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19. Electric Forces and Electric Fields19. Electric Forces and Electric Fields

• Newton’s Law0=F

r⇒ 0=∆vr

amF rr=

원인

1. 관성의법칙

2. 운동의법칙

3. 작용반작용의법칙결과

2112 FFrr

=

• Force in NatureMass1. Gravitational Force

2. Electromagnetic Force

3. Strong Force

4. Weak Force

Charge

19-1. Historical Overview

• Ancient Period ---- Charge

---- Current vs Magnetic Field • Oested

---- Electromagnetism --- Phenomenology • Faraday

---- Electromagnetism 의완성• Maxwell

Maxell’s Equations

cf) Newton’s Law

19-2. Properties of Electric Charges

• Two kinds of charges (positive and negative) in Nature

+ + : Repulsive– – : Repulsive+ – : Attractive– + : Attractive

• The force : proportional to the inverse square of the separation.

21 rF ∝ ⇒ Coulomb’s Law

• Charge : Conserved & Quantized

Sum of Charges = 0

Charge:

⋅⋅⋅⋅±±±= ,e,e,eQ 32

C.e 191061 −×=Electron has charge of -e.

19-3. Insulators and Conductors

• Materials

Metals (Conductors) : free movement of electric chargese.g. Fe, Cu, Au, Ag, ···

Insulators : no free movement of electric chargese.g. Rubber, Ceramic, Cotton, ···

Semiconductor : Insulator but the electric charges can be created either by the thermal energy or by dopings

e.g. Si, GaAs, ···

• Charging by Induction

Grounded

Conductor

Charges are uniformly distributed at the surface of conductors.

Induced Charge due to polarization in insulators.

19-4. Coulomb’s Law

• Gravitational Force : between two massive objects

221

rmmGFG −=2

21

rmmFG −∝ G = 6.67×10-11N·m2/kg2⇒

Always attractive

• Electrical Force (Coulomb’s Law) : between two charged particles

221

rqqFC ∝ 2

21

rqqkF eC =⇒ ke= 8.99×10-11N·m2/C2

041πε

=ek2212

0 1085428 mNC. ⋅×=ε −

Permittivity of vacuum

Charge of an electron or proton : |e| = 1.602 × 10-19 C= (6.25 × 1018)-1C

• Vector notation of Coulomb’s Law

r̂rqqkF eC 2

21=r

12221

21 r̂rqqkF e=

r

21221

12 r̂rqqkF e=

r

1221 r̂r̂ −=

1221 FFrr

−= : Newton’s third law

The magnitude is the same but the direction is opposite.

• Coulomb Force due to many charges

+ –

+−

q1 q2

q3q4

12Fr13F

r

14Fr

Vector Sum

∑=++=i

iFFFFF 11413121

rrrrr

Example 19.2 Hydrogen Atom

+- e

p

Average separation r = a0 = 5.3 × 10-11m|e| = 1.60 × 10-19Cmp = 1.67 × 10-27kg , me = 9.11 × 10-31kg

( )( )211

219

2

29

221

10351060110998

m.C.

CmN.

rqqkF ee −

−

×

×⋅×==

Electrical force

N. 81028 −×=

Gravitational force

( )211

2731

2

211

221

103510671101191076

m.kg.kg.

kgmN.

rmmGFG −

−−−

×

×⋅×⋅×==

N. 471063 −×=

19-5. Electric Fields

Electric Field : Electric force per unit charge acting on a test charge

0qFEr

r≡

SI unit : N/C

For a point charge q:

Electric force on a test charge

r̂r

qqkF e 20=

r

r̂rqkE e 2=

rElectric Field:

Force ⇒ Field : Often much easier to describe the physical phenomena

For a group of charges : Vector sum

∑=i

ii

ie r̂

rqkE 2

rElectric Field:

Example 19.3: Electric Field of a dipole

221221 r̂rqkr̂

rqkEEE ee −=+=

rrr

21 EE =

( )θ−+θ= sinEsinEEy 21

011 =θ−θ= sinEsinE

θ=θ+θ== cosEcosEcosEEEx 121 2

θ= cosrqkE e 22

( ) ( )

( ) 2322

212222

2

2

ayqak

aya

ayqk

e

e

+=

+⋅

+=

For r >> a:

3

2yqakE e=

3

1r

∝

• Electric Field Due to Continuous Charge Distributions

Electric field at P due to charge ∆q

r̂rqkE e 2

∆=∆

r

Total Electric field at P

∑∆

=i

ii

ie r̂

rqkE 2

r

∆qi → 0 for the continuous charge distributions:

r̂rdqkr̂

rqkE e

ii

i

i

qe lim

i

∫∑ =∆

=→∆

220

r

• Charge Density for uniform distributions

- Charge desity , ρ : Charge per unit volumes

r̂rdVkE e ∫

ρ= 2

r

VQtot≡ρ ,

- Surface charge density, σ : Charge per unit area

r̂rdSkE e ∫

σ= 2

r

AQtot≡σ ,

- Linear charge density, λ : Charge per unit length

r̂rdlkE e ∫

λ= 2

r

LQtot≡λ ,

Example 19.4 Electric Field due to a Charged Rod

Total Charge Q

Linear charge density

lQ

=λ

∫∫+ λ

=λ

=ld

dee x

dxkrdlkE 22

( )dldlk

dldk

xk e

e

dlx

dxe +

λ=

+−λ=

−λ=

+=

=

111

( )dldQkE e

+=

2dQkE e≈If d >> l, : like a point charge

Example 19.5 Electric Field due to an Uniform Ring Charge

aQπ

=λ2

Total Charge Q Linear charge density

∫∫λ

== r̂rdlkEdE e 2

rr

xx dEdE 21 =,dEdE ⊥⊥ −= 21

( ) ( ) ( ) 23222122222 axxdqk

axx

axdqkcos

rdqkdE ee

ex+

=+

⋅+

=θ=

( ) ( ) Qaxxkdq

axxkE ee

x 23222322 +=

+= ∫

If x = 0, E = 0.

2xQkE e≈If x >> a, : like a point charge

19-6. Electric Field Lines

• Electric Field lines : Visualize the electric field patterns

1. E is tangent to the electric field lines

Er E

r

2. Strength of E ∝ the number of lines per unit area

• Rules for drawing Electric field lines

1. Begin on positive charges and terminate on negative charges.

2. The number of lines ∝ the magnitude of charges

3. No two lines cross each otherQAE ∝⋅

19-7. Electric FluxElectric Flux Φ

AE ⋅=Φ

: ∝ the number of the electric field lines through a surface

Electric Flux Φ is defined as

θ⋅=⋅≡Φ cosAEAErr

Ar

:Directed to the surface normalA

r

Electric flux, ∆Φi, through a small surface ∆Ai

iii AErr

∆⋅=∆Φ

A general definition of Φ :

∫∑ ⋅=∆⋅≡Φ→∆ Surfacei

iiA

AdEAElimi

rrrr

0

19-8. Gauss Law

• Electric Flux through a closed surface

∫ ⋅=Φ AdEC

rr

(i) No Charge inside

Er

Er

2Ar

1Ar 21 AEAEC

rrrr⋅+⋅=Φ

0=⋅+⋅−= AEAE

0=⋅=Φ ∫ AdEC

rr

(i) A charge inside

∫∫ ⋅=⋅=Φ21 SS

C AdEAdErrrr

S1

S2

qkrrqk ee π=π⋅= 44 2

2

041πε

=ek

0ε=Φ

qC

For an empirical surface, ΦC is the same. i.e. ΦC is independent of the surface contour.

• Gauss’s Law

0ε=⋅=Φ ∫ inc

CqAdE

rr

The net flux through any closed surface is proportional to the net charge inside of the surface.

19-9. Applications of Gauss’s Law

• Gauss’s law is very useful to determine the electric field due to charge distributions with high degree of symmetries, such as spherical, cylindrical, or long plane shapes.

Example 19.7 Electrical Field due to a point charge

0ε=⋅=Φ ∫

qAdEC

rr

0

24ε

=⋅πqEr

204

1rqE

πε= r̂

rqE 2

041πε

=⇒r

Example 19.8 A Spherically Symmetric Charge Distribution

Charge density ρ

343

aQ

π=ρ⇒Qa =ρπ 3

34

(a) r >a

00 ε=

ε=⋅=Φ ∫

QqAdE incC

rr

0

24ε

=⋅πQEr

2204

1rQk

rQE e=

πε=⇒ for r > a

(b) r <a

0ε=⋅=Φ ∫ inc

CqAdE

rr

3

0

0

34

1

r

dV

περ

=

⋅ρε

= ∫

3

3

0

24arQEr

ε=π 33

04 aQrk

arQE e=

πε=⇒

π=ρ 34

3aQ for r < a

Example 19.9 A Cylindrically Symmetric Charge Distribution+++++++++

xy

dy

θ

( ) 232222 yxxdykcos

yxdykdE e

ex+

λ=θ

+λ

=

( )∫∫∞+

∞−

∞+

∞− +

λ== dy

yxxkdyEE e

x 2322

0ε=⋅=Φ ∫ inc

CqAdE

rr

0

2ελ

=⋅πlErl

rk

rE e

λ=

λπε

= 22

1

0

Example 19.10 A Non-conducting Plane Sheet of Charge

0ε=⋅=Φ ∫ inc

CqAdE

rr

0

2ε⋅σ

=⋅AEA

02εσ

=E Constant i.e. Uniform field

19-10. Conductors in Electrostatic Equilibrium

• Charges freely move inside Conductors. • Excess charges entirely on its surface • E = 0 inside of Conductors

0εσ

=⋅=⋅=Φ ∫AAEAdEC

rr

0εσ

=E

For r < R, + +

++

++

+

++

++

++

+ ++

+++

R

r

Gauss Surface

0=⋅=Φ ∫ AdEC

rr

0=⇒ E

For r > R,

0ε=⋅=Φ ∫

QAdEC

rr

0

24ε

=π⋅QrE 2

041

rQE

πε=⇒