBaKr

105.2100.410)2.64.5(1085.84

)106.1(3656

)(4

⋅=⋅=⋅+⋅⋅⋅

⋅⋅⋅≈

=+

⋅=Φ=Φ=

−−−

−

π

πε

typical energies in chemical reactions: a few eV

consider three charges:

++=

23

2

13

1

0

3

12

1

0

2

44 r

q

r

qq

r

qqW

πεπε

and many charges:

∑∑ ∑

∑ ∑

Φ==

==

≠

<

iii

i ij ij

ji

i ij ij

ji

qr

qq

r

qqW

2

1

4

1

2

1

4

1

0

0

πε

πε

potential energy for a system of charges: ∑ Φ=i

iiqW2

1


Capacitors

macroscopic charge distributions - caution!

∑∑∑∑ Φ+Φ+Φ+Φ=≠+2

2121

1212

221

11212

1

2

1

2

1

2

1

iii

iii

iii

iii qqqqWWW

simplest case: conductor in vacuum

a pair of conductors: capacitor

E ≈ 0 far away → q1 = - q2

∫ Φ= dSW )()(2

1rrσ

Gauss

QUQ

dSdSW

QQ

2

1)(

2

1

)(2

1)(

2

1

=Φ−Φ=

=Φ+Φ=

−+

−

−−++ ∫∫ 4342143421rr σσ

field in the capacitor: E ∝ Q

QUdU ∝→⋅−= ∫ 2

1

lE → capacitance of a capacitor C: Q = CU

C

QCUW

22

2

1

2

1 ==→

[ ] FV

CC :==

usually µF, pF

Parallel plate capacitor:ignore the edges:

from Gauss:

dUE =

E0εσ =

with σ = Q/Ad

AC

d

UE

A

Q 000

εεε =→==→

A = 1 cm2, d = 1mm

→ C = 0.88 .10-12F


The energy of the electric field

for a parallel plate capacitor:2

02

0222

2

1

2

1

2

1

2

1VEAdEdCECUW εε ====

for any field:

∫=→⋅∆=∆ dVEWEVW2

0212

02

1εε

→ the energy density of the electric field:

202

1 Ew ε=

31 Electromagnetism WS 07/[email protected] II-31


I.2 The electrostatic field in isolators


Polarisationconductor in field → electromagnetic induction

E = 0 within conductor + surface charges

isolator in field → polarisationmodified field + displaced charges

→ “dielectrics”

polarisation of an atom in an electric field

consider the centres of massof the nucleus and electron cloud

→ dipole p = Zea

EPpP || ,N=

→ no net charge!

→ a net polarisation per unit volume[ ]

22m

As

m

CP ==

(in most materials)

small displacements → linear: Eαp 0ε=

αααα (atomic polarisability) can be a tensor

insulator in homogeneous field(with N atoms per unit volume)


Polarisation charges

The situation is quite different at the surfaces:

PNpS

SaNZe

S

qp ==

∆

∆⋅=

∆=σ Pp −=σPolarisation charge σp

SP∆=∆Spσ

SP∆=∆⋅= SaNZeq cosθ

→ the two charges on both sideshave always the same magnitude

polarisation charges are real, but immobile and they disappear for E = 0

polarisation charges are opposed to free charges


The dielectric constant

the dielectric reduces the voltage, i.e. the field in a capacitor

by a factor εr

consider a capacitor completely filled with dielectric:

QUCUCQ DDDD === ε/0d

ACC r

rD0

0 εε

ε ==→

air 1.00058water 81glass ~4(SrBi)TiO2 ~1000

→ the capacitance is increased by εr

εr: dielectric constant, relative permittivity

← the polarisation charges cause a “depolarisation field”


The electric susceptibility

Gauss:

without dielectric: σ0 = ε0 E0

with dielectric:

with σD = ε0 ED

PEpD −=+= 000 εσσσ

000

εε

ε

PE

PEE −=−=→ DrD EP 0)1( εε −=→ r

define electric susceptibility χE := εr – 1 → EP 0εχE=

generally: response of material = susceptibility . intensive quantity

S

Caution!

with P = Np and p = αε0 Elocal “→” χE = Nα – for gases only!

in crystals and liquids: Elocal = Emacroscopic + ECoulomb(neighbours) + Edipol(neighbours)


The Clausius-Mosotti formula

consider a non-polar liquiddetermine Elocal for a molecule in a “continuous medium”

gedanken experiment: remove the molecule → cavity

polarisation charge:

EdSdddS EEp cos)( 00 θεχεχθσ −=== SESP

at the centre:2

2

20 4

coscos

4

)()(

R

EdS

R

dSE Eppol

zπ

θχθ

πε

θσθ =−=

3sincos

2

sin4

cos

4

cos

32

0

2

22

0 02

2

2

2

Ed

E

ddRR

EdS

R

EEE

EE

ES

Epolz

pol

χθθθ

χ

ϕθθπ

θχ

π

θχ

π

π π

==

===

∫

∫ ∫∫

44 344 21

EEEE )1(31

Epol

local χ+=+≈→

)1( )1(31

31

EEE NN χαχχα +≈→+≈→ EP

3/1 α

αχ

N

NE

−≈→ Clausius-Mosotti formula

(α from χE,gas or solutions)


Polarisation charges in the volume

if E ≠ const → ρp ≠ 0

( ) zyxx

PzyxxPxxPqqq x

xxoutinx δδδδδδδ∂

∂−≈+−+−=−= )()(""

PP −∇=−=∂

∂−

∂

∂−

∂

∂−=→ div

z

P

y

P

x

P zyxpρ

total polarisation charge:

0=∇−=+= ∫∫∫∫ VSV pS pp dVddVdSQ PSPρσ

Gauss’ theorem

total charge density: pf ρρρ +=

000 ε

ρ

ε

ρρ

ε

ρ ffp +∇−=

+==∇

PEGauss’ law: )(

:

00 43421D

PEPE

=

+∇==∇+∇→ ερε f

fρ=∇D→ with )()()( 0 rPrErD += ε electric displacement

[ ]23

m

As

m

CmD ==


The electric displacement

is an “artificial” vector field

DEEEPED ==+=+= 0000 εεεχεε rE from fρ=∇D ∫∫ =→SV fS

dVd ρSD

consider a coaxial cable:

rr

arE

r

arDarrDd

rr

fffS

00 2)( )( 2)(2

επε

λ

εε

σσσππ ==→=→==∫ SD

compare with charged rod: 0 2

)(επ

λ

rrE =

homogeneous dielectric: ε0 → εrε0 or0εε

ρ

r

f=∇E

E is reduced by factor εr

potential: )/ln(22 00

abdrr

dUr

b

a r

b

a επε

λ

επε

λ∫∫ =−=−= lE

capacitanceper length )/ln(

2 0

abUC rεπελ

==

typical: 75 pF/m

inner conductor

insulatorouter conductor

outer sheath


Poisson’s equation and Laplace’s equation

general case:0

)()(

ε

ρ rrE =∇ and Φ−∇=E

0

2 )()(

ε

ρ rrE −=∆Φ−=Φ−∇=∇→

Laplacian operator2

2

2

2

2

2

zyx ∂

∂+

∂

∂+

∂

∂=∆

0

2 )(

ε

ρ r−=∆Φ=Φ∇ Poisson’s equation

if there is no charge: 02 =∆Φ=Φ∇ Laplace’s equation

→ calculate potential and field for any given charge distribution(at least in principle, or numerically)

still another way to express Gauss’ or Coulomb’s law

with homogeneous dielectric: →=∇ )(

)(0εε

ρ

r

f rrE

0

2 )(

εε

ρ

r

f r−=∆Φ=Φ∇

pf ρρρ +=

usually experimental boundary conditions: Φ not ρ


Once more: the coaxial cable

there are no free charges: 0=∆Φ

symmetry: cylinder coordinates: r, z, ϕ

2

2

2

2

2

2

zyx ∂

∂+

∂

∂+

∂

∂=∆

with ϕcosrx = and ϕsinry =

ϕϕ sincosyxr

y

yr

x

xr ∂

∂+

∂

∂=

∂

∂

∂

∂+

∂

∂

∂

∂=

∂

∂

we need

consider: etc.

2

2

2

2

2

11

zrrr

rr ∂

∂+

∂

∂+

∂

∂

∂

∂=∆→

ϕ

no dependence on ϕ and z: 0)(1

)( =Φ

∂

∂

∂

∂=∆Φ r

rr

rrr

BrArAr

r +=Φ→∫=∂

Φ∂→ )ln()( with Φ(a) = U and Φ(b) = 0

)/ln()()/ln(

)/ln()(

abr

U

rrEbr

ab

Ur =

∂

Φ∂−=→

−=Φ→ as before


Inhomogeneous dielectric

shape of E changes when ε not homogeneous!

Drvacp

Dvac EES

SESE εε

σ=→=+−

0

000Gauss:

Drp EP 0)1( εεσ −−=−=

but ρf = 0 → Dvac = DD

potential: vacr

Dvac Eb

babEEbaU

+−=+−=

ε)(


Boundary conditions

consider any boundary between two dielectrics

line integral:

→=+−=∫ 0||22||11 EdlEdld lEl

||2||1 EE =S

Gauss for D:

→=−= ⊥⊥∫ 02211 dSDdSDdS

SD ⊥⊥ = 21 DD

but E is discontinuous

but D is discontinuous

Snell’s law: E continuous:

D continuous:

βα sinsin 21 EE =

βεαε coscos 2211 EE =

βεαε cotcot 21 =→

with →= 0ED εεr2

||2

1

||1

εε

DD= and ⊥⊥ = 2211 EE εε


Electric field energy in a dielectricremember: for a parallel plate capacitor DEVEVdE

d

ACUW r

r

2

1

2

1

2

1

2

1 20

2202 ==== εεεε

for any field: ∫= DEdVWD 21 DEwD 2

1 =→

consider insulated capacitor → Q = const

without dielectric:

with dielectric: E → E0/εr and D → D0 :

the missing energy → kinetic energy of the dielectric

→ the dielectric is sucked into the capacitor

VEDW 0021

0 =

rr

WV

EDDEVWW

εε00

021

21 ===→

consider capacitor connected to voltage supply → U = const

with dielectric: E → E0 and D → εrD0 :

energy supplied by power supply Q → εrQ0 (on each plate!): Wps= ∆Q U

energy needed for field:

the rest → kinetic energy of the dielectric

→ the dielectric is sucked into the capacitor

0WW rε→

QUW ∆=21

the force on a dielectric hanging into a capacitor:

)2/()1()1(

22

11 20

02

2dbU

d

zb

z

UCU

zz

WF r

r εεεε

−=∆−

∆=∆

∆=

∆

∆=∆

→ determine εr


The field equations of the electrostatic field

0=×∇ E The electrostatic field has no rotation

fρ=∇DThe charges are the sources of the electric field

0/ερ=∇E

EPED 00 εεε r=+= The dielectric displacement

Electromagnetism - Universität Luxemburg · Grant, Phillips “Electromagnetism”, Wiley...

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Transcript of Electromagnetism - Universität Luxemburg · Grant, Phillips “Electromagnetism”, Wiley...