Mitschrift Physik III

8/7/2019 Mitschrift Physik III

http://slidepdf.com/reader/full/mitschrift-physik-iii 1/90

Physik III

Vorlesungs-Skript

Prof. Dr. Simon Lilly

Mitschrift

Marc Maetz

HS 2010





Contents

Contents 10.1 General Information . . . . . . . . . . . . . . . . . . . . . . . . 3

0.1.1 Testat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30.1.2 Exam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30.1.3 books . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30.1.4 webpage . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1 Electrostatics: Electric charge 5

1.1 empirical fact: charges exist . . . . . . . . . . . . . . . . . . . . 51.2 Coulomb’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3 Potential energy of a system of charges . . . . . . . . . . . . . 61.4 Concept of electric field . . . . . . . . . . . . . . . . . . . . . . 61.5 Charge distributions . . . . . . . . . . . . . . . . . . . . . . . . 71.6 Concept of Flux and Gauss’ Law . . . . . . . . . . . . . . . . . 71.7 Applications of Gauss’ Law . . . . . . . . . . . . . . . . . . . . 81.8 Energy associated with E-field . . . . . . . . . . . . . . . . . . 91.9 Concept of electrical Potential . . . . . . . . . . . . . . . . . . . 101.10 Experiment: charging two spheres equally . . . . . . . . . . . 101.11 Other useful operators . . . . . . . . . . . . . . . . . . . . . . . 11

1.12 summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2 Conductors 13

2.1 Conductors + insulators . . . . . . . . . . . . . . . . . . . . . . 132.2 Conditions for conductor . . . . . . . . . . . . . . . . . . . . . 142.3 The general electrostatic problem . . . . . . . . . . . . . . . . . 142.4 Uniqueness Theorem . . . . . . . . . . . . . . . . . . . . . . . . 152.5 Some interesting cases . . . . . . . . . . . . . . . . . . . . . . . 15

2.5.1 Enclosed cavity . . . . . . . . . . . . . . . . . . . . . . . 152.5.2 Faraday cage . . . . . . . . . . . . . . . . . . . . . . . . 16

1



Contents

2.6 Some tricks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.6.1 Mirror charges . . . . . . . . . . . . . . . . . . . . . . . 172.7 Capacitance and capacitors . . . . . . . . . . . . . . . . . . . . 172.8 Energy stored in capacitor . . . . . . . . . . . . . . . . . . . . . 18

2.8.1 Parallel plate capacitor . . . . . . . . . . . . . . . . . . . 192.9 General System of conductors . . . . . . . . . . . . . . . . . . . 19

3 Elecric Currents 21

3.1 Electric current . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.2 Charge conservation . . . . . . . . . . . . . . . . . . . . . . . . 223.3 Ohm’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.4 Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3.4.1 Kirchoff’s Rules . . . . . . . . . . . . . . . . . . . . . . . 233.5 Energy dissipation in a resistor . . . . . . . . . . . . . . . . . . 243.6 Sources of energy . . . . . . . . . . . . . . . . . . . . . . . . . . 253.7 Circuits with capacitors . . . . . . . . . . . . . . . . . . . . . . 26

4 Fields of moving charges 28

4.0 Magnetic Fields or Forces . . . . . . . . . . . . . . . . . . . . . 284.1 Reminder about special Relativity . . . . . . . . . . . . . . . . 294.2 Invariance of charge . . . . . . . . . . . . . . . . . . . . . . . . 294.3 Transformation of E-fields . . . . . . . . . . . . . . . . . . . . . 304.4 Field of accelerated charge . . . . . . . . . . . . . . . . . . . . . 31

4.5 Force on moving charge . . . . . . . . . . . . . . . . . . . . . . 334.6 Interaction of moving charge with other moning charges . . . 33

5 Magnetic Fields 35

5.1 Simplest case . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355.1.1 Force between two ∞ long wires . . . . . . . . . . . . . 36

5.2 Properties of magnetic fields . . . . . . . . . . . . . . . . . . . 365.2.1 Uniqueness theorem . . . . . . . . . . . . . . . . . . . . 37

5.3 Vector Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . 375.4 Fields of coils and solenoids . . . . . . . . . . . . . . . . . . . . 38

5.4.1 Biot-Savart law . . . . . . . . . . . . . . . . . . . . . . . 39

5.5 summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405.6 Change in B-field across a current sheet . . . . . . . . . . . . . 405.7 How do E and B transform . . . . . . . . . . . . . . . . . . . . 425.8 Hall Effect (1879) . . . . . . . . . . . . . . . . . . . . . . . . . . 44

6 Magnetic Induction 45

6.1 Magnetic Induction . . . . . . . . . . . . . . . . . . . . . . . . . 456.1.1 Wire in circuit . . . . . . . . . . . . . . . . . . . . . . . . 47

6.2 Faraday’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . 486.3 Lenz’ Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

2



Contents

6.4 Mutual inductance . . . . . . . . . . . . . . . . . . . . . . . . . 50

6.5 Self-inductance . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

7 Alternating current circuits 53

7.1 The RLC circuit as damped oscillator . . . . . . . . . . . . . . 537.2 Circuits driven by alternating voltage . . . . . . . . . . . . . . 54

7.2.1 Solenoid pushing . . . . . . . . . . . . . . . . . . . . . . 55

7.2.2 Some circuits . . . . . . . . . . . . . . . . . . . . . . . . 567.3 Power Consumption . . . . . . . . . . . . . . . . . . . . . . . . 60

8 Maxwell’s Equations 61

8.1 Wave equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

8.2 Superposition of two opposite directions . . . . . . . . . . . . 628.3 Standing wave . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

8.4 Energy Transport of E and M waves . . . . . . . . . . . . . . . 638.5 Lorentz transformation of waves . . . . . . . . . . . . . . . . . 638.6 summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

9 Dielectric materials 65

9.1 Introdiuction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 659.2 Electric dipoles . . . . . . . . . . . . . . . . . . . . . . . . . . . 669.3 Atomic and molecular dipoles . . . . . . . . . . . . . . . . . . 67

9.3.1 Permanent dipoles . . . . . . . . . . . . . . . . . . . . . 67

9.3.2 Induced dipoles . . . . . . . . . . . . . . . . . . . . . . . 679.4 Electric fields from polarized matter . . . . . . . . . . . . . . . 68

9.4.1 Gauss’ Law in medium and vector field D . . . . . . . 709.5 Currents in dielectrics and Maxwell’s equations . . . . . . . . 71

9.6 Eloctromagnetic waves in dielectric . . . . . . . . . . . . . . . 719.7 Example: Electric field around dielectric sphere . . . . . . . . 72

10 Magnetic phenomena in matter 76

10.1 Phenomenology . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

10.2 Magnetic dipoles . . . . . . . . . . . . . . . . . . . . . . . . . . 7610.3 Force on m in external field . . . . . . . . . . . . . . . . . . . . 77

10.4 Current in loop in atom . . . . . . . . . . . . . . . . . . . . . . 7810.5 Electron spin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7910.6 Magnetic fields of magnetized matter . . . . . . . . . . . . . . 79

10.7 Maxwell’s equations . . . . . . . . . . . . . . . . . . . . . . . . 8010.8 Ferromagnetism (Fe,Ni) . . . . . . . . . . . . . . . . . . . . . . 82

11 Generation of electromagnetic waves 84

11.1 Potentials and potential wave equations . . . . . . . . . . . . . 84

11.2 Delayed potentials . . . . . . . . . . . . . . . . . . . . . . . . . 8511.2.1 Hertzian Dipole . . . . . . . . . . . . . . . . . . . . . . . 86

3



Contents

0.1 General Information

If you find any mistake, feel free to send me an e-mail to ’[email protected]’.

0.1.1 Testat

To receive the Testat, you must make a serious attempt and hand-in at least3/4 of the questions on at least 3/4 of the Exercise Sheets.

0.1.2 Exam

• 2-hour written

• calculator, ”translation” dictionary English-German and 10 sides of notes in their original own handwriting

• all questions in German and English and can be answered in Germanand English

0.1.3 books

english deutsch

Purcell Electricity and Magnetism Elektrizitat and Magnetismus

Jackson Classical Electrodynamics Klassische Elektrodynamik

Tipler & Mosca Physik Physics for Scientists and Engine

Kanzig Elektrizitat und Magnetismus (v—d—f)

0.1.4 webpage

http://www.exp-astro.phys.ethz.ch/PhysikIII

4



Chapter 1

Electrostatics: Electric

charge

1.1 empirical fact: charges exist

t0 explain observed forces

a) + and − charges - like repel opposite attract

b) Net charge∑ q is conserved, and is relativistically invariant

c) Charge is apparently quantized qe = 1.602 · 10−19C

d) Charges appear ∼ pointlike r < 10−15m

1.2 Coulomb’s Law

Fig.1

F21 = kq1q2

r221

r21

cgs: F =q1q2

r2SI: F =

1

4πε 0

q1q2

r2

ε0 = permittivity of free space = 8.854 · 10−12Fm−1

a)

5



1. Electrostatics: Electric charge

b)

c) effect of charges is additive

Fx = ∑ i=x

1

4πε 0

qxqi

r2xi

rX I → superposition Principle

”Force experienced by one charge from another is not affected by thepresence of a third ”

1.3 Potential energy of a system of charges

Work done on q2 W 2 =

F21 · ds. If radial path:

W 2 = − r2

∞

1

4πε 0

q1q2

r2dr =

1

4πε 0

q1q2

r12= W 1

Independent of path3rd charge brought up

F3 = F31 + F32

W 3 = W 31 + W 32

Total of all charges W 31 + W 32 + W 21

W j =1

4πε 0

j−1

∑ i=1

q jqi

rij

Total W = ∑ j

W j =1

4πε 0∑

j

j−1

∑ i=1

qiq j

rij=

1

4πε 0∑

i∑ i= j

qiq j

rij· 1

2

e.g. NaCl

W energy per ion

=e2

8πε 0

−1.748

a

1.4 Concept of electric field

Fq =1

4πε 0q∑

qi

r2i

ri

Can consider this due to a field

Fq = qE(x, y, z) with E(x, y, z) =1

4πε 0∑

i

qi

r2i

ri

key concept: E describes locally the effects of distant charges

6



1.5. Charge distributions

1.5 Charge distributions

Often useful to introduce charge density, ρ(x, y, z)

dq = ρdV per unit volume

→ E(x, y, z) =1

4πε 0

all

space

ρ(x, y, z)r2

r dx dydz r =

x − x

y − y

z − z

also:

σ -charge per unit area in 2-d system dq = σ da (1.1)

λ-charge per unit system in 1-d system dq = λdb (1.2)

1.6 Concept of Flux and Gauss’ Law

da is ⊥ to surface and oriented outwards. Define Flux Φ

dΦ = E · da.

Φ =1

4πε 0

q

r2· 4π r2 =

q

ε0

→ Φ =∑ i qi

ε0for any closed surface =

1

ε0

V

ρdV

7




1.7 Applications of Gauss’ Law

a) Uniformly charged spherical surface

+

+

++

+

+

+

+

++

+

+

S1

σ

E = 0

Φ = 0

E = Q4πε0r2 as if all Q at center!

R

Q = 4π R2δ (1.3)

Inside surface E = 0 (1.4)

Outside surface 4π r2E(r) =1

ε04π R2σ =

1

ε0Q → E(r) =

Q

4π r2ε0=

σ

ε0at the surface R = r

(1.5)

b) Infinite line charge, constant λ

missing picture

2π r lE(r) =lλ

ε0(1.6)

E(r) =λ

2πε 0r(1.7)

c) Infinite sheet charge

2π a2E(r) =π a2σ

ε0(1.8)

E(r) =σ

2ε0(1.9)

8



1.8. Energy associated with E-field

1.8 Energy associated with E-field

+

+

++

+

+

+

+

++

+

+

E =σ

ε0

R2

r2(1.10)

→P.E. of charge at surface dq =

1

4πε 0

Q

r

d (1.11)

Total P.E. W =1

2

Q2

4πε 0r2(1.12)

dW

dr= −1

2

Q2

4πε 0r2(1.13)

dW

dr= −UEdV = −UE4π r2dr (1.14)

E =Q

4πε 0r2→ UE =

ε0

2E2 (1.15)

9




1.9 Concept of electrical Potential

Charge moves in E-field from A to B. Work by E-field = q B

A Eds Indepen-dent of path

C Eds = 0

Define electric Petential ϕ so that ∆ϕ AB = − B A E · ds → ∆ϕ(x, y, z) +

fix zero-point→ ϕ(x, y, z) usually ϕ(∞) = 0→ P.E of individual charge = qϕ(x, y, z) e.g E-field from single charge

ϕ =

1

4πε 0

1

R

+superposition

ϕ =1

4πε 0∑

i

qi

ri=

1

4πε 0

all

space

ρ(x, y, z)r

dV

P.E of ” last charge” = qϕ P.E. of system of charges = 12 ∑ i qi ϕi

dϕ = −E · ds = − Exdx + E ydy + Ezdz

(1.16)

always dϕ =∂ϕ

∂xdx +

∂ϕ

∂ ydy +

∂ϕ

∂zdz (1.17)

l.e.Ex = −∂ϕ

∂x, . . . (1.18)

E = −

ϕ (1.19)

A =

∂ A

∂xx +

∂ A

∂ yy . . . (1.20)

→E is always ⊥ to surfaces of constant ϕ (1.21)

1.10 Experiment: charging two spheres equally

r force 1 force 2

10 55 44

14 31 25

20 14 13

40 7

charge 1 charge 2

10



1.11. Other useful operators

Gauss’ Law

S

E · da =

V

ρ

ε0dV = Φs (1.22)

for general F(x, y, z)

S

S · da =

S1

F · a +

S2

F · da (1.23)

= ∑ i

Si

Fda = ∑ i

V i

S

Fda

V i(1.24)

(1.25)

take limit V i → 0 ∑ i V i → dV

S

Fda =

V

limV →0

S

F · da

V

divE

dV

SF · da =

FdV Gauss Theorem

(1.26)

⇒ ρ

ε0= divE Gauss’ Law

(1.27)

→ divF =∂Fx

∂x+

∂F y

∂ y+

∂Fz

∂z(1.28)

1.11 Other useful operators

•

E =

−

ϕ (1.29)

divE =ρε0

(1.30)

→ div

ϕ

= − ρ

ε0(1.31)

2ϕ = − ρ

ε0Poisson’s equation (1.32)

2=

∂2

∂x2+

∂2

∂ y2+

∂2

∂z2

Laplacian operator (1.33)

(1.34)

11




• curl Remember ”circulation”

Γ =

cF · ds = ∑

i

Γi = ∑ i

ai

ci

F · ds

ai(1.35)

curl F = lima→0

C

F · ds

a(1.36)

→

CF · ds =

A

curlF · da Stokes’ The

(1.37)

curl F =

∂Fz

∂ y− ∂F y

∂z

x +

∂Fz

∂z− ∂Fz

∂x

y +

∂F y

∂x− ∂Fx

∂ y

z =

×F

(1.38) E · ds = 0 in Electrostatics → curl E = 0 + . . . (1.39)

1.12 summary

E = −

ϕ (1.40)

ϕ = − Eds (1.41)

ρ = −ε0

2ϕ (1.42)

ϕ =

ρ

4πε 0rdV (1.43)

ρ = ε div E (1.44)

E =

ρ

4πε 0r2 rdV (1.45)

12



Chapter 2

Conductors

A conductor can’t have an electric field inside because the charge can move.

+

2.1 Conductors + insulators

Difference is mobility of charges. ≈ 1020

→ Key point: E (x, y, z) inside conductor = 0

13



2. Conductors

E

2.2 Conditions for conductor

a) E = 0 inside conductor

b) π = constant on surface and throughout conductor. ϕ:

c) At surface, E is perpendicular to surface, E = σ ε0

(Gauss’ Law)

d) Total charge on conductor Q =

S σ da = ε0

S Eda

2.3 The general electrostatic problem

General set of conductors k in Vacuum. ϕK, Qk

→ Solve2 ϕ = 0 subject to .b.

define ϕK Dirichlet problem (2.1)

Qk Neumann bonday conditions (2.2)

mixture of ϕK + QK but don’t overconstrain (2.3)

14



2.4. Uniqueness Theorem

2 ϕ ≡ 0

2.4 Uniqueness Theorem

There is at most one unique solution.

Proof: Let ϕ1 (x, y, z) be a solution to2 ϕ = 0 + ϕK . . .

ϕ2 (x, y, z) be another solution to 2 ϕ = 0 + ϕK . . .

→ ψ = ϕ1 − ϕ2 is a solution to

2 ϕ = 0 and ϕ = 0 on all surfaces→ ψ = 0 everywhere ⇒ ϕ1 = ϕ2

2.5 Some interesting cases

2.5.1 Enclosed cavity

2 ϕ = 0

ϕ ≡constant

⇒ ϕ = constant throughout cavity

→ E = 0

15



2. Conductors

2.5.2 Faraday cage

Farraday cage

+

+

+

+++

+

+

+

+

+

+

++ +

+

+

+

−

−

−−

−−−

−

−

−

−−

−− −

−−

−

+Q

E = 0

−Q

+Q

+

+

+

+++

+

+

+

+

+

+

++ +

+

+

+

+Q

16



2.6. Some tricks

2.6 Some tricks

2.6.1 Mirror charges

ϕ

r

ϕ ≡constant

The both following cases match the same conditions. (Mirror trick)q q

−q

2.7 Capacitance and capacitors

Single conductor carrying a charge Q and some potential ϕ

Q

ϕ

from superposition Q&ϕ Q = Cϕ

C = ”capacitance” = Qϕ units

17



2. Conductors

Coulombs Volt−1 = ”FARAD” Normally, two conductors close together,

+Q on one (2.4)

−Q on the other (2.5)

V = ∆ϕ between them (2.6)

+Q

−Q

ϕ1

ϕ2s

Area A

Uniform E between plates

E =ϕ1 − ϕ2

s=

V

SV = ϕ1 − ϕ2 (2.7)

σ = ε0E = ε0V

S(2.8)

Q = Aσ =Aε0V

S(2.9)

→ C =A

Sε0 (2.10)

2.8 Energy stored in capacitor

+Q −QE

V = ϕ1 − ϕ2

V = QC

dW = Vdq (2.11)

W = Q

0

q

cdq =

1

2

Q2

C=

1

2CV 2 =

1

2QV (2.12)

(2.13)

18



2.9. General System of conductors

2.8.1 Parallel plate capacitor

Note:

C =A

Sε0 (2.14)

Q = Aδ = AEε0 δ = Eε0 (2.15)

W = A2E2ε20 · 1

2

S

Aε0=

1

2ε0E2 (Volume) (2.16)

2.9 General System of conductors

Q1ϕ1

∞

ϕ∞ = 0

Q3

ϕ3

Q2

ϕ2

2 ϕ = 0

If we define all ϕi → E (x, y, z) → σ (x, y, z) (2.17)

→ Qi (2.18)

or define all Qi → E (x, y, z) → ϕi (2.19)

(2.20)

19



2. Conductors

Set ϕ = 0 on all except i = 1 (2.21)

Q1 = C11 ϕ1 (2.22)

Q1 = C21 ϕ1 (2.23)

Q1 = C31 ϕ1 (2.24)

Now ϕ2 varies, all other ϕ = 0

Q1 = C12 ϕ2 (2.25)

Q2 = C22 ϕ2 (2.26)

Q3 = C32 ϕ2 (2.27). . . (2.28)

→ By superposition for general ϕi

Q1 = C11 ϕ1 + C12 ϕ2 + C13 ϕ3 + . . . (2.29)

Q2 = C21 ϕ1 + C22 ϕ22 . . . (2.30)

→ Qi = Cij ϕ j with Cij matrix capacitance (2.31)

20



Chapter 3

Elecric Currents

3.1 Electric current

Net non-zero charge in net motion

a

a

Charges q

number density n

mean velocity u

Define current through a, I a=rate of net motion of charges through a.

I a = nqu · a (3.1)

For multiple species I a = ∑ i

(niqiui) J : current density

·a (3.2)

J = ∑ Ji J = ρu 3-d

I = λv 1-d

(3.3)

21



3. Elecric Currents

3.2 Charge conservation

S

net flow of charge rate of charge into surface of enclosed charge

S J · da = − d

dt V ρdV (3.4)

applying Gauss’ Theorem

div J = −d ρ

dt(3.5)

3.3 Ohm’s Law

Motion of charge due to E-field in non-perfect conductor. Conductors are

not perfect. The charges are not infinitely mobile.Expect J (x, y, z) = σ E Ohm’s Lawσ is the conductivity. In practice the σ can be a Tensor dependend on itsposition. That is just an approximation. It’s not an as fundamental Law asCoulomb’s law or Gauss’ law.Consider a simple cylindrical ”component”.

A J

e−

L

I =

Jda = JA (3.6)

Potential Voltage differenc =

Eds = EL (3.7)

Define Resistance R =V

I =

EL

J A=

1

σ = ρ =

1

δ⇒ R = ρ

resistivity

L

A

(3.8)

22



3.4. Circuits

Aside: In steady state

div J = − dϕ

dt= 0 (3.9)

→ div E = 0 (Ohm’s Law) (3.10)

Charge density ρ = 0

div E =

ρ

ε0

(3.11)

3.4 Circuits

Circuit = discrete ”component” linked together by ideal conducting wires,may join in ”nodes”

eg

A B

C D

E F

ϕ A = ϕB = ϕC = ϕD

ϕC > ϕE

ϕD > ϕF

ϕ

For each component I i current through it, V i potential difference acress it.For any arbitrary circuit, we want to know I i, V i for all component:→ Commonsense rules (≡ )Kirchoff’s Rules

3.4.1 Kirchoff’s Rules

• For every element i I i = V iR (Ohm’s Law)

• ∑ I in = ∑ I out or ∑ all I in = 0 since I out = −I in(charge conservation in steady state)

• Around any ”loop” ∑ V = 0→ Can usually break down a circuit into simple equivalent elements

23



3. Elecric Currents

eg series connection

R1 R2

I 1 = I 2 = I (3.12)

V 1 + V 2 = V (3.13)

R =V

I =

V 1I

+V 2I

=V 1I

1

+V 2I

2

= R1 + R2 (3.14)

In parallel

V 1 = V 2 = V

I = I 1 + I 2

→ 1

R=

1

R1+

1

R2(3.15)

R1

R2

3.5 Energy dissipation in a resistor

ϕ1 ϕ2IV = ϕ1 − ϕ2

P.E. released when Q flows through V = QV .

Energy per second = V I = V 2

R = I 2R appear as heat.

24



3.6. Sources of energy

3.6 Sources of energy

A

B

I Rgives energy to charges

”electro-motiveforce” e.m.f -units of Volt

eg chemical reactions:

H +

H 2

SO

HSO−4

PB

PbO2

2e− 2e−

+ −

1. Pb + HSO−4 → PbSO4 + H + + 2e− + energy

2. PbO2 + HSO−4 + 3 H + + 2e− → PbSO4 + 2 H 2O + energy

3. combined effect Pb + PbO2 + 2 HSO−4 − 2 H + → 2PbSO4 + 2 H 2O +

energy+2e− used on LHS reapon on RHS, +2 H + cross from right to left

Some energy wasted as heat

25



3. Elecric Currents

$I$

$R$

$+\phi$

$-\phi$$R i$

internal resistanc

I = εR+Ri

useful V = ε

−IRi Thevenin’s Theorem any system of emf &

resistors equivalent to a single emf and intend resistance.

3.7 Circuits with capacitors

Q = CV (3.16)

I =V

R= −dQ

dt(3.17)

−dQ

dt=

Q

CR(3.18)

dQ

dQ= − dt

CR(3.19)

In Q = −t

RC + constant (3.20)

Q = Q0e− tRC (3.21)

I =V 0R

e−t tRC (3.22)

RC time

FΩv

cs−1· cv (3.23)

Also charging of CPic 3.2

26



3.7. Circuits with capacitors

27



Chapter 4

Fields of moving charges

4.0 Magnetic Fields or Forces

F = q E + v × B Lorentz Force

I

B

q v

F

28



4.1. Reminder about special Relativity

4.1 Reminder about special Relativity

F F

∆x = γ∆x − βγc∆t Lenght/Lorentz contraction (4.1)

∆ y = y β =

v

c , γ =

1 1 − β2 (4.2)

∆z = ∆z (4.3)

c∆t = γc∆t − βγ∆x time dilation (4.4)

(4.5)

ux =

ux − v

1 − uxvc2

ux =u

x + v

1 + uxvc2

(4.6)

u y =

u y

γ 1 −uxv

c2 u y =u

y

γ 1 +uxv

c2 (4.7)

cpx = γcpx − βrE (4.8)

cp y = cp y (4.9)

cpz = cpz (4.10)

E = γE − βγcpx (4.11)

Forces? If particle at rest in F

f x = f x (4.12)

f y = γ f y (4.13) f z = γ−1 f z (4.14)

4.2 Invariance of charge

S(t)

E(t)da =

S(t)Eda =

Q

ε0(4.15)

29



4. Fields of moving charges

4.3 Transformation of E-fields

V

+σ −σ

E = σ ε0

At rest in F. In F, moving related to F ⊥to plates

σ = σ (4.16)

E =σ

ε0=

σ

ε0= E (4.17)

Now v plates

V

⇒ must be general result:

E = E

E⊥ = γE⊥

Now consider point charge moving with velocity v

qvx

In F, rest-frame of particle

Ex =q

4πε 0

x

(x2 + z2)32

(4.18)

Ez =q

4πε 0

z

(x2 + z2)32

(4.19)

sety = 0 xz plane (4.20)

30



4.4. Field of accelerated charge

Now consider F moving −vx (our lab!) swap primes and use

x = γx − γβct (4.21)

z = z (4.22)

t = γt − γβx

c2(4.23)

Set t = 0 when x = 0, look at field at t = 0

(4.24)

Note:

a) Field is radial: Ex

Ez

= xz , field points at origin

x

at origin

at t = 0

radial but not isotropic

q v v

→ E · ds = 0 ⇒ curl E = 0

4.4 Field of accelerated charge

Start at rest, then accelerate

31




q

Look at deceleration

q 0v0

12 v0t

a = v0τ

Look at this time T later. Assume 12 v0τ v0T cT

origin Where it would have been

θv0T sin θ

ErcT Eθ

Er

cτ

Er

→ Eθ

Er=

v0T sin θ

cτ (4.25)

Er =q

4πε 0R2=

q

4πε 0c2T 2(4.26)

⇒ Eθ =v0T sin θ

cτ · q

4πε 0c2T 2=

q

4πε 0

v0

τ a

·sin θ

Rc2(4.27)

→ Eθ =qa sin θ

4πε 0c2R(4.28)

32



4.5. Force on moving charge

4.5 Force on moving charge

q vxF our lab, Fq at rest.

Ex = Ex

E y = γE y

→dp

x

dt= E

xq = Exq

dp y

dt= E

yq = γE yq

(4.29)

• Particle is at rest in F, so flip primes in above eqn for F, F

dpx

dt= dpx

dt(4.30)

dp y

dt=

1

γ

dp y

dt(4.31)

back in F,

dpx

dt= Exq and

dp y

dt= E yq = E yq (4.32)

→ i.e. Force on q due to E does not change with vq

4.6 Interaction of moving charge with other moning

charges

q vx

I Infinite wire is uncharged

λ+0 = −λ−

0

−ve → v0

+ve 0

Negative charge moves, positive doesn’t → I = −λ0v0

Transform to F = rest-frame of q

λ+= γλ0 (4.33)

λ−= γ

γ due tovelocity less

than V

λ0 (4.34)

⇒wire looks positively charged in F (4.35)

33




Algebra work out λ− in rest of negative charges

net λ = γβ F→F

β0 v0

λ0

q

E

λ(> 0)

Er = λ

2πε 0r = γββ0λ0

2πε 0r (4.36)

New force is radial from wire and perpendicular to motion.

F y = qE y = −qγββ0λ0

4πε 0r for y = −r (4.37)

Transform to F y =F

y

γ= − q ββ0λ0

2πε 0r(4.38)

but I = −λ0v0 = −λ0 β0c (4.39)

F y = I 2πε 0c2

qvr

looklike magnetic force

∝ v

∝ I in wire

⊥ to velocity

(4.40)

v0

I

+ + + + + + + + +− − − − − − − − −

34



Chapter 5

Magnetic Fields

Total force on electric charge

F = qE + qv × B Lorentzforce (5.1)

B come from current and act on currentE come from charge and act on charge

dF = dqv × B = (λdl)v × B (5.2)

dF = I × Bdl (5.3)

5.1 Simplest case

B =µ0 I

2π r

µ0 = Permeability of Force Space (5.4)

Units of B held is Tesla. 1 Tesla excerts a force of 1Nm−1 on a current of 1Amp.

I

B

vF

⊗I

B

35



5. Magnetic Fields

5.1.1 Force between two ∞ long wires

B1

FI1

B2

FI2

d

B1 =µ0 I 12π d B2 =

µ0 I 22π d (5.5)

F1 =µ0 I 22π d

I 1 = F2 (5.6)

µ0 = 4π × 10−7Tm−1 A−1 (5.7)

F = 2 × 10−7 Nm−1 S.I. convention for defining an amp (5.8)

5.2 Properties of magnetic fields

Consider simple case of ∞ wire

B =µ0 I

2π r(5.9)

36



5.3. Vector Potential

B · ds = 0

B · ds = µ0 I

2π r · 2π r = µ0 I

General rule: Amperes Law B · ds = µ0 I = µ0

A

J0da (5.10)

5.2.1 Uniqueness theorem

For a given J(x, y, z), there is a unique B(x, y, z).

Proof:Suupose there are two solutions B1&B2 for J(x, y, z).→ D = B1 − B2 is a solution

div D = div B1 − div B2 = 0 (5.11)

curl D = curl B1 − curl B2 = µ0 J − µ0 J = 0 (5.12)

5.3 Vector Potential

ϕ = 14πε 0

ρr

dV (5.13)

⇒ E = −

ϕ (5.14)

Imagine

B =

×A (5.15)

div B = 0 = div(curl A) = 0 (5.16)

curl B = µ0 J ⇒ curl(curl A) = µ0 J (5.17)

37



5. Magnetic Fields

Expand× (

×A)

5.4 Fields of coils and solenoids

x-component:

∂

∂ y

∂ A y

∂x− ∂ Ax

∂ y

− ∂

∂z

∂ Ax

∂z− ∂ Az

∂x

= µ J x (5.18)

→∂2 AY

∂ y∂x− ∂2 Ax

∂ y2− ∂2 Ax

∂z2+ ∂2 Az

∂x∂z= µ J x

(5.19)

−∂2 Az

∂ y2− ∂2 Ax

∂z2 −∂2 Ax

∂x2 2 Ax

+∂2 A y

∂ y∂x+

∂ Az

∂x∂z +

∂2 Ax

∂x2 ∂

∂x(div A)=µ0 J x

= µ0 J x (5.20)

Trick: div A = f (x, y, z)

Then we can find another F, div F = f and curl F = 0Can add −F to A ⇒ div F = 0 and the same B

2 Ax = −µ0 J x

c.f.

ϕ = − ρ

ε0

for y, z − comp.analogous (5.21)

We had

ϕ =1

4πε 0

ρ

rdV (5.22)

A =µ0

4π

J

rdV Vectorpotential (5.23)

Apply this to general current carrying loop.

dA =µ2

4π r JdV =

µ0

4π rIdldB =

×dA =

I µ0

4π

×1

rdl

(5.24)

=I µ0

4π

−dl ×

1

r

=

I µ0

4π dl × r

r2(5.25)

(5.26)

38



5.4. Fields of coils and solenoids

5.4.1 Biot-Savart law

dB = Idl × r · µ0

4π r2= JdV × r

µ0

4π r2Biot-Savart law (5.27)

c.f

dE = ρ r1

4πε 2r2dV (5.28)

dB = ( J × r)µ0

4π r2dV (5.29)

usually Biot-Savart is easier to work with then calculating A and the takingits curl.on axis Bx = B y = 0

Bz =µ0

4π (b2 + z2)· I · 2π b

sin θ

b

(b2 + z2) 12

(5.30)

@z = 0 (θ = 90).

Bz =µ0 I

2b(5.31)

Bz = µ0 I 2

b2

(b2 + z2) 32

(5.32)

z

x

Bz

rθ

I

b

Now consider long Solenoid n turns per unit lenght.

current =rdθ

sin θ· n · I (5.33)

On the axis

dBz =µ0

2

rdϑ nI

sin ϑ

b sin ϑ

r2=

µ0

2nI sin ϑ (5.34)

39



5. Magnetic Fields

Integrating over ϑ

Bz =µ0

2nI

ϑ 2

ϑ 1sin ϑ 2dϑ =

µ0

2nI [cos ϑ 2 − cos ϑ 1] (5.35)

For an infinitely long solenoid

Bz = µ0nI (5.36)

Applying Ampere’s Law, it is clear that B cannot depend on position withinthe solenoid, for an infinite solenoid.

5.5 summary

GaussLaw AmperesLaw

S E · da = Q

ε0= 1

ε0

ρdV

B · ds = µ0 I = µ0 Jdacurl B = µ0 J

div E = ρε0

div B = 0

ϕ = 14πε0

ρr dV A = µ0

4π

Jr dV

E = − ϕ B = curl A

dE = 14πε0

ρdV rr2 dB = µ0 Idl× r

4π r2 = JdV × rµ0

4π r2

5.6 Change in B-field across a current sheet

σ

40



5.6. Change in B-field across a current sheet

∆E⊥ = σ ε0

Gauss’ Law (5.37)

∆E = 0

E · ds = 0 (5.38)

→ Pressure on sheet = σ

E1 + E2

2

= σ E⊥ (= 0 if no extend E-fields)

(5.39)

Now look at the current sheet

J = σ v Am−1 (5.40)

x

z

y

σ

J

B1B2 ∆B = (B1 − B2)

B · ds = µ0 J xl

I l

(5.41)

∆Bz = µ0J x (5.42)

∆Bx = 0 (5.43)

∆B y = 0 (div B = 0) (5.44)

∆B = µ0J (5.45)

∆B = 0 (5.46)

∆B⊥ = 0 (5.47)

∆B parallel to sheet perpendicular to current∆B parallel to sheet and parallel to current.Exactly as in electrostatics

Pressure =1

2µ

B2

2 − B21

(5.48)

+energy density in B-field =1

2µ0

c f

1

2εE2

(5.49)

41



5. Magnetic Fields

5.7 How do E and B transform

x

z

y

F

v

F

v0 v0

−σ σ

E y =σ

ε0(5.50)

Bz = µ0J = µ0σ v0 (5.51)

v0 = v0 − v

1 − vv0

c2

= c ( β0 − β)(1 − ββ0)

(5.52)

σ = ϕ0

charge densityof charges

in F

σ

γ0

charge densityof charges

in rest-frame

(5.53)

γ0 =

1 − v

02

c2

(5.54)

42



5.7. How do E and B transform

σ = σγ (1 − ββ0) (5.55)

and J = σ v0 = σγ (1 − ββ0) c ( β0 − β)

(1 − ββ0)= σγ (v0 − v) (5.56)

E y =

σ

ε0=

σγ

ε0− σγββ0

ε0(σ V 0µ0 = B2) (5.57)

= γE y − γβ

ε0µ0cBz (5.58)

εµ0= 1c2

= γ

E y − βc

=v

Bz

= γ

E y − vBz

(5.59)

Bz = µ0

J (5.60)

= µ0σγv0 − µ0σγv (5.61)

= γ

Bz − εγ0 βE y

→ Bz = γ

Bz − β

cE y

(5.62)

Other components

Ex = Ex (5.63)

E y = γ

E y − βcBz

(5.64)

Ez = γ

Ez + βcB y

(5.65)

Bx = Bx (5.66)

B y = γ

B y +

β

cEz

(5.67)

Bz = γ

Bz − β

cE y

(5.68)

If represent + ⊥ components to v (motion of frames)

E = E B

= B (5.69)

E⊥ = γ E⊥ + cfi × B⊥ B

⊥ = γ

B⊥ − 1

cfi × E⊥

(5.70)

43



5. Magnetic Fields

5.8 Hall Effect (1879)

x

z

y

J

+

−

B

F = qE + qv × B (5.71)

If +ve charge carrier

If −ve charge carrier

+

− E

Eint = vBSign of Ez tells us +ve or −ve charge carriers.Answer: −ve!

44



Chapter 6

Magnetic Induction

6.1 Magnetic Induction

F = qE + qv × B (6.1)

Charges in moveing conductor in a B-field experience Lorentz force.

x

z

y

Bz

v

F−

F+

E

Eint =

−v

×B

45



6. Magnetic Induction

Eint = −v × B B

− − − − − − −

+ + + + + + +

E

−

+

F−

F+

Aside: Look is rest-frame of conductor

E y = γ

E y

=0

− βcBz

(6.2)

yz

x

Eint

E y

Eint, y = γvBz (6.3)

Eint = vBz (6.4)

46



6.1. Magnetic Induction

6.1.1 Wire in circuit

A

B

lRF+

F−

B

+ charges moving A → Bgains energy (6.5)

+ charges moving B→

Agains energy (6.6)

different Situations

B1 B2 B(t)

Energy gain = B

Aq (v × B) ds = qvlB = qE

(6.7)

i.e.E = vlB = rate of charge of ” enclosed b” =Φ

t(6.8)

In general:E =

(v × B) ds (6.9)

47




6.2 Faraday’s Law

Φ(t) =

SB · da (6.10)

Φ (t + dt) =

S+dSB · da = Φ(t) +

dS

B · da dΦ

(6.11)

On rim of loop

da = vdt

×ds (6.12)

⇒ dΦ =

dSB · da =

C

B · (vd × ds) (6.13)

dΦ

dr=

CB · (v × ds) = −

c

(v × B) · ds = −E from before (6.14)

i.e.E = −dΦ

dtFaraday’s Law (6.15)

Note

Φ = S

B · da (6.16)

div B = 0 (6.17)

What if loop fixed and B is changing

→ look in rest-frame of loop

B1 B2

v v fixed

E⊥ = γ E⊥ + v × B⊥ (6.18)

E = γv × B (6.19)

E = γv × B1 (6.20)

E2 = γv × B2 (6.21)

energy gain per q = E = γv (B1 − B2) l (6.22)

= −γdΦ

dt= −dΦ

dt(6.23)

48



6.3. Lenz’ Law

Consider a static loop Faraday’s Law

E = − d

dt Ψ = − d

dt

S

B · da (6.24)

moving loop E =

(v × B) · ds (6.25)

E =

Eds = − d

dt

S

Bda (6.26) S

curl Eda =? (6.27)

(6.28)

curl E = −∂B

∂t(6.29)

note: this does not uniquely specify E (x, y, z), - can add any E (x, y, z) fromstatic ρ (x, y, z), which has curl E = 0 as before.

6.3 Lenz’ Law

B1 B2

v

I

F1+ F2+F−

Binducted

I B

E

emf drives current around loop

B1 > B2, F1 > F2 anddΦ

dtis negative

→I is in direction to reduce change in Φ

(6.30)

also force on loop ← F resists motion

dF = I (ds × B) E =

(v × B) · ds (6.31)

dW = dF · v I =E R

(6.32)

W =

I (ds × B) · v (6.33)

49




6.4 Mutual inductance

Change in IU in one circuit may produce a change in Φ in a second circuit(mutual ind.) or a change in Φ in itself (self ind.)

I 1C1

Let

Φ21 = flux though C2due to I , in C1 (6.34)

=

S2

B1 · da = kI 1 (6.35)

E 21 = emf in C2due to change in I 1 (6.36)

= − ddtΦ21 (6.37)

E 21 = −kdI 1dt

k = “mutual inductance“ M21

(6.38)

units = VA−1s = Ωs = “Henry“ (6.39)

⇒ M21 =Φ21

I 1(6.40)

Use A

50



6.5. Self-inductance

Φ =

SB · da (6.41)

=

Scurl A · da (6.42)

Stoke’s=

C

A · ds (6.43)

A1from Chap 4?

=µ0

4π

all

space

J

rdV =

µ0

4π I 1

C

ds1

r(6.44)

Φ21 = C2

A1 · ds2 = C2

µ04π I 1

C1ds1

r ds2 (6.45)

Φ12 =

C1

µ0

4π I 2

C2

ds2

rds1 (6.46)

Since

C1 C2

ds2 · ds1

r= C2 C1

ds1 · ds2

r(6.47)

→ Φ12

I 2=Φ21

I 1(6.48)

6.5 Self-inductance

Pic 6.1

Changing I → change in Φ → E to oppose change in I Clase switch at t = 0

I =ε0

R

1 − e− R

C t

(6.49)

51




I

t

ε0R

Pic 6.3

UC =1

2

CV 2 (6.50)

UL =1

2LI 20 (6.51)

52



Chapter 7

Alternating current circuits

7.1 The RLC circuit as damped oscillator

Pic 7.1R

LC+Q

−Q

Q = CV (7.1)

I = −dQ

dt= −C

dV

dt(7.2)

V − LdI

dt− RI = 0 Kirchoff nr.2∑ v = 0 (7.3)

d2V

dt2 +R

L

dV

dt +1

LC V = 0 → damped harmonic oscillator (7.4)

Try solution V ∝ eλt

λ2 +R

Lλ +

1

LC= 0 (7.5)

λ =− R

L ±

R2

L24

LC

2=

−R

2L±

R2

4L2− 1

LC(7.6)

53



7. Alternating current circuits

a) R2<

4Lc

V (t) = Ae−R2L t cos ωt ω2 =

1

LC− R2

4L2

(7.7)

pic7.2 (7.8)

b) R2>

4LC

V (t) = Ae− β1t + Be− β2t β =R

2L±

R2

4L2− 1

LC(7.9)

pic 7.3

c) R2 = 4LC

V (t) = Ae−R2L t (1 + Bt) (7.10)

pic 7.4

7.2 Circuits driven by alternating voltage

Easy to generate voltage source that varies sinusordiallyPic 7.5

R

LVAC

Calculate amplitude and phase of I ,which must have same ω

I = I 0 cos (ωt + α) (7.11)

Apply Kirchoff’s nr 2

ε0 cos ωt = LdI

dt+ RI (7.12)

Try solution as above

ε0 cos ωt = −LI 0ω sin (ωt + α) + RI 0 cos (ωt + α) (7.13)

= −LI 0ω sin ωt cos α + cos ωt sin α + RI 0 cos ωt cos α − sin ωt sin α(7.14)

54



7.2. Circuits driven by alternating voltage

Balance sin ωt, cos ωt

−I 0 Lω cos α − RI 0 sin α = 0 → tan α =−ωL

R(7.15)

→I peaks after V , I “lags“ V − LI 0ω sin α + RI 0 cos α = ε0

(7.16)

→ I 0 =ε0

R cos α − ωL sin α(7.17)

=ε0

R (cos α + sin α tan α)(7.18)

=ε0

Rcos α (7.19)

tan α =ωL

R→ cos α =

R

(R2 + ω2L2)12

→ I 0 =ε0√

R2 + ω2C2I is reduced in amplitude

(7.20)

E

current in loop lags

I in solenoid

I s, I lis negative (7.21)

E = −dΦ

dt(7.22)

Φin ring (7.23)

7.2.1 Solenoid pushing

Some description would be nice. . .

55




7.2.2 Some circuits

pic 7.7

V = ε0 cos ωt (7.24)

ε0 cos ωt = − Q

C+ IR (7.25)

I = I 0 cos(ωt + ϕ0) (7.26)

I = −dQ

dt(7.27)

Q = −

Idt = − I

ωsin(ωt + ϕ) (7.28)

ε0 cos ωt =I 0

ωCsin(ωt + ϕ) + RI 0 cos(ωt + ϕ) (7.29)

(7.30)

steady-state

tan ϕ =1

RωC, I 0 =

ε0 R2 +

1

ωC

2(7.31)

I = I 0 cos ωt (7.32)

V L = LdI

dt= −LI 0ω sin (ωt + ϕ) (7.33)

V C = − Q

C=

1

C

Idt =

I 0ωC

sin (ωt + ϕ) (7.34)

V = V L + V C = −

ωL − 1

ωC

I 0 sin (ωt + ϕ) (7.35)

ωL = ωL − 1

ωC(7.36)

56




pic 7.9

ωL > 1ωC ; tan ϕ = − ωL

R − 1ωRC I 0 =

ε0 R2 +

ωL − 1

ωC

2 (7.37)

ωL =1

ωC⇒ ω ≡ ω0 =

1√LC

(7.38)

I = I 0 =ε0

R; (7.39)

P(ω0) =ε2

0

R2; ω; P(ω) =

1

2P(ω0) (7.40)

I

ω/ω0

RR> R

Q f = ω

energy stored

energy power dissipated =

ωL

R (7.41)

ω = ω0 + ∆ω, ω20 =

1

LC⇒ 1

ω0C= ω0 L (7.42)

ωL − 1

ωC= (ω0 + ∆ω) L − 1

ω0 + ∆ω

1

C= ω0L

1 +

∆ω

ω0

−

1 + ∆ωω0

−1

ω0C(7.43)

= ω0L

1 +

∆ω

ω0−

1 − ∆ω

ω0

= ω0L

2∆ω

ω0

+ O

∆ω

ω0

(7.44)

pic 7.10

I 2 = I 02 cos (ωt + ϕ2) (7.45)

V 2 = V 02 cos (ωt + ϑ 2) (7.46)

eiϑ = cos ϑ + i sin ϑ , i2 = −1 (7.47)

1) I = I 0 cos (ωt + ϕ) → I 0eiϕ (7.48)

2) z = x + iy Re

zeiωt

(7.49)

57




I 1 = I 01 cos (ωtϕ1) (7.50)

I 2 = I 12 cos (ωt + ϕ2) (7.51)

I 1 = I 01eiϕ1 (7.52)

I 2 = I 02eiϕ2 (7.53)

I 1 + I 2 = Re

I 01eiϕ1 + I 02eiϕ2

eiωt

(7.54)

Kirchoff’s rules

a) ∑ I in = 0

b) ∑ ∆V = ∑ emt

V = ε0 (7.55)

I = I 0eiϕ (7.56)

admittance (7.57)

Y =eiϕ

(R2 + ω2 L2);tan =

−ωL

R(7.58)

impendance (7.59)

z =1

y(7.60)

I = YV , V = ZI (7.61)

(7.62)

R → 0

C → 0, ϕ = −π

2; y =

−i

ωL, V = IR (7.63)

Im

Re

V

I

π 2

L → 0, ϕ = 0; ϕ =1

R; z = R, C = 0, V = IR (7.64)

58




Im

Re

V

L → 0, R → 0,tan ϕ =1

ωRC, ϕ =

π

2; y = iωC; z =

−i

ωC(7.65)

Im

Re

V

I

π 2

pic 7.11

I = YV (7.66)

V = ε0 (7.67)Y = YC + YR + YL = iωC +

1

R− i

1

ωL(7.68)

I = ε0

1

R+ i

ωC − 1

ωL

= I 0eiϕ (7.69)

I 0 = ε0

1

R2+

ωC − 1

ωL

2

(7.70)

tan ϕ =

RωC − R

ωL

(7.71)

59




7.3 Power Consumption

P = V I = I 2R =V 20R

cos2 ωt (7.72)

V = V 0 cos ωt (7.73)

P =v2

0

R

cos2 ωt

=

1

2

V 20R

=1

2I 2R (7.74)

P = V I = V 0 cos ωtI 0 cos (ωt + ϕ) (7.75)

= V 0 I 0 cos ωt (cos ωt cos ϕ−

sin ωt sin ϕ) (7.76)

= V 0 I 0

cos2 ωt cos ϕ − /cosωt sin ωt sin ϕ

(7.77)

(7.78)

P = V 0 I 01

2cos ϕ (7.79)

cos ωt sin ωt =1

2sin2ωt (7.80)

V rms =

1

√2 V 0 (7.81)

I rms =1√

2I 0 (7.82)

P = V rms I rms cos ϕ (7.83)

60



Chapter 8

Maxwell’s Equations

8.1 Wave equation

E & B

×E =

∂B

∂t(8.1)

×B = µ0ε0

∂E

∂t(8.2)

·E =

·B = 0 (8.3)

×

×B

= ε0µ0

×∂E

∂t(8.4)

×B

0

−2

B = −ε0µ0∂2B

∂t2(8.5)

⇒2

B − ε0µ0∂2B

∂t2= 0 (8.6)

2E − ε0µ0

∂2E

∂t2= 0 (8.7)

61



8. Maxwell’s Equations

x

z

y

E

B

v

Direction of travel = E × B and E · B = 0

Note: Wave doesn’t have to be a sine wave. Any function of (K · r − ωt) will

do so long spatial and temporal derivatives match and E, B, v are perpen-dicular.

8.2 Superposition of two opposite directions

Maxwell equations linear in E and B → Sum should also be a solution.evt ein Fehler hier. . .

E = z · E0 (sin (Ky − ωt) + sin (Kyωt)) (8.8)

= z · 2E0 sin (Ky) cos (ωt) (8.9)

B = xB0 (sin (Ky − ωt) − sin (Ky − ωt)) (8.10)

= x2B0 cos (Ky) sin(ωt) (8.11)

(?Standing wave)

8.3 Standing wave

E = 0 at certain points at all times.Satisfies the boundary conditions for conducting surface.⇒ conductor to reflect lightB is changing rapidly at the surface and is zero inside the conductor.⇒ surface currents

62



8.4. Energy Transport of E and M waves

8.4 Energy Transport of E and M waves

Compute the energy density for aur travelling wave.

dU =

ε0E2

2+

1

2µ0B2

dV = ε0E2dV =

1

µ0B2dV (8.12)

=1

cµ0EBdv (8.13)

from before: ε0E2 = B2

µ0(8.14)

Energy transport per unit area

S =1

µ0EB (8.15)

Pointing Flux

S =1

µ0E × B (8.16)

8.5 Lorentz transformation of waves

Ex = Ex E

y = γ

E y − βcBz

E

z = γ (Ez + βcBy) (8.17)

Bx = Bx B

y = γ

By +

β

cEz

B

z = γ

Bz − β

cE y

(8.18)

E · B = 0 (8.19)

ε0E2 − B2

µ0= 0 (8.20)

E · B = Ex B

x + E y B

y + EzB

z (8.21)

= Ex Bx + γ2 (Ey − βcBz)

B y +

β

cEz

+ γ2

Ez + βcB y

Bz − β

cE y

(8.22)

= Ex Bx + γ2

1 − β2

1

E y B y + EzBz

= E · B (8.23)

63



8. Maxwell’s Equations

ε0E2 − B2

µ0= ε0E2 − B2

µ0(8.24)

→ em wave in one Frame appear as an em in all frame including ωK = c

What about a frame moving at c?Consider E y = E0 and Bz = E0

c

E y = γ

E0 − βc

E0

c

= E0γ (1 − β) (8.25)

Bz = γE0

c −β

c E0 =

E0

c γ (1 − β) (8.26)

as v → c, β → 1, E&B → 0

8.6 summary

f (k · r − ωt) curl E = − ∂B∂t I = YV = 1

Z V

ωk = v = 1√ε0µ0

= c curl B = µ0ε0 ∂E∂t + µ0 J

B0 = E0c div E = ρ

ε0YR = 1

R ZR = R

E · B = 0 div B = 0 YL = −iωL ZL = iωL

E × B = direction of travel div J = − ∂ρ∂t YC = iωC ZC = −i

ωC

(8.27)

⇒ (10)Φ =

S B · da = BS cos θ

64



Chapter 9

Dielectric materials

9.1 Introdiuction

When I have a material in a capacitor, capacitance changes C = QV → C =

εCVac , ε ≥ 1

For fixed V (= ∆ϕ) Q = εQVac (9.1)

For fixed Q V =V Vac

ε(9.2)

(9.3)

ε = 1.0 for vacuum (9.4)

≈ 1.0006 for air (9.5)

1.01 typical gases (9.6)

≈ 2 − 10 typical solids (9.7)

≈ 20 − 100 typical liquids (9.8)

empirically, related to density or mobility

65



9. Dielectric materials

9.2 Electric dipoles

Recall p = ql r from − to +

p =

allspace

r ρdV = “first moment“ of p (9.9)

→ p in E

a) Torque N = p × E → align p with E

b) Work to align dipole with E ⊥→ ω = pE dω = − ρE sin θdθ = averageω for randomly oriented dipole

c) potential energy: E pot = −p · E

d) Net force in inhomogeneous E-field

Fx = p ·

Ex (9.10)

F y = p ·

E y (9.11)

Fz = p ·

Ez (9.12)

(9.13)

V ≡ ∆ϕ

Q

−Q−σ

σ

V =Q

εCVacε ≤ 1.0 (9.14)

E =EVac

ε(9.15)

66



9.3. Atomic and molecular dipoles

9.3 Atomic and molecular dipoles

9.3.1 Permanent dipoles

H

+

Cl

−

H +

H +

O−−

→ external E-field will pref-erentially align permanent dipoles.

9.3.2 Induced dipoles

+

−

F+F−

E

⇒

−

+p

Expect p ∝ E (9.16)

= E4πε 0α α : dimensions = Volume (9.17)

Compare E-field insied atom to external one

Inside E =1

4πε 0

e

a20

for Hydrogenexpect∆z

a0≈ E pot

14πε0

ea2

0

(9.18)

p = e∆z (9.19)

α ∝ a30 (9.20)

actually α =q

2a3

0 for H (9.21)

67




9.4 Electric fields from polarized matter

Consider material composed of dipoles number density N .

Total dipole moment

in volume dV = p · N · dV

P

P = density of polarization (9.22)

P

dz

da

−σ

+σ

E p

i.e.σ = P (9.23)

E p =σ

ε0= − P

ε0(9.24)

H P is result of external Eext

→ E p is opposite to Eext

→ E final = Eext + E p is recuced → look at capacitor

+Q

−Q −σ

σ

Evac = σ ε0

+Q

−Q−σ

σ

P E p

E final ≡ Evac − pε0

68



9.4. Electric fields from polarized matter

1 =Evac

E final− p

E finalε0(9.25)

! E =Evac

E final= 1 +

p

ε0E final(9.26)

→ E = 1 +p

E finalε0 X =“permittivity“

(9.27)

Note: Final E-field makes sence! Will drop “final“ from now on.

E = 1 +p

ε0E final X susceptibility

(9.28)

Look again at Capacitor with E (≡ ∆ϕ) constant

V E

+ + + + + + + + +

− − − − − − − − −

+ + + + + + +

− − − − − − −

E from po-larization P = (E − 1) Eε0 from →

top surface has σ = −P (9.29)

= (1 − ε) ε0E (9.30)

(9.31)

top plate of capacitor needs extra charge

total charge = ε0E (9.32)

ε0E = δC + (1 − ε) ε0 top dielectric

E (9.33)

→ σ C = εε0∃ = εσ 0 σ : capacitance ↑ ×ε (9.34)

69




9.4.1 Gauss’ Law in medium and vector field D

q

E

dilectric ε

E reduced by factor 1ε

E =q

4πεε0r2(9.35)

Now introduc concept ρ free (what we control)and ρbound what nature controls due to polarization Always true

div E =1

ε0 ρ =

1

ε0

ρ free + ρbound

(2) (9.36)

div EVac =ρ free

ε0 = ε div E +ρbound

ε0(9.37)

→ (ε − 1) div E = − ρbound

ε0(3) (9.38)

Aside: we had before

P

E= (ε − 1) ε0 (9.39)

P = ε0 (ε − 1) E (9.40)

(3) → ρbound = − div PReturn to (3)

div E = ε div E − 1

ε0div P (9.41)

div

E +

P

ε0

D/ε0

=ρ free

ε0(9.42)

div D = ρ free (9.43)

70



9.5. Currents in dielectrics and Maxwell’s equations

D = E1 +p

Eε0 ε0 ε

Electric Displacement (9.44)

D = εε0E (9.45)

i.e If we deal with D, things are as in vacuum.→ get E from D

9.5 Currents in dielectrics and Maxwell’s equations

ρbound = − div P (9.46)

div J = −∂ρ

∂t(9.47)

→ Jbound =∂P

∂t(9.48)

Maxwell

curl B = µ0ε0

∂E

∂t + µ0 J free +

∂P

∂t (9.49)

and E +P

ε0=

D

ε0(9.50)

“boxed“

curl B = µ0∂D

∂t+ µ0 J free (9.51)

= µεε 0∂E

∂t+ µ0 J free (9.52)

9.6 Eloctromagnetic waves in dielectric

ρ free = 0 J free = 0 (9.53)

curl E = −∂B

∂tdiv E =

1

εε0div D = 0 (9.54)

curl B = εε0µ0∂E

∂tdiv = 0 (9.55)

71




→ Same wave-like solutions but now

ω

k=

1√εε0µ0

→ c =c

ε12

(9.56)

ε12 = refractive index n

B0 = ε12

E0

c=

E0

c (9.57)

9.7 Example: Electric field around dielectric sphere

Consider a polarized sphere

r 0 p− ⇔

outside looks like

+

−

+

−

p

Total dipole is the same

p0 =4

3

π pr30 (9.58)

→ use standard expressions for ϕ ∝ E outside of sphere

ϕ =1

4πε 0 p0

cos θ

r2=

1

3ε0

r30

r2p cos θ (9.59)

On surface sphere, r = r0

ϕ =1

3ε0r0 P cos θ

z

=P

3ε0z (9.60)

72



9.7. Example: Electric field around dielectric sphere

Inside2 ϕ = 0. By inspection and uniqueness

Ez = − p3ε0

(9.61)

ϕ =pz

3ε0(9.62)

p

Field at poles:

outside E = − ∂

∂r

Pr30 cos θ

3ε0r2=

2P

3ε0

(r = r0,cos θ = 1) (9.63)

D ≡ 2P3

73




Eout = 2P3ε0

(9.64)

Ein =−P

3ε0(9.65)

∆E⊥ =P

ε0(9.66)

∆D⊥ = 0 (9.67)

What about E?

→ General bondary conditions for dielectric surface without dfree charges

∆E = 0 (9.68)

∆D⊥ = 0 (9.69)

What if polarization produced by extend E-field (E0)?

E = E0 + E p (9.70)

also P = (ε − 1) χ

ε0Eint (9.71)

Eint = E0 − P

3ε0(9.72)

= E0 − (ε − 1)

3Eint (9.73)

Eint =3

2 + εE0 (9.74)

P =3 (ε − 1)

ε + 2ε0E0 (9.75)

74



9.7. Example: Electric field around dielectric sphere

vacuum dielectric

curl E = 0 curl E = 0

div E = ρε0

div D = ρ free

E = − ϕ D = εε0E2 ϕ = 0

∆E

= 0 ∆⊥

= 0 (9.76)

75



Chapter 10

Magnetic phenomena in

matter

10.1 Phenomenology

• Ferromagneticsm (Fe,Ni, permanent magnets)

• Para-magnetism (few, eg Bi)

• Dia-magnetism (almost all meterials)

PARA DIA

Force ∝ B dBdx

10.2 Magnetic dipoles

B-field from current loop

76



10.3. Force on m in external field

I

is (for from loop) exactly same as E-field from dipole.

+

−

f

p = ql

E

I M

Some response in external fields: eg

N = m × B c.f.N = p × E (10.1)

10.3 Force on m in external field

As with electric dipole

Fx = m ·

Bx (10.2)

m

BrBr

I

77



10. Magnetic phenomena in matter

F = l × B (10.3)

Fz = −2π b Oa of loop

IBr (10.4)

Using div B = 0

→ π b2

dBz

dz

∆z + 2π b∆zBr = 0 (10.5)

→ Fz = 2π bI ·b

2dBz

dz (10.6)

= π b2 I ∂Bz

∂z(10.7)

= m · ∂Bz

∂z= m ·

Bz

(10.8)

10.4 Current in loop in atom

mev2

r=

Ze2

4πε 0r2(10.9)

e−

+Ze

Change B → emf in loop → E-field

E

·ds

2π rE

=

−dΦ

dt

= π r2 ∂B

∂t

(10.10)

E =r

2

dB

dt(10.11)

medv

dt= e

r

2

dB

dt

(10.12)

∆v =e

me· r

2∆B (10.13)

Change in v → change in I = ev2π r

78



10.5. Electron spin

dI =e

2π rdv =

e2

me4π (10.14)

dm = − e2r2

4medB from Lenz’s Law (10.15)

• not dependent on sense of v.Applying external B, all electron orbitals aqcuire opposite ∆m. Because ∆mis anti-parallel to B → diamagnetism

10.5 Electron spin

“Spin“ of electron → mangetic moment (Quantum mechanical effect)→ External B can align m of electron spin → paramagnetismNot in all atoms because electrons paired with opposite spin, ∑ m = 0 -noeffect

10.6 Magnetic fields of magnetized matter

As before for p

M = N m (10.16)

J

M dz

da

MdV = (J dz) da = M (dadz) dV

(10.17)

M = J c.f. P = σ (10.18)

curl M = Jbound (10.19)

79




10.7 Maxwell’s equations

curl B = µ0ε0∂E

∂t+ µ0

J free + J bound

curl M

(10.20)

curl (B − µM)

H

= µε0∂E

∂t+ µ0 J free (10.21)

B = µ (H + M) (10.22)

i.e. Amperes Law:

C

H · ds = I enclosed free

(10.23)

cf

SD · da = ρ free (10.24)

Usually M ∝ H (10.25)

B = µµ0H (10.26)D = εε0E (10.27)

B = µµ0H (10.28)

D = εε0E (10.29)

¡++¿

∆E = 0 curl E = 0 always when∂

∂t= 0 (10.30)

∆D⊥ = 0 div D = 0 when ρ free = 0 (10.31)

∆ H = 0 curl H = 0 when J free = 0 (10.32)

∆B⊥ = 0 div B = 0 always (10.33)

80



10.7. Maxwell’s equations

[surfaces with ρ free = J free = 0]

I q

filed lines will bend at surfaces

81




E

θ

E

θ

E⊥

E⊥

E

E⊥ =

E⊥ε

E = E (10.34)

θ> θ E

= E⊥

2 + E2

< E (10.35)

¡++¿

¡++¿

10.8 Ferromagnetism (Fe,Ni)

• magnetic moment m even in absence of Bext

• suggests that M due to stable alignment of atomic m

Iron permanent magnet M ≈ 1.8 · 106 JT −1m−3

M ≡ every atom with 2 aligned electron spins. QM effect, enerycally favor-able to align spins

82



10.8. Ferromagnetism (Fe,Ni)

83



Chapter 11

Generation ofelectromagnetic waves

11.1 Potentials and potential wave equations

We had before B = curl A+Maxwell-equation

curl E = −∂B

∂t= − curl

∂ A

∂t(11.1)

E = −∂ A

∂t+ anything with curl = 0 (11.2)

E = −∂ A

∂t−

ϕ (11.3)

Remember

div E =ρ

ε0curl E = −∂B

∂t(11.4)

div B = 0 curl B = µ0 J + ε0µ0∂E

∂t(11.5)

84



11.2. Delayed potentials

div E = ρε0

(11.6)

+E=− ∂ A∂t − ϕ−−−−−−−−→ div

−∂A

∂t−

ϕ

=

ρ

ε0(11.7)

−2

ϕ − ∂

∂t(div A) =

ρ

ε0(11.8)

Earlier, we imposed div A = 0 because we can find a field, F, such that

curl F = 0 div F = anything (11.9)

Now, generalize this to force div A = −ε0µ0∂ϕ∂t “Lorentz condition“

−2

ϕ + ε0µ0∂2 ϕ

∂t2=

ρ

ε0in vacuum (11.10)

curl B = µ0 J + ε0µ0∂E

∂t(11.11)

+ fact E = −∂A

∂t−

ϕ (11.12)

+ fact B = curl A (11.13)

curl (curl A) div A−2A

−ε0µ0

∂2 A

∂t2− ∂ϕ

∂t div A

ε0µ0

= µ0 J (11.14)

−2

A + ε0µ0∂2A

∂t2= µ0 J = 0 in vacuum (11.15)

11.2 Delayed potentials

Solution to −2 ϕ = ρε0

is from earlier

ϕ =1

4πε 0

V

ρ(r)|r − r|dV (11.16)

Solution to full equation

85



11. Generation of electromagnetic waves

ϕ(r, t) =1

4πε 0

V

ρ

r, t − |r−r|c

|r − r| dV (11.17)

“delayed“ or “retarded“ potentials. Likewise:

A(r, t) =m0

4π

V

J r, t

−|r−r |

c |r − r| (11.18)

11.2.1 Hertzian Dipole

+

− Hertzian dipolelI

q = q0 sin ωt (11.19)

I =dq

dt= I 0 cos ωt I 0 = ωq0 (11.20)

p = ql = p0 sin ωt p0 = l · q0 =I 0l

ω(11.21)

Calculate A

Az(r, t) =µ0

4π

l2

− l2

dzI

z, t − |r − z|

c

· 1

|r − z| =µ0

4π

I

t − rc

l

r

(11.22)

Ax = A y = 0 because I z(11.23)

Now use

86



11.2. Delayed potentials

div A = −ε0µ0∂ϕ∂t

Lorentz condition (11.24)

∂ A

∂t= . . . (11.25)

−ε0µ0∂ϕ

∂t=

µ0l

4π

∂

∂z

I

t − rc

r

(11.26)

=µ0l

4π

− I

t − r

c z

r3− ∂I

t − r

c

∂

t − rc

τ

cr2

(11.27)

∂ϕ

∂t=

l

4πε 0 z

r3

I t−

r

c +z

cr2

∂I t − rc∂ t − rc (11.28)

→ ϕ =l

4πε 0

z

r3q

t − r

c

Delayed potentialfrom ϕ

→ ignore at large r

+z

cr2I

t − r

c

∂B∂t effect

(11.29)

→ E&B from ϕ&A

Er = 0 Eϕ = 0 Eθ = −ωl I 0 sin θ sin ω t −r

c4πε 0c2 · r

(11.30)

Br = 0 Bθ = 0 Bϕ =−µ0ωl I 0 sin ω

t − r

c

r

(11.31)

rθ Eθ

Bϕ

v = c

ϕ

Amplitude of E&B

|E|

θ → maximum in plane of dipole

87



11. Generation of electromagnetic waves

Instanteous power

W =

S

1

µ0(E × B) da =

ω2l2 I 206πε 0c3

sin2

ω

t − r

c

(11.32)

W =ω2l2 I 20

12πε 0c3=

p20ω4

12πε 0c3not related to r (11.33)

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Mitschrift Physik III