電 磁 學

Textbook : Field and Wave Electromagnetics

Author: David K. Cheng (偉明圖書)

References :
1. Applied Electromagnetics: Early Transmission Lines Approach ( Stuart M. Wentworth)
2. Introductory Electromagnetics ( Zoya Popovic’ and Branko D. Popovic’)
3. Engineering Electromagnetic ( Umran S. Inan and Aziz S. Inan)

Contents
期 中 考 期 末 考
1. The Electromagnetic Model 3. Static Electric Fields (cont.)
9. Theory of Transmission Lines and 4. Solution of Electrostatic Problems
Smith Chart 5. Steady Electric Currents
2. Vector Analysis 6. Static Magnetic Fields
3. Static Electric Fields
Office hours: Mon. periods 5&6; Tue. periods 5&6; Wen. periods 5&6
Chapter 1 The Electromagnetic Model

# Electromagnetic: the study of the effects of electric charges at rest

and in motion.

# Field: a spatial distribution of a scalar or vector quantity

# Basic conception :
1. Charges (positive or negative) at rest  static electric fields
2. Charges in motion  a time-varying electric field
accompanied by a magnetic field
3. Time-dependent electromagnetic fields electromagnetic waves
(i.e. time-dependent electric field and time-dependent magnetic field)
1 /4
# Definitions of charge densities:

1.  (or v )  im qv ( c m3 ) : volume charge density

v  0

where q is the amount of charge in a very small volume v .

2.  s  im qs ( c m 2 ) : surface charge density

s  0

where q is the amount of charge in a very small element of

surface s .

q
3.   im  ( c m) : line charge density
 0

where q is the amount of charge in a very short element of line  .

2 /4
# Definition of current densities:
*A current must flow through a finite area; hence it is not a point function.*

 I A
1. J S  im ( ) : surface current density.
l  0 
m

  I
2. J (or J v )  im ( A 2 ) : volume current density
s  0 s m
# Four fundamental vector field quantities in electromagnetics :
Symbols and units Field quantity Symbol Unit
for Field quantities
Electric Electric field intensity  V/m
E
Electric flux density  C/m2
(Electric displacement) D
Magnetic Magnetic field intensity  A/m
H
Magnetic flux density  T or
B Wb/m2
3 /4
# In free space (or vacum):
1. C : velocity of electromagnetic wave  C = 3  108 m / sec
2.  0 : permittivity of free space  0 
10 9
36
( F/m )

3.  : 0

permeability of free space  0  4  10 7
(H/m)

from 1, 2 and 3, we obtain:

1
C
0 0

 
Meanwhile, in free space (or vacum) relationship between D and E is
 
D   0E ,
   
whereas B and H is B  0 H

# In a medium (i.e. not a free space or vacum):

 
D   E and B  H , here    r 0 and   r 0
 

4 /4
 Chapter 9 Theory and Application of Transmission Lines (T.L.)
9.1 Introduction
For effective point-to-point transmission of power and information the source
energy must be directed or guided. Usually, the transmission line (TL) is used to
guide the transverse electromagnetic (TEM) wave.
◆ Types of Transmission Lines

(a) parallel-plate TL (b) Two-wire TL (c) coaxial TL (d) microstrip-line TL (e) microstrip-line TL
(microwave) (low frequency) (high frequency) (microwave) (microwave)

9.3 General Transmission Line Equations

z z+∆z
AC

Transmission line

R: Ω /m
G: Ω-1 /m
L: H /m
C: F /m

Transmission-line circuit model

from KVL :
∂i ( z , t ) ∂i ( z , t )
v( z , t ) = L∆z + R∆zi ( z , t ) + v( z + ∆z , t ) ⇒ v( z + ∆z , t ) − v( z , t ) = −∆z[ L + Ri ( z , t )]
∂t ∂t
v( z + ∆z , t ) − v( z , t ) ∂i ( z , t ) ∂v( z , t ) ∂i ( z , t )
⇒ = −[ L + Ri ( z , t )], if ∆z → 0 ⇒ = −[ L + Ri ( z , t )].........(1)
∆z ∂t ∂z ∂t
∂v( z + ∆z , t )
from KCL : i ( z , t ) = C∆z + G∆zv ( z + ∆z , t ) + i ( z + ∆z , t )
∂t
∂v( z + ∆z , t )
⇒ i ( z + ∆z , t ) − i ( z , t ) = −∆z[C + Gv( z + ∆z , t )]
∂t
i ( z + ∆z , t ) − i ( z , t ) ∂v ( z + ∆z , t )
⇒ = −[C + Gv( z + ∆z , t )],
∆z ∂t
1
∂i ( z , t ) ∂v ( z , t )
if ∆z → 0 ⇒ = −[C + Gv( z , t )].........(2)
∂z ∂t
◆ Definition of Phasor:
if v(t)=V cos (ωω+ φ)=Re[Ve j (ωt +φ ) ] = Re[Ve jφ e jωt ], thus, Ve jφ is called the " phasor" of v(t).
dv(t ) dv(t ) d
⇒ For the " phasor" of , it is jωVe jφ . ( = Re[Ve jφ e jωt ] = Re[jωVe jφ e ] jωt ])
dt dt dt

◆ Derivations of V(z) and I(z) in phasor domain:

∂v( z, t ) ∂i( z, t ) dV ( z )
for eq. (1) : = −[ L + Ri( z, t )] ⇒ = −[ LjωI ( z ) + RI ( z )]
∂z ∂t dz
dV ( z )
⇒ = −( R + jωL) I ( z ) ..........(3)
dz
∂i( z, t ) ∂v( z, t ) dI ( z )
for eq. (2) : = −[C + Gv( z, t )] ⇒ = −[CjωV ( z ) + GV ( z )]
∂z ∂t dz
dI ( z )
⇒ = −(G + jωC )V ( z )...........(4)
dz
d dV ( z ) d d 2V ( z ) dI ( z )
for eq. (3) : [ ] = [ −( R + jωL) I ( z )] ⇒ 2
= −( R + jωL)
dz dz dz dz dz
2
d V ( z) dI ( z )
⇒ 2
= −( R + jωL)[−(G + jωC )V ( z )] ( = −(G + jωC )V ( z ).)
dz dz
d 2V ( z )
⇒ − ( R + jωL)(G + jωC )V ( z ) = 0
dz 2
d 2V ( z )
Let r = ( R + jωL)(G + jωC ) ⇒
2
2
− r 2V ( z ) = 0 ........(5)
dz
2
d V ( z) d 2V ( z )
for eq. (5) : Let V(z) = V0 e ⇒ sz
2
= S V0 e ⇒
2 sz
2
− r 2V ( z ) = ( S 2 − r 2 )V0 e sz = 0
dz dz
+ − rz − rz
⇒ S = − r and + r ⇒ V ( z ) = V0 e + V0 e
here, V0+ e − rz : forward part of V(z) ; V0− e rz : backward part of V(z) ; r : propagation constant of wave
Similarily,
d dI ( z ) d d 2 I ( z) dV ( z )
for eq. (4) : [ ] = [ −(G + jωC )V ( z )] ⇒ 2
= −(G + jωC )
dz dz dz dz dz
2
d I ( z) dV ( z )
⇒ 2
= −(G + jωC )[−( R + jωL) I ( z )] ( = −( R + jωL) I ( z ).)
dz dz
d 2 I ( z) d 2 I ( z)
⇒ 2
− (G + j ω C )( R + j ω L ) I ( z ) = 0 ⇒ 2
− r 2 I ( z ) = 0 ........(6)
dz dz
2 2
d I ( z) d I ( z)
for eq. (6) : Let I(z) = I 0 e sz ⇒ 2
= S 2 I 0 e sz ⇒ 2
− r 2 I ( z ) = ( S 2 − r 2 ) I 0 e sz = 0
dz dz
+ − rz − rz
⇒ S = − r and + r ⇒ I ( z ) = I 0 e + I 0 e
here, I 0+ e − rz : forward part of I(z) ; I 0− e rz : backward part of I(z)

2
整理 正向波

V ( z ) = V0+ e− rz + V0−erz V(z)

逆向波
I ( z ) = I 0+ e− rz + I 0−erz
here, r = ( R + jωL)(G + jωC ) = α + jβ Z

r : propagation constant (傳播常數) (unit : m-1 ) V(z)= +

α : attenuation constant (衰減常數) (unit : Np/m) 正向波 逆向波
+ − (α + j β ) z + − α z − jβ z
β : phase constant (相位常數) (unit : rad/m) 說明： V e 0
+ − rz
=V e 0 =V e e
0

meanwhile, 影響振幅 影響相位

dV ( z ) d
 = [V0+ e −rz + V0− e rz ] = -rV0+ e −rz + rV0− e rz .... (A)
dz dz
and from eq. (3) :
dV ( z )
= −( R + jωL) I ( z ) = −( R + jωL)[ I 0+ e − rz + I 0− e rz ] = −( R + jωL) I 0+ e − rz − ( R + jωL) I 0− e rz .....(B)
dz
Comparing (A) with (B), we obtain
V0+ R + jωL R + jωL R + jω L
-rV0+ = −( R + jωL) I 0+ ⇒ = = = ≡ Z0
I 0+ r ( R + jωL)(G + jωC ) G + jω C
V0− R + jωL R + jωL V0+
rV0− = −( R + jωL) I 0− ⇒ − = = = Z ( = ) 注意：電流有方向性
I 0− G + jωC I 0+
0
r
here, Z 0 is defined as the " characteristic impedance" of the transmission line.
V0+ V0− R + jωL
整理： Z0 ≡ = − = (電流有方向性)
I0+
I0−
G + jωC

9.4 Wave Characteristic on Finite Transmission Lines

z
A steady-state,
single-frequency, Z0 ZL
time-harmonic
source
Z=0
If Z L is the load of the TL with the location of z = 0,
VL = V ( z = 0) = V0+ e − r ( 0 ) + V0− e r ( 0 ) = V0+ + V0−
I L = I ( z = 0) = I 0+ e − r ( 0 ) + I 0− e r ( 0 ) = I 0+ + I 0−
VL V0+ + V0− V0+ 1 + V0− / V0+ 1+ Γ Z L − Z0
and, Z L = = + −
= + ⋅ − +
= Z0 or Γ=
IL I0 + I0 I0 1+ I0 / I0 1- Γ Z L + Z0
here, Γ = V0− / V0+ = −( I 0− / I 0+ ) is the reflection coefficiency (反射係常數) at " load".
Case discussion :
Z L − Z0 0 − Z0
Case 1 : if Z L = 0 (i.e. shorted load) ⇒ Γ = = = -1 : 反相完全反射
Z L + Z0 0 + Z0
Z L − Z0 ∞ − Z0
Case 2 : if Z L → ∞ (i.e. open load) ⇒ Γ = = = 1 :同相完全反射
Z L + Z0 ∞ + Z0
Z L − Z0 Z0 − Z0
Case 3 : if Z L = Z 0 (i.e. matching load) ⇒ Γ = = = 0 : 匹配(無反射) ∴ 0 ≤ Γ ≤ 1 3
Z L + Z0 Z0 + Z0
◆ For any point on the transmission line

+ I
V  Z0 ZL
-
Z= −  Z=0
For any point on the TL, z = −,
V = V ( z = −) = V0+ e − r ( −  ) + V0− e r ( −  ) = V0+ e r + V0− e − r
I  = I ( z = −) = I 0+ e − r ( −  ) + I 0− e r ( −  ) = I 0+ e r + I 0− e − r
1 1
(Note : average power : ( Pav )  = Re[V ⋅ I  ]; ( Pav ) L = Re[V L ⋅ I L ])
* *

2 2
−
V
+ r − − r + r
1 + 0+ e − 2 r
V V V e + V0 e V0 e V0 1 + Γe − 2 r 1 + Γ
then, Z  = = ( ) z =− = 0+ r − − r
= + r
⋅ −
= Z − 2 r
= Z0
I0 e + I0 e 1 − Γe 1 - Γ
0
I I I0 e I
1 + 0+ e − 2 r
I0
Z  - Z0
here, Γ = Γe − 2 r = is the reflection coefficiency (反射係數) at z = -
Z  + Z0

◆ Another form for Z 

For any point on the TL, z = −,
V0− − r Z - Z 0 − r
e r +
e e r + L e
V V0+ e r + V0− e − r V0+
V0 + r
e + Γe − r
Z L + Z0
Z = = + r = + ⋅ = Z 0 r = Z0
I I 0 e + I 0− e − r I0 I 0− − r e − Γe − r Z - Z 0 − r
r
e + +e e r − L e
I0 Z L + Z0
( Z L + Z 0 )e r + ( Z L - Z 0 )e − r Z L (e r + e − r ) + Z 0 (e r − e − r )
= Z0 = Z0
( Z L + Z 0 )e r − ( Z L - Z 0 )e − r Z L (e r − e − r ) + Z 0 (e r + e − r )
sinh( r)
ZL + Z0
Z [2 cosh(r)] + Z 0 [2 sinh( r)] cosh(r) Z + Z 0 tanh(r)
= Z0 L = Z0 = Z0 L ....(9.103)
Z L [2 sinh( r)] + Z 0 [2 cosh(r)] sinh( r) Z L tanh(r) + Z 0
ZL + Z0
cosh(r)
e x − e -x e x + e -x sinh( x)
( Note : sinh (x) = ; cosh (x) = , tanh(x) = )
2 2 cosh( x)
Case discussion :
For a lossless (無損耗) transmission line, the R and G are both equal to 0. Therefore,
r = (R + jωL)(G + jωC) = (jωL)(jωC) = jω LC = α + jβ
ω 2πf 2π
⇒ α = 0; β = ω LC = = = (背) (Note : u = 1/ LC : velocity nof wave)
u fλ λ
e jβ − e -jβ
sinh( r) sinh( jβ) jsin( β)
for this case (lossess TL) : tanh(r) = = = jβ 2 -jβ  = = j tan( β)
cosh(r) cosh( jβ) e +e cos( β)
2
Z + Z 0 tanh(r) Z + jZ 0 tan( β) Z cos( β) + jZ 0 sin( β)
∴ Z = Z0 L = Z0 L = Z0 L . 4
Z L tanh(r) + Z 0 jZ L tan( β) + Z 0 jZ L sin( β) + Z 0 cos( β)
Example: For a lossless TL with a characteristic impedance (Z 0 ) of 50 Ω and the load (Z L )
of 25-j25 Ω, please find (1) Γ at load; (2) Z  at =λ/4; and (3) Γ at  =λ/4
 =λ/4
+ I
V  Z0=50Ω ZL =25-j25Ω
-
Z= −  Z=0

(Sol.)
For a lossless TL, r = α + jβ = jβ (  α = 0)
Z L - Z0 (25 - j25) - 50 - 25 - j25 - 1 - j − 1 2 5
(1). Γ == = = = = −j = ∠ tan −1 2
Z L + Z 0 (25 - j25) + 50 75 - j25 3- j 5 5 5
2π λ
(25 − j 25) + j 50 tan(
⋅ )
Z L + jZ 0 tan( β) λ 4 ≅ 50 50
(2). Z  ( = λ / 4) = Z 0 = 50
jZ L tan( β) + Z 0 2π λ 25 − j 25
j (25 − j 25) tan( ⋅ ) + 50
λ 4
100
= = 50 + j 50 (note : β = 2π / λ ; tan(π/2) → ∞)
1− j
Z  − Z 0 (50 + j 50) − 50 j 1 + j2 5
(3). Γ ( = λ / 4) = = = = = ∠ tan −1 2
Z  + Z 0 (50 + j 50) + 50 2 + j 5 5
−1 2 −1 2 −2 j 2π ⋅ λ
another method : Γ ( = λ / 4) = Γe − 2 r = ( − j ) e − 2 jβ  = ( − j ) e λ 4
5 5 5 5
− 1 − j 2 − jπ 1 + j 2 5
= e = = ∠ tan −1 2
5 5 5

Case discussion for a lossless TL:

Case 1: Short-circuit termination (Z L =0)

+ I Z + Z 0 j tan( β)
V  Z0 ZL =0 Z S = Z  ( Z L = 0) = Z 0 L = jZ 0 tan( β)......(A)
- Z L j tan( β) + Z 0
Z= −  Z=0

Case 2: Open-circuit termination (Z L →∞)


+ I Z L + Z 0 j tan( β) Z0
ZL →∞ Z  = Z  ( Z L → ∞) = Z 0 =
O
V  Z0 ......(B)
- Z L j tan( β  ) + Z 0 j tan( β  )
Z= −  Z=0
Z0
From (A) and (B) ⇒ Z S ⋅ Z O = jZ 0 tan( β) ⋅ = Z0 or Z 0 = Z S ⋅ Z O
2

j tan( β)
Case 3: For a length of λ/4 (quarter-wave line):
 =λ/4 Z + Z 0 j tan( β)
Z in = Z  ( = λ / 4) = Z 0 L
+ I Z L j tan( β) + Z 0
V  Z0 ZL
- 2π λ
Z L + Z 0 j tan( ⋅ ) 2
Z= −  Z=0 λ 4 ≅ Z 0 or Z ⋅ Z = Z 2
= Z0
2π λ ZL
in L 0
Z L j tan( ⋅ ) + Z 0 5
λ 4
Sandwitch rule
 Case 4: For a length of λ/2 (half-wave line): Z + Z 0 j tan( β)
 =λ/2 Z in = Z  ( = λ / 2) = Z 0 L
Z L j tan( β) + Z 0
+ I
V  Z0 ZL 2π λ
Z L + Z 0 j tan( ⋅ )
-
= Z0 λ 2 =Z
Z= −  Z=0
2π λ L
Z L j tan( ⋅ ) + Z 0
λ 2
Case 5: For a length of nλ/2 : Z + Z 0 j tan( β)
 =nλ/2 Z in = Z  ( = nλ / 2) = Z 0 L
Z L j tan( β) + Z 0
+ I
V  Z0 ZL 2π nλ
Z L + Z 0 j tan( ⋅ )
-
= Z0 λ 2 =Z
Z= −  Z=0
2π nλ L
Z L j tan( ⋅ ) + Z 0
λ 2
Example 9.6 The open-circuit and short-circuit impedances measured at the input
terminals of a lossless transmission line of length 1.5 m, which is less than a quarter
wavelength, are –j54.6 Ω and j103 Ω, respectively.
(a) Find Z0 and r of the line.
(b) Without changing the oprtating frequency, find the input impedance of a
short-circuited line that is twice the given length.
(c) How long should the short-circuited line be in order for it to appear as an open
circuit at the input terminals ?
(Sol.)
For a lossless TL, r = α + jβ = jβ (  α = 0); tanh(r) = j tan( β)
Z + jZ 0 tan( β) Z0
(a). Z o = Z  ( Z L → ∞) = Z 0 L = = − j 54.6(Ω)...( A)
jZ L tan( β) + Z 0 j tan( β)
Z L + jZ 0 tan( β)
Z S = Z  ( Z L = 0) = Z 0 = jZ 0 tan( β) = j103(Ω)......( B)
jZ L tan( β) + Z 0
∴ Z 0 = Z o ⋅ Z S = (− j 54.6)( j103) ⇒ Z 0 = 54.6 × 103 ≈ 75(Ω)
2

103
 Z S = jZ 0 tan( β) = j103(Ω) ⇒ j 75 tan( β × 1.5) ⇒ β × 1.5 = tan −1 ( ) ≈ 0.941
75
∴ β = 0.627 ⇒ r = jβ = j 0.627
Z L + jZ 0 tan( β)
(b). Z S = Z  =3 ( Z L = 0) = Z 0 = jZ 0 tan( β) = j 75 tan(0.628 × 3) = − j 231(Ω)
jZ L tan( β) + Z 0
Z L + jZ 0 tan( β)
(c) Z S = Z  ( Z L = 0) = Z 0 = jZ 0 tan( β) = j tan(0.628) → ∞
jZ L tan( β) + Z 0
∴ 0.628 = nπ − π / 2, n = 1, 2, 3, ..... ⇒  = (nπ − π / 2) / 0.628 = 5n − 2.5(m) = 2.5(m), 7.5(m), 12.5(m)....

◆ Quality factor Q of a circuit

VL
Vmax The Q of a shorted lossy line having a length equal to an odd multiple of λ/4 is :
f or β or ωL
Vmax Q= 0 = =
2 ∆f 2α R + GL / C
Δf
f 0 : resonant frequency; ∆f : half - power bandwidth
α : attenuation constant; β : phase constant; 6

f
Example 9.7 The measured attenuation of an air-dielectric coaxial transmission line at
400 (MHz) is 0.01 (dB/m). Determine the Q and the half-power bandwidth of a
quarter-wavelength section of the line with a short-circuit termination.
(Sol.)
This is a shorted lossy line with quarter - wavelength length.
c 3 × 10 8 2π 2 × 3.14159
λ= = = 0.75(m) ∴ β= = ≈ 8.38
f 400 × 10 6
λ 0.75
and, for z = 1m : 20 log e −α = −0.01 ⇒ log e −α = −0.0005
0.0005 0.0005
⇒α = = = 0.0001151
log e 0.43429
β 8.38 8.38 × 0.43429
Q= = = ≈ 3641
2α 0.0005 0.001
2×
0.43429
f 0 400 × 10 6
∆f = = ≈ 0.11 × 10 6 = 110 × 10 3 (Hz)
Q 3641

◆ Standing Wave Ratio (SWR or S)

 Review :
+ I I L+
V ZL V L V V Z − Z0 Z − Z0
-
Z0
Z  =  , Z L = L ⇒ Γ =  = Γ e jθ  ; Γ = L = Γ e jθ Γ
- I IL Z + Z 0 ZL + Z0
Z= −  Z=0

+ − + − - r V0− -2r
V = V ( z = -) = V e 0
- r(-  )
+V e 0
r(-  )
= V e +V e
0
r
0 = V e (1 + + e ) = V0+ e r (1 + Γe -2r )...(A)
0
+ r

V0
+ − I 0− -2r
I  = I ( z = -) = I e0
- r(-  )
+I e = I e + I e = I e (1 + + e ) = I 0+ e r (1 − Γe -2r )............(B)
0
r(-  ) +
0
r − - r
0
+
0
r

I0
−
+ - r(0) − r(0) + − + V0 + 2 Z LV0+
VL = V ( z = 0) = V0 e + V0 e = V0 + V0 = V0 (1 + + ) = V0 (1 + Γ) = ........... ........(C)
V0 Z L + Z0
I 0− 2 Z 0 I 0+
I L = I ( z = 0) = I 0+ e -r(0) + I 0− e r(0) = I 0+ + I 0− = I 0+ (1 + ) = I +
(1 − Γ ) = ............................(D)
I 0+ Z L + Z0
0

Therefore,
VL I
(C) → (A) ⇒ V = V0+ e r (1 + Γe -2r ) = ( Z L + Z 0 )e r (1 + Γe -2r ) = L ( Z L + Z 0 )e r (1 + Γ e jθ Γ e -2r )
2Z L 2
IL
For a lossless TL : r = jβ ⇒ V = ( Z L + Z 0 )e jβ (1 + Γ e jθ Γ e - j2 β )
2
I
= L ( Z L + Z 0 )e jβ (1 + Γ e j (θ Γ − 2 β ) )..................9.135a
2
I
if θ Γ − 2β = −2nπ , n = 0, 1, 2,.... V = L ( Z L + Z 0 ) (1 + Γ ) = V max
2
2nπ + θ Γ
⇒  max = , n = 0, 1, 2,........ max : 在TL上最大電壓發生的位置(與Z L的距離)
2β
I I
(D) → (B) ⇒ I  = I 0+ e r (1 − Γe -2r ) = L ( Z L + Z 0 )e r (1 − Γe -2r ) = L ( Z L + Z 0 )e r (1 − Γ e jθ Γ e -2r7 )
2Z 0 2Z 0
 IL
For a lossless TL : r = jβ ⇒ I  = ( Z L + Z 0 )e jβ (1 − Γ e jθ Γ e - j2 β )
2Z 0
IL
= ( Z L + Z 0 )e jβ (1 − Γ e j (θ Γ − 2 β ) )...............9.135b
2Z 0
IL
if θ Γ − 2β = −2nπ − π , n = 0, 1, 2,.... V = ( Z L + Z 0 ) (1 − Γ ) = Vmin
2
(2n + 1)π + θ Γ
⇒  min = , n = 0, 1, 2,........ min : 在TL上最低電壓發生的位置(與Z L的距離)
2β
Define "Standing Wave Ratio" , SWR, as the ratio of the maximum to the minimum voltages along the TL.
IL
( Z L + Z 0 ) (1 + Γ )
V max
2 1+ Γ
SWR (or simply by S) = = = (dimensionless) ⇒ 1 ≤ S < ∞
Vmin IL 1− Γ
( Z L + Z 0 ) (1 − Γ )
2
A high SWR on a line is undesirable because it results in a large power loss.
1+ Γ S −1
S == ⇒Γ =
1− Γ S +1
Case discussion :
1+ 0
(1) if Γ = 0(i.e.Z L = Z 0 ) ⇒ S = =1
1− 0
1+1
(2) if Γ = 1(i.e.Z L → ∞) ⇒ S = →∞
1 −1
1+1
(3) if Γ = −1(i.e.Z L = 0) ⇒ S = →∞
1−1

Example 9.9 The standing-wave ratio on a lossless 50(Ω) transmission line terminated
in an unknown load impedance is found to be 3.0. The distance between successive
voltage minima is 20(cm), and the first minimum is located at 5(cm) from the load.
Determine (a) the reflection coefficient Γ, and (b) the load impedance ZL. In addition,
find (c) the equivalent length and terminating resistance of a line such that the input
impedance is equal to ZL.
(Sol.)
1- Γ S −1 2
(a)  S = = 3.0 ⇒ Γ = = = 0.5
1+ Γ S +1 4
(Note : distance between successive voltage minima ⇐ 表示相鄰兩極小電壓的距離 = λ / 2)
θ + (2n + 1)π =1 =0 θ + (2 × 1 + 1)π θ Γ + (2 × 0 + 1)π 2π 2π λ
  min = Γ ∴  nmin −  nmin = Γ - = = =
2β 2β 2β 2 β 2 ⋅ 2π / λ 2
λ 2π 2π
⇒ = 20cm ⇒ λ = 40 cm = 0.4 m ⇒ β = = = 5π
2 λ 0.4
θ + (2 × 0 + 1)π
and, 0.05 = Γ ⇒ θ Γ = 0.5π − π = −0.5π
2 × 5π
∴ Γ = Γ e jθ Γ = 0.5e - j0.5π = 0.5[cos(−0.5π ) + j sin( −0.5π )] = − j 0.5 #
ZL − Z0 Z − 50 50 − j 25
(b)  Γ = ⇒ − j 0.5 = L ⇐ ZL = = 30 − j 40 # 8
ZL + Z0 Z L + 50 1 + j 0.5
 V V0+ e r + V0− e − r 1 + Γe −2 r 1 + (− j 0.5)e − j 2×5π
(c) Z  = = + r = Z = Z ⇒ 50 = 30 − j 40
I 0 e + I 0− e − r 1 − Γ e − 2 r 1 − (− j 0.5)e − j 2×5π
0 L
I

◆ Transmission-line circuits
 Vi = V g -I i Z g , and

Zg I L+ IL
+ + Ii From 9.135a : Vi (V ) = ( Z L + Z 0 )e r (1 + Γe − 2 r ).
Vg Vi Z0 ZL V L 2
- - I
Z= −  Z=0 From 9.135b : I i ( I  ) = L ( Z L + Z 0 )e r (1 − Γe − 2 r )
2Z 0
IL I
∴ ( Z L + Z 0 )e r (1 + Γe − 2 r ). = V g − L ( Z L + Z 0 )e r (1 − Γe − 2 ) Z g (對照 Vi = V g -I i Z g )
2 2Z 0
IL Z 0V g 1 IL Z 0V g 1
⇒ ( Z L + Z 0 )e r = [ ]= ( Z L + Z 0 )e r = [ ]
2 Z0 + Z g Zg − Z0 2 Z 0 + Z g 1 - Γg Γe − 2 r
1- Γe − 2 r
Zg + Z0
Zg − Z0
here, define Γg = (在generator端的反射係數)
Zg + Z0
 求與訊號源相距 z的位置的電壓及的位置的電壓及電流大小為：
Zg I L+
+ + Ii
Vg Vi Z0 ZL V L 將式子V = I L ( Z + Z )e r (1 + Γe − 2 r ).中的改以z′帶入
 L 0
- - 2
Z= −  Z=0
IL IL
rz ′ − 2 rz ′
z ( ⇒ V (z′) =
Z L + Z 0 ) e (1 + Γ e ) = ( Z L + Z 0 )e r( -z) (1 + Γe − 2 rz′ )
z′ 2 2
Z 0V g e -rz 1 + Γe − 2 rz′
= ⋅( ) ..........9.153a (Note : z′ =  − z )
Similarily, Z 0 + Z g 1 - Γg Γe − 2 r
I I
I (z′) = L ( Z L + Z 0 )e rz′ (1 − Γe − 2 rz′ ) = L ( Z L + Z 0 )e r( -z) (1 − Γe − 2 rz′ )
2Z 0 2Z 0
V g e -rz 1 − Γe − 2 rz′
= ⋅( ) ...............9.153b
Z 0 + Z g 1 - Γg Γe − 2 r
1 + Γe − 2 rz′
Z 0V g e -rz
Z 0V g e -rz
From. 9.153a ⇒ V (z′or z) = ⋅( − 2 r
)= (1 + Γe − 2 rz′ )(1 - Γg Γe − 2 r ) −1
Z 0 + Z g 1 - Γg Γe Z0 + Z g
Z 0V g e -rz
= (1 + Γe − 2 rz′ ) ⋅ (1 + Γg Γe − 2 r + Γg Γ 2 e − 4 r + Γg Γ 3 e −6 r + ............)
2 3

Z0 + Z g
Z 0V g
= [e -rz + (Γe − r )e − rz′ + Γg (Γe − 2 r )e − rz + .............]
Z0 + Z g
= V1+ + V1− + V2+ + V2− + V3+ + V3− + ............
and V1+ = VM e -rz , V1− = Γ(VM e -r )e − rz′ , V2+ = Γg (ΓVM e -2r )e − rz , V2− = (Γg Γ 2VM e -3r )e − rz′ , ..........
Z 0V g
here, VM =
Z0 + Z g
Discussions :
(1) If Z L = Z 0 and Z g ≠ Z 0 (i.e. load matching) ⇒ Γ = 0 ⇒ V (z′or z) = V1+
9
(1) If Z g = Z 0 and Z L ≠ Z 0 (i.e. line) ⇒ Γg = 0 ⇒ V (z′or z) = V1+ + V1−
Similarily, for I (z′or z)
1 + Γe − 2 rz′
V g e -rz Z 0V g e -rz
I (z′or z) = ⋅( − 2 r
)= (1 + Γe − 2 rz′ )(1 + Γg Γe − 2 r ) −1
Z 0 + Z g 1 + Γg Γe Z0 + Z g
V g e -rz
= (1 + Γe − 2 rz′ ) ⋅ (1 − Γg Γe − 2 r + Γg Γ 2 e − 4 r − Γg Γ 3 e −6 r + ............)
2 3

Z0 + Z g
Vg
= [e -rz − (Γe − r )e − rz′ + Γg (Γe − 2 r )e − rz − .............]
Z0 + Z g
= I 1+ − I 1− + I 2+ − I 2− + I 3+ − I 3− + ............
and I 1+ = I M e -rz , I 1− = Γ( I M e -r )e − rz′ , I 2+ = Γg (ΓI M e -2r )e − rz , I 2− = (Γg Γ 2 I M e -3r )e − rz′ , ..........
Vg
here, I M =
Z0 + Z g

Example 9.10 A 100 (MHz) generator with Vg=10∠0° (V) and internal resistance 50 (Ω)
is connected to a lossless 50 (Ω) air line that is 3.6 (m) long and terminated in a 25+j25
(Ω) load. Find,
(a) V(z) at a location z from the generator,
(b) Vi at the input terminals and VL at the load,
(c) The voltage standing-wave ratio on the line, and
(d) The average power delivered to the load.
Sol.
Known : f = 100 MHz; V g = 10∠0  = 10 (V); Z g = 50 (Ω); Z 0 = 50 (Ω); Z L = 25 + j 25 (Ω);  = 3.6 (m)
c 3 × 10 8 2π 2π Z − Z 0 (25 + j 25) − 50 - 1 + 2j
∴ λ= = = 3 (m) ⇒ β = = ; Γ= L = =
f 100 × 10 6
λ 3 Z L + Z 0 (25 + j 25) + 50 5
Z g − Z0 50 − 50 Z 0V g 50 × 10
Γg = = = 0 , VM = = = 5 (V)
Z g + Z0 50 + 50 Z 0 + Z g 50 + 50
(a ) V (z) = V1+ + V1− + V2+ + V2− + V3+ + V3− + .............
= VM e − rz + ΓVM e − r e − r (  − z ) + Γg ΓVM e − 2 r e − rz + Γg Γ 2VM e −3r e − r (  − z) + .....
2π 2π 2π 2π 2π 2π
- 1 + 2j
−j z − j ×3.6 − j ×( 3.6- z) −j z −j ×7.2 j z
= 5e + 3
⋅ 5e 3 e 3 = 5e 3 + (-1 + 2j) e 3
e 3
5
(b) Vi = V (z = 0) = V1 + V1− + V2+ + V2− + V3+ + V3− + .............
+

= VM e − rz + ΓVM e − r e − r (  − z ) + Γg ΓVM e − 2 r e − rz + Γg Γ 2VM e −3r e − r (  − z) + .....

2π 2π 2π 2π
−j - 1 + 2j
⋅0 − j ×3.6 − j ×( 3.6-0) − j ×7.2
= 5e 3
+
⋅ 5e 3 e 3 = 5 + (-1 + 2j) e 3
5
VL = V (z = 3.6m) = V1+ + V1− + V2+ + V2− + V3+ + V3− + .............
= VM e − rz + ΓVM e − r e − r (  − z ) + Γg ΓVM e − 2 r e − rz + Γg Γ 2VM e −3r e − r (  − z) + .....
2π 2π 2π 2π 2π
- 1 + 2j
−j ×3.6 − j ×3.6 − j ×( 3.6-3.6) − j ×3.6 −j ×3.6
= 5e + 3
⋅ 5e 3 e 3 = 5e 3 + (-1 + 2j) e 3
= 3.137 − j 3.186 #
5
− 1 + j2 5
1+ 1 +
1+ Γ 5 5 = 1 + 0.447 ≈ 2.62
(c) SWR = = = 10
1− Γ − 1 + j2 5 1 − 0.447
1− 1−
5 5
 3.137 − j 3.186
2 2
1 1 V 1 VL 1
= VL I L = Re[VL ( L ) * ] = Re[ * ] = ⋅ Re[ ] ≈ 0.2 (w)
av *
(d) P
25 + j 25
L
2 2 ZL 2 ZL 2

9.5 Transients (瞬變) on Transmission Lines

If the applied source is not steady state or not time harmonic, for example a DC source,
the behaviour on the line is called “transient”.
(1) Load-matching case (i.e. ZL=Z0)
For a lossless transmission line : R = 0; G = 0; u = 1/ LC
u : propagation velocity of wave on the TL
R0
Z L -Z 0 R -R Z g -Z 0 R g -R0
Γ= = 0 0 = 0; Γ g = =
Z L + Z 0 R0 + R0 Z g + Z 0 R g + R0
As the switch closed at t = 0, a voltage wave of magnitude V +1
travels down the line in the + z - direction with a velocity u.
R0V0 V+ V0
V +1= ; I +1= 1 = ⇒
R g + R0 R0 R g + R0

(2) Load-dismatching case (i.e. ZL≠Z0)

R0  Z 0V0 V0
T= ; VM = ; IM =
u Z g + Z0 Z g + Z0
Z L -Z 0 R -R Z g -Z 0 R g -R0
ΓL = = L 0 ; Γg = =
(1) For voltage on the line Z L + Z 0 R L + R0 Z g + Z 0 R g + R0
VL = V1+ + V1− + V2+ + V2− + V3+ + V3− + V4+ + V4− + ..................
here, V1+ = VM ; V1− = V1+ Γ L = VM Γ L
V2+ = V1− Γ g ; V2− = V2+ Γ L
V3+ = V2− Γ g ; V3− = V3+ Γ L
..............................................
∴ VL = V1+ + V1− + V2+ + V2− + V3+ + V3− + V4+ + V4− + .................
= V1+ + V1+ Γ L + (V1+ Γ L ) Γ g + (V1+ Γ L Γ g ) Γ L + (V1+ Γ L Γ g ) Γ g + .............
2

= V1+ (1 + Γ L + Γ L Γ g + Γ L Γ g + Γ L Γ g + ..............)
2 2 2

= V1+ [(1 + Γ L Γ g + Γ L Γ g + ...........) + Γ L (1 + Γ L Γ g + Γ L Γ g + ...........)]

2 2 2 2

= V1+ (1 + Γ L )(1 + Γ L Γ g + Γ L Γ g + ...........)

2 2

1
= V1+ (1 + Γ L ) (if Γ L < 1 and Γ g < 1)
1− ΓL Γg
(2) For current on the line
I L = I 1+ − I 1− + I 2+ − I 2− + I 3+ − I 3− + I 4+ − I 4− + .................
here, I 1+ = I M ; I 1− = I 1+ Γ L = I M Γ L
11
I 2+ = I 1− Γ g ; I 2− = I 2+ Γ L
 I 3+ = I 2− Γ g ; I 3− = I 3+ Γ L
..............................................
∴ I L = I 1+ − I 1− + I 2+ − I 2− + I 3+ − I 3− + I 4+ − I 4− + .................
= I 1+ − I 1+ Γ L + ( I 1+ Γ L ) Γ g − ( I 1+ Γ L Γ g ) Γ L + ( I 1+ Γ L Γ g ) Γ g − .............
2

= I 1+ (1 − Γ L + Γ L Γ g − Γ L Γ g + Γ L Γ g − ..............)
2 2 2

= I 1+ [(1 + Γ L Γ g + Γ L Γ g + ...........) − Γ L (1 + Γ L Γ g + Γ L Γ g + ...........)]

2 2 2 2

= I 1+ (1 − Γ L )(1 + Γ L Γ g + Γ L Γ g + ...........)
2 2

1
= V1+ (1 − Γ L ) (if Γ L < 1 and Γ g < 1)
1− ΓL Γg

R0

For example : R L = 3R0 , R g = 2 R0

3R0 -R0 1 2 R0 -R0 1 R0V0 1 V V0 V
ΓL = = ; Γg = = , VM = = V0 ; I M = M = = 0
3R0 + R0 2 2 R0 + R0 3 2 R0 + R0 3 R0 2 R0 + R0 3R0
1 1 1 1
(1) For voltage ⇒ V1+ = VM = V0 ; V1− = V1+ Γ L = V0 ⋅ = V0
3 3 2 6
1 1 1 1 1 1
V2+ = V1− Γ g = V0 ⋅ = V0 ; V2− = V2+ Γ L = V0 ⋅ = V0
6 3 18 18 2 36
....................................... .................................................
1 1 1 1 3
∴V L = V1+ (1 + Γ L ) = V0 (1 + ) = V0
1− ΓL Γg 3 2 1 1 5
1− ⋅
2 3

Z' = u ⋅ t Z ' = 2 − u ⋅ t Z ' = u ⋅ t - 2

Z’ Z’ Z’

1 1 1 1
(1) For voltage ⇒ V1+ = VM = V0 ; V1− = V1+ Γ L = V0 ⋅ = V0
3 3 2 6
1 1 1 1 1 1
V2+ = V1− Γ g = V0 ⋅ = V0 ; V2− = V2+ Γ L = V0 ⋅ = V0
6 3 18 18 2 36
....................................... .................................................
1 1 1 1 3
∴V L = V1+ (1 + Γ L ) = V0 (1 + ) = V0
1− ΓL Γg 3 2 1 1 5
1− ⋅
2 3 12
 V0 V0 1 V0
(2) For current ⇒ I1+ = I M = ; I1− = I1+ Γ L = ⋅ =
3R0 3R0 2 6 R0
V0 1 V V0 1 V
I 2+ = I1− Γ g = ⋅ = 0 ; I 2− = I 2+ Γ L = ⋅ = 0
6 R0 3 18 R0 18 R0 2 36 R0
.......................................
.................................................
1 V 1 1 V
I L = I1+ (1 − Γ L ) = 0 (1 − ) = 0
1 − Γ L Γ g 3R0 2 1 − 1 ⋅ 1 5 R0
2 3

Z' = u ⋅ t Z ' = 2 − u ⋅ t Z ' = u ⋅ t - 2

Z’ Z’ Z’

◆ Reflection Diagrams

# Reflection diagram
Z1 
Zg I L+
+ + IM
Vg VM Z0 ZL V L
- -
Z=0 Z= 

Time Range Voltage Voltage Discon

0 ≦ t＜ t 1 (t 1 = z 1 /u) 0 0
t 1 ≦ t＜ t 2 (t 2 = 2T - t 1 ) V1+ V1+ at t1
t 2 ≦ t＜ t 3 (t 3 = 2T + t 1 ) V1+ + ΓLV1+ ΓLV1+ at t 2
t 3 ≦ t＜ t 4 (t 4 = 4T - t 1 ) V1+ + ΓLV1+ + Γg (ΓLV1+ ) Γg ΓLV1+ at t 3
t 4 ≦ t＜ t 5 (t 5 = 4T + t 1 ) V1+ + ΓLV1+ + Γg ΓLV1+ + ΓL (Γg ΓLV1+ ) Γg ΓL V1+ at t 4
2

13
 Z1 
Zg I L+
+ + IM
Vg VM Z0 ZL V L
- -
Z=0 Z=  Time Range Current Voltage Discon
0 ≦ t＜ t 1 (t 1 = z 1 /u) 0 0
t 1 ≦ t＜ t 2 (t 2 = 2T - t 1 ) I 1+ I 1+ at t1
t 2 ≦ t＜ t 3 (t 3 = 2T + t 1 ) I 1+ − ΓL I 1+ - ΓL I 1+ at t 2
t 3 ≦ t＜ t 4 (t 4 = 4T - t 1 ) I 1+ − ΓL I 1+ + Γg (ΓL I 1+ ) Γg ΓL I 1+ at t 3
t 4 ≦ t＜ t 5 (t 5 = 4T + t 1 ) I 1+ − ΓL I 1+ + Γg ΓL I 1+ − ΓL (Γg ΓL I 1+ ) - Γg ΓL I 1+ at t 4
2

Example 9.11 A rectangular pulse of an amplitude 15 (V) and a duration 1 (µs)is applied
through a series resistance of 25 (Ω) to the input terminals of a 50 (Ω) lossless coaxial
transmission line. The line is 400 (m) long and is short-circuited at the far end.
Determine the voltage response at the midpoint of the line as a function of time up to 8
(µs). The dielectric constant of the insulating material in the cable is 2.25.

Pulse Excitation

sol.
Vg1 (t ) Vg2 (t )
15
Vg= + 1
0 t ( µs ) t ( µs )

-15
known : R0 = 50(Ω), R g = 25(Ω), R L = 0(Ω),  = 400 (m), ε r = 2.25, µ r = 1
R L -R0 0-50 R g -R0 1
⇒ ΓL = = = −1; Γ g = =− ,
R L + R0 0 + 50 R g + R0 3
c  400
µ= = 2 × 10 8 (m/s) ⇒ T = = = 2 ( µs )
µγ ε γ µ 2 × 10 8
(1) For source V g1 :
R0 50
V1+ = V g1 ⋅ = 15 ⋅ = 10 (V)
R g + R0 25 + 50
∴ v a = V1+ + Γ LV1+ + Γ g Γ LV1+ + Γ g Γ L V1+
2

14
+ Γ g Γ L V1+ + Γ g Γ L V1+ + ......( Please see the red curve shown in the following reflection diagram.
2 2 2 3
)
(2) For source Vg2 :
R0 50
V1+ = Vg2 ⋅ = −15 ⋅ = −10 (V)
Rg + R0 25 + 50
∴ vb = V1+ + Γ LV1+ + Γ g Γ LV1+ + Γ g Γ L V1+ + Γ g Γ L V1+ + Γ g Γ L V1+ + ......(Please see the black curve
2 2 2 2 3

shown in the following reflection diagram.) ⇒ vm = va + vb

At the midpoint of the line (200m), the voltage response of time up to 8 µs is :

vm

15
 9.6 The Smith Chart
◆ Developed in 1939 by P. W. Smith as a graphical tool to
characterize the performance of microwave circuits, and analyze
as well as design transmission-line circuits.

# Z-Smith chart (Impedance Smith chart)

ZL − Z0
Γ= = Γ e jθ L = Γr + jΓi
Z0 ZL + Z0
ZL
1+ Γ Z 1 + Γ 1 + (Γr + jΓ)
Z L = Z0 ⇒ zL = L = = = r + jx
z = − z=0 1+ Γ Z 0 1 − Γ 1 − (Γr + jΓ)
z L : normalization of Z L by Z0
1 + (Γr + jΓ) (1 + Γr + jΓi )(1 − Γr + jΓi ) 1 − Γr − Γi 2Γi
2 2

∴ r + jx = = = + j
1 − (Γr + jΓ) (1 − Γr − jΓi )(1 − Γr + jΓi ) (1 − Γr ) 2 + Γi 2
(1 − Γr ) 2 + Γi
2

 1 − Γr − Γi
2 2

 r = ........(1)
 (1 − Γr ) 2 + Γi
2

⇒
x = 2Γi
.......(2)
 (1 − Γr ) 2 + Γi
2

2r 1− r
from (1) ⇒ [ (1 − Γr ) 2 + Γi ] ⋅ r = 1 − Γr − Γi ⇒ Γr − Γr + Γi =
2 2 2 2 2

1+ r 1+ r
2r r 2 1− r r 2 r 2 1 2
Γr − Γr + ( ) + Γi = +( ) ⇒ (Γr − ) + Γi = (
2 2 2
)
1+ r 1+ r 1+ r 1+ r 1+ r 1+ r
Considering :
r = 0 ⇒ Γr + Γi = 12 ⇒ curve A
2 2

1 1 2 x
r= ⇒ (Γr − ) 2 + Γi = ( ) 2 ⇒ curve B
2

2 3 3 x x
1 1
r = 1 ⇒ (Γr − ) 2 + Γi = ( ) 2 ⇒ curve C
2

2 2 Psc Poc
2 2 1 2 ․-1 x
D ․1
r = 2 ⇒ (Γr − ) + Γi = ( ) ⇒ curve D
2

3 3 r C x r
r → ∞ ⇒ (Γr − 1) 2 + Γi = 0 ⇒ (Γr , Γi ) = (1, 0) ⇒ 點 POC
2

r x B r
x
from (2) ⇒ [ (1 − Γr ) + Γi ] ⋅ x = 2Γi ⇒ xΓr − 2 xΓr + x + xΓi = 2Γi
2 2 2 2
A
2 1 1
⇒ Γr − 2Γr + 1 + Γi = Γi ⇒ (Γr − 1) 2 + (Γi − ) 2 = ( ) 2
2 2

x x x

Considering : x = 0 ⇒ (Γr − 1) 2 + (Γi − ∞) 2 = ∞ 2 ⇒ Γr 軸

x = 0.5 ⇒ (Γr − 1) 2 + (Γi − 2) 2 = 2 2 and x = −0.5 ⇒ (Γr − 1) 2 + (Γi + 2) 2 = (−2) 2
x = 1 ⇒ (Γr − 1) 2 + (Γi − 1) 2 = 12 and x = −1 ⇒ (Γr − 1) 2 + (Γi + 1) 2 = (−1) 2
x = 2 ⇒ (Γr − 1) 2 + (Γi − 1 ) 2 = ( 1 ) 2 and x = −2 ⇒ (Γr − 1) 2 + (Γi + 1 ) 2 = (− 1 ) 2
2 2 2 2
16
x → ∞ ⇒ (Γr − 1) + Γi = 0 ⇒ (Γr , Γi ) = (1, 0) ⇒ 點 POC
2 2
 Example: For a T.L. with Z 0 =50 Ω, Z L =100+j50 Ω, please determine Γ at load using Smith chart.

x
Z0 ZL
Γ x x
circle
z = − z=0
A
Psc Poc
․-1 x
O
․1
r
x r

r x r

x
Γ circle

sol :
Z L 100 + j50
Step1 : z L = = = 2 + j1 = r + jx ⇒ r = 2 ; x = 1 ⇒ 找出A點, 並延伸OA找出與
Z0 50
最外圓的交點後，讀出位置(角度)的值 ⇐目的在定出負載在傳輸線上的位置
Γi
求出 Γ (= Γr2 + Γi2 = OA (以OPOC = 1為基準)) 及 θ Γ (= tan -1 為OA與Γr 軸的夾角)
Γr
Step2 :以O點為圓心，OP為半徑畫出Γ - circle ⇒ 傳輸線上最大電壓會在 Γ - circle與正Γr 軸的
交點處(Pmax )；最小電壓則在與負Γr 軸的交點處(Pmin 。
)
最大電壓與負載的距離則為d max (參考右上圖); 最小電壓與負載的距離則為d min (參考右上圖)
說明：On the T.L. the maximum voltage, Vmax is V0+ (1 + Γ ), whereas the minimum voltage isV0+ [1 + (- Γ )]
Step 3 : 讀出Pmax 點處的r值，即為此傳輸線的S (VSWR)值。
1+ Γ 1 + Γr
說明：S (VSWR) = = ( Γ = Γ r + j Γ i , 但在Pmax 處，Γ r > 0 且 Γ i = 0 ........(a)
1- Γ 1 - Γr
r 2 1 2 r 2 1 2
另外，(Γr − ) + Γi = ( ) , 在Pmax 處 Γ i = 0, 所以(Γr − ) =(
2
)
1+ r 1+ r 1+ r 1+ r
r -1 1 r -1 1 + Γr
⇒ Γr − = (or : 不合) ⇒ Γr = ⇒r= ..........(b)
1+ r 1+ r 1+ r 1+ r 1 - Γr
17
比較(a) 與 (b) ⇒ S (VSWR) = r
 Step4 : 求傳輸線上任意位置( z = -)的Z℘及Γ 方法： Z0 ZL
在 Γ - circle上，由負載的位置順時針轉長度後，
z = − z=0
找出Z℘所在位置，讀出 r與x值 ⇒ Z  = (r + jx) ⋅ Z 0
V V ( z = −) 1 + Γe − 2 r 1 + Γ
說明：Z  = = = Z0 − 2 r
= Z ⇒ Γ = Γe − 2 r
I ( z = −) 1 − Γe 1 − Γ
0
I
for a lossless T.L. ⇒ Γ = Γe − j2 β = Γ e jθ Γ ⋅ e − j2 β = Γ e j(θ Γ − 2 β )
2π
⇒ Γ = Γ θ Γ = θ Γ − 2β  = θ Γ - 2 
℘
λ

Example 9.13 For a lossless T.L. with Z 0 =50 Ω, Z L =0 Ω (i.e. short-circuit load), please using Smith
chart to determine the input impedance Z i and Γ i of this line that is 0.1 wavelength
long.
P1

Z0 ZL

z = −0.1λ z=0

Psc
O
sol :
ZL 0
zL = = = 0 + j0 = r + jx
Z 0 50
⇒ r = 0 ; x = 0 ⇒ 找出P點 ⇒ 圖上之點PSC
Γ = Γr2 + Γi2 = OPSC = 1 ⇐ 即 Γi = Γ = 1 #
以O為圓心；OPSC為半徑順時針轉2 β 
2π
⇒ 2⋅ ⋅ 0.1λ = 0.4π = 72  ⇒ 由180  轉至108  (或由0λ順時針轉至0.1λ處)
λ
到達圖上之P1點 ⇒ 讀出 (ri , x i ) ≈ (0, 0.725)
Zi
∴ zi = = 0 + j0.725 ⇒ Z i = (0 + j0.725) ⋅ Z 0 = j0.725 × 50 = j36.25 (Ω) #
Z0

Use the transmission line theory :

2π
- j2 ⋅0.1λ
- j2 β
1 + Γe 1 + (-1)e λ 1 - e - j0.4πλ
Zi = Z0 = 50 = 50 = 50 × j tan(0.2π )
1 - Γe - j2 β 2π
- j2 ⋅0.1λ
λ
1 + e - j0.4πλ
1 - (-1)e
j50tan(36  ) ≈ j 36.3 (Ω) #

18
 Example 9.14 For a lossless T.L. with Z 0 =100 Ω, Z L =260+j180 Ω, please using Smith chart to
determine the Γ, S(VSWR), input impedance Z i, and the location d max of a voltage
maximum on the line of length 0.434 wavelength.
sol :
Z L 260 + j180
Step1 : z L = = = 2.6 + j1.8 = r + jx ⇒ r = 2 .6; x = 1 .8 ⇒ 找出P2點, 並延伸OP2 找出與
Z0 100
最外圓的交點P2' 後，讀出位置(角度)的值為0.22λ ⇐ 此為負載在傳輸線上的位置 .
Γi
求出 Γ ( = Γr2 + Γi2 = OP2 長度 (以OPOC = 1 為基準)) = 0.6 及 θ Γ ( = tan -1 為OA與Γr 軸的夾角) =
Γr
(0.25λ - 0.22λ ) × 4π ≈ 21 或用21 - 0  (正Γr 軸) ⇒ Γ = 0.6∠21
Step2 :以O點為圓心，OP2為半徑畫出Γ - circle ⇒ 傳輸線上最大電壓會在 Γ圓與正Γr 軸的
'
交點PM 處, 延伸OPM 找出與最外圓的交點POC 後, 最大電壓與負載的距離d max = 0.25λ - 0.22λ =
0.03λ
Step 3 : 讀出PM 點處的r值 = 4，此值即為此傳輸線的S(VSWR)值 ⇒ S = 4
Step4 : Z i處為  = 0.434λ：
由負載的位置順時針轉0.434λ後到達點P3' ( 0.434λ = 0.5λ - 0.22λ + 0.154λ，找出
) OP3' 與
Γ - circle 的交點P3後，讀出r = 0.69與x = 1.2 ⇒ Z i = (r + jx)Z 0
= (0.69 + j1.2) × 100 = 69 + j120 (Ω)

Example 9.15 For a lossless T.L. with Z 0 =50 Ω, S=3.0, λ=0.4 m, and first voltage minimum has a
distance of 0.05 m from the load. Please find Γ and Z L
(Homework)
19
 9.7 Transmission line impedance matching
◆ Impedance matching by quarter-wave transformer

lossless
R0 R0′’ RL ⇒ R ′'0 = RL R0

λ/4
transformer

Example 9.17 A signal generator is to feed equal power through a lossless air transmission line with
a characteristic impedance 50 (Ω) to two separate resistive loads, 64 (Ω) and 25 (Ω),
depicted as figure. Determine the required characteristic impedances and standing
wave ratios of the two quarter-wave lines.
sol :
Let Ri1 = Ri 2 and Ri1 // Ri 2 = R0
∴ Ri1 = Ri 2 = 50 (Ω) × 2 = 100 (Ω)
 R × R = R ' 2 ⇒ R ' = R × R = 100 × 64 = 80 (Ω) #
 i1
⇒
L1 01 01 i1 L1
2
 Ri 2 × R L 2 = R02
'
⇒ R02
'
= Ri 2 × R L 2 = 100 × 25 = 50 (Ω) #
 R L1 − R01
'
64 − 80 1 + Γ1 1 + 0.11
Γ
 1 = = = −0.11 ⇒ S1 = = = 1.25 #
 R −
+ R '
64 + 80 1 − Γ 1 − 0 . 11
⇒
L1 01 1

Γ = R L 2 − R02 = 25 − 50 = −0.33 ⇒ S = 1 + Γ2 = 1 + 0.33 = 1.99 #

'

 2 R +− R ' 25 + 50
2
1 − Γ2 1 − 0.33
 L2 02

# Y-Smith chart (Admittance Smith chart)

Z L − Z0
Γ= = Γ e jθ L = Γr + jΓi
Z L + Z0
1+ Γ 1 1 1− Γ 1− Γ Y 1 − Γ 1 − (Γr + jΓ)
Z L = Z0 ⇒ = ⇒ YL = Y0 ⇒ yL = L = = = g + jb
1 +− Γ ZL Z0 1+ Γ 1+ Γ Y0 1 + Γ 1 + (Γr + jΓ)
1 − (Γr + jΓ) (1 − Γr − jΓi )(1 + Γr − jΓi ) 1 − Γr − Γi 2Γi
2 2

∴ g + jb = = = −j
1 + (Γr + jΓ) (1 + Γr + jΓi )(1 + Γr − jΓi ) (1 + Γr ) + Γi
2 2
(1 + Γr ) 2 + Γi
2

 1 − Γr − Γi
2 2

g = ........(1)
 (1 + Γr ) 2 + Γi
2

⇒
b = − 2Γi
.......(2)
 + Γ 2
+ Γ
2
 (1 r ) i

2g 1− g
from (1) ⇒ [ (1 + Γr ) 2 + Γi ] ⋅ g = 1 − Γr − Γi ⇒ Γr +
Γr + Γi =
2 2 2 2 2

1+ g 1+ g
2g g 2 1− g g 2 g 2 1 2
Γr + Γr + ( ) + Γi = +( ) ⇒ (Γr + ) + Γi = (
2 2 2
) ....(A)
1+ g 1+ g 1+ g 1+ g 1+ g 1+ g 20
 from (2) ⇒ [ (1 + Γr ) 2 + Γi ] ⋅ b = - 2Γi ⇒ bΓr + 2bΓr + b + bΓi = −2Γi
2 2 2

−2 1 1
⇒ Γr + 2Γr + 1 + Γi = Γi ⇒ (Γr + 1) 2 + (Γi + ) 2 = ( ) 2 ........(B)
2 2

b b b

# Exchange between Impedance and Admittance on Smith chart

On the Smith chart we need only move the point representing z L along the Γ-circle by a
quarter-wavelength to located the point representing y L . (亦即將 z L 點沿著Γ-circle 轉 180°，便
可得到 y L 的位置)
Example 9.18: Given Z L =95+j20, find Y L .

Sol :
General method :
1 95 − j 20
YL = 1 / Z L = ⇒ 2 ≈ 0.01 − j 0.002 (Ω -1 )
95 + j 20 95 + 20 2

Z 95 + j 20
Smith − chart method : z L = L = = 1.9 + j 0.4
Z0 50
⇒ Find z L as point P1 on the Smith chart. ⇒ Plot the Γ - circle
⇒ Move point P1 along the Γ - circle by 180  (i.e. the opposite
position with respect to the center pount O) to point P2
⇒ Read g (r ) = 0.5, b ( x) = −0.1 ∴ y L = 0.5 − j 0.1
1
⇒ YL = y L ⋅ Y0 = (0.5 − j 0.1) × = 0.01 − j 0.002 (Ω -1 )
50 21
 Example 9.19: Find the input admittance of an open-circuited line of characteristic impedance 300 Ω
and length 0.04 λ.

Sol :
Smith − chart method 1 :
For an open - circuited line, z L is located at point POC ⇒ Plot Γ - circle ⇒ Move point POC
along Γ - circle in clockwise direction by a length of 0.04λ to point P3 , which is the position
of z i ⇒ Move point P3 along Γ - circle in clockwise direction by 180  to point P3' ⇒ Read
1
r (g) = 0 and x (b) = 0.26 ∴ y i = 0 + j 0.26 ⇒ Yi = y i ⋅ Y0 = (0 + j 0.26) × = j 0.00087 (Ω -1 ) #
300

Smith − chart method 2 :

For an open - circuited line, z L is located at point POC ⇒ Plot Γ - circle ⇒ Move point POC
along Γ - circle by 180  to point PSC ⇒ Move point PSC along Γ - circle in clockwise
direction by a length of 0.04λ to point P3' ⇒ Read r (g) = 0 and x (b) = 0.26 ∴ y i = 0 + j 0.26
1
⇒ Yi = y i ⋅ Y0 = (0 + j 0.26) × = j 0.00087 (Ω -1 ) #
300
Method 1 Method 2

◆ Single-stub matching
1 1 1
If circuit matched ⇒ Z s //Z B = R 0 ⇒ + = ⇒ Ys + YB = Y0
Z s Z B R0
Ys YB Y0
⇒ + = ⇒ ys + y B = 1
Y0 Y0 Y0
Becauset the stub is shorted, its input impedance Z s is purely imaginary.
1 + (-1)e - j2 β 
(說明：Z s = R 0 = j R 0 tan( β ) , note Γ = -1 for short - load)
1 - (-1)e - j2 β 
Assume y s = -jb B , then y B = 1 + jb B .
(Here, b B can be either positive or negative. )
IfTherefore, for matching this circuit, it is necessary to ;
1) find d such that y B has a unity real part, and
22
2) find  such that the stub can cancel the imaginary part of y B .
 Example 9.20: A 50 (Ω) transmission line is connected to a load Z L =35-j47.5 (Ω). Find the position
and length of a short-circuited stub required to match the line.

=50 Ω =35-j47.5 Ω

PSC POC

Sol :
Z L 35 - j47.5
Step 1 : z L = = = 0.7 - j0.95 ⇒ Find z L on the Smith chart → point P1
R0 50
Step 2 : Plot the Γ - circle (以O為圓心, OP1為半徑畫圓)
Step 3 : Move P1 along Γ - circle by 180  to the point shown as P2 . ⇒ Point P2 reprents the y L .
Step 4 : Plot the unit circle of r = 1 and then find out the two points of intersection with Γ - circle
P3 ⇒ Read at P3 : y B1 = 1 + j1.2 = 1 + jb B1
⇒
P4 ⇒ Read at P4 : y B2 = 1 - j1.2 = 1 + jb B2
Step 5 : Solutions for the position of the stub :
Extend OP2 , OP3 and OP4 to the outer circle with the points of P2' , P3' and P4' , respectively.
⇒ Read the position parameters : P2' = 0.109 λ ; P3' = 0.168 λ ; and P4' = 0.332 λ
⇒ 傳輸線上並接stub的位置可有兩處，與負載的距離分別為：
For P3 (from P2' to P3' ) : d 1 = 0.168 λ - 0.109 λ = 0.059 λ #

For P4 (from P2' to P4' ) : d 2 = 0.332 λ - 0.109 λ = 0.223 λ #
Step 6 : Solutions for the length of short - circuited stubs :
1) For P3 its imaginary part, b B1 , has to be canceled by adding the short - circuited stub. Therefore
the y s1 for this stub is 0 - j1.2 ⇒ y s1 = -j1.2 ⇒ position of the input terminal of this stub is P3'' (= 0.361 λ ).
⇒ length of the stub is :  B1 = 0.361 λ - 0.250 λ (因為負載短路，所以導抗為開路 ⇒ POC ) = 0.111 λ #
2) For P4 its imaginary part, b B2 , has to be canceled by adding the short - circuited stub. Therefore
the y s2 for this stub is 0 + j1.2 ⇒ y s2 = j1.2 ⇒ position of the input terminal of this stub is P4'' (= 0.139 λ ).
⇒ length of the stub is :  B2 = 0.5λ - 0.250 λ + 0.139 λ = 0.389 λ # 23
 Cheapter 2 Vector Analysis
2-1~2-3 Vector algebra and formula
Scalar : completely specified by a magnitude  charge、current、
(有極性) (有極性)
energy.

Vector : completely specified by both a magnitude and a direction.

 
 electric field intensity E 、magnetic field intensity H
(1) Unit Vector
 
    A A 
A  aA A , here A= A aA    : the unit vector of A
A A

 A
A A
 
a
1 A
  
 A   a A A  A(  a A ) ․
Start point

-A
(2) Vector Addition

B
 
 A B
  B
 A B
A  
A

Parallelogram rule

(3) Vector Subtraction

 
B B
   
 A
A  B  A  ( B )


A  B  
A-B

1
(4) Products of Vectors

(a). k A = ka A A  a A (kA) , here k is a constant

(b). Scalar (or dot) product

 
A  B ABcos  AB  : 定義的意思  AB 取較小的那個角,  AB  
 
(1) if  AB  (2) if  AB 
2 2
 
B B
 AB

 AB A
B cos 
 B cos
A

hence,
   
(a). A  A =AAcos 0  = A 2  A= A  A
   
(b). A  B = B  A  commutative law (交換率)

      
(c). A  ( B + C )= A  B + A  C  distributive law (分配)

(4) Vector (or Cross) Product

  
A  B ABsin  AB  a n


B

B sin  AB
 AB 
A
  
an : the unit vector perpendicular to the plane containing A and B

and its direction follow the right-hand rule.

2
Hence,
   
a. B  A = - A  B  not commutative
      
b. A  ( B + C ) = A  B + A  C  distributive law
           
c. A  ( B  C ) = B ( A  C )- C ( A  B )  ( A  B)  C  not associative (結合)

(6) Product of three vectors

(a). Scalar triple vectors

(ref. Fig.2.8 in textbook)

        
A  ( B  C )  B  (C  A)  C  ( A  B) = magnitude egual to the volume of the
 
parallelepiped formed by the three vectors A , B ,and


C.

           
or A  ( B  C )   A  (C  B )   B  ( A  C )  C  ( B  A)

(b).Vector triple product

        
A  ( B  C )  B( A  C )  C ( A  B) : known as the “back-cab”rule. (記)

BAC - CAB

3
2-4 Orthogonal Coordinate Systems

(1) General specifications of vectors

  
Let au , au ,and
1 2
a u3 be the unit vectors in the three coordinate

directions.

They are called the base vectors.

  
Chain rule : au  au  au
1 2 3

        
 au1  au2  au3 au2  au3  au1 au3  au1  au2

     
au1  au2  au2  au3  au3  au1  0

     
au1  au1  au2  au2  au3  au3  1

Any vector A can be written as the sum of its components in
the three orthogonal directions, as follows:
4
    
A  au1 Au1  au 2 Au 2  au 3 Au 3
 1
A  A  ( Au1  Au2  Au3 )
2 2 2 2



 A Au1  Au  Au 
aA   au1  2 au2  3 au3 )
A A A A

  
Given vectors A, B and C, and let
   
A  au1 Au1  au2 Au2  au3 Au3
   
B  au1 Bu1  au2 Bu2  au3 Bu3
   
C  au1 Cu1  au2 Cu2  au3 Cu3

therefore,
        
(i) A  B  ( A  au Au  au Au  au Au )  (au1 Bu1  au 2 Bu 2  au 3 Bu 3 )  Au1 Bu1  Au 2 Bu 2  Au 3 Bu 3
1 1 2 2 3 3
 
A  C  Au1C u1  Au 2 Cu 2  Au 3C u 3

 
B  C  Bu1C u1  Bu 2 Cu 2  Bu 3C u 3

       
(ii) A  B  (au1 Au1  au 2 Au 2  au 3 Au 3 )  (au1 Bu1  au 2 Bu 2  au 3 Bu 3 )
  
= au1 ( Au 2 Bu 3  Au 3 Bu 2 )  au 2 ( Au 3 Bu1  Au1 Bu 3 )  au 3 ( Au1 Bu 2  Au 2 Bu1 )

formula :
base vectors
  
au1 au 2 au 3
  
A  B  Au1 Au 2 Au 3 components of A

Bu1 Bu 2 Bu 3 components of B

5
(iii)
     
a u1 au 2 au 3 a u1 au 2 au 3
      
C  ( A  B)  C  Au1 Au 2 Au 3  (au1C u1  au 2 C u 2  au 3 C u 3 )  Au1 Au 2 Au 3
Bu1 Bu 2 Bu 3 Bu1 Bu 2 Bu 3
C u1 Cu 2 Cu 3
 Au1 Au 2 Au 3
Bu1 Bu 2 Bu 3

(iv) Because in some of the coordinates , ui (i  1,2, or 3) may not be a length; a

conversion factor is needed to convert a differential change dui into a
change in length d  i

r 通式
d i  r d i d  i  hi d ui
d i

Here, hi is called a metric coefficient and therefore,

a. differential length
      
d  au1d1  au2d 2  au3d 3  au1 (h1du1 )  au2 (h2du2 )  au3 (h3du3 )
1 1
d  (d 1  d 2  d 3
2 2 2 2
)  [(h1du1 )  (h2 du 2 )  (h3 du3
2 2 2 2
) ]

b. differential area ds
 
ds  a n ds

an : the unit vector normal to the surface of dS

 
here, ds 3  d 1 d 2  h1 du1  h2 du 2  h1 h2 du1 du 2  ds 3  a u3 h1 h2 du1 du 2
 
ds1  d 2 d 3  h2 h3 du 2 du 3  ds1  a u1 h2 h3 du 2 du 3
 
ds 2  d 3 d 1  h3 h1 du 3 du1  ds 2  a u 2 h1 h3 du1 du 3

c. differential volume dv
dv  d 1  d 2  d 3  h1du1  h2 du2  h3du3  h1h2 h3du1du2 du3

6
(2) The three most common and useful orthogonal coordinate
systems  Cartesian coordinate、 Cylindrical coordinates

and spherical coordinates.

(i) Cartesian coordinates ( 卡迪遜座標系統 or 直角座標系統 )
 
Z Y u1 = x u2 = y u3 = z ax ay

az
X A vector in Cartesian coordinates is written as
   
A  a x Ax  a y A y  a z Az
   
 d   a x dx  a y dy  a z dz

 h1  1, h2  1, h3  1
 
 dS x  dydz  dS x  a x dydz
 
dS y  dxdz  dS y  a y dxdz dv  dxdydz
 
dS z  dxdy  dS z a z dxdy
    
Example 2 - 5 Given A  ax 5-a y 2  az , find a unit vector B such that
    
(a) B//A (b) B  A, if B lies in the xy - plane.
(sol)

(ii) Cylindrical coordinates

u1  r , u 2   , u3  z
7
  
a a

az

A vector in cylindrical coordinates is written as

   
A  a r Ar  a A  a z Az
   
d   ar d r  a rd  a z d z  h1  1, h2  r , h3  1

 
dSr  rddz  dS r a r rd dz
  dv  dr  rd  dz
dS  drdz  dS a  dr dz
 rdrddz
 
dS z  rddr  dS z a z rd dr

The relations between the components of a vector in carteslan and

cylindrical coordinates :
   
let A  a r Ar  a A  a z Az
       
 ar  cos  a x  sin  a y a   sin  a x  cos  a y az  az
8
      
 A  Ar (cos a x sin a y )  A ( sin a x cos a y )  Az a z
  
= ( Ar cos   A sin  ) a x  ( Ar sin   A cos  ) a y  Az a z

 Ax  cos   sin  0  Ar 
   0  A 
  A y    sin  cos 
 Az   0 0 1  Az 
here, x  r cos  , y  r sin  , z  z
and, r  x 2  y 2 ,  tan 1 t , z  z
   
Example 2 - 9 Express the vector A  ar (3 cos   a 2r  a z 5 in Cartesian coordinate s.

(sol):

 Ax  cos   sin  0 3 cos  

 A    sin  cos  0   2r   a x (3 cos 2   2r sin  )  a y (3 sin  cos   2r cos  )  a z 5
  
 y 
 Az   0 0 1  5 

x y
and, cos   , sin  
x y
2 2
x  y2
2

  3x 2 y  y x x 
 A  ax ( 2  2 x2  y2  )  a y (3   2 x2  y2  )  az 5
x y 2
x y2 2
x y
2 2
x y
2 2
x y 2 2

 3x 2  3 xy 
 ax ( 2  2 y)  a y ( 2  2 x)  a z 5
x y 2
x y 2

(iii) Spherical coordinates u1  R , u 2   , u 3  

 
Ref. Fig. 2.18 aR a

a
A vector in spherical coordinates is written as
   
A  a R AR  a A  a A
   
 d   a R dR  a Rd  a R sin d  h1  1, h2  R, h3  R sin 
Therefore,

9
  
dS R  Rd sin d  R 2 sin dd  dS R  a R R 2 sin dd
 
dS  dR  R sin  d  R sin dRd  dS  a R sin dRd
 
dS  dR  Rd  RdRd  dS  a R RdRd

dv  (dR)( Rd )( R sin d )  R 2 sin dRdd

x  r cos   R sin  cos  R  x2  y2  z2

1 x2  y2
y  r sin   R sin  sin    tan
 z
y
z  R cos    tan 1
x

2-5 Integrals Containing Vector Functions

       
F  a x Fx  a y Fy  a z Fz A  a x Ax  a y Ay  a z Az
     
 a r Fr  a F  a z Fz  ar Ar  a A  a z Az
let      
 a R FR  a F  a F  a R AR  a A  a A

then
   
 dv 
v
F  (a
v
x F x  a y F y  a z F z ) dxdydz (for Cartesian coordinate )
  
  (a
v
r Fr  a  F  a z F z ) rdrd  dz (for cylindrica l coordinate )
  
  (a
v
R F R  a  F  a  F ) R 2 sin  dRd  d  (for spherical coordinate )

   

c
Vd    V
c
( x , y , z )( a x dx  a y dy  a z dz ) (for Cartesian coordinate )

  
  V ( r ,  , z )(a r dr  a rd   a z dz ) (for cylindrica l coordinate )
c
  
  V ( R ,  ,  )(a R dR  a Rd   a R sin  d  ) (for spherical coordinate )
c

10
        
  d    ( a x Fx a y Fy  a z Fz )  ( a x dx  a y dy  a z dz ) (for Cartesian coordinate )
c
F
c
     
  ( a r Fr a F  a z Fz )  ( a r dr  a rd   a z dz ) (for cylindrica l coordinate )
c
     
  ( a R FR a F  a F )  ( a R dR  a Rd   a R sin d ) (for spherical coordinate )
c

     

s
A  d s  s
( a x Ax a y Ay  a z Az )  a n ds (for Cartesian coordinate )

   
  ( ar Ar a A  a z Az )  an ds (for cylindrical coordinate )
s
   
  ( a R AR a A  a A )  an ds (for spherical coordinate )
s

Here, is the unit vector of the surface ds
an
 
a
** How to decide the positive direction of n or ds ?

a. If S is a closed surface enclosing a volume, then the positive direction for a n

is always in the outward direction from the volume .

b. If S is an open surface, apply the right-hand rule.

11
    B  
Example 2 - 14 Given F  a x xy  a y 2 x, evaluate  F  dl , along the quarter - circle
A
y shown as following.
B   B     
B (sol)  F  dl   ( a x xy  a y 2 x )  ( a x dx  a y dy  a z dz )
A A
r =3 B 0 3
o A
x   ( xydx  2 xdy )   x 32  x 2 dx  2  32  y 2 dy
A 3 0

1 0 3 
 (9  x 2 ) 3 / 2  [ y 9  y 2  9 sin 1 3y ]  9(1  )
3 3 0 2
Example 2-15
    
F  ar k1 / r  a z k 2 z evaluate the scalar surface integral  F  ds
s

over the surface of a closed clinder about the Z-axis specified by

Z=  3 and r = 2

(Sol)
z        
 F d S 
S
  an ds 
F
top
  an ds 
F
bottom
  an ds
F
side
3 face face face
2
 k    k  
 
top
( ar 1  a z k 2 z )  a z ( rdrd )   ( ar 1  a z k 2 z )  (  a z )( rdrd )
r bottom
r
y face face

 k1  
x -3  side r
( a r  a z k 2 z )  ( a r rddz )

face

2 2 2
2 2 3
  k zrdrd
0
0
2 z 3 
0  (k zr )drd
0
2 z 3 
3  k ddz
0
1

 3k 2  2  2  3k 2  2  2  k1  2  6
 24k 2  12k1

2-6 Gradient of a Scalar Field

Ref. Fig. 2-24
  dV
(V )  grad V  a n : represent both the magnitude and direction of the
dn

maximum space rate of increase of a scalar.

12
meanwhile,
dV dV dn dV dV   
    cos  an  a  V  a
d dn d dn dn
 
 dV  V  a d  V  d

  V  V  V  V  V  V
V  au1  au 2  a u3  a u1  au 2  au 3
 1  2  3 h1u1 h2 u 2 h3 u 3

     
   au1  au2  a u3
h1u1 h2 u 2 h3 u 3

 
a r  a 
*  a ,  a r (自行證明)
 
  
Example 2 - 16 For E  V , determine E at the point (1, 1, 0), if
y
(a) V  V0 e  x  sin( ) (b) V  V0 R cos θ (課本有誤)
4

(sol)
   V  V  V
(a ) E  V  (a x  ay  az )
h1x h2 y h3 z
y y y
[V0 e  x  sin( )] [V0 e  x  sin( )] [V0 e  x  sin( )]
 4  4  4
 ( a x  ay  az )
x y z
 y   y 
 [a x (V0 e  x  sin( ) )  a y ( V0 e  x  cos( ))  a z (0)]
4 4 4 ( x, y, z )  (1,1,0)
 2   2 2   
 a xV0 e 1   a y V0 e 1   V0 e 1  (a x  a y )
2 4 2 2 4
   V  V  V
(b) E  V  (a R  a  a )
h1R h2  h3 
  (V0 R cos θ )   (V0 R cos θ )   (V0 R cos θ )
 ( a R  a  a )
R R R sin 
  
 [a RV0 cos ) )  a V0 sin  a (0)]
( x, y, z )  (1,1,0)  ( R, ,  )  ( 2 ,  2,  4)
   
 a V0  -a zV0 (在  /2的位置，a  -a z )

13
2-7 Divergence of a Vector Field
 
 
div A    A  im
 A  ds
s 
v : the divergence of a vector field A at a
v  0

point as the net outward flux of A per unit volume as the volume about the

point tends to zero.

 1   
 A  [ (h2 h3 A1 )  ( h1h3 A2 )  (h1h2 A3 )] (背法：2/3 法則)
h1h2 h3 u1 u 2 u 3

Example 2-17 Find the divergence of the position vector to an arbitrary point.
(Sol)
     
OP  a x x  a y y  a z z  a r r  a z z  a R R

a. Cartesian coordinates
        
 OP  (a x  ay  a z )  (a x x  a y y  a z z )
x y z
x y z
   3
x y z

b. Cylindrical coordinates

1    1
 OP  [ (r  1 r]  (1 1 0)  (1 r  z)]   3r  3
1 r  1 r  z r

c. Spherical coordinates
1   
 OP  [ (R  R sin  R]  (1 R sin  0)  (1 R  0)]
1 R  R sin R  
1
  3R 2 sin  3
R sin
2

14
 2-8 Divergence Theorem
  
   A dv   A  ds (背)
v s
 

  A  im
 A  ds
s
hint: v
v0
  2  
Example 2 - 19 Given A  a x x  a y xy  a z yz , verify divergence
theorem over a cube with one unit on each side.

      
(sol) Divergence theorem :    A dv   A  ds , 已知 A  a x x 2  a y xy  a z yz
v s
 1  (1 1  x )  (1 1  xy )  (1 1  yz )
2
(a )    A dv   [   ]dxdydz
v v1 1 1 x y z
1 1 1 1 1 1 1 1 1 3 1
    (3 x  y )dxdydz     3 xdxdydz     ydxdydz    2
0 0 0 0 0 0 0 0 0 2 2
             
(b)  A  ds   A  ds   A  ds   A  ds   A  ds   A  ds   A  ds
s front face back face left face right face top face bottom face

1 1     1 1    
  (a x x 2  a y xy  a z yz )  a x dydz    (a x  a y xy  a z yz )  (  a x )dydz
2
x
0 0 x 1 0 0 x0
1 1     1 1    
  (a x x 2  a y xy  a z yz ) ( a y ) dxdz    (a x x
2
 a y xy  a z yz )  a y dxdz
0 0 y0 0 0 y 1
1 1     1 1    
  (a x x 2  a y xy  a z yz )  a z )dxdy   (a x x 2  a y xy  a z yz )  (- a z )dxdy
0 0 z 1 0 0 z0
1 1
 1 0  0    0  2 (a)  (b) 故得證
2 2
 
Example 2 - 20 Given F  a R kR, verify divergence theorem for the
shell region enclosed by spherical surface at R  R 1 and R  R 2
(R 2  R 1 ) centered at the origin. (自己試看看)
15
2-9 Curl of a Vector Field
  
  [an  A  d]max
curl A    A  im c

S 0 S
     
The component of   A in other direction au is au  (  A)
     1  
 (  A) u  a u  (  A)  Sim
u  0 S
( 
u cu
A  de )

Where the direction of the line integral around the contour Cu bounding


S u and the direction au follow the right-hand rule.

ax ay az
   
 A 
行列式法 x y z
Ax Ay Az

for general orthogonal curvilinear coordinates (u1 , u 2 , u3 ) :

  
au1 h1 au 2 h2 au3 h3 直接乘，不作微分
 1   
 A  (背法：2/3 法則)
h1h2 h3 u1 u 2 u3
h1 A1 h2 A2 h3 A3 進行微分運算

2-10 Stoke’s Theorem

  

aj  A d
cj
 (  A ) j   im S j
S j  0
   
 (  A ) j   S j   A  d  .......( 1)
cj

(A) (B)

16
    
im  j 1 
N
from (A)  (  A ) j  S j  (  A )  S ..........(2)
S j  0 S

   
im  j 1    d ............(3)
N
from (B)  A  d   A
S j 0 cj c

combing eqs. (2) and (3),we obtain

   

S
(  A )  dS    d :
A
c
Stoke’s theorem (背)
 
if surface S is a closed one   (  A )  dS  0 (because closed path c =0)
S
  
Example2 - 22 Given F  ax xy  a y 2 x, please verifyStoke's Theoremover a
quarter - circulardisk with a radius 3 in the first quadrent.
(Sol.)    
Stoke' s theorem 
S
(   F )  d S    d
F
c
  
y ax ay az
    
B (a )   F   a z (2  x)
x y z
xy  2 x 0
r =3
O x   3 9 y 2   
A
S   F  ds  0 3  a z (2  x)  a z dxdy  9(1  2 ) ........(A )
        
(b )  F  d    F  d    ( a x xy  a y 2 x )  ( a x dx  a y dy  a z dz )
C OABO OABO

 
OABO
xydx 
OABO
 2 xdy   xydx   xydx   xydx  (  2 xdy   2 xdy )   2 xdy )
OA AB BO OA AB BO
3 0 0 0 3 0
  x  0 dx   x 32  x 2 dx   0  ydx  (  2  xdy   2 32  y 2 dy   2  0 dy )
0 3 0 0 0 3

 
 0  ( 9)  0  ( 0  9   0)  9(1  ) .......(B)  (A)  (B) 故得證
2 2 17
2-11 Two Null Identities
(1) Identity I : the curl of the gradient of any scalar field is identically zero.
   (V ) 0 (記)

Proof: from Stoke’s theorem 

 
   (V )  ds   V  d 
S c


here,  V  d    dV  0
c

(2) Identity II : the divergence of the curl of any vector field is identically
zero.

  (  A)  0 (記)

Proof: from divergence theorem 

  
   (  A)dv   (  A)  ds
v s

here,
    
 (  A )  ds   (  A)  a n1 ds   (  A)  a n 2 ds
S S1 S2
 
  A  d    d  0...( C1與C 2 相同迴路但是方向相反)
A
C1 C2

2-12 Helmholtz’s Theorem

# Definitions of solenodial and irrotational fields


(i) A vector field F is solenodial when the divergence of this field is

egual to zero    F  0

18
 
(ii) A vector field F is irrotational when the curl of this field is egual to

zero    F  0

Therefore, a vector field F is
 
1. Solenooidal and irrotational if   F  0 and   F  0

 
2. Solenooidal but not irrotational if   F  0 and   F  0

 
3. Irrotational but not solenooidal and if   F  0 and   F  0

 
4. Neither solenooidal nor irrotational if   F  0 and   F  0

Helmholt’s Theorem : A vector field (vector point function) is determined to

within an additive constant if both its divergence and
its curl are specified everywhere.

Proof: A general vector field F can be decomposed into an irrotational part
 
Fi and a solenoidal part Fs

i.e.
  
F  Fi  Fs

 
  Fi  g   Fs  0
with   
  Fi  0   Fs  G


where g and G are assumed to be known. We have
    
  F    ( Fi  Fs )    Fi    Fs  g
and      
  Fi    ( Fi  Fs )    Fi    Fs  G
 
in view of identity 1,   Fi  0  Fi  V

    
in view of identity 2,   Fs  0  Fs    A Thus F  V    A

19
    
Example 2 - 23 Given F  a x (3 y  c1 z )  a y (c2 x  2 z )  a z (c3 y  z ),

(a) If F isirrotational, please determine c1 , c2 , and c3
 
(b) If F  V , please find V .

(sol)
  
1 a x 1 a y 1 a z
  1   
(a)    F  0 
1  1 1 x y z
1  (3 y  c1 z ) 1  (c2 x-2 z ) 1  (c3 y  z )
     
 c3 a x - c1a y  c 2 a z - 3a z  2a x  0a y  0  c1  0 ; c 2  3 ; c 3  2

      V  V  V
(b)  F  V  3 ya x  (3x - 2z)a y  (2 y  z ) a z  -(a x  ay  ax )
1  x 1  y 1  z
 V
 x  3 y  V  3 xy  k1 ( y, z )

 V
  3 x  2 z  V  3 xy  2 yz  k 2 ( x, z )
 y
 V 1
  2 y  z  V  2 yz  z 2  k 3 ( x, y )
 z 2
1
 V  3 xy  2 yz  z 2  k , k is a constant, which can be determinted
2
by the boundary condition such as V  0 at infinity (無窮遠處).

20
 Chapter 3 Static Electric Fields
3-1 Introduction
# In electrostatics, electric charges (or electric sources) are at rest, and
electric fields do not change with time . Therefore there are no magnetic
fields.
# For two fixed positioned point charge q1 and q 2 , from Coulomb’s law the

vector force F12 exerted by q1 on q 2 can be written mathematically as

q1   R12
q2
  qq
F12  a R12 K 1 22 .......( 1 )
R12

 R12
a
here R12 R 
12

   V
F  E 
 D

3-2 Fundamental Postulates of Electrostatics in Free Space


Electric field intensity E : the force per unit charge that a very small
stationary test charge experiences when it is placed in a region where
an electric field exists.

 F
That is: E  im ( N ) or (V )......( 2 )
q0 q
c m

An inversion relation of eq. (2) give the force F on a stationary charge q
  
in an electric field E to be F  qE (N)
# Two fundamental postulates of electrostatics in free space :

(i.e. divergence and curl of E )
1
  
  E  ......( 3 ) and
0

  E  0........( 4 )  means that static electric fields are irrotational.
here,  : the volume charge density of free charge. ( c m3 )
 0 : the permittivity of free space.
Eqs. (3) and (4) are point relations and this holds at every point in space.

# For an arbitrary volume V, the total field can be obtained by

 
V

v
  E dv   dv.........(5)
v
0
S   Q
Q   E  ds  .............(6)
l s
0
 Gauss Law

here, Q is the total charge contained in volume V bounded by surface S.

from (4)
   from Stokes theorem  
  E  0........( 4)   (  E )  ds  0    d   0  Kirchhoff ' s Law( KVL)
E
S c

Postulates of Electrostatics in Free Space

Differential Form Integral Form

    Q
E 
0 
S
E  ds 
0
  
 E  0
  d  0
c
E

3-3 Coulomb’s Law

(1). Electric Field Due to a Simple Point Charge
Case 1: The charge q is positioned at the origin.
Case 2: The charge q is not positioned at the origin.
2
     q
  ds   (aR ER )  aR dS 
S
E
S
0
  q
q E p  aqp
 E R  ds   2
S
0 4 0 R  R'
 
 E R  4R 2 
q  R  R'
0 and aqp   
R  R'
q  
 ER   q( R  R' )
4 0 R 2  Ep    3 (V / m)
   q 4 0 R  R'
 E  aR ER  aR ( V / m)
4 0 R 2

(We can veerify   E  0 for this.)
Example 3-1 Determine the electric field intensity at P(-0.2, 0, -2.3) due
to a point charge of +5 (nC) at Q(0.2, 0.1, -2.5) in air. (unit: m)

Example 3-2 A total charge Q is put on a thin sphrical shell of radius b.

Determine the electric field intensity at any arbitrary point inside the shell.
(自行參考)

(2). Electric Field Due to a System of Discrete Charges

q2
。P      
z  q1 ( R  R1 ' ) q 2 ( R  R2 ' ) q n ( R  Rn ' )
E   3   3      3
q1 
' R
q3 4 0 R  R1 ' 4 0 R  R2 ' 4 0 R  Rn '
 R2  q4  
R1' R3'  ' .. q k ( R  Rk ' )
n

R4 .    3 (V / m )
' . k 1 4 R  R '
0 k
O R n qn
y
3
x
Example : There is an electric dipole with two charges, +q and –q, and a
distance d between them. Find E at point P if R>>d.
 
 d  d
  q (R  2 ) q
(R  )
2
E  3   3
4 0  d 4 0  d
R R
2 2
 
 d  d
(R  ) (R  )
q 2 2
 { 3   3}
4 0  d  d
R R
2 2
Consider if R>>d,
 3   3 3
 d   d  d  2  2   d 2
2
R  ( R  )  ( R  )    R  R  d  
2  2 2   4
   
R  d  3 3 R d
 R 3 [1  2 ] 2  R 3 [1  ]
R 2R 2
Similarly,
 3   3 3
 d   d  d  2    d 2 2
R  ( R  )  ( R  )    R 2  R  d  
2  2 2   4
   
Rd  3 3R  d
 R 3 [1  2 ] 2  R 3 [1  ]
R 2R 2
     
 q 1  d 3R  d  d 3R  d
E  [( R  )(1  )  ( R  )(1  )]
4 0 R 3 2 2R 2 2 2R 2
 
q 3R  d  
 [ R  d ] (V / m)
4 0 R 3 R 2

# Electric Dipole Moment

. q
 
   1 RP  

d P  qd Thus, E  4 R 3 [3 R 2 R  P]
0
q
4
If the dipole lies along the z-axis
   
 P  a z P  P (aR cos  a sin  )
  
 1 RaR  P ( aR cos  a sin  )   
 E [3   R a  P ( a cos   a sin  )]
4 0 R 3
R R 
R2
P  
 ( a 2 cos   a sin  ) (V / m)
4 0 R 3 R 

(3). Electric Field Due to a Continuous Distribution of Charge

(a) . If the charge is distributed in a volume v’ with a

volume charge density  (c / m ) :

3

 1  
4 0 v ' R 2
E a R dv'

or 1 R
  
4 0 v ' R 3
dv'

(b) . If the charge is distributed on a surface S with a

volume charge density  s (c/m2) :

 1  s
4 0  R 2
E a R ds' (V / m)
S'

(c). If the charge is distributed along a line L’ with a

line charge density   (c/m2) :
 1  
4 0 L' R 2
E a R d ' (V / m)

5
 3-4 Gauss’s Law and Applications
  
Gauss’s Law: E  : differential form
 0
total outward flux of the electric-field intensity
  Q
or 
S
E  ds  : integral form
0
Example 3-5 :
Known: (1). infinite long line (2)uniform charge density  
r (Sol):
from Gauss’s Law: (先取一長度為 L 的圓柱高斯面 S)
L
L
   Q 2 2
  
 E  ds  
s
0 
L 0
 a r E r  a r rd dz 
0
L

Cylindrical Gaussian surface 2


infinite long line  E r 2rL  L
0
   
 E  Er ar  a r (V / m)
2 0 r

Example 3-6 : Known (1) infinite plannar charge

(2) uniform surface charge density  s

(Sol.)
  Q
from  E  ds   (先取高斯面 S，長方體、立方體或圓柱體皆可)
0
   
E  E z a z , ds  a z ds
(i)on the top face :   
E  ds  E z a z  a z ds  Eds
   
E   E z a z , ds  a z ds
(ii)on the bottom face :    
E  ds  ( E z a z )  (a z ds )  E z ds

6
 
(iii)on the side faces, because the plane is infinite  E  0
Summarize (i)(ii)and (iii) to obtain :
  Q
 E  ds  Ez ds
s
top
face
 E z ds bottom 
face 0
, and ds  A, Q   s A

s A 
 2Ez A   Ez  s
0 2 0
     s 
 E  s az , z  0 ; E az , z  0
2 0 2 0

Example 3.7 Known: volume charge density =-0 for 0Rb, and =0

for R>b; find E ?

(sol)

(a). for 0  R  b
  Q
        : Gauss Law and, Q    dv
2
E d s E a a dS E 4 R
s Si
R R R R
0 v

4 3  
    0 dv    0 R  E  E R a R   a R 0 R (V / m)
v
3 3 0
(b). for R  b
  4 3
s   S R R R    v   v 0
    
2
E d s E a a dS E R 4 R and, Q dv dv 0 b
i
3
  0b 3
 E  E R aR  aR (V / m)
3 0 R 2

7
3-5 Electric Potential
  
# For a curl–free vector E    E  0 , from Identity I 

E can be defined as the grdient of an arbitrary scalar quantity like：
 
EV ……….(A)
Here, V is a scalar quantity and called electric potential.
  p2
W
p2
 
W    qE  d      E d
p1
q p1
W : 反抗電場所作的功 (與能量相同)

  p2

here, W  V2  V1  V2  V1    E  d ......( B ) (必背)

q p1

“potential difference” or “electrostatic voltage”

  q
Example 1: from Gauss Law :  E  ds   E R a R  a R dS  E R 4R 2 
s Si
0
q  q
E  E  ER aR  aR (V / m )
4 0 R 2
4 0 R 2
R2 q q R2 q 1 1
V21  Vp 2 - Vp1    a R  a R dR   (  )
R1 4 0 R
4 0 R R1
2
4 0 R2 R1

* * for n discrete point charges (q k ) located at R 'k ,
n
1 qk
V
4 0
 R  R
k 1
'
k

Example 2 : For two exist charges, +q and –q, with a distance of d, the
potential at point P can be written as V, thus

q 1 1
V (  )
4 0 R R

if R>>d, we have :
8
 d 1 d d
R  ( R  cos )   ( R  cos ) 1  R 1 (1  cos )
2 R 2 2R
d 1 d d
R  ( R  cos )   ( R  cos ) 1  R 1 (1  cos )
2 R 2 2R
q d d
V  [ R 1 (1  cos  )  R 1 (1  cos  )]
4 0 2R 2R
   
qd cos  qd a  a P  aR
  z R
 (V ) ..........................................( D )
4 0 R 2 4 0 R 2 4 0 R 2
   V  V  V
E   V   ( a R  a  a )
h1R h2  h3
  qd cos    qd cos    qd cos 
 aR ( )  a ( )  a ( )
R 4 0 R 2
R 4 0 R 2
R sin   4 0 R 2
2qd cos   qd sin   p  
 a  a  ( 2 cos  a  sin  a) (V / m )
4 0 R 3 4 0 R 3 4 0 R 3
R R

(無法訂出高斯面的例子)

# The electric potential due to a continuous distribution of

charge confined in a given region:
(1) for a line charge
1 
V  d' (V )
4 0 L' R

(2) for a surface charge distribution

1 s
V  ds ' (V )
4 0 s' R

(3) for a volume charge distribution

1 v
V  dv' (V )
4 0 v' R

9
Example 3-9 There is a disk with radius b and carying a uniform

surface charge density  s , please find E on the axis.
(無法訂出高斯面的例子)
( Sol ) 先假設要求電場強度的點位在在disk中心上方Z的位置)
R  Z 2  r ' 2 , ds '  r ' ddr '
2 b
1 s  1
V 
4 0  R
ds '  s
4 0  1
r ' dr ' d
s' 0 0
(Z  r ' )
2 2 2

s 1
 [( Z 2  b 2 )  Z ]
2
(V )
2 0
  s 1

a [1  z ( z  b ) ], z  0
2 2 2
    V  z 2 0
E   V   a z  1
z    s
a [1  z ( z  b ) ], z  0
2 2 2
 z 2 0
1 1
b2 2 b2 b2
and z ( z  b )  (1  2 )  1  2 , if z  0 or   1  2 , if z  0
2 2 2

z 2z 2z
thus
  s b2   sb2  b 2  s  Q
 az [1  (1  2 )]  a z  az  az , z0
  2 0 2z 4 0 z 2
4 0 z 2
4 0 z 2
E
 a  s [1  ( 1  b )]   a  s b   a b  s   a
2 2 2
Q
, z0
 z
2 0 2z 2 z
4 0 z 2 z
4 0 z 2 z
4 0 z 2

Example 3-10 : (無法訂出高斯面的例子)

( sol )
L
 R  z  z' , z 
2
L L
 dz '2
 dz ' 2
V  [ ]  
4 0  L R 4 0  L z  z '
2 2

L
z
 2 ], z  L
  ln[
4 0 z
L 2
2
   dV  s L L
 E   V   a z  az ,z 
dz L 2
4 0 [ z 2  ( ) 2 ]
2
10
3.6 Conductors in Static Electric Field
According to their electrical properties, the material can generally be
classified into three types: conductors , semiconductors, and insulators (or
dielectrics)

說明:
Introduce some positive (or negative) charge in
the interior of a conductor, three will be no
charge in the interior of a conductor (  =0) and
the all charge will reach the conductor surface to
produce an electric field normal to the surface .

Note: Inside a conductor, under static

 0
conditions : 
E0

 
  E  0
 
  E  d   Et  w  0  Et  0 (E t : tangential component)
abcda

      s  s 
  E    E  ds  E n  s  0  s  s  E n  s  s
0 S
0  0 s  0
 s  
 En  a n , (a n : unit vector normal to the surface)
0
Summary:
Boundary conditions at a conductor /free space interface:
(1) Tangent component Et=0 (2) Normal component E n  0
Example 3-11 (重要)

Ro
Q Ri
air

air Conducting
shell 11
3.7 Dielectrics in Static Electric Field
# For a dielectric material, an external electric field
will polarise the material and creat electric dipoles.

Define the volume density of electric dipole moment P as :
nv 

  Pk k
P  im k 1
(c / m 2 )
v  0 v
 
P  aR
from eq. (D) (see page 9)  dV  dv'
4 0 R 2
 
1 P  aR
V  ' 2 dv'
4 0 R
v

 R  ( x  x' )  ( y  y' )  ( z  z ' )

2 2 2 2

and   1 aR
  ' ( ) 
R R2
1   1
V   P  ' ( )dv'
4 0 v ' R

  '   '  P 1     1
[Note:  ' ( f A )  f   A  A   f   ' ( )  'P  P  ' ( ) ]
R R R

1  P 1 1  
4 0 v ' 4 0 v ' R
V  ' ( )dv'  ( )'P
R
  
1 P  1 ('P)
4 0 s ' R 4 0 v ' R
V   a n ' ds' dv'

Summary :
 
(1) P  a n '   ps : polarization surface charge density
 
(2)    P   p : polarization volume charge density

(3) Total charge in dv' is:

   
  ps ds'   p dv'   P  an ' ds'    Pdv'
s' v' s' v'
   
    Pdv'    Pdv'  0 12
3.8 Electric Flux Density and Dielectric Constant
In a dielectric the total volume charge density is decomposed by a free
volume charge density  and a polarization volume charge density
 
p  (Note: Referring   E   /  0 in free space)
  1
  E  (   p )
0
   
   ( 0 E )    (  P)
  
   ( 0 E  P )  

Define the electric flux density (or electric displacement ) D , such that
  
D  0E  P (c/m2 ) (要熟記)
 
 D  
 
    D dv    dv
v v
 
  D  ds  Q : Gauss Law
s

If the dielectric properties of the material are linear and isotropic

 
 P   0 xe E , where xe is electric susceptibility .
     
thus, D   0 E   0 xe E   0 (1  xe ) E   0 r E  E (要熟記)


where  r  1  xe  : the relative permittivity(or the dielectric constant)
0
 : the absolute permittivity
If  r is independent of position, the medium is said to be homogenous.

And, a linear、homogenous and isotropic medium is called a

simple medium.

13
For an anisotropic material:
 Dx  11 12 13   Ex 
 D      E 
  
y 21 22 23  y 
 Dz   31  32  33   Ez 

The maximum electric field density beyond causing the dielectric material
becoming conducting and resulting large current (i.e. dielectric breakdown )
is called the dielectric strength.

Example 3-12 :
Ro (sol)
Q Ri
 then  then   
air (提示) Find E  V ( E  V )  D ( D  E )
air Dielectric then    
shell  P ( D  P   0 E )

(a) for R>R0     (i.e. in air)0

(i) from Gaussian’s law

  Q 2 
  Q
 E  ds    RE a  R 2
sin d da 
s 0 0 0
R
0
Q  Q 
 4R 2 E  E aR
0 4R  0
2

 R R
Q  
(ii) V R  V    E  d   V R    a  dR ' a
 2 R R
  4 0 R '
Q R Q Q
  (V )  V R0  (V )
4 0 R'  4 0 R 4 0 R0
14
   Q 
(iii) D   0 E  a R (c / m 2 )
4R 2

  
(iv) P  D   0 E  0 (c / m 2 )
(b) for R  R  R      (i.e. in dielectric shell)
i 0 r 0

(i) from Gaussion’s law

  Q 2 
  Q
 E  ds     EaR  R sin ddaR 
s  0 0  0 r
Q
 E 4R 2 
 0 r
 Q 
E aR (V / m)
4 0 r R 2

  R R
Q  
(ii) VR  VR0    E  d   VR  VR0   4  a R  dR' a R
R0 R0 r 0 R'2
Q R Q Q 1 1 Q 1 1 1
 VR0    (  ) [(1  )  ] (V )
4 r  0 R ' R0 4 0 Ro 4 r  0 R R0 4 0  r R0  r R
Q 1 1 1
and VRi  VR ( R  Ri )  [(1  )  ]
4 0 r R0  r Ri
  Q  Q 
(iii) D   r  0 E   r  0 a  aR (c / m 2 )
4 r  0 R 2 4R 2
R

   Q  Q  Q 1 
(iv) P  D   0 E  aR   0 aR  (1  )aR (c/m 2 )
4R 2
4 r  0 R 2
4R 2
r

(c) for R  R     (i.e. in air)

i 0

  Q  Q 
(i)  E  ds   E  aR (V / m)
s 0 4 0 R 2

15
 R  
(ii) VR  VR    E  d  
i
Ri

R
Q  
VR  VR   a  dR ' a
4 0 R '2
i R R
Ri

Q 1 1 1 Q 1 1
 [(1  )  ] (  )
4 0  r R0  r Ri 4 0 R Ri
Q 1 1 1 1 1
 [(1  )  (1  )  ] (V )
4 0  r R0  r Ri R
  Q  Q 
(iii) D   0 E   0 aR  aR (c / m 2 )
4 0 R
2
4R 2
   Q  Q 
(iv) P  D   0 E  a   aR  0
4R 2 4 0 R 2
R 0

(d) Surface charge density:

(i)on the inner shell surface :

  1 1  
 sp  P ( R  Ri )  an  (1  ) a  (  a )
4Ri r R
R  Ri 2 R

1 1
 2 (1  ) (c / m 2 )
4Ri r
(ii)on the outer shell surface
  1 1  
 sp  P ( R  R0 )  an  2 (1  )a  a
4R0 r R R
R  R0

1 1
 (1  ) (c / m 2 )
4R0 r
2

16
(e)Volume charge density:
   Q  1
 p    P    [(1  aR ] )
 r 4R 2
1  1 Q
 2 [ R 2 sin   (1  ) ]0
R sin  R  r 4R 2

Example 3-13: Two spherical conductors:

Known (1) total charge: Q

(2) d  b or b
1 2

Find (1) the charge on the two surface


(2) E at the sphere surface
(sol) (提示：先假設有 Q1 在 b1 球;有 Q2 在 b2 球,且 Q1+Q2= Q)
 
b1

(i) V1  V    E  d 


b1
Q1   Q1 b1 Q1
 V1    aR  dRaR  
 4 0 R 2 4 0 R  4 0b1
b2
Q2  
V2  V    aR  dRaR
 4 0 R 2
b2
Q2 Q2 b2 Q2
V2    dR  
 4 0 R 2
4 0 R  4 0b2
Q1 Q2 Q1 Q2
V1  V2    
4 0b1 4 0b2 b1 b2
b1 b2
and Q1  Q2  Q  Q1  Q, Q2  Q
b1  b2 b1  b2

17
  Q1  1 b1Q  Q 
(ii) E1  a  a  a
4 0b1 4 0b1 b1  b2 4 0 (b1  b2 )b1
2 R 2 R R

 Q2  1 b2Q  Q 
E2  2 aR  aR  aR
4 0b2 4 0b2 b1  b2 4 0 (b1  b2 )b2
2

3.9 Boundary Condition for Electrostatic Fields

(a) for path abcda

let bc=da= h and approach zero (i.e. h 0)

           
then
abcda
 E  d   E1  w  E 2  (  w )  ( E1t  E1n )  w  ( E 2 t  E 2 n )  (  w )

 E1t w  E 2t w  0
 
 E1t  E 2t ; E1t , E 2t : the tangential components of E1 and E 2

 This means that the tangential component of an E field is continuous
across an interface.
(b) for a small pillbox
let h be vanishingly small (i.e. h 0)
     
then s
D  ds  D1  a n2 s  D2 an1s   s s

    
 an 2  ( D1  D2 )   s ( an1  - an 2 )
    
 an 2  ( D1t  D1n  D2t  D2 n )   D1n  D2 n   s   1 E1n   2 E2 n   s

此處D1的方向是射入medium 2; 與課本3.121b所定的方向相反
18
Boundary conditions :
tangential components  E1t  E2 t
  
normal components  a n 2  ( D1  D2 )   s

Cases:

(1) if medium 2 is a conductor :


 D1n   1 E1n   s (or  D1n   1 E1n   s , 視D2 的方向而定 )
 D2  0  
 E1t  E 2t

(2) if no free charges at the interface

  s  0  D1n  D2 n
  
Example 3-14: In the following structure, determine Ei , D i and Pi
inside the lucite.
   
E o  axE 0 Ei Eo
   
D o  a x 0 E 0 Di Do

Free Lucite Free

space =3.2 space

Example 3-15:

19
3.10 Capacitance and Capacitors

# Introduction
Inside the conduction :

free space E0 ;  0

En At the interface :
Et s
Et  0 ; E n 
conductor 0

說明：對一導電體(conductor)而言，導體內部維持等電位(因此內部沒

有電場)，自由電荷將累積在導體的表面上而對外形成電場。外加電

荷增加則電場變大 ( kE  (k s ) a n ) ；電場變大則電位增加


0
  Q
( kE  (kV ) )，所以電荷與電位維持常數變化關係  C (F)
V

C: called the capacitance of the isolated conducting body.

# Consider two conductors separated by free space or a dielectric

medium, for example :

Q
C
V12

Generally , capacitance C can be determined by the following

steps:

20
1. Choose an appropriate coordinate system for the given
geometry;
2.Assume charges +Q and –Q on the conductors :

3.Find E from Q (use Gauss law);
 1  
4.Find V from E (use V    E  d  );
2

Q
5. C .
V

Example 3-17 : Find capacitance for the following parallel-plate capacitor

Conductor, Area S y

d Dielectric,  +Q Conductor, Area S

d
(sol) 
Conductor, Area S E Dielectric, 
0 -Q x
Conductor, Area S

(1) Choose Cartesian coordinate

(2) Assume  Q and - Q on the upper and lower plate conductors , respective ly,
(3) on the upper plate :  s   Q/S
  Q   Q
for 0  y  d, from Gauss law  E  d s    a y E  (  a y ) S 
s
 
Q   Q
E  E  a y
S S
  d
 Q 
d
Q
(4)  V  Vupper - Vlower    E d y    (  a y )  a y dy  d
0 0
 S  S
Q S
(5)  Q  CV  C   (F)
V d
Example 3-18 : Find capacitance for the following cylindrical capacitor

b a

L
21
 (Sol)
Put  Q and - Q on the inner and outer conductors , respective ly,
-Q

b a
for a  r  b +Q
r
  Q 
From Gauss law   d s   E  2r  L
s
E
 L
Q    Q   a

a
Q  Q b
  E  ar E  ar and, Vab    E d r    a r a r dr  ln
 2r  L b b
2r  L 2 L a
2 L
for Q  CV ab  C  (F)
b
ln
a

Example 3-19 : Find capacitance for the following spherical capacitor.

Ro
Ro Ri
Ri
R
+Q

 -Q
(Sol)
Put  Q and - Q on the inner and outer spherical conductors , respective ly,
for R i  R  R o
  Q   Q
From Gauss law s
E  d s    R R
a E

 a R  Rsin
s
 d  d  

Q    Q   Ri

 E  4R  2
 E  aR E  aR and, Vio    E d R
 4 R 2 Ro
Ri
 Q  Q 1 1
   aR  a R dR  (  )
Ro
4 R 2
4 R i R o
4
for Q  CV io  C  (F)
1 1

Ri Ro

3-11 Electrostatic Energy and Forces


free space Q2 ＊ At beginning, to bring the charge Q1 from
infinity to a set position requires no work since

R12 neither the charge nor the electric field existed in free space.

22
Q1
 ＊To bring the charge Q from infinity against the field of the charge Q
2 1

in free space to a distance R12 , the amount of work required is :

Q1 or Q2
W2  Q2V2  Q2  Q1  Q1V1  Q1V1  Q2V2
4 0 R12 4 0 R12
1
 W2  (Q1V1  Q2V2 )
2
＊Suppose another charge Q3 is brought from infinity to a point that is
R13 from Q1 and R23 from Q2 ; an additional amount of work is
required that equals:
Q1 W  Q3V3  Q3 (
Q1

Q2
)
 4 0 R13 4 0 R23
1 Q1Q2 Q1Q3 Q2Q3
W3  W2  W  (   )
R13 4 0 R12 R13 R23
R12
1 QQ QQ Q1Q3 QQ Q2Q3 Q2Q3
 [ 1 2  1 2   1 3   ]
2 4 0 R12 4 0 R12 4 0 R13 4 0 R13 4 0 R23 4 0 R23
Q2   Q3 1
 [Q1 (
Q2

Q3
)  Q2 (
Q1

Q3
)  Q3 (
Q1

Q2
)]
R23 2 4 0 R12 4 0 R13 4 0 R12 4 0 R23 4 0 R13 4 0 R23
1
 [Q1V1  Q2V2  Q3V3 ]
2

# The general expression for the potential energy of a group of N discrete

point charges at rest is:
1 N
We   QkVk ( J ) or (eV ) ,
2 k 1
N Qj
where Vk  
j 1 4 R
jk 0 jk

1 '

2 v '
# For a continuous charge distribution of density  : We  Vdv

Example 3-22 :
dR
R
b


23
(sol)
R 2 
QR 1  R3
  R sin dddR    2 ( cos ) 0
2
VR   ' '

4 0 R 4 0 R 0 0 0
4 0 R 3
R 3  2  2 R 2
 
4 0 R  3 3 0

The differential charge in a spherical layer of thickness dR is :

4 4 4
dQR   ( R  dR) 3   R 3   [ R 3  3R 2 dR  3RdR 2  dR 3  R 3 ]
3 3 3
4 4
 [3R 2 dR  3RdR 2  dR 3 ]    3R 2 dR  4R 2 dR
3 3
R 2
4 2 R 4
dW  VR  dQR   4R dR 
2
dR
3 0 3 0
4 2 R 4 4 2 R 5 4 2 b 5
b
W   dW   dR  b
 (J )
 3 0 5 15 0
0
0
3 0

1 '

2 v '
** Example 3-23 同 Example 3-22,但改用 We  Vdv 法來求.

# Electrostatic Energy in Terms of Field Quantities

1 1   ' 1   ' 1   '
We   Vdv     D Vdv     (VD ) dv   D  Vdv
'

2 v' 2 v' 2 v' 2 v'

1  ' 1   '      
  VD  ds   D  Edv (   ( DV )  V  D  D  V )
2 s' 2 v'

Consider: 1
 V
R  1
  1   V D  ds 
D, E  2 s R
 
R
if R  , then  VD  ds  0
S  R2

1   ' 1   '
2 v '
 We  D  E dv   E  E dv ( if for a linear medium)
2 v'

1 or 1  D ' 1 D2 '
  E  dv   D  dv  
2 '
dv
2 v' 2 v'  2 v' 
24
 1   1 2 D2
Let We   we dv  we  D  E  E 
'
( J/m 3 )
v'
2 2 2
we : eletrostat ic energy density
Example 3-24: Find the electrostatic energy stored in the folloing
Area S
parallel-plate capacitor.
(sol＞ +Q
V  d
 V  -Q
E  (a Z )
d
1  
d
1 V 1 S 1
We   E  Edv    ( ) 2 Sdz  ( )V 2  CV 2
2 v' 20 d 2 d 2
1
or or Q 2
 QV 
2 2C

電路方法：
dw e (t ) dq (t )
p (t )  v(t )  i (t )   v(t )   dwe (t )  v(t )  dq (t )
dt dt
1 Q2
  dwe (t )   v(t )  dq (t )  We  VQ 
2 2C

Example 3-25: Find capacitance of the the following cylindrical

capacitor by using the electrostatic energy formulas.

b a


(Sol) L -Q

b a +Q
r

for a  r  b
L
  Q Q    Q
From Gauss law  E  d s   E  2r  L   E  a r E  a r
s
  2r  L
1  
b
1 1 Q
and, W e    E  E dv    E dv    (
2 '
) 2  2r  Ldr
2 v' 2 v' '
2 a 2r  L
2 L
b
Q2 1 Q2 b Q2
 
4 L a r
dr  ln ,
4 L a
Compared to W e 
2C
 C
b
ln 25
a
Chapter 4 Solution of Electrostatic Problems
4.2 Poisson’s and Laplace’s Equations
# For electrostatics :  
  D   ....(1)
 
  E  0 ....(2) : irrotational
 
E  V ......(3)
 
In a linear and isotropic medium  D   E ... ( 4)
Substitution of eq.(3) and (4) into eq.(1)
 
   ( E )  
 
   ( V )  

for a medium that is also homogeneous 

      
  ( V )   becomes    (V )       V  


  2V  : the Poisson’s equation

 
and here  2     ：the Laplcian operator.

(1) in Cartesian coordinates

                 
 2V    [( a x  a y  a z )V ]  ( a x  a y  a z )  [( a x  a y  a z )V ]
x y z x y z x y z
 2V  2V  2V  2V  2V  2V  
 2  2  2  V  2  2  2 
2

x y z x y z 
     
( Note :   au1  au2  au
h1u1 h2 u 2 h3 u 3 3
1 h h A h h A h h A
 A  [ 2 3 1  1 3 2  1 2 3] )
h1 h2 h3 u1 u 2 u 3

(2) in cylindrical coordinates

      1  V 1  2V  2V
 V    [( a r 
2
a  a z )V ]  (r ) 2 
r r z r r r r  2 z 2
1  V 1  2V  2V  
 V 
2
(r ) 2  
r r r r  2 z 2 
1
 (3) in spherical coordinates
    
 2V    [( aR  a  )V ]
R R R sin 
1  2 V 1  V 1  2V
 ( R )  (sin  ) 
R 2 R R R 2 sin    R 2 sin 2   2
1  2 V 1  V 1  2V  
 V  2
2
(R ) 2 (sin  ) 2 
R R R R sin    R sin 2   2 

For charge free medium (i.e.   0 ) (or no free charge in a simple medium)

 2V  0 : called the Laplace’s equation

Example 4-1: Two plates of a parallel-plate capacitor, with a distance d

and a voltage difference of V0. Assuming negligible fringing effect at the
edges, determine (a) potential inside the capacitor, and (b) the surface
charge density on the plates.
y
(Sol) no free charge in this capacitor (i.e.   0 )
V0
 d d 2V dV
 V 0 2 0
2
 c1  V  c1 y  c2
dy dy
Here, c1 and c2 are constants under being determined by satisfying the
boundary conditions.
Boundary conditions:
V0    V  V 
at y=0, V=0 ; at y=d, V=V0 ;  V  y and E  V   ( 0 y )a y   0 a y
d y d d
(i) at the lower plate:
     V 0  V0
s D  ds  s   s E  (a y )s  s s  d s  s s   s  d
(ii) at the upper plate:
     V0 V
s D  ds  s   su E  (a y ) s  s su 
d
( s )  s su   su  0
d

2
 Example 4-2 Known: volume charge density =-0 for 0Rb, and =0 for
R>b; Find V by solving Poisson’s and Laplace’s equations.
free space =0

free space (sol)

b 
(a ) 0  R  b, ρ  -ρ0 , from  2Vi 
=-0 0
1  2 Vi 1  Vi 1  2Vi  0
 2 (R ) 2 (sin  ) 2 
R R R R sin    R sin 2   2  0
Since Vi is independant with the  and  , the above equation can be simplified as
1 d dV  d dV  dV 
(R 2 i )  0  ( R 2 i )  0 R 2  R 2 i  0 R 3  C1
2
R dR dR 0 dR dR 0 dR 3 0
dV  C    dV   C
 i  0 R  12  Consider Ei  Vi   a R i   a R ( 0 R  12 ),
dR 3 0 R dR 3 0 R
C 
at R  0, 12   ( Ei   unreasonable !) Therefore, C1  0
R
   dVi  
 Ei   a R 0 R, and  0 R  Vi  0 R 2  C1' ..............(A)
3 0 dR 3 0 6 0

(b) R  b, ρ  0 , from  2Vi  0
0
1  2 Vo 1  Vo 1  2Vo
 2 (R ) 2 (sin  ) 2 0
R R R R sin    R sin 2   2
Since Vo is independant with the  and  , the above equation can be simplified as
1 d dV d dV dV
2
(R 2 o )  0  ( R 2 o )  0  R 2 o  C2
R dR dR dR dR dR
dV C    dV  C
 o  22  Consider Eo  Vo   a R o   a R 22 ,
dR R dR R
 
因為球內外為同物質， 所以電場應有連續性， 亦即 Ei ( R  b)  Eo ( R  b)
   C     b3  b3
  a R 0 b   a R 22  C 2  0 b 3  Eo   a R 0 2 , Vo  - 0  C '2
3 0 b 3 0 3 0 R 3 0 R
 0b 3
當 R    Vo  C '2 應該為0  C '2  0  Vo  - ..............(B)
3 0 R
0 2  b3
因為電位具有連續性 , 所以 Vi ( R  b)  Vo ( R  b)  b  C1'  - 0
6 0 3 0 b
0 2    b3
 C1'   b  Vi  0 R 2  0 b 2 , .0  R  b ; Vo  - 0 , R  b
2 0 6 0 2 0 3 0 R

3
4.3 Uniqueness of Electrostatic Solutions
Uniqueness theorem: A solution of Poisson’s equation that satisfies the given
boundary conditions is a unique solution.

4.4 Method of Images

Method of Images : The method of replacing bounding surfaces by appropriate
image charges in lieu of a formal solution of Poisson’s or
Laplace’s equation.

(1). Point charge and conducting planes

Solve such a problem by using another method – image method.

 The formal procedure :

In Cartesian coordinates :
 2V  2V  2V
 2V     0 , for y>0 except at the point charge
x 2 y 2 z 2

and the solution V(x,y,z) should satisfy the following conditions:

1. V(x, 0, z)=0 (potetial on the grounded plane conductor)
Q
2. V  as R  0 (R: distance between field point and source Q)
4 0 R

3. V  0 as R  ( x   , y , z    )
4. V(x, y, z)= V(-x, y, z) ; V(x,y,z)= V(x, y, -z)

 Apply the image method

Step1: Put image charges
(0,d,0)

(0,-d,0)
4
 Step2: Find V
Q Q Q 1 1
V(x, y, z)=   (  )
4 0 R 4 0 R 4 0 R R

R   ( x  0) 2  ( y  d ) 2  ( z  0) 2
here,
R   ( x  0) 2  ( y  d ) 2  ( z  0) 2

Step3: Find E
  V  V  V 
E   V =  ax  ay  az
x y z
Note: In this problem the point charge +Q and –Q cannot be used to calculate the V

or E in the y<0 region.

Example 4-3

   
For the net force on Q  F  F1  F2  F3

 Q2 
where F1  (a y )
4 0 (2d 2 ) 2

 Q2 
F2  (a x )
4 0 (2d1 ) 2

 
 Q2 2d 1 a x 2d 2 a y
F3  (  )
4 0 (4d1  4d 2 )
2 2 2 2 2 2
4d 1  4d 2 4d 1  4d 2

 Q2  d1 1  d2 1
Therefore, F  [a x (  )  ay (  )
16 0 2 2
3
d1
2
2 2
3
d2
2
(d1  d 2 ) 2
(d1  d 2 ) 2

(2). Line charge and Parallel Conductor Cylinder

5
 The image line charge must lie somewhere along op say at point Pi ,
which is at a distance d i from the axis.
Let us assume :  i    

The E and V of point M produced by the source line charge   are :
  
E   a r , where r is the distance from line charge of density  
2 0 r
r   r 1  r
V      E  a r dr    dr   ln 0
r0
2 0 r0 r 2 0 r

r0 : the reference point


The E and V of point M porduced by the image line charge  i are :
   
E i  ar , where r is the distance from the line charge of density  i
2 0 r
ri r
   (   ) i 1  r
V  i    E  i  a r dr   dr   ln i
r0
2 0 ro r 2 0 r0

 r  r  r
VM  V    V i  ln 0   ln i   ln i
2 0 r 2 0 r0 2 0 r
Because the surfaces of the conducting cylinder should be equipotential, therefore
ri
choosing =constant such that the located point Pi has to be chosen to make
r
OMPi  OPM .
Pi M OPi OM r d a a2
  or i  i   cons tan t  d i  (image charge 位置選定公式)
PM OM OP r a d d

Example 4-4 Determine the capacitamce per unit length between the
two long, parallel, circular conducting wires of radius a. The axes of the
wire are separated by a distance D.
＜sol＞
from the method of images

a a
-l l

d  
i
di
d
D
 a   a
 V2  ln , V1  ln ,
2 0 d 2 0 d

6
 a2 D  D 2  4a 2
Where d  D - d i  D  d 
d 2
   0  0
C    
V1  V2   a  a d D  D 2  4a 2
ln   ln ln( ) ln[ ]
2 0 d 2 0 d a 2a
 0  0
  (  ln[ x  x 2  1]  cosh 1 x )
D D D
ln[  ( ) 2  1] cosh 1 ( )
2a 2a 2a

Another example:

The potential at any point P(x,y) due to    and    is,

等電位及電場曲線圖
 r    r0  r
V p  V    V    ln 0  ln   ln 2
2 0 r1 2 0 r2 2 0 r1
r ( x  b) 2  y 2
if 2   k (cons tan t ), 表示等電位
r1 ( x  b) 2  y 2
k 2 1 2 2k 2k
 (x  2 b)  y 2  ( 2 b ) 2 : a family of circles in the xy - plane with radii a  2 b
k 1 k 1 k 1
k 2 1 k 2 1
and centered at ( 2 b, 0) , if let c  2 b, thus c 2  a 2  b 2
k 1 k 1

Advanced examples:
Example1 : y

a1 a2
- O l
  l   x
b b
c1 c2
D

7
 2 2
we have b 2  c1  a1
2 2 1
and b 2  c 2  a 2  c1  ( D 2  a12  a 22 )
2D
1
But c1  c 2  D c2  ( D 2  a 22  a12 )
2D
Example 2 :

2 2
here b 2  c1  a1
2 2 1
b 2  c2  a2  c1  (a 22  a12  D 2 )
2D
1
and c 2  c1  D c2  (a 22  a12  D 2 )
2D

(3). Point Charge and Conducting Sphere


The E and V of point M produced by the point charge Q are ,
 Q 
EQ  a r where r is the distance from Q
4 0 r 2

8
 r   r
Q 1 Q 1 1 Q 1
VQ    EQ  a r dr  r dr  (  ) ,
r0
4 0 r0
2
4 0 r r0 4 0 r

r0 : the reference point and let r0  


Similarly, the E and V of point M produced by the point image charge Qi are,
r    Qi r
1 Qi 1 1 Qi 1
VQI i    E Qi  a r dr  r dr  (  )
r0
4 0 r0
2
4 0 r r0 4 0 r
1 Q Qi Q Q
and  VM  VQ  VQi  (  )  0 ( grounding)   i  0,
4 0 r ri r rI
ri Qi ri a d i a2 a
   constant,     di   Qi   Q
r Q r d a d d

Example 4-5
(4). Charge Sphere and Grounded Plane

image


a a a a2 a2
Q1  Q0  Q0  kQ 0 ( k  ), di  
d 2c 2c d 2c
2 2
a a k a a2
Q 2  Q1  Q1  Q 0 , d i '  
d' a2 1 k 2 d' a2
2c  2c 
2c 2c
a a aQ 2 k k2
Q3  Q2  Q 2    Q0
di '' 2c  d i' a2 k3 1 k 2
2c  1
a2 1 k3
( 2c  )
2c
3
k
 Q0
k3
(1  k )(1 
2
)
1 k3
.......... .......

k2
 The total charge on the sphere is Q  Q 0  Q1  Q 2         Q 0 (1  k       )
1 k 2
Since the charge pairs (  Q0 , Q1 ),(  Q1 , Q2 ),(  Q2 , Q3 )….yield a zero potential on the
9
sphere, only the original Q0 contributes to the potential of the sphere , which is
k2
Q0 (1  k       )
Q0 Q 1 k 2 k2
Hence, V0  , hence C    4 0 a(1  k       )
4 0 a V0 Q0 1 k 2
4 0 a

4.5 Boundary-Value Problem in Cartesian Coordinates

Boundary-value problems: problems governed by partial differential equation
with prescribed boundary conditions.
Three types of boundary-value problems:
(1). Dirichlet problem : the value of the potential is specified everywhere on the
boundary.
(2). Neumann problems : the normal derivative of the potential is specified
everywhere on the boundary.
(3). Mixed boundary-value problems : the potential is specified over some
boundaries and the normal derivative of the potential is
specified over the remaining ones.
In Cartesian coordinates, Laplace’s equation for scalar electric potential V is
 2V  2V  2V
  0
x 2 y 2 z 2
Assume the solution V(x,y,z) can be expressed as a product in the following from:

V ( x, y, z )  X ( x)Y ( y ) Z ( z )
d 2 X ( x) d 2Y ( y ) d 2 Z ( z)
Thus  Y ( y ) Z ( z )  X ( x ) Z ( z )  X ( x )Y ( y ) 0
dx 2 dy 2 dz 2
1 d 2 X ( x) 1 d 2Y ( y ) 1 d 2 Z ( z)
   0
X ( x) dx 2 Y ( y ) dy 2 Z ( z ) dz 2
1 d 2 X ( x) 2 1 d 2Y ( y ) 1 d 2 Z ( z) 2
Now let 2
  k x , 2
  k y
2
, 2
 k z
X ( x) dx Y ( y ) dy Z ( z ) dz
1 d 2 X ( x) d 2 X ( x)
then k x  k y  k z  0 , and
2 2 2
2
2
 k x  2
2
 k x X ( x)  0
X ( x) dx dx
d 2 X ( x)
The possible solution of 2
 k x2 X ( x)  0 is shown as following.
dx
kx2 kx X(x) Exponential forms of X(x)
0 0 A0x+B0
+ k A1sinkx+B1coskx C1ejkx+D1e-jkx
- jk A2sinhkx+B2coshkx C2ekx+D2e-kx

d 2Y ( y ) d 2 Z ( z)
Similar method for solving 2
 k y Y ( y )  0 and
2
2
 k z2 Z ( z )  0
dy dz

10
 Example 4-6 Determine the potential distribution in the region enclosed
by the three electrodes.

(sol)
In this structure, V is independent of z.  k z  0  Z ( z )  B0  V(x, y, z)  B0V(x, y)  B0 X ( x)Y ( y )
All boundary conditions :
For x - direction : V( 0 , y)  V0 , V(, y)  0 ; For y - direction : V(x, 0 )  V(x, b)  0
 k  k  k  0  k  k
2
x
2
y
2
z
2
y
2
x ( k z  0) let k  k 2  k x2   k 2
2
y

d 2 X ( x)
 2
 k 2 X ( x)  0  X ( x)  C 2 e kx  D2 e  kx (C 2  0, as X( x  )  0 )
dx
d 2Y ( y )
and 2
 k 2 X ( x)  0  Y ( y )  A1 sin( ky )  B1 cos( ky ) ( B1  0, as Y(y  0)  0)
dy
 V(x, y)  B0  D2 e -kx  A1 sin (ky)  Ce -kx sin (ky ), C  B0 D2 A1
from boundary condition : V ( x, b)  0  sin( kb)  0  kb  n  k  n / b, n  1, 2 , 3,......
 
π 2π nπ
sin ( y )  C 2 e sin ( y )  ......   C n e sin ( y )   Vn ( x,y )
- πb x - 2bπ x - nπ x
 V(x, y)  C1e b

b b n 1 b n 1
nπ  
 x nπ nπ
Here, Vn(x, y)  C n e b
sin ( y), Meanwhile, V (0 , y )   C n   C n sin ( y )  V0 , 0  y  b
b n 1 n 1 b

nπ
here, C
n 1
n sin (
b
y )  V0 , 0  y  b is essentially a Fourier - series expansion,

mπ nπ mπ
y)   C n sin ( y )dy   V0 sin (
b b
Therefore, 
0
sin (
b n 1 b 0 b
y)d y

Cn  4V0
b mπ 
nπ  b, if m  n mπ  , if n is odd
y)   C n sin ( y )dy   2
b
  sin ( , 0 V0 sin ( b y)dy   n
0 b n 1 b 0 , if m  n 0 , if n is even
 
nπ 4V0 - nπb x nπ
 V(x, y)   C n e y)  
- nπ x
b
sin ( e sin ( y ), x  0 and 0  y  b
n 1 b n  odd n b

Similar example: Determine the potential distribution in the region enclosed

by the two electrodes.

V=0

a 11
 (sol)
In this structure, V is independent of z.  k z  0  Z ( z )  B0  V(x, y, z)  B0V(x, y)  B0 X ( x)Y ( y )
All boundary conditions :
For x - direction : V( 0 , y)  V0 , V(a, y)  0 ; For y - direction : V(x, 0 )  V(x, b)  0
 k x2  k y2  k z2  0  k y2   k x2 ( k z2  0) let k y2  k 2  k x2   k 2
d 2 X ( x) d 2 X ( x)
from 2
 k 2
x X ( x )  0  2
 k 2 X ( x)  0  X ( x)  C 2 e kx  D2 e  kx
dx dx
or
 A2 sinh (kx)  B2 cosh (kx)
sinh (ka)
as X ( x  a )  0  A2 sinh (ka)  B2 cosh (ka)  0  B2   A2  X ( x)  A2 sinh (kx)
cosh (ka)
sinh (ka) 1
 A2 cosh (kx)  X ( x)  A2 [cosh (ka) sinh (kx)  sinh (ka) cosh (kx)]
cosh (ka) cosh (ka)
 A3 sinh ( k ( x  a ))
d 2Y ( y )
from 2
 k 2 X ( x)  0  Y ( y )  A1 sin( ky )  B1 cos( ky ) ( B1  0, as Y(y  0)  0)
dy
 V(x, y)  B0  A3 sinh (k ( x  a ))  A1 sin (ky)  C sinh (k ( x  a )) sin (ky ), C  B0 A3 A1
from boundary condition : V ( x, b)  0  sin( kb)  0  kb  n  k  n / b, n  1, 2 , 3,......
 
nπ nπ
 V(x, y)   C n' sinh ( ( x  a )) sin ( y )   Vn ( x,y )
n 1 b b n 1

nπ nπ nπ nπ
Here, Vn(x, y)  C sinh ( ( x  a )) sin ( y), Meanwhile, V (0 , y )   C n' sinh (- a ) sin ( y )
'
n
b b n 1 b b
 
nπ nπ nπ
  C n sin ( y )  V0 , 0  y  b, here, C n  C n' sinh (- a )   C n sin ( y )  V0 , 0  y  b
n 1 b b n 1 b
is essentially a Fourier - series expansion,

mπ nπ mπ
y)   C n sin ( y )dy   V0 sin (
b b
Therefore, 0
sin (
b n 1 b 0 b
y)d y

 Cn  4V0
mπ 
nπ  b, if m  n mπ  , if n is odd
y)   C n sin ( y )dy   2
b b
  sin ( , 0 V0 sin ( b y)dy   n
0 b n 1 b 0 , if m  n 0 , if n is even
4V0 nπ 4V0 4V0
Cn   C n' sinh (- a )  C n'   , if n is odd
n b nπ nπ
n sinh (- a ) n sinh ( a )
b b

nπ nπ
 V(x, y)   C n' sinh[ ( x  a )] sin ( y )
n 1 b b

4V0 nπ nπ
 sinh[ ( x  a )] sin ( y ), 0  x  a and 0  y  b
nπ b b
n  odd
n sinh ( a )
b
12
 4.6 Boundary-Value Problem in Cylindrical Coordinates

In cylindrical coordinates, Laplace’s equation for scalar electric potential V is

1  V 1  2V  2V
(r ) 2  0
r r r r  2 z 2
Consider the lengthwise dimension of the cylindrical geometry is large in comparison to
 2V
its radius  field quantity is approximately independent of z  0
z 2
1  V 1  2V
(r ) 2 0
r r r r  2
Applying the method of separation of variables, thus the solution V(x,y,z) can be
expressed as a product in the following from:

r d dR ( r ) 1 d 2  ( )
V (r ,  , , z )  Z 0 R (r ) ( )  [r ] 0
R ( r ) dr dr  ( ) d 2
r d dR ( r ) 1 d 2  ( ) d 2  ( )
Now let [r ]  k , then
2
 k 
2
 k 2  ( )  0
R ( r ) dr dr  ( ) d 2
d 2

d dR ( r ) d 2 R(r ) dR ( r )
r [r ]  k 2 R(r )  r 2 2
r  k 2 R (r )  0  R ( r )  Ar r k  Br r  k
dr dr dr dr
1 d  ( )
2
d  ( )
2
Similarily for  :   k 2
  k 2  ( )  0   ( )  A sin( k )  B cos( k )
 ( ) d 2
d 2

Therefore, V ( r ,  , z )  Z 0 R ( r ) ( )  Z 0 ( Ar r k  Br r  k )[ A sin( k )  B cos( k )]

 r k [ A sin( k )  B cos( k )]  r  k [ A sin( k )  B cos( k )]
Consider : (1) for case at the cylindrical axis (i.e. r  0) the r  k factor can not exist.
(2) for case at infinity (i.e. r   ) the r k factor can not exist.
For the simplest case : k  0
d dR ( r ) 1 d 2  ( )
r [r ]  0  R ( r )  C 0 ln r  D0  0   ( )  A0  B0
dr dr  ( ) d 2
If there is no circumferential variation   ( )  B0  V (r ,  , z )  Z 0 (C 0 ln r  D0 ) B0  C1 ln r  C 2
Example 4-8 For the following cylindrical structure, determine the potntial
distribution in the space between the conductors.
(sol)
If the lengthwise dimension is large in comparison with
b its radus and there is no circumferential variation 
a
V0 V ( r ,  , z)  V ( r )  C1 ln r  C 2
Boundary conditions : V ( r  a )  V0 ; V ( r  b)  0
 C1 ln a  C 2  V0 ......(A) C1 ln b  C 2  0 ......(B)
V0 13
from (A) and (B)  V ( r )  ln(b / r )
ln(b / a )
 Example 4-9 For the following cylindrical structure, determine the potntial
distribution both inside and outside the tune.
V0
(Sol)
V(b, )
b
V0

 
0 2
-V0 -V0

Because the lengthwise dimension is large in comparison with its radus

 V (r ,  , z )  V (r ,  )  r k [ A sin( k )  B cos( k )]  r  k [ A sin( k )  B cos( k )]
V0 for 0    
Boundary conditions : V (b,  )  
 V0 for     2
(1) Inside the tube ( r  b)
Because the region includes r  0, the r  k factor can not exist.
 V (r ,  )  r k [ A sin( k )  B cos( k )] Moreover, since V ( r ,  ) is an odd function of  ,
the cos( k ) factor can not exist. Therefore, V (r ,  )  A r k sin( k ), k is integer

For this case, V (r ,  ) should be composed by a lot of such terms  V (r ,  )   An r n sin( n )
n 1

 4V0

V0 for 0      if n is odd
 V (b,  )   An b sin( n )  
n
 An   nb n
n 1  V0 for     2  0 if n is even
4V  r n
 V (r ,  )  0  sin( n ), r  b
 n odd nb n
(2) Inside the tube ( r  b)
Because the region includes r  , the r k factor can not exist.
 V (r ,  )  r  k [ A sin( k )  B cos( k )] Moreover, since V (r ,  ) is an odd function of  ,
the cos( k ) factor can not exist. Therefore, V (r ,  )  A r  k sin( k ), k is integer

For this case, V (r ,  ) should be composed by a lot of such terms  V (r ,  )   An r  n sin( n )
n 1

 4V0 b n

V for 0      if n is odd
 V (b,  )   An b  n sin( n )  
0
 An   n
n 1  V0 for     2  0 if n is even

 n
4V b
 V (r ,  )  0  sin( n ), r  b
 n odd nr n

14
 4.7 Boundary-Value Problem in Spherical Coordinates
Laplace's equation in spherical coordinates :
1  2 V 1  V 1  2V
(R ) 2 (sin  ) 2 2  0 (   0)
R 2 R R R sin    R sin   2
For simplifying the problem, we limit discussion to cases in which potential V is independent of  .
1  V 1  V
 (R2 ) 2 (sin  )  0 Assume V ( R, θ )  Γ( R )  Θ(θ )
R R
2
R R sin   
1 d dΓ( R) 1 d dΘ(θ )
 (R2 ) (sin  )0
Γ( R) dR dR Θ(θ ) sin  d d
1 d dΓ( R ) 1 d dΘ(θ )
Let (R2 )  k 2 , then (sin  )  k 2
Γ( R ) dR dR Θ(θ ) sin  d d
d 2 Γ( R) dΓ( R)
 (1) R 2 2
 2R  k 2 Γ( R)  0  Γ n ( R)  An R n  Bn R -(n1 ) , here n(n  1 )  k 2 ,
dR dR
n  0,1,2,3,...positive integer
1 d dΘ(θ ) d dΘ(θ )
(2) (sin  )  n(n  1)  (sin  )  n(n  1)Θ(θ ) sin   0
Θ(θ ) sin  d d d d
It' s a form of Legendre's equation. The solutions for this eq. are called Legendre function.
Θ n (θ )  Pn (cos ) : Legendre polynomials
Finally,Vn ( R, θ )  Γ n ( R)  Θ n (θ )  [ An R n  Bn R -(n1 ) ]Pn (cos )
 
V ( R, θ )   Vn ( R, θ )   [ An R n  Bn R -(n1 ) ]Pn (cos )
n 1 n 1

Example 4 - 10

15
 Chapter 5 Steady Electric Currents
5.1 Introduction
Conduction currents : caused by drift motion of conduction electrons and/or
holes in conductors and semiconductors.
Electrolytic currents : the result of migration of positive and negative ions.
Convention currents : result from motion of electrons and/or ions in a vacuum..

5.2 Current Density and Ohm’s Law

 
Q  Nqu  a n st (c)

Q    
 I   Nqu  a n s  Nqu  s
t
 
( here, s  a n s )

Let J : volume current density (or simply current density)

Define
 J : in amperes per square meter
 
 J  Nqu  u ( here,   Nq : volume charge density )
   
 I  J  s  I   J  ds
s

Example 5-1

Neglecting fringing effects.

1
   dV
E (0)  a y E ( y ) y 0  a y y 0 0
dy
  
J  a y J  a y  ( y )u ( y ) (Note : (y) is negative )

from Newtown’s law of motion

du ( y ) dv( y )
m  eE ( y )  e
dt dy

here, m  9.11  10 31 kg ; e  1.6  10 19 c

Noting that :

du du dy du d 1
m m  mu  ( mu 2 ) (Note : dy/dt=u )
dt dy dt dy dy 2

1
d 1 dV 1 2eV 2
 ( mu 2 )  e  mu 2  eV  u  ( )
dy 2 dy 2 m

1
J m
now, J   u       J V 2
u 2e

from Poisson’s equation :

1
d 2V   J m
  V 2
dy 2 0 0 2e
1
dV dV d 2V J m dV
兩邊乘 2 2  2  2 V 2
dy dy dy 0 2e dy
1
d dV 4J m d
 [( ) 2 ]    [V 2 ]
dy dy  0 2e dy
1
dV 4J m 2
 ( )2  V c
dy 0 2e

c : an arbitrary constant
1
dV 4J m 2
at y  0  V  0, and 00 0 c c  0
dy 0 2e

2
 1 1 1
dV 4J m
  ( )2 ( )4V 4
dy 0 2e
1 1 1
m J
 V dV  2( ) ( ) 4 dy
4 2
 0 2e
3 1 1
4 J m
 V 4  2( ) 2 ( ) 4 y  c '
3  0 2e

 (1) y = 0, V=0  c' 0

(2) y = d , V=V 0
4 3 J m
1
4 2e 3
1

 V0 4  2( ) 2 ( ) 4 d  J  02 V0 2
3 0 2e 9d m
Child-Langmuir Law
For more than one kind of charge carriers (electrons, holes and ions)
 
J   N i qi u i
i (A / m2 )
For metallic conductors the main kind of charge carriers is electrons.
 
 J  N e qe u
 
由分析可知: u  u e E  亦即電荷移動速度與所受電場強度成正比

u e : the electron mobility

   
 J   N e q e u e E    e u e E  E
 
 J  E ,     e u e : conductivity

For semiconductor, the charge carries are electrons and holes.

    e u e   h u h

## Verify Ohm’s law :

3
   2
Because V12    E  d   V12  E
1

The total current is :

   
I   J  ds   E  ds  ES
s s

I
E
S
I 
V12    V12  I  R ; R  ( )
S S

1 S
Conductance : G   (S )
R 

Example 5-2
Copper (radius =1 mm) Aluminum (radius=1 mm)
(a). (b).
Find R ?
1 km 1 km
＜sol＞


(a) from R  and  copper  5.8  10 (查表B4,Page 675)
S
10 3
R  5.49()
5.8  10 7   (10 3 ) 2


(b) from R  and  alu min um  3.54  107 (查表B4,Page 675)
S
10 3
R  8.99()
3.54  10 7   (10 3 ) 2

5.3 Eelctromotive Force Kirchoff’s Voltage Law

4
5.4 Equation of Continuity and Kirchoff’s Current law
Consider : S, V
Q
I

   
I   J  ds  I     Jdv...(1)
s r
5
  dQ  d
dt v
and I   dv...(2)
dt
  d 
    Jdv    dv    dv
v
dt v v
t
  
J  ( A / m 3 ) : the equation of continuity
t

For stendy currents charge density does not very with time   0,
t
or for no flow source    0

or for no flow source   0

 
   J  0  divergenceless or solenoidal
   
    Jdv  0   J  ds  0
v s

  I j  0 : Kirchhoffs current law (通過電路中任一節點的 電流代數和  0)

j

From the equation

of continuity

  
 J  
t
  
   E  
t

  
in a simple medium    E 
0

6
  
 
 t

  t
    0     0e  where  0 is the initial charge density at t  0 ,
t 
 

As t      0 e     0 e 1  0.368  0


   , : called the relaxation time


5.5 Power dissipation and Joule’s law


The work w done by an electric field E in moving a charge q a
distance  in a small time interval of t is:
 
w  qE   

w     
  im   im qE   P  q E  u , u :the drift velocity.
t  0 t t  0 t

The total power delivered to all the charge carrier in a volume dv is :

 
dP   pi  ( N i qi E  u i )dv
i i
 
 E  ( N i qi  u i )dv
i
  dP  
 dP  E  Jdv   EJ
dv
 
 P   E  J dv (w) : Joules law
v

In a conductor of a constant cross section, dv  ds  d

7
  
 P   E  Jdv
v
   
  E  d   J  ds
L s

 VI  I 2 R

5.6 Boundary Condition for Current Density

For Steady Current Density:

   

 
J  0    J  0
   J  ds  0.....( A)
  J  
  or  J 
   E  0     0   d   0.....( B )
  c 

from (A)   J1n S  J 2 n S  0  J1n  J 2 n : normal component

J1t  J 2t  J1t J 2t J1t  1

from (B)   0    : tangential component
1 2 1 2 J 2t  2

Example 5-3

8
<sol>
J 1n  J 2 n
 J1n  J1 cos1  J1t  J1 sin 1
  J 1t  1
 J 2 n  J 2 cos 2  J 2t  J 2 sin  2 
, and
J 2t  2

 J 1 cos  1  J 2 cos  2 cot 1 cot  2 tan  2  2

   
 2 J 1 sin  1   1 J 2 sin  2 2 1 tan 1  1

2
J2  J 2t  J 2 n  ( J1 sin 1 ) 2  ( J1 cos1 ) 2
2 2

1
2
 J1 ( sin 1 ) 2  cos1
2

1

** for a homogeneous conducting medium (i.e.  is a constant to the spatial)


 J  
   ( )  0    J  0 : a curl - free vector

    
 J   ( J  E   (V )   V )
    
and   J  0    (  )  0   2  0
  
以此求出  , 再以J   求出J

# Consider the boundary with two different lossy dielectrics ( 1 ,  2 ,  1 ,  2 )

9
  J 1n  J 2 n   1 E1n   2 E 2 n
D1n  D2 n   s   1 E1n   2 E 2 n   s
 2 E 2n 
  s  1   2 E 2n  ( 1 2   2 ) E 2n
1 1
 E  
or  s   1 E1n   2 1 1n  ( 1  2 1 ) E1n
2 2

Consider, if medium 2 is a much better conductor than medium 1 (i.e.  2   1 )

 21
  s  (1  )E1n  1 E1n  D1n
2
Example 5-4

＜sol＞

(a). let J is the normal component of the current density.

d1 d
V  ( R1  R 2 ) I  (  2 )I
 1S  2 S
I V
J  (A / m2 )
S d d
( 1  2)
1 2

(b). V  E1 d 1  E 2 d 2

and  1 E1   2 E2 ( J 1n  J 2 n )
 2V  1V
 E1  (V / m) , E 2  (V / m)
 2 d1   1d 2  2 d1   1 d 2

(c). on the upper and lower plates

10
 1 2V
 s1  1E1  (c / m 2 )
 2 d1   1d 2
  2 1V
 s 2   2 E 2  (c / m 2 )
 2 d1   1d 2
at the interface of the dielectrics

  1 2V  2 1V
 si   1 E1   2 E 2  
 2 d1   1 d 2  2 d1   1 d 2
 2 1  2V
(  1 )
2  2 d1   1 d 2

Sum :

 s1   s 2
 s1   s 2   si  0

5.7 Resistance Calculations

11
    
Q
  ds
D  E  ds
C  s   s  
V   E  d   E  d
L L
   
  E  d   E  d
V
R     L 
L

I  J  ds  E  ds
s s
   
V Q Q
  E  d
  E  ds
 RC     L   s  
I V I  E  ds   E  d  
s L

C 
 RC  
G 

Example 5-5

＜sol＞

(a).
  
from Gausss Law :  E  ds    d
s

   
 E  2r       E   ar
 2r
a  a
     b
Vba    E  dr     a r a r  dr   ln
b b
2r 2 a
 2
   c1 ( F / m)
Vba b
ln
a

12
 and

b b
ln ln
  
R1C1   R1    a  a ( / m )
 c1  2 2

## The resistance between the conductors for a length  of the coaxial

cable is R1 /  , not R1 ##

(b). In example 4-4 the analysis results show the capacitance per unit
length for the parallel-wire transmission line is :

C1 ' 
cosh 1 ( D / 2a)

 1  cosh 1 ( D / 2a ) 1
therefore R1 '      cosh 1 ( D / 2a )
 C1 '   
1 D D
 ln[  ( ) 2 1 ( / m )
 2a 2a

The leakage resistance of a length  of the parallel-wire transmission line

is R1 ' /  , not R1 '

Summary:
The procedure for computing the resistance of a piece of conducting
material between specified equipotential surface ( or terminal ) is as
follows:
1. Choose an appropriate coordinate system for the given geometry.
2. Assume a potential difference V0 between conductor terminals.
 2  
3. Find E ( from  V  0  V  E   V)
13
    
4. Find I ( from I   J  ds   E  ds )
s s

5. Find resistance R by taking the ratio V0 / I

Example 5-6

14
 Chapter 6 Static Magnetic Fields
6.1 Introduction

For electrostatic field: For static Magnetic field: Note:

B =magnetic flux density

D   B  0

 E  0  B  J H ：magnetic field intensity

and D   E and B   H
Fe  q E Fm  qu  B

Thus, the total electromagnetic force on a charge q is:

 
F  Fe  Fm  q E  qu  B  q E  u  B ：called the Lorentz’s force equation

6.2 Fundamental Postulates of Magnetostatics in Free

Spaces

In free space , the divergence and cure of B are ：

B  0
here,  0  4  10 7 H / m 
  B  0 J

   Bdv   B  d S  0
 
v s

    B   0   J  0    J  0 : from Identity II.

 B ds   
s s
0 J  d s   B d l   0  J  d s   0 I : Amphere' s circuital law
c s

Summary:
Postulates of Magnetostatics in Free Space
Differential Form Integral Form
B  0  Bds  0
s

6- 1
   B  0 J  B  dl  
C
0 I

Example 6-1

<Sol>
(a) Inside the conductor

 r1 2
B1  a B1 ， d l  a r1 d ， I1  I
b 2
r1 2
  B1  d l   0 I 1   a B 1  a r1 d   2 I ， r1  b
C C1 b
2 2
2 r  r 
 B 1 r1 d   0  1  I  B 1 r1 2   0  1  I
0
b b

 0 r1 I  rI
 B1   B1  0 12 a
2b 2
2b
(b) Outside the conductor

B2  a B 2 ， d e  a r2 d ， I 2  I

  B2  d l   0 I 2   a B 2  a r2 d   0 I ， r2  b
c2

2 0 I
  B 2 r2 d   0 I  B 2 r2 2   0 I  B 2 
0 2r2
 0 r1 I
B a , rb
2b 2
0 I
a , r b
2r2
Example 6.2
(Sol)
For b  a  r  b  a

 B  d   0 NI

  B a  r a d   0 NI
C

6- 2
 2
  B  rd   0 NI
0

 0 NI  NI
 B 2r   0 NI  B   B  0 a
2r 2r

Note : B  0 for r  (b  a) and r  (b  a)

Example 6.3

<Sol>
Applying Ampere’s circuital law :

 B d    I
0 total

 BL   0 ILn  B   0 NI
From right-hand rule,the direction of B goes from right to left.

6.3 Vector Magnetic Potential

   B  0 and from Identity II   (  A)  0

 B can be expressed as the curl of another rector field , say A  B    A

A : defined and called the vector magnetic potential Wb / m 

How do we choose  、 A ?

2
Because :   B   0 J    (  A)   0 J  ( A)   A   0 J
2  
Comparing to  V  ，we can choose   A  0 such that
0
2
 A   0 J

6- 3
  2
 AX    0 J X

2
 Ay    0 J y

2
 Az    0 J z
2 
The solution form of  V  is
0
1 
V 
4 0 
V' R
dv '

2 0 J '
 The solution form of  A    0 J is A 
4 
V '
R
dv

Define  as the magnetic flux     B d s     A  d s   A  d s

s s c

6.4 The Biot-Savart Law and Applications

For a thin wire with cross-sectional area S.

J dv '  JSd '  I d  '
0 J ' 0 Id l  0 I dl '
 A
4 
V' R
dv 
4 c' R

4 C ' R (Wb / m)

 I d l '  0 I dl '
then, B    A     0
 4
C ' R   4 C '

R
........( A)

Note:
1. The unprimed curl operation implies differentiations with respect to the
space coordinates of the field point.
2. The integrand operation is with respect to the primed source coordinates.

0 I d l ' 0 I 1 1 '
Continue ( A)  B       R   d l  ( R )  d l 
'

4 C '
R 4 C'
 

 '
'
 I  aR 0 I d l  aR
 0
4 C  R 2
 0   d l  C ' R 2 (T)：Biot-Savart Law
 4
'

( Applying   ( f G )  f   G   f  G )

6- 4
 Sometimes,it is convenient to write Biot-Savart Law in two steps:

B   ' dB
C

 0 I  d l  a R   I  dl '  R 
'

dB   0  
4  R 2  4  R 3 
   

Example 6.4

<Sol>
'
A typical element on the wire is d l  a Z dz '

'
 I dl 0 I L a Z dz ' az 0 I ln( z '  z ' 2  r 2 ) | L
(a) A  0
4 C ' R  4 L '2

4    L
z  r2

 0 I  L2  r 2  L 
 az ln  a z Az
4  L2  r 2  L 

1 Az A
Therefore, B    A    (a z AZ )  a r  a Z
r  r

  0 I L2  r 2  L   0 IL
 a  ln( )  a .......( A)
r  4 L  r  L 
2 2
2r L  r
2 2

 0 IL 0 I
if r  L  B  a  a
2r L  r 2 2 2r
(b) By applying Biot-Savart Law:

 R  a1 r  a z z '

'
 d l  R  a z dz '  (a r r  a z z ' )  a rdz '

6- 5
 ' 
 I dl  R
L  I L rdz '
 B  dB   0  a 0 
 L 4 3
4 L 2
R (z '  r 2 )3/ 2
 0 IL
 a .......( B)
2r L2  r 2
Finally, (B)=(A)

Example 6-5 、 Example 6-6  Homeworks

6- 6