field and wave electromagneticsocw.snu.ac.kr/sites/default/files/note/3276.pdf · 2018. 1. 30. ·...

Field and Wave Electromagnetics

Chapter 3 Static Electric FieldsChapter 3. Static Electric Fields

3 1 Introduction3-1 Introduction

Coulomb’s law Experimental lawCoulomb’s law Experimental law

1 2ˆ q qF k12

1 212 2

12

ˆRq qF a kR

=

Radio Technology Laboratory 2

3 2 Fundamental Postulates of Electrostatics in Free Space3.2 Fundamental Postulates of Electrostatics in Free Space

Force : (N)FForce :

El i fi ld i i (전계 강도)

(N)F

Electric field intensity (전계 강도)

F0

lim [ V m ] or [ N C]q

FEq→

=

why? - Not to disturb the charge distributionwhy? Not to disturb the charge distribution of the source


3 2 Fundamental Postulates of Electrostatics in Free Space

Two fundamental postulates of electrostatics

3.2 Fundamental Postulates of Electrostatics in Free Space

Two fundamental postulates of electrostatics

E ρε

⎧ ∇ =⎪⎨

(MKS)i ―① 4E πρ∇ =note (cgs)i0

0 E

ε⎨⎪ ∇× = →⎩ irrotational ―②

0

1 v v

Edv dv

Q

ρε

→ ∇ =∫ ∫

∫

i①

0

S

QE d sε

= =∫ Gauss's lawi

0E dl→ =∫ i② 0

( ) 0c

S

E dl

E ds E dl

→ =

∇× = =

∫∫ ∫

i

∵ i i

②


Kirchhoff’s voltage law

3 2 Fundamental Postulates of Electrostatics in Free Space

Conservative field

3.2 Fundamental Postulates of Electrostatics in Free Space

Conservative field

1 2

0C C

E dl E dl+ =∫ ∫i i1 2

2 1

1 2

1 2Along Along

P P

P PC C

E dl E dl= −∫ ∫i i1 2

2

1

2

g g

Along

P

PC

E dl= ∫ i2g

Why?

E dl

E∫

Why?

(Usually depends on integral path)

cf) is irrotational!!

i

∵


Ecf) is irrotational!!∵

3 3 Coulomb’s Law3-3 Coulomb s Law

Coulomb’s lawCoulomb’s law

ˆ ˆ( ) RS S

qE d s R E R ds= =∫ ∫i i0

20

( )

4

RS S

RqE

R

ε

πε=∴

∫ ∫

0

20

4

ˆ ˆ [V m ] or [ N C]4R

RqE R E R

R

πε

πε= =

General Expression

2 3

0 0

( )ˆ4 4

p qpq q R RE aR R R Rπε πε

′−= =

′ ′− −1 2

20

ˆ [N ]4

q qF RRπε

=cf)


3 3 Coulomb’s Law

Charged Shell

3-3 Coulomb s Law

Charged Shell

1 2s ds dsdE ρ ⎛ ⎞= −⎜ ⎟2 2

0 1 2

2 21 2

4

cos

dEr r

d r d r

πε

α

= ⎜ ⎟⎝ ⎠

Ω ⋅ Ω⋅= =solid angle

1 2

cos

0ds ds

dE

α= =

∴ =

solid angle

Discrete ChargesDiscrete Charges

31

( )1 [V m]4

nk k

k

q R RER Rπε

′−=

′∑


104 kkR Rπε = ′−

3 3 1 Electric Field due to a System of Discrete Charges3-3.1 Electric Field due to a System of Discrete Charges

Electric dipolep

2 2d dR Rq

⎧ ⎫⎪ ⎪− +⎪ ⎪

3 30

2 24

2 2

qEd dR R

πε⎪ ⎪

= −⎨ ⎬⎪ ⎪

− +⎪ ⎪⎩ ⎭ 3 333 22 2

2

cos3

,2 2 2 4

Rd

d d d dR R R R R d R d

θ

−− −⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎡ ⎤− = − ⋅ − = − ⋅ +⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎢ ⎥

⎢ ⎥ ⎣ ⎦⎝ ⎠ ⎝ ⎠⎣ ⎦⎡ ⎤3

23 3

2 2

31 1 (binomial expansion)2

R d R dR RR R

−

− −

⎡ ⎤⎡ ⎤ ⎢ ⎥⋅ ⋅

≅ − ≅ +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎢ ⎥

⎣ ⎦2 3( 1) ( 1)( 2)(1 ) 1

2! 3!n n n n n nx nx x x

⎣ ⎦− − −

± = ± ± ±note

33 q R dd R d

−

±

⎡ ⎤ ⎡ ⎤


02 3 2

3 3 312 2 4

q R dE R dR R

d R dR RR πε

− ⎡ ⎤⋅+ ≅ − ∴

⎡ ⎤⋅≅ ⋅ −⎢ ⎥

⎣⎢⎣ ⎦

⎥⎦

3 3 1 Electric Field due to a System of Discrete Charges3-3.1 Electric Field due to a System of Discrete Charges

3 2

( )

1 ˆˆˆ3 , cos sin4

p qd electric dipole moment vector

R pE R p z RR R

θ θ θ

=

⎡ ⎤= ⋅ − = −⎢ ⎥

⎣ ⎦

i

Define

then,

( )( )

3 204

cos

ˆˆ i

R R

R p Rp

R

πε

θ

θ θ θ

⎢ ⎥⎣ ⎦

=i zzθ R

( )( )3

0

cos sin

ˆˆ 2cos sin [V/4

p p R

pE RR

θ θ θ

θ θ θπε

= −

= + m]θ

θ

y

x


3 3 2 Electric field due to a continuous distribution of charge3-3.2 Electric field due to a continuous distribution of charge

Continuous distributionContinuous distribution

2 2'0 0

' 1ˆ ˆ ˆ', where 4 4 v

dv Rd E R E R dv RR R R

ρ ρπε πε

= ⎯→ = =⎯ ∫0 0

3'0

4 4

1 ' [V/m]4 v

R R R

RE dvR

πε πε

ρπε

=∴ ∫

) Surface & line charge density casecf

0

' 20

) Surface & line charge density case1 ˆ ' [V/m]

4s

S

cf

E R dsRρ

πε= ∫

' 20

1 ˆ ' [V/m]4

lL

E R dlRρ

πε= ∫


Ex 3 4 Infinitely long line chargeEx 3-4 Infinitely long line charge

Determine the electric field intensity of an infinitely long,

ˆ ˆ sol) ', ' 'l l

l

R r r z z dl dzρ ρ

ρ= − =

Determine the electric field intensity of an infinitely long, straight, line charge of a uniform density in air

30 0

ˆ ˆ' ' ' 4 4

l ldz dzR r r z zd ER

ρ ρπε πε

−= = 3

2 2 2

ˆ ˆ( ' )

'

r zr dE zdEr z

d

= ++

32 2 2

0

' where 4 ( ' )

' '

lr

r dzdEr z

z dz

ρ

περ

=+

32 2 2

0

( ' )

'

lz

l

z dzdEr z

r dzdE

ρ

περ

= −4 +

⎧⎪ 3

2 2 20

3

4 ( ' ) at ',

' '

lr

l

dEr z

z zz dzdE

ρ

περ

=⎪⎪ +⎪= ⎨ −⎪ =⎪


3'2 2 2

0 ( ' )z z z

dEr zπε

=⎪4 +⎪⎩

Ex 3 4 Infinitely long line chargeEx 3-4 Infinitely long line charge

⎧3

2 2 20

'

4 ( ' ) at ',

' '

lr

rdzdEr z

z zd

ρ

πε

⎧ =⎪⎪ +⎪= − ⎨⎪

3' '2 2 2

0

' '

( ' )'

lz zz z z z

z dzdE dEr z

r dz

ρ

περ ρ

=− =

∞

⎪ = = −⎪

4 +⎪⎩

∫ 32 20 02

ˆ ˆ ˆ [V/m]2

( ' ) l l

rr dzE r E r r

rr z

ρ ρπε πε

∞

−∞∴ = = =

4+

∫

HW. Integrate to have the above result cf) Symmetry Gauss law→


3 4 Gauss’s Law and Applications3-4 Gauss s Law and Applications

C lind ical s mmetCylindrical symmetry

( ) QE E dvρ∇ = ⇒ ∇ =∫i i( )'

0 0

By Divergence theorem

0

v

s

QE ds

ε ε

ε←⎯⎯⎯⎯⎯⎯→ =

∫

∫ i0zE =

0ε

22 ,

L l LE ds E r d dz rLEπ ρφ π= = =∫ ∫ ∫i

rE

0 00

2 ,

where, , 0

r rS

r Z

E ds E r d dz rLE

E rE E

φ πε

= =

∫ ∫ ∫

0zE =

0

2

lrE

rρπε

∴ =


Ex 3 7 Spherical electron cloudEx 3-7 Spherical electron cloud

for 0 R bρ ρ= ≤ ≤⎧ ˆSpherical symmetry i e E r E→ =0 , for 00 , for

R bR b

ρ ρρ=− ≤ ≤⎧

⎨ = >⎩2

Spherical symmetry i.e. i) 0 b

4

r

R R

E r ER

E ds E ds E Rπ

→ =≤ ≤

⋅ = =∫ ∫iS

30 0

4 3

iR RS

v vQ dv dv R

π

πρ ρ ρ= = − = − ⋅

∫ ∫

∫ ∫0

0

ˆ , 03

E R R R bρε

= − ⋅ ≤ ≤∴

3

ii) b4 (fixed)

R

Q bπρ

≥

= 0

30

(fixed)3

ˆ 3

Q b

bE R

ρ

ρε

=

∴

−

−= 2 , R bR

≥


3ε0R

3 5 Electric Potential3-5 Electric Potential

2

1

0 (Vector Identity)

[J/C or V] (Work must be done against the field)P

P

E E VW E dlq

∇× = → =−∇

=− ⋅∫2 2 2 2

1 1 1 12 1

ˆ [V] ( ) ( )P P P P

P P P P

q

V V E dl E dl V l dl dV− =− ⋅ − ⋅ =− −∇ =∫ ∫ ∫ ∫∵ i

Remember when we define gradient

ˆ ˆˆ cos ( )dV dV dn dV dV n l V ldl dn dl dn dn

α= = = = ∇i i i

( )dV V dl=∴ ∇ i

) Refernece point of potential: the zero potential point is usually at infinitycf ) Refernece point of potential: the zero potential point is usually at infinity.cf


3 5 1 Electric potential due to a charge distribution3-5.1 Electric potential due to a charge distribution

Point charge at origin potential at R w r t that at infinityPoint charge at origin, potential at R w.r.t. that at infinity

20

ˆ4

qE RRπε

=

20 0

ˆ ˆ( ) [V]4 4

R q qV R R dRR Rπε πε∞

⎛ ⎞=− =⎜

⎠∴ ⎟

⎝∫ i

V21 : Potential difference between point P1 and P2

2 121 P P0 2 1

1 14

qV V VR Rπε

⎛ ⎞= − = −⎜ ⎟

⎝ ⎠


3 5 1 Electric potential due to a charge distribution3-5.1 Electric potential due to a charge distribution

No work ˆ

ˆ

r

dl R d

F F r

φ φ=

=

iNo work

0F dl =∴ i

1 [V]n

kqV ∑ '10

[V]4

k

k k

VR Rπε =

=−

∑


Ex 3 8 An Electric dipoleEx 3-8 An Electric dipole

1 1q ⎛ ⎞

0

11

1 14

1 cos 1 cos2 2

qVR R

d dR RR R

πε

θ θ

+ −

−−

⎛ ⎞= −⎜ ⎟

⎝ ⎠⎧ ⎛ ⎞ ⎛ ⎞− ≅ +⎪ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎪1

1

2 2 ,

1 cos 1 cos2 2

R Rd R

d dR RR R

θ θ

+

−−

−

⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎪

⎨⎛ ⎞ ⎛ ⎞⎪ + ≅ −⎜ ⎟ ⎜ ⎟⎪ ⎝ ⎠ ⎝ ⎠⎩

2 20 0

cos 1 ˆ , where : dipole moment vector4 4

qdV p R p qdR R

E

θπε πε

= = ⋅ =

= −∇ ( )3ˆ ˆˆ ˆ2cos sin

4V V pV R RR R R

θ θ θ θθ

∂ ∂= − − = +

∂ ∂ ( )30

1 2 31 1 2 2 3 3

4

ˆ ˆ ˆ )

R R RV V Vcf V u u u

h u h u h u

θ πε∂ ∂∂ ∂ ∂

∇ = + +∂ ∂ ∂

2

Note

1. Equipotential line : cos

2 fi ld li iVR c

E R

θ

θ

=


2 2. field line : sinEE R c θ− =

Potential due to continuous charge

'

1 '4 v

V dvRρ

πε= ∫ [V] : volume charge distribution

0

'0

41 '

4

v

s

R

V dsR

περ

πε=

∫

∫ [V] : surface charge distribution

'

0

1 '4

lL

V dlRρ

πε= ∫ [V] : line charge distribution


Ex 3 9 A uniformly charged diskEx 3-9 A uniformly charged disk

Obtain a formula for the electric field intensity2 2

2

1

y

sol) ' ' ' ' and '' ' '

bs

ds r dr d R z rrV dr d

π

φρ φ

= = +

= ∫ ∫( )

( )

10 02 20 2

12 2 2

'

'b

s

z r

z r z

φπε

ρ

4+

⎡ ⎤= +⎢ ⎥

∫ ∫

( )0 0

2

z r zε

= + −⎢ ⎥⎣ ⎦

'

( )2 2 2

12 2 2

2 2

'cf) ', substitute ''

' 0

r dr z r kz r

k k

+ =+

⎫

∫

( )( )

( )1

2 2 2

2 2

2 2 2 2 2 21

2 2

' 0,

' ,

2 ' ' 2

z b

z

r k z k z

k dkr b k z b k z b dk kkd k dk

1+

⎫= = → =⎪⎪= = + → = + ⇒ = =⎬⎪⎪

∴ ∫ ∫


( ) 2 ' ' 2 kr dr k dk ⎪= ⎪⎭

Ex 3 9 A uniformly charged diskEx 3-9 A uniformly charged disk

1⎧ ⎡ ⎤( )1

2 2 2

0

1

ˆ 1 , 02

ˆ

sz z z b zVE V zz

ρε

−⎧ ⎡ ⎤− + >⎪ ⎢ ⎥

∂ ⎪ ⎣ ⎦= −∇ = − = ⎨∂ ⎡ ⎤⎪ ( )1

2 2 2

0

ˆ 1 , 02

sz

z z z b zρε

−∂ ⎡ ⎤⎪− + + <⎢ ⎥⎪ ⎣ ⎦⎩

220

ˆ , 04

ˆFor very large s

NoteQz z

zbz E zπεπ ρ

⎧ >⎪⎪⎨2

02

0

For very large , 4 ˆ , 0

4

sz E zQz z z

zπε

πε

= = ⎨⎪ − <⎪⎩


3 6 Conductors in Static Electric Field3-6 Conductors in Static Electric Field

Conductor : Free ElectronsThe electrons in the outermost shells of the atoms of conductors are free to move

Insulator or dielectricsThe electrons in the atoms of insulators are confined to their orbits: no free electronselectrons

SemiconductorsA relatively small number of freely movable charge carriersA relatively small number of freely movable charge carriers→ Allowed energy bands for electrons. → Forbidden band → Band gap


3 6 Conductors in Static Electric Field

Charge introduced → Field will be set up → Charges will move away

3-6 Conductors in Static Electric Field

Charge introduced Field will be set up Charges will move away from each other to reach conductor’s surface until the charge and the fieldinside vanish

: Inside a conductor 0, 0Eρ = =

Under static conditions, the E field on a conductor surface is everywhere normal to the surface → Equipotential surface (No tangential component)

0 ( E inside conductor is zero)tabcdaE dl E w= Δ =∫ i ∵

0abcda

t

sn

EsE ds E s ρ

∴ =

Δ= Δ =

∫

∫ i0

0

ns

snE

ερε

=∴

∫


3 6 Conductors in Static Electric Field

Boundary conditions at a conductor / Free space interface

3-6 Conductors in Static Electric Field

Boundary conditions at a conductor / Free space interface

0tEρ

=⎧⎪⎨ ( 0 )E field insideconductor=∵

0

snE ρ

ε⎨ =⎪⎩

( 0 )E field insideconductor=∵

Uncharged conductor in a static fieldElectrons move in a direction opposite to that of the field- Electrons move in a direction opposite to that of the field

- Positive charges move in the direction of the field- So that induced free charge will distribute on the conductor surface and create an

induced field in such a way to cancel the external field inside the conductor andinduced field in such a way to cancel the external field inside the conductor and tangent to its surface


Ex 3 11 A point charge +Q at the center of a conducting shellEx 3-11 A point charge +Q at the center of a conducting shell

Determine and as functions of the radial distanceE V R

0

Determine and as functions of the radial distance "Ungrounded"

ˆ (a) Spherical Symmetry R

E V R

R R E E R> =∵

( )

21

0

1 1 12

4

,

RS

R

R R

QE ds E R

Q QE V E dR

πε

⋅ = =

= = −∴ =

∫

∫ ( )

( )

1 1 120 0

0

,4 4

b , Ins

R R

i

R RR R R

πε πε∞

<<

∫

2

ide a conductor

0 0RE E= ∴ =

0

2

2 10 0

, whole conducting shell is an equipotential body4

R

R R

QV VRπε=

= =

An amount of negative charge equal to must be induced on the inner shell surface at in order to cancel out the field induced by positive charge at center. i

NoteQ

R R Q−

= +


This implies again that + 0must be induced on the outer shell surface at .Q R R=

Ex 3 11 A point charge +Q at the center of a conducting shellEx 3-11 A point charge +Q at the center of a conducting shell

( ) R R<

3 3 320 0

(c)

,4 4

i

R R

R RQ QE V E dR C C

R Rπε πε

<

= = − + = +∫3 2 0

30 0 0 0

where, at is equal to at

1 1 1 1 1 4 4

i

i i

V R R V R R

Q QC VR R R R Rπε πε

= =

⎛ ⎞ ⎛ ⎞= − ∴ = + −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠0 0 0 0i i⎝ ⎠ ⎝ ⎠

Electric field intensity and potential variations of A point charge +Q at the center of a conducting shell


field and wave electromagneticsocw.snu.ac.kr/sites/default/files/note/3276.pdf · 2018. 1. 30. ·...

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