Electromagnetic Fields & Wave
Jeong Phill Kim
phill@cau.ac.kr
010‐8918‐5344
School of EE
Chung‐Ang University,
Seoul 156‐756, Korea
Required Knowledges
Calculus
‐ Coordinate system
‐ Vector calculus
‐ Derivative & Integral
‐ Differential equations
Physics
‐ Mechanics
‐ Basic Electromagnetics
Programming
‐ Matlab
Chung‐Ang University
Plan of Study
1st semester
‐ Coordinate System
‐ Vector Calculus
‐ Electrostatics
‐ Magnetostatics
2nd semester
‐ Time‐varying Electromagnetics
‐ Maxwell’s Equations
‐ Plane Wave Propagation
‐ Transmission Line
‐ Impedance Matching
Senior courses
‐ RF / Microwave Engineering
‐ Radar Engineering
‐ Antenna Engineering
Graduate courses
‐ Advanced Electromagnetics
‐ RF/Microwave Circuit Design
‐ Antenna Theory & Design
‐ Wave propagation & Scattering
‐ Radar System Design
‐ Microwave Imaging
‐…
References
• D. K. Cheng, Field & Wave Electromagnetics,
• U.S.Inan & A. S. Inan, Engineering Elctromagnetics
• Markus Zahn, Electromagnetic Field Theory: a problem
solving approach
• Lecture notes : will be uploaded in e‐class
² How can we develope these things?
of each one?
Ch. 1 Intro
Electronic enginnering around us:
² Now a days, so many electric or electronic devices and apparatus are
around us as follows:
{
{
{
{
{
{
{
{
{
{
{
{
{
{
Radio and TV
Microphone and Speaker
Smart phone
Bluetus and WiFi connected Devices
Computer connected to LAN
T-money (simple contact or contacless trraffic card)
Electric motor
Dispalys (ex, Cathode-ray tube (CRT)
Touch panel
Maglev (Magnetic Levitation) Train
Light (one of EM wave)
Smart highway and car
Induction heating (IH) and microwave oven
Radar and Laser
What is the theory of operations
{ Students studying electrical engieering generally meet voltages
and currents in electric circuits at first. They study the
characteristics of circuit elements (such as resistor, capacitor,
and inductors) and the Kirchhoff's voltage and current laws, and
use them to analize the circuit characteristics.
{ Some questions ???
1) What are the voltage and current physically? In addition, how
can we explain the operation of capacitor and inductor clearly?
2) The source signal may be direct current (DC) or alternating
current (AC) one.
¡ fDC = 0,
fAC 6= 0
¡ Low frequency ! Circuit theory can be used for analysis
¡ Increasing frquency ! Electric circuit ???
3) Let us consider
1 cm Resistor and Wavelength of 30 GHz
¡ In this case, is the current in resistor unform ???
4) Now the wireless connection becomes common way data
communication. Can you explaine the related phenomena ???
Permanent
Magnet
Electric
signal
+
–
Cone
² The related phenomena is strongly related with the physical
phenomena of electromagnetic fields.
! This is the reason why we have to study electromagnetics.
Coil
Vibration
Speaker
4H
+
–
Chung‐Ang University
5V
10 
8
1F
8H
15 
2A
λ
Branches of EE department
Electromagnetics
² Physics based
{ Electromagnetics is strongly based on the physics.
¡ Electric circuit and energy
{ Therefore it is needed for you to have interests in physics.
¡ Semiconductor
¡ Electromagnetics, Microwave, and optics
{ However, the problem is that most student think that physics is
one of most difficult subjects to study.
¡ High speed circuit design
{ Then How ???
¡ Wireless communication (HW, but SW ???)
¡ Remote sensing
Electromagnetics is the core and prerequisite subject for:
{ Electric circuit and energy
² Non-physics based
{ Semiconductor
¡ Software-based communication (Data compression and coding)
{ High speed circuit design
¡ Artificial Inteligence (AI) and Neural network (NN)
{ Microwave, optics, and RemoteSensing
¡ Internet of things (IOT)
{ EMI and EMC
¡ Algorithm and programming
² How about the autonomous navigation ???
EM is a Basic and Major Area of EE Department
Communication &
Signal processing
Electric
energy
RF, Microwave &
Optics
Electromagnetics
Integrated
circuits
Semiconductor
Control &
System
Computer
Chung‐Ang University
History of Electromagnetics
² Electricity
{ Long ago ancient people were aware of thunder (lightening) and
shocks from electric fishes. The record of electric fishes dates
back to 2750 BC, where these fish were referred to as the Thunderer of the Nile.
{ Ancient traders around the Mediterranean had known rods of amber
could be rubbed with cloth like to attract feathers. In about
600 BC, the ancient Greek philosopher Thales (his living city was
Miletus) first wrote about the attraction phenomena of amber with
a series of observations.
{ Electricity remained relatively unknown until 1600 AD, when the
English scientist William Gilbert made a careful study about it
using amber, and denominated it as the latin word electricus, meaning
like amber. Now electricity becomes the representing terminology
instead of electricus.
Black
Sea
Balkan
Peninsula
Miletus
Magnesia
² Magnetism
Mediterranean
Sea
{ It is known that Ancient Greeks living in Magnesia (about 20
km from Miletus) probably first observed that iron ore in mine
attracted one another and also attracted small iron objects. The
word magnet comes from Magnesia, the name of the place where this
ore was found.
Africa
Wireless
Transmission
Maxwell’s
Eq.
{ From around 1400 AD, the magnetic loadstone has been used as a
compass for safe sailing in sea.
Coulomb’s
Law
Ampere’s
Circuital Law
Marconi
Maxwell
Coulomb
Ampere
1820
1861
1897
1866
1831
1784
Savart
Bio‐Savart
Biot Law
Chung‐Ang University
Faraday
EM
Induction
Hertz
Dipole
Antenna
Radiation
Ch. 1: Electrostatics
1.1 Gravitational Force and Field
² Phenomena related to the electric charge, specially for ststic
chrage
² Gravitational Force (Newton / Apple tree)
{ Earth environment
² Statics and Dynamics
(
=0
@(Physical quantity)
@t
6= 0
m me
= mg
R2
k g me
Gravitational
! g = 2 = 9:8 m/s2 acceleration
R
8
¡11
>
N m2 /kg2
<kg = 6:67 £ 10
R = Re + h ' Re = 6; 400 km
>
:
me = £1024 kg
Fg = kg
:
Static
:
Dynamic, Time-varying
² What is charge ???
{ Charge is formed by collection of a lot of atoms with their
electrons gained or lost.
h
{ An atom consists of proton and some number of electrons
¡ Simple model (the following Figure)
¡ An electron is known to have a charge (negative).
e = ¡1:6 £ 10
¡19
mt
{ Meaning of g
Fg
Earth
me ms
¡ Gravitational force for m = 1 kg
¡ Gravitational field
R≃ae
(C)
² Important concept of souce mass and test mass
Total charge of electrons = Multiple of e (quantized)
For most matters
R
me ! ms ;
m ! mt
mt ms
Fg = kg 2 = mt g
R
jTotal charges of protonj = jTotal charge of electronsj
{ An ion is an atom (or group of atoms) that has lost or gained one
or more electrons, giving it a net positive or negative charge,
respectively (the following Figure).
g!a
¡¡¡!
m1 Fg
Fg m2
Fg = mt a
² Gravitational energy (Wg ) and potential (Vg )
Neutral charge
Neutral charge
{ Energy
Z
Wg = ¡
9+
h:
Negative charge
9+
Fg ¢ dl ' Fg l ! mt g h :
Joule (J)
4+
1 electron
gained
Positive charge
4+
{ Potential:
Energy for mt = 1 kg
Vg =
1 electron
lost
² Negative mass ?
Chung‐Ang University
vertical height from Earth surface
Electromagnetics
Wg
= gh
mt
(Now only mathematically ?
5
1.2 Expemiment on Electric Force
² Analogy:
Torsion
Balance
apparatus
² Experiment (Charles-Augustin de Coulomb)
³q q ´
1 2
Fe = ke
R2
ke = 8:9877 £ 109
Gravitational
ms
Force (N) Fg = mt kg 2
R
= mt g
ms
Field
g = kg 2
R
Energy
Wg = mt gh
Wg
= gh
Potential Vg =
mt
² Torsion balance
{ The torsion balance consists of a bar
suspended from its middle by a thin fiber
of a very weak torsion spring.
R
Fe
{ A known charge is loaded at the end of
the bar. If a test charge is located with
a distance R from the charge on the bar
(normal direction to the bar axis).
q1
Fe
= qt E
qs
E = ke 2
R
We = qt El
We
Ve =
= El
qt
Example Comparison of related forces in atom
² Parameters proton and electron in hydrogen atom
qp = ¡qe = 1:6 £ 10¡19 C
R = 0:53 £ 10¡10 m
(Bohr radius of H atom)
Charles A. de Coulomb
² Gravitational and electric forces
me m
= 3:6 £ 10¡47
R2
qp qe
Fe = ke 2 = 8:2 £ 10¡8
R
Fg = kg
² With the concept of source and test charges
³ q ´
s
Fe = qt ke 2
R
(N)
(N)
8:2 £ 10¡8
Fe
=
= 2:3 £ 1039
Fg
3:6 £ 10¡47
² Thus we can ignore the gravitational force comapred to the electric
force.
² Corresponding one to graviational field g ?
Chung‐Ang University
qs
R2
me = 9:11 £ 10¡31 kg
{ The sensitivity of the instrument comes
from the weak spring constant of the
fiber, so a very weak force causes a large
rotation of the bar.
E = ke
Fe = qt ke
mp = 1:67 £ 10¡27 kg
{ The equilibrium sate means that the
twisting force (calculated from the
measured rotated angle) equals to the
applied force.
!
Electric
q2
{ The bar will rotate, twisting the fiber,
until it reaches an equilibrium.
Fe = qt E
Gravitational and Electric
qs
R2
Electromagnetics
6
² Constant ke and ²0
1.3 Electric Force and Field in Free Space
{ You might have already known the following
relation between ke and ²0
² What is the meaning of Electrostatics ?
1) Charges are not moving.
2) Quanties of charges are not varying in
time.
Source charge
at the origin
1.3.1 Discrete Source Charges
F
z
E
² Single point charge at origin
³ q ´
s
Fe = qt ke 2 ^
r = qt E
r
qt = 2 C
qt = 1 C
E
qt = 1 C
r
{
y
qs
x
³ q ´
s
E = ke 2 ^
r
r
1
4¼²0
1
²0 =
= 8:854 £ 10¡12
4¼ ke
ke =
qt = 1 C
E
² There is no change in E field itself with
the coordinate chosen, but its expression
may be different depending on the chosen
coordinate systems.
² Concept of source- and field-points:
8
>
: Field-point vector
<r = (x; y; z)
0
0 0 0
r = (x ; y ; z ) : Source-point vector
>
:
R = r ¡ r0
: Source-to-field point vector
qs ^ ´
Fe = qt ke 2 R
R
³ q
´
s ^
E = ke 2 R
R
8
R = r ¡ r0
>
>
>
<
r = jRj
>
>
>
:R
^ =R
R
qs1 R1
qs2
N
X
E=
z
r 1’
En
n=1
Source charge
at arbitrary point
Field
point E
qt=1 C
z
² Multiple point charges
Force on qt = 1 C
³
{ This relation can be clearly understood
after studying a electric flux density D.
{ From now ke and 1=(4¼²0 ) will be used
case by case.
² Single point charge at an arbitrary point
r0 = (x0 ; y0 ; z0 ) ???
² Electric field:
(unit charge)
Where does this expression come ?
En = ke
μ
r 2’
qn
R2n
¶
^
Rn
x
0
R2
r
r n’
rN’
Rn
qsn
RN
qsN
Rn = r ¡ r0n
E
qt = 1 C
r
x
0 r’
qs
R
y
² Practical measurement (Caution !!!)
Fe
qs !0 qt
E , lim
Chung‐Ang University
Electromagnetics
7
y
1.3.2 Continuous Distribution
Example Electric Force and Field
² There are two point charges 10 nC at (1,0,2)
and 5 nC at (-2,0,3).
{ Calculate E at (1,3,2).
R1 = r ¡ r01 = (1; 3; 2) ¡ (1; 0; 2)
= (0; 3; 0)
R1 = 3
p
p
R2 = 32 + 32 + 12 = 19
R2
3
qs1 2
= 10 nC R1
‐2
r02
R2 = r ¡ = (1; 3; 2) ¡ (¡2; 0; 3)
= (3; 3; ¡1)
qs2 = 5 nC
z
0
x
1
3
y
² Expressions of charge distributions
8
>
<½L ¢ln : line charge (C)
qn = ½S ¢Sn : surface charge (C)
>
:
½V ¢vn : volume charge (C)
8
>
<½L : line charge density (C/m)
½ = ½S : surface charge density (C/m2 )
>
:
½V : volume charge density (C/m3 )
vn
r n’
r
y
Line charge
z
E = E1 + E2 = (¡1:63; ¡11:62; 0:54)
jEj = 11:74 (V/m)
dq
En = ke
qn
R2n
¶
y
where qn = ½V ¢vn
Surface charge
² This vector calculation is a quite tedious
and time-consuming work !!!
dE
z
qn ! dq = ½ dv; Rn ! R
μ
¶
½ dv ^
En ! ke
R
R2
μ^¶
Z
R
0
!
E = ke
½(r )
dv
R2
V0
(nN)
0
x
^ n;
R
{ Exact expression with
{ Calculate F to the charge of ¡10 nC at
(1,3,2).
r
ρL r’
En
μ
dE
R
N
X
i=1
qt = 1 C
0
x
{ Approximation formula of total E
E'
Rn
qn
² Case study (volume charge distribution)
Therefore the related E in V/m are
μ
¶
R1
E1 = ke qs1
= (0; ¡9:99; 0)
R31
μ
¶
R2
E2 = ke qs2
= (¡1:63; ¡1:63; 0:54)
R32
F = qt E = (16:3; 116:2; ¡5:4)
jFj = 117:4 (nN)
Field
point En
z
r
R
s
y
r’
dq
x
Volume charge
{ Any simpler methods to find the E ???
dE
z
¡ Gauss' Law
¡ Electric potential approaches
r
R
ρv
r’
y
dq
x
Chung‐Ang University
Electromagnetics
8
² Therefore dE½ and the resulting E½ become
Example Finite Length Line Charge
^ ½ from
² Differential electric field at r0 = ½
small charge element ½L dz0 at z = z0 :
Finite length
μ ¶
z
R
0
dE = ke ½L
dz
R3
L
8 0
0
zz
r =^
>
>
>
<r = ½
^½
dq = L dz’
RL
0
0
>R = r ¡ r = ½
^½¡^
zz
>
>
p
:
r′
R = jRj = ½2 + z02
L
² dEÁ = 0 by axial symmetry.
Then dE½ and dEz are
(
^=½
R¢½
R¢^
z = ¡z0
μ
¶
^
R¢½
^ = ke ½L
dz0
dE½ = dE ¢ ½
R3
0
dE½ =
μ
= ke ½L
0
¡z
dz0
R3
R¢^
z
R3
ke ½L
½
E½ =
Z
¯L
cos ¯ d¯
0
ke ½L
sin ¯L
½
L
L
=p
RL
½2 + L2
² Ez can be found similarly as
ke ½L
Ez =
½
dE

dEz
!
dE
!
Z
¯L
sin ¯ d¯
0
ke ½L
(cos ¯L ¡ 1)
½
½
½
cos ¯L =
=p
RL
½2 + L2
Ez =
² Sum of results from upper and lower wings
ke ½L
(sin ¯1 + sin ¯2 )
½
ke ½L
Ez = Ez1 + Ez2 =
(cos ¯1 ¡ cos ¯2 )
½
8
8
L1
>sin ¯ = p L2
>
>
>
2
<
<sin ¯1 = p 2
2
½ + L1
½2 + L22
½
½
>
>
>
>
:cos ¯2 = p 2
:cos ¯1 = p 2
2
½ + L1
½ + L22
E½ = E½;1 + E½;2 =
μ
E½ =
sin ¯L =
³½
´
0
= ke ½L
dz
R3
dEz = dE ¢ ^
z = ke ½L
!
!
R
ΒL β
r
ke ½L
cos ¯ d¯
½
¶
dz0
¶
L1
L
E = ??
?

1
2
L2
² With a change of variables z0 = ½ tan ¯ (Fig.
XXX)
(
dz0 = ½ sec2 ¯ d¯
R = ½ sec ¯
cos ¯ d¯
½ dz0
½2 sec2 ¯ d¯
=
=
3
R
½3 sec3 ¯
½
0
0
2
2
sin ¯ d¯
½ tan ¯ sec ¯ d¯
z dz
=
=
3
3
3
R
½ sec ¯
½
Chung‐Ang University
Electromagnetics
9
Case a
² For a ¯eld point on the z-axis (jzj > L=2), ¯1 = ¯2 =
¼=2, Ri = jz § L=2j
E½ = 0;
d
E
P
{ Field point for a maximum Ez can be found by
@ ³z´
@Ez
=0
=0 !
@z
@z R3
Ez =???
² For a ¯eld point with ¯2 = ¯1
L
§a
! a2 ¡ 2z2 = 0 ! zm = p
2
L
{ This can be solved with N-polygon approximation (Appendix).
Ez = 0
ke ½L
sin ¯1
½
L=2
ke ½L
p
=2
2
½
½ + (L=2)2
E½ = 2
² Surface Charge Distribution
Circular surface
Case b
L/2
{ Ez along the z-axis can be found as
L
P
Example Uniform Line Charge on In¯nite Line
E

L/2
² With L ! 1
E½ = 2
ke ½L
½
Case c
The integrand is not a function of Á.
∞
Z
L

{ Along the z-axis, E = ^
zEz with
I ^
R¢^
z
dl
Ez = ke ½L
2
R
dl = adÁ
z
z
^ ¢^
R
z = cos ° = = p
2
R
a + z2
az
Cc z
= k e ½L 3
3
R
R
Cc = 2¼a (circle's circumference)
Chung‐Ang University
E
∞
½z
d½
3
0 R
½ = z tan °;
d½ = z sec2 ° d°
¸
Z °0 · 3
z tan °
sec2 ° d°
Ez = ke ½S (2¼)
z3 sec3 °
0
Z °0
= ke ½S (2¼)
sin ° d°
(0,0,z)

s
Ez
0
b
0
y
dq
x
0
z
dE

x
= ke ½S (2¼) (1 ¡ cos °0 )
z
cos °0 = p
b2 + z2

(0,0,z)
{ Ez is simpli¯ed
(Due to the axial symmetry)
Ez = ke ½L (2¼)
Circular disk charge
b
Ez = ke ½S (2¼)
Example Uniform line charge on Circular line
² Calculate the electric ¯eld on the z-axis.
dq = ½S dS = ½S ½ d½ dÁ
¶
Z b Z 2¼ μ ^
R¢^
z
Ez = ke ½S
½ d½ dÁ
R2
0
0
z
 R
0
′ r′
y
L dq
Infinite planar charge
{ For an infinite plane (b ! 1)
zm
0
Ez
(0,0,z)
Ez
Ez = ke ½s (2¼)
‐zm
{ For a ring surface charge with inner and
outer radius a and b, respectively,
s
y
0
x
(Appendix)
Electromagnetics
10
Example Spherical cloud of charge
² Charges are collected (uniformly
distributed) in a form of spherical cloud
whose radius is a.
Spherical shell ???
Inside and outside
z
z=z0
v
a
y
{ E for z0 < a and z0 > a ???
x
Ezi
Q1 =
4¼z30 ½v
3
i=
1;
2;
Q2 =
4¼a ½v
3

{ Two conducting hemispheres are clamped
together.
{ The outer conductor is momentarily short.
{ Unclamping the outer conducting
hemispheres, and then measure the induced
charge Qout . The observed experimental
result is

Flux line



 
 

 




z
D
!


x
D»E
² Proportional constant ke ???
{ Comparison of the expression of ²0 and ke
1
= 8:854 £ 10¡12
4¼ ke
1
ke =
4¼²0
²0 =


Flux line




 





Gaussian surface
{ With replacing ke with 1=(4¼²0 ), we can
obtain the familiar formula:
μ
¶
qt qs
^
F=
R
4¼²0 R2
μ
¶
qs
^
E=
R
4¼²0 R2
Electromagnetics
E
Same
direction
y
qs
{ A proportional constant ²0 is introduced
and called permittivity of the air.
Gaussian surface
Qout = ¡Qin
Chung‐Ang University
8
qs
<E = ke 2 ^r
r
:D = qs ^r
4¼ r2
D = ²0 E
D
Gaussian surface
² Experiment by Michael Faraday
{ Inner conducting sphere is charged (Qin )
E
{ For a point charge at the origin in a free
space


1.4 Electric Flux and Gauss' Law
{ For validating his idea on flux concept

 


 

1.4.1 Electric Flux and its Density
Flux line


How and Why ???
² Introducing a flux density D, the
following mathematical formulation called
Gauss' Law for flux can is valid:
I
D ¢ dS = Q
(all charges inside S)
² Relation between D and E
for z0 < a
for z0 > a
3
;
qs
S
(
Qi
= ke 2 ;
z0
Flux lines
ª = Qin
Ez
² Let us consider the case of circular
disk carrying uniform charge distribution
(Appendix)
E=^
zEz
{ Concept of flux line ª was suggested to be
equal to total charge in the inner sphere.
11
1.4.2 Applications of Gauss' Law
² Why is this formula called the Gauss' Law ?
Gauss is known to provide a mathematica
formula, which can also be applied to
the many physical flux quantities such
as light, wave, etc.. Therefore our case
is one special one.
Example Uniform planar charge distribution
S2
+2 nC
+4 nC +3 nC
‐1 nC
S3
‐3 nC
+1 nC
S1
² Charge is assumed to be uniformly
distributed on an infinite plane xz-plane.
Then D = ^
yDy due to the symmetry.
² If the surace of centered cube
the Gaussian surface S,
8
>
<Dy =
Dy 2S = ½s S !
>
:Ey =
² Since the Gauss' law is valid for any
closed surface,
I
D ¢ dS = 2 nC
S1
I
D ¢ dS = ¡1 nC
S
I2
D ¢ dS = 0
½s
2
½s
2²0
y
x
d
Dy
s/2
z
4¼r3
½v
Dr 4¼r =
3
2
{ Outside the cloud,
8
<Dr 4¼r2 = Q
3
:Q = 4¼a ½v
3
z
ρ
ρL
L/2
S
S
S
2
S
1
y
a
x
{ Inside the cloud,
Ex. Uniform line charge distribution
y
0
² There are two cases of a field point; the
inside and the other outside of the cloud.
{ Constant En on S
(some part En = 0 is allowed)
Chung‐Ang University
is chosen as
² The charge is uniformly distributed, then
D=^
rDr due to the symmetry.
{ Symmetric charge distribution
(cube / cylinder / sphere)
{ D½ , E½ are found to be proportional to
1=½.
Area
A
Example Uniform spherical cloud of electron
² For meaningful use of Gauss' Law, S should
be chosen appropriately:
{ Charge is assumed to be uniformly
distributed along z-direction. Then
D = ½^ D½ due to the symmetry. If the
surface of the centered cylinder is
chosen as the Gausian surace S,
8
½L
>
<D½ =
2¼½
D½ 2¼½ L = ½L L !
½L
>
:E½ =
2¼²0 ½
ρs
² It is very interesting that Dy and Ey become
constant along y.
S3
S
z
!
½v r
Dr =
3
D
r
D0
a
!
Dr =
Q
4¼r2
y
x
{ Continuity of Dr at r = a,
and if it is D0 ,
‐L/2
D0 =
D
!
0

Electromagnetics
Q
a
= ½v
2
4¼a
3
8 μ 2¶
a
>
<D0
r2
Dr =
³
´
>
:D r
0
a
12
r
1.5 Curl and Divergence of Electric Field
² Divergence of E field (r ¢ E)
Vb surface
z
² As we know, in order to charaterize the E
field thoroughly, its divergence and curl
should be known.
qt=1 C
b
C=C1+C2
a
C2
y
C1
qs
² Curl of E field (r £ E)
{ First way: No electric energy is
required to move a charge along any closed
contour C.
I
W = F ¢ dl = 0
x
Va surface
C
Because F = q E,
I
Z
E ¢ dl = 0 !
(r £ E) ¢ dS = 0
C
!
r£E=0
S
where S is the open surface formed by
closed contour.
{ Second way:
E for a single point source charge at r0
{ r ¢ E is expressed as
· Z
μ^¶ ¸
R
0
r ¢ E = r ¢ ke
dv
½(r )
R2
V0
·
μ ^ ¶¸
Z
R
0
= ke
½(r ) r ¢
dv
R2
V0
E=?
r1
r1
qs
y
x
{ Therefore
Z
½(r0 )±(r ¡ r0 ) dv = 4¼ke ½(r)
r ¢ E = 4¼ke
V0
r¢E =
Using the following prperty and null identity
8
μ ¶
μ ¶
μ^¶
^
1
R
>
>
1
R
=
¡r
>
< R2
= ¡r £ r
=0
! r£
R
2
R
R
r £ (rf) = 0
>
>
>
:
r £ E1 = 0
!
f : scalar function
z
^ 2 can be expressed as
{ Because R=R
(Appendix)
μ ¶
^
R
1
= ¡r
R2
R
μ ¶
μ^¶
1
R
2
=
¡r
r¢
R2
R
μ ¶
1
2
r
= ¡4¼ ±(R)
R
μ^¶
R
0
E1 = ke qs (r )
R2
Because r is an operator to the field point variable,
·
μ ^ ¶¸
R
0
r £ E1 = ke qs (r ) r £
R2
E=?
½(r)
²0
r ¢ E for a point charge
{ For a point charge at the origin, how is
r ¢ E at some specific points ?
For all charges,
N
X
i=1
Chung‐Ang University
N
X
Ei =
r£E =r£
(r £ Ei ) = 0
i=1
Electromagnetics
13
z
1.6 Electric Energy and Potential
W
² E for a point charge qs (source) at the
origin
1 ³ qs ´
Er =
E = ^r Er ;
4¼²0 r2
If a test charge qt is at infinity, there is
no energy.
r→∞
qt
y
r
x
N
X
V=
1C
V
qs
² Multiple point charges (not vector sum but scalar sum)
Vn ;
where
Vn =
n=1
qn
4¼²0 Rn
² Continuous charge distribution
1
V=
4¼²0
Z
v
0
½(r0 )
dv
R
We (r = 1) = 0
² Relation between V and E
Required energy to move this charge from 1
to r
Z r
Z r
F ¢ dl = ¡qt
E ¢ dl
We (r) = ¡
1
Z
Vba = ¡
1
We (r) = qt
² Electric potential is Electric energy for
qt = 1 C.
z
C4
qt
² Electric potential (V) for a point charge
qs ,
qt
x
C3
C2
Vb surface
C5
qs
@V
@V
@V
¢x +
¢y +
¢z
¢V =
@x
@y
@z
¶
μ
@V
@V
@V
^
^ ¢y + ^
^
+y
+^
z
¢ (^
x ¢x + y
z ¢z)
= x
@x
@y
@z
qt
C6
{ From these two expressions
y
E = ¡r V
x
qt
Va surface
b
E2
E
E1 E
2
Δl
E1
Vba=‐E1⋅l
r£E=0
!
E » rf
!
E = ¡rV
² Since only scalar quantity is associated,
it is easier to handle the problem in
calculation of E via V .
Electromagnetics
E
E
a
² Mathematical validation
(
r £ (rf) = 0; (Vector identity)
independent of integration path(r £ E = 0)
Chung‐Ang University
y
a 1C
C1
a
:
b
{ Expression 2 (Gradient)
1.6.1 Electric Potential
² Potential difference
Z b
E ¢ dl
Vba = Vb ¡ Va = ¡
qs
E ¢ dl
¢V = ¡E ¢ ¢l
^ ¢y + ^
¢l = ¢x + y
z ¢z)
qs
4¼² r
qs
W(r)
=
qt
4¼² r
b
a
With dl = ^
r dr
V(r) =
z
{ Expression 1
14
{ At z = 0, R1 = R2 = R10 =
z
Example Calculation of E via V
L/2
² Electric potential V for a finite line
charge
Z L=2
1
½L 0
dz
V=
4¼²0 ¡L=2 R
p
R = ½2 + (z ¡ z0 )2
Z L=2
1
½L
p
V=
dz0
4¼²0 ¡L=2 ½2 + (z ¡ z0 )2
dq
β2
β
½
E½ = A
R10
R
A=
ρ
ρL
z § L=2
½
where where + for i = 1, and ¡ for i = 2
¶
μ
R1 + z + L=2
½
V = L ln
4¼²0
R2 + z ¡ L=2
{ Resulting electric field
8
¡@ V
½L 1
>
>
E
=
=
(sin ¯1 ¡ sin ¯2 ))
½
>
<
@½
4¼²0 ½
E = ¡rV !
μ
¶
>
¡@ V
1
½L
1
>
>
=
¡
:Ez =
@z
4¼²0 R2
R1
1
1
¡
(¡L=2 + R10 ) (L=2 + R10 )
L
½ R10
¸
¶
(Same expression)
Example V and E for a circular line charge
z
dV (0,0,z)
² Find V and E for a circular ring charge on
the z-axis.
{ Differential potential dV
μ ¶
1
½L
dl
dV =
4¼²0 R
p
R = b2 + z2 ;
{ The fact V = 0 when z ! 1 or ½ ! 1 implies that the constant C =
0. In addition
tan ¯i =
½2 + (L=2)2
{ Along the z-axis (jzj > L=2), we can obtain
the same results before.
β1
(z ¡ z0 ) = ½ tan ¯
(
dz0 = ¡½ sec2 ¯ d¯
p
!
R = ½2 + (z ¡ z0 )2 = ½ sec ¯
Z ¯2
¡½L
V=
sec ¯ d¯
4¼²0 ¯1
¤¯2
¡½L £
=
ln (sec ¯ + tan ¯) ¯ + C
1
4¼²0
Ri
;
½
·
½L
4¼²0
μ
½
E½ = L
4¼²0
R1
{ Introducing a angle ¯ from the ½-axis as
sec ¯i =
Ez = 0
V = ???
r
r′
0
‐L/2
R2
p

b
dl = bdÁ
x
R
0
y
′
L
dq
{ Electric potential (2 times of the
contribution from the half circle)
¶
μ
½L b 1
V=
2²0 R
{ Similarly we can find V off the axis,
and we can know that the derivatives of
V along x and y are zero. Therefore the
resulting electric field on the z-axis has
only the z component as
μ
¶
½L b
¡@ V
z
=
Ez =
@z
2²0 R3
! Same expression.
! Same expressions (See Appendix)
Chung‐Ang University
Electromagnetics
15
1.6.2 Electrostatic Energy
² Considering all charges
² Situation 1 (only one charge Q1 )
N
1X
Qk Vk
We =
2 k=1
E-field is formed by source charge Q1
Q2 at r = 1
!
W=0
Vk =
1
4¼²0
² Situation 2 (moving Q2 to a point R21 )
Required energy
Q1
W2 = ¢W = Q2 V21 = Q2
4¼²0 R21
¡ Subscripts
(
1-st subsript : field point
2-nd subscrip : source point
Here the issue may be the contribution due to ½v at the field point
where V is evaluated. Because V is known to be finite over the
region of interest, and the infinitesimal integration volume ±v ! 0,
the contribution of ½v is zero anyway.
Therefore
W2 = Q1
² Situation 3 (moving Q3 to R31 )
W3 = W2 + ¢W
¢W = (Q3 V31 + Q3 V32 )
(Q3 V31 + Q1 V13 ) + (Q3 V32 + Q2 V23 )
=
2
(Q2 V21 + Q1 V12 )
W3 =
2
(Q3 V31 + Q1 V13 ) + (Q3 V32 + Q2 V23 )
+
2
1
= [Q1 (V12 + V13 ) + Q2 (V21 + V23 ) + Q3 (V31 + V32 )]
2
1
= (Q1 V1 + Q2 V2 + Q3 V3 )
2
Chung‐Ang University
j=1(j6=k)
Qj
Rjk
² Generalized to a continuous charge distribution
Z
1
½v (r0 )
V(r) =
dv
4¼²0 r0 6=r R
Z
1
½v (r0 ) V(r) dv
We =
2 V
¢W :
Q2
= Q1 V12
4¼²0 R12
Q2 V21 + Q1 V12
! W2 =
2
1
! W2 = (Q1 V1 + Q2 V2 )
2
Vi : potential due to charges except Qi
N
X
ROI: Region of Interest
Q1
ROI
r→∞
² Therefore we can express We with ½v = r ¢ D as
Z
Z
1
1
½v V(r) dv =
(r ¢ D) V(r) dv
We =
2 V
2 V
Using vector identity
Q1
R12
Q1 R12
R13
r→∞
r ¢ (V D) = V (r ¢ D) + D ¢ (rV)
Q2
Q2
R23
Q3
r→∞
Z
Z
1
1
We =
[r ¢ (V D)] dv ¡
D ¢ (rV) dv
2 V
2 V
I
Z
1
1
(V D) ¢ dS +
(D ¢ E) dv
=
2 S
2 V
|
{z
}
! 0 as r ! 1
¶
Z μ
Z
D¢E
=
we dv
dv =
2
V
V0
we =
Electromagnetics
D¢E
2
:
electric energy density
16
² Different definition of energy desity
8
½V
>
: defined only the source region
>
<we1 = 2
>
>
:we = D ¢ E
2
:
{ Method 2 (with we1 = ½v V=2)
The potential V0 at r = a is calculated as
Z
Z a
¡½v 3 a ¡2
½v a2
Er dr =
a
r dr =
V0 = ¡
3²0
3²0
¡1
¡1
defined over all problem space
and the potential for r < a is
Z r
½v
½v (a2 ¡ r2 )
V = V0 ¡
» d» = V0 +
3²0 a
6²0
¢
½v ¡ 2
3a ¡ r2
=
6²0
Example Energy required to forming a cloud charge
² Calculate We required to assemble a uniform cloud of charge of
radius a and volume charge density ½v .
{ Method 1
z
The total charges inside S and the
resulting electric flux density are
¶
8μ
4¼r3
>
>
½v ; r < a
<
3
¶
Qi = μ
>
4¼a3
>
:
½v ; r > a
3
a
Therefore the stored We is calculated as
ρv
¢
½v V
½2 ¡
= v 3a2 ¡ r2
2
12²0
Z 2¼ Z ¼ Z a
We =
we1 r2 sin μdrdμdÁ
we1 =
y
x
0
0
μ
¶
4¼½2v
a5
4¼½2v a5
5
=
a ¡
=
12²0
5
15²0
Therefore, Dr is
8³ ½ ´
v
>
r;
<
Qi
3
Dr =
= ³ ´ 3
4¼r2 >
: ½v a ;
3 r2
0
Same results !!!
r<a
r>a
Now we and We are
8μ 2 ¶
½v
>
>
r2 ; r < a
<
2
18²0
Dr
we =
= μ 2 ¶ 6
2²0 >
a
½v
>
:
; r>a
18²0 r4
Z 2¼ Z ¼ Z 1
we r2 sin μdrdμdÁ
We =
0
=
=
4¼ ½2v
0
Z
a
4
a5
+ a5
5
r
dr
a
0
μ
¶
1
¡2
6
r dr + a
18²0
2¼ ½2v
9²0
0
μZ
¶
=
Chung‐Ang University
4¼½2v a5
15²0
Electromagnetics
17
² Distribution process of charges
inserted into the conductor
1) Charge moves to the conductor
surface due to repulsive force
2) Finally, ½ = 0 and E = 0 inside the
conductor (equilibrium state)
I
D ¢ dS = 0
Electron energy
² Material in which electrons can move
freely, but totally neutralized.
0 just outside a
4) If not, Et makes the charges to move
until Et = 0.
² Various electrostatic problems realted
to conductors can be solved with
2nd-order differential equations and
the boundary-value problems (This will
be covered later).
E
² What to be noteworthy:
{ If we can find V for a charge
distribution, the E can be found
by simple mathematical operation.
{ Generally it is easier to find V
than the direct calculation of E
because of scalar operation.
Chung‐Ang University
Point charge at the center of conducting
shell

E 
Conductor






E
E
Inserting
positive
charges
+ +
+++ +
+ 4V
‐ ‐ ‐ ‐
E
+ 4V
2V
‐ ‐ ‐
0V
² A point charge Q is at the center of
neutralized conducting shell. Calculate
E and V with inner and outer radius a
and b, respectively.
{ Because the given conductor is neutral,
and the electrons in conductor can move
freely.
{ Due to the effect of the point charge
+Q at center, the electrons and
positive charges move to the inner
(by attaction) and outer (by repucsion)
surfaces, respectively, as shown in
Fig. XXX .
{ Since the electric field is zero inside
conductor in steady state (equilibrium
status), the net charge at inner and
outer surfaces of the conductor should
be ¡Q and +Q respectively.
² Electric field can be found by applying
the Gauss' law:
Conducting shell
b
Q
a
+ + +
+ –– – ––+
+–
Q –+
–+
+– b
a
Er
r
V
V0
r
{ By the Gauss' law, the electric field
becomes E = ^r with
8
< Q ; for r < a and r > b
Er = 4¼²0 r2
:0;
for a < r < b
Electromagnetics
E
⊝ ⊕
⊕
⊝
⊕
⊝
⊕
⊝
⊝ ⊕
Region where a conductor
is to be placed
E
– E
After
Eext
Constant V lines
Conductor
5) Finaly, only En can exist on a
conductor surface.
Before
Bandgap
Semiconductor
Insulator
any closed S inside the conductor
=
E ? Equipotential lines
|
{z
}
Band
overlap
S
3) In addition, Et
conductor.
{ E is always perpendicular to the
constant V lines:
Conduction band
Valance band
1.7.1 Conductor
+ + +
+ – –– +
–
+ ––
– +
1.7 Conductor and Dielectrics
18
{ Now the potential V can be found for r > b at first as
Z r
Q
V=¡
Er dr =
; for r > b
4
¼²
0r
¡1
The potential V0 at r = a can be found and the resulting We
becomes
Z a
Q
V0 = ¡
Er dr =
4
¼²
0a
1
¸
·
Z
a+
4¼
Q2
Q2
2
±(
r
¡
a
)
r
dr
=
We =
2 (4¼)2 ²0 a3 a¡
8¼²0 a
: Same results !!!
Inside the conducting shell, the electric field is zero, and
therfore the electric potential becomes
Q
;
4¼²0 b
for a < r < b
6.2 Dielectrics and Electric Dipole
Finally the potential for r < a is
μ
¶
Z r
Q
1 1
V=¡
Er dr + V0 =
+ V0
¡
4¼²0 r a
a
² A dielectric material consists of specific
atoms, which can be modeled by a lot of
electric dipole (pair of + and ¡ charge),
which are bounded to some region, and the
directions from ¡ to + charge of the pairs
are randomly distributed without E.
The plots of Er and V are depicted in Figure.
Example Energy of a conducting sphere
² Our concern is what happens when the
external E is applied.
² A conducting sphere having a radius a is
charged with Q. Calculate the stored electric
energy.
{ Method 1: Calculation using Dr ,
8
< Q ; r>a
Dr = 4¼r2
:0;
r<a
Q2
Dr Er
=
;
r>a
2
2 (4¼)2 ²0 r4
¸Z 1
·
Z
Q2
2
We = we r sin μdrdμdÁ = 4¼
r¡2 dr
2²
2
(4¼)
0
a
v
Q2
=
8¼²0 a
we =
{ Method 2:
Calculation with ½ and V
The stored electric energy is we1 = ½v V=2 with
½v = ½s ±(r ¡ a)
Q
½s =
4¼a2
Chung‐Ang University
z
Q
a
y
x
Conducting
sphere
+
‐
V0 =
² The electron cloud model, (all electron
charges are distributed uniformly in a
spherical cloud with a radius of its orbit)
was suggested and has been used for a simple
study the phenomena of an atom under an
applied external electric field.
E0 = 0
Atom in external electric field
{ Let us consider an atom whose electron
charge is ¡q under an applied external
electric field E0 as shown in Fig. By
the electron cloud model, calculate the
equivalent electron charge density and
relative separation d=r0 , where d is the
distance between the center of electron
cloud and a point positive charge.
{ The equivalent electron charge density is
calculated as
½c =
Electromagnetics
3q
4¼r30
–
–
–
q
–
–
–
–
–
E0 ≠ 0
–
–
–
q
–
–
–
–
–
19
{ For an atom with one electron, even a
relatively high external field E0 = 106
V/m is applied, d=r0 becomes
{ When an external electric field E0 is applied,
let us assume that the positive point charge is
displaced while the electron cloud is fixed.
d
' 6:95 £ 10¡6
r0
{ Since two kinds of force are exerted on the
positive point charge; one by the electron
cloud, and the other by the external electric
field, and equating these two yields waht we
want to calculate.
{ At first let us consider the force on the
positive charge by the electron cloud. The
electric field Ec at the point of the positive
charge q by the electron cloud can be found by
the Gauss' law as
μ ¶3
¡qi
d
;
q
=
q
Ec =
i
4¼²0 d2
r0
{ The magnitudes of two electric fields Ec and
E0 should be same for equating the two related
forces. Therefore
qd
= E0
4¼²0 r30
{ Now d=r0 can be found as
μ
¶
d
4¼²0 r20
=
E0
r0
q
This implies that the separation d is very
small comapred to the radius of orbit.
² Many molecules (dipoles) in dielectrics are
are aligned if the external electric field
is applied.
Model of
Polarized atom
–
–
–
q
–
–
q –
–
qd
d
–
Chung‐Ang University
R1
q
–
r
R2
θ
p = qd
d 0
+ q
where d is a vector from negative to
positive charges.
z
{ Electric potentail at a far range (d ¿ r)
μ
¶
μ
¶
1
R1 ¡ R2
1
q
q
¡
=
V=
4¼²0 R1
R2
4¼²0
R1 R2
R1 R2 ' r2
p ¢ ^r
;
V=
4¼²0 r2
r
θ
R2
R1 ¡ R2 ' d ¢ ^r
d p
0
x
p¢^
r = p cos μ
d cosθ
z
V(r)
+
‐
+
‐
+
‐
^ Eμ
E = ¡r V = ^r Er + μ
¡@V
p cos μ
=
@r
2¼²0 r3
¡@V
p sin μ
Eμ =
=
r@μ
4¼²0 r3
Field
point
r≫d
R1
{ Electric field at a far range (d ¿ r)
Er =
Electromagnetics
x
‐q
² Electric potential and electric field
e = 1:6 £ 10¡19
4¼²0 £ 10¡20
d
E0
'
r0
q
6:95 £ 10¡12
'
E0
Ne
Field
point
² Dipole moment p for a positive charge q and
separation d is defined as
{ With the approximate value r0 ' 10¡10 (m) and Ne
the number of electrons
q = Ne e;
z
a pair of charges of
² Electric dipole:
same magnitude but opposite sign.
r′
+
‐
+
‐
+
‐
+
‐
+
‐
+
‐
+
‐
+
‐
+
‐
R
r
y
x
20
² Continuous distributed aligned dipoles
total dipole
E0
pk
¢v !0
k=1
2
(C/m , Same dimension of D)
¢v
(
N
N¢v
:
:
No.
No.
of dipoles per unit volume
of total dipoles in ¢v
{ Electric potential for an infinitesimal
volume dv
μ
¶ μ^¶
^
P dv
(P dv) ¢ R
R
dV =
=
¢
2
4¼²0 R
4¼²0
R2
· μ
¶¸
1
1
=
P ¢ r0
dv
4¼²0
R
·
μ
¶
¸
1
1
1
0
0
r ¢
P ¡ (r ¢ P) dv
=
4¼²0
R
R
{ Polarization surface and volume charge
density
(
^0;
½ps = P ¢ n
on S0
½pv = ¡r0 ¢ P; in V0
I
Z
½ps
½pv
V(r) =
dS +
dv
4¼²
R
4¼²
0
0
0R
S
V
– + – +
– +
– + – +
– +
– +
– + – +
– +
– +
– +
– +
– +
– + – +
– +
– +
– + – + – +
– +
– + – + – + – +
² Based on this result, a dielectric material
under the external electric field can be
modeled by an air-filled parallel plates
carrying charges Qp with surface S, and
the equivalent electric flux density Dp and
field Ep are
Q
½ps =
– +
– +
{ The final expression can be obtained by
using the following vector identity:
Q
– +
– +
– +
Chung‐Ang University
Applying external field results
in net charge transition
Qp
– +
– +
Parallel
plate model
d
–
–
–
–
–
–
Qp
^
=P¢n
S
The net charge remaining within the volume V
(inside S) is the negative of this integral
I
Z
Qv = ¡ P ¢ dS = (¡r ¢ P) dv
V
Z S
Qv = ½pv dv
P=Nqd/(Sd)
=Nq/S
Qp=Nq
+
+
+
+
+
+
E0
P
Ep
– +
r ¢ (fA) = rf ¢ A + f(r ¢ A)
1
A = P; f =
R μ
μ
¶
¶
1
1
1
0
0
P ¡ (r0 ¢ P)
=r ¢
P¢ r
R
R
R
E0=0
P=0
+ –
N
¢v
X
P = lim
– +
S
+ –
+ –+ –
{ Polarization vector P :
moment per unit volume
Applying the divergence theorem to the
first term
I
Z
^0
P¢n
¡r0 ¢ P
V(r) =
dS +
dv
S 4¼²0 R
V0 4¼²0 R
– +
–
1
r3
–+ –
–
+ –
–+ –
E»
–
1
;
r2
– +
V»
Before applying
external field
+ –+
+ –+ –
+
+ –
+ –+
+ –+ –
+
+ –
+ –+
+ –+ –
+ –
{ Total electric potential
μ
¶
¸
Z ·
1
1
1
0
0
r ¢
V(r) =
P ¡ (r ¢ P) dv
4¼²0 V0
R
R
{ Dependency of V and E on r
Hence, when r ¢ P 6= 0, the bulk of the polarized dielectric
appears to be charged. However, originally since the
dielectric body is electrically neutral dielectric body, thc
total charge of the body after polarization must remain zero
because of the bound charge characteristics. This can be
readily verified by noting that
Electromagnetics
21
I
Total charge =
Z
½ps dS +
S
V
I
=
½pv dv
Z
(r ¢ P) dv = 0
P ¢ dS ¡
S
V
² In case of free space (no dielectrics), the E and free (not
bounded) charge density ½ are related as
r¢E=
½
²0
Now the presence of the dielectric is known to be modeled
by polarization charges in free space. Therefore the above
should be modified as
½ + ½pv
²0
r ¢ (²0 E + P) = ½
r¢E=
As we have studied that the concept of electric flux density
comes from ther free charge as ½ = r ¢ D, we can find the
following relation:
² Case A (Parallel configuration):
We have studied that the
E inside the dielectric
becomes smaller than the
applied E-field. Therefore
we may think that E-field in
dielectric region is smaller
than that in air region.
However, this is not the
case because the charges
on the conductor moves to
the left due to the calling
from the electric dipoles of
dielectrics. Therefore this
different charge distribution
on the conductor results in
P = ²0 Âe E
Now D can be written as
8
>
² = ²0 ²r
>
>
>
²
<
= 1 + Âe
²r =
D = ²E
²0
>
>
>
>
:Â
e
D1z = ²r1 D2z
!
+ + + + + +
D0
x
Metal
electrode
– – –
Case A
+
air
– – – –
– –
Case B
Dielectric



–
Dielectric
     +
– – – – –
+ + + + +
– – – – –
+Region
+ + + 2+
– – – – –
+ + + + +
– – – – –
+ + + + +
+ +
– – – – – – –
– –
Region 1
(air)
–
and finally
+ +
z
D1z > D2z
D = ²0 E + P
P describing the properties of dielectrics is known to
be a function of E. If we introduce the proportionality
condition  as
Metal
electrode
Ex. Distribution of E and D
+ + + +
– – –
+ + +
– – – –
+ + + +
–
+
– –
+ + +
–
+ + + +
–
+
Region 2
– – – –
– – – –
E1z = E2z
This can be verified by the
boundary conditions which will
be studied later.
² Case B (Series configuration)
Physically, the electric flux density becomes same in both
regions:
(Permittivity)
D1z = D2z
(Relative permitivity), or
(dielectric constant)
!
!
(Electric susceptibility)
²r1 E1z = Ez2
E2z
E1z =
²r1
For the free space, it is found that
² = ²0
²
²r =
=1
²0
Chung‐Ang University
Electromagnetics
+ + +
Region 1
(air)
22
–
– –
7. Boundary Conditions
Condition 1
² Tangential component of electric field
Z
(r £ E) ¢ dS = 0
r£E= 0 !
S
I
E ¢ dl = 0
!
Region 1 (1)
w
Region 2 (2)
w!0
² Normal component of electric field
Z
r¢D =½ !
(r ¢ D) dv = Q
V
I
D ¢ dS = Q
!
Condition 2
h
Region 1 (1)
D1n dS S
s
n̂
D2n dS
Region 2 (2)
S
h!0
¡¡¡! D1n S ¡ D2n S = ½s S
!
D1n ¡ D2n = ½s
² Concept of superposition
8
>
No source at the boundary:
>
>
<
>
Source only with ½s :
>
>
:
Combining the results
D1n ¡ D2n = ½s
(D1n )0 = (D2n )0
8
½
<(D1n )s = s
2
:(D2n ) = ¡½s
s
2
x
² A dielectric sheet with a dielectric
constant ²r is placed perpendicularly in
^ E0 in free
a uniform electric field E = x
space. Determine E, D, and P.
{ In this case, Et = 0. By the boundary
condition on Dn for ½ = 0,
C
¡¡¡¡! E2t L ¡ E1t L = 0 ! E2t = E1t
Ex. Dielectric sheet in unform E
Decomposition
Superposition
No surface
charge density Region 1 (1)
D1n0
D2n0
Region 2 (2)
Surface charge
density only
s
Region 1 (1)
D1ns
D2ns
Region 2 (2)
Dx0 = Dxd
!
E0
Ed
E0
D0
Dd
D0
Air
Dielectrics
r
Ex1 = ²r Exd
The subscripts 0 and d stand for air and
dielectric regions, respectively. Since
Ex0 = E0 ,
8
<Ex0 = E0
! Dx = ²0 E0
:Exd = E0
²r
8
>
in air
<0;
¶
μ
! Px = Dx ¡ ²0 Ex =
1
>
²0 Ex ; in dielec.
: 1¡
²r
Ex. E, D = ??? for a point charge Q at the
center of dielectric shell
² The dielectric shell with inner and outer
radius a and b, respectively, has a dielectric
constant ²r .
{ D has only a radial components.
of normal D,
By the B.C.
r
Q
b
a
Dielectric
shell
Dr
0Er
Dr0 = Drd
{ Now Dr and Er can be found as
Q
4¼r2
8
Dr
Q
>
>
;
< =
²0
4¼ ²0
Er (r) =
Dr
Q
>
>
: =
;
²r
4¼²0 ²r
0
r
Dr (r) =
Chung‐Ang University
Air
Electromagnetics
in air
in dielelctrics
23
Ex. Boundary conditions at dielectric-air interface
² Case B (Series configuration):
becomes same in both regions:
z
² A dielectric region having a dielectric constant ²r is
placed in air as shown in Fig. XXX, and E in air
is assumed to have magnitude E0 and make an angle
μ1 with the air-dielectric boundary. Determine the
magnitude of E and the angle μ2 .
1 E1
Air
D1z = D2z
Physically, the electric flux density
!
²1 E1z = ²2 E2z
x
r
2
² In a similar way, we can find the related boundary conditions for the
following structures (See Appendix).
E2
Conducting
sphere
{ In order to apply the boundary conditions, we need
to ¯nd the tangential and normal components of
electric ¯eld. From the given magnitude and angle
information, they are
2
2
1
2
1
E1t = E1x = E0 sin μ1
E1n = E1z = E0 cos μ1
E1=E2
1
1
D1=D2
Er1=Er2
2
Dr1=Dr2
{ From the boundary conditions, E2t and E2n become
8. Capacitors and Capacitance
E2t = E1t
D2n = D1n
! E2x = E1x = E0 sin μ1
E1z
E0 cos μ1
! E2z =
=
²r
²r
² Capacitor:
Electronic device formed by two conducting plates
preserving charges
{ Therefore the magnitude of E2 and angle μ2 are
found as
s
cos2 μ1
jE2 j = E0 sin2 μ1 +
²2r
V=0
V = V1
V = 2V1
¢¢¢
E2x
²r sin μ1
tan μ2 =
=
! μ2 = tan¡1 (²r tan μ1 )
E2z
cos μ1
² Capacitance:
Let us assume that the electric field has only
the z-component, and there is no fringing field.
From the continuity of the tangential component
of the electric field,
E1z = E2z
!
D1z
D2z
=
²1
²2
     

‐Q
Q=0
Q = Q1
Q = 2 Q1
Q
!
Q = kV
!
C = slope
+
+
Q = CV
² Capacitance C depends
² Case A (Parallel configuration):
V
Proportional constant of Q w.r.t V
Q/V
Ex. Parallel conducting plates
Q

     
+
+
0
V
{ not on Q and V
1
2
E1z=E2z
z
x
{ but on the geometry and medium.
² Capacitance of various configurations of
capacitor
Circuit symbol
C
V
1
2
D1z=D2z
Chung‐Ang University
+
Electromagnetics
24
Ex. Parallel plate capacitor
² When a voltage V is applied between
two conductors with an assumption of no
fringing field, electric field and flux
density become
¡V
Ez =
d
Parallel plate capacitor
!
² Let a charge Q is assumed to be uniformly distributed
on the surface of the center conductor. Then D and
E are found from the Gauss' law as
I
Q
D ¢ dS = Q ! D½ =
2¼½L
S
Q
! E½ =
2¼²½L
z
S
r
d
Dz = ²Ez
The an induced positive charge on the
upper conductor is found from the Gauss'
law as
I
Q = D ¢ dS = Dz S
S
Therefore the capaciatance becomes
Metal Foil
Plates
Q
²S
C= =
V
d
It is found that the capaciatnce increase
as dielectric constant and surface area
S, but deceases as spacing d, and this is
a general trend for any other capacitors.
{ The other is to use multilayer
structure with very high dielectric
constant material.
Multilayer ceramic capacitor (lower
figure), which is widely used in
small-scale circuit design.
Now the V0 between conductors and the resulting C
can be found as
μ ¶
Z b
b
Q
V=
ln
E ¢ dl =
2¼²L
a
a
2¼²L
Q
C= = μ ¶
b
V
ln
a
Coaxial‐line capacitor
L
ϵr
a
b
Spherical capacitor
Ex. Spherical-shell capacitor
Lead
Paper Lead
Electrolytic capacitors
–
100 uF / 10 V
² Based on the previous result, we can man
make the capacitor structure yielding
large capacitance. Fig. XXX shows two
main approaches:
{ Increasing S with keeping d to be very
small
Eelectrolithic capacitor (upper figure)
Ex. Coaxial capacitor
Multi‐layer Ceramic chip capacitor
1005:
1.0 mm
0.5 mm
² Let a charge Q is assumed to be uniformly distributed
on the surface of the inner conducting sphere. Then
resulting D and C can be found as
I
Q
D ¢ dS = Q ! Dr =
2
4¼r
S
Q
! Er =
4¼²r2
b
a
ϵr
Now V0 between two conductors C are
μ
¶
Z b
1 1
Q
¡
E ¢ dl =
V=
4¼² a b
a
Q
4¼²
¶
C= = μ
1 1
V
¡
a b
Multilayer
configuration
Chung‐Ang University
Electromagnetics
25
² Connection of capacitors
Series connected capacitor
{ Series connection of capacitor
Q
Q
Q
V = V1 + V2 + ¢ ¢ ¢ + VN =
+
+ ¢¢¢ +
C1 C2
CN
¶
μ
1
1
1
=Q
+
+ ¢¢¢ +
C1 C2
CN
Q
=
Cs
+ VN ‐
CN
+ V2 ‐
C2
+ V1 ‐
C1
+Q ‐Q +Q ‐Q +Q ‐Q
+V ‐
Cs
+Q ‐Q
Parallel connected capacitor
1
1
1
1
=
+
+ ¢¢¢
Cs
C1 C2
CN
Q = Q1 + Q2 + ¢ ¢ ¢ + QN = C1 V + C2 V + ¢ ¢ ¢ + CN V
= (C1 + C2 + ¢ ¢ ¢ + CN )V
= Cp V
Cp = C1 + C2 + ¢ ¢ ¢ CN
S1
ϵ1
S2
ϵ2
{ Constant V:
battery.
Conductors are connected to
{ Constant Q:
Conductors are kept isolated.
d
d2
d1
² Coaxial and Spherical capacitors (See Appendix)
b1
ϵ2
b
b
b2
a
a ϵ1
ϵ2
a
ϵ1
L
L
Chung‐Ang University
dWm = Fdz;
ϵ2
ϵ1
b2
b1
XXX)
dz
V
dQ:
a ϵ2
F
z
dWm + dWb = dWe
8
dWm : Additionally supplied mechanical
>
>
>
>
>
energy by the force F
>
>
>
<dW : Additionally supplied electric
b
>
energy by the battery
>
>
>
>
>
dWe : Increased stored electric energy
>
>
:
in the capacitor
S
ϵ2
ϵ1
F2=‐F1
F1
z=0
² Considering two kinds of supplied energy
(dependong on the approaches), the above
equation can be expressed as
‐Q
+Q
Parallel connection Series Connection
1
1
1
=
C = C1 + C2
+
C
C1 C2
8
8
Q
²1 S
²1 S1
Q1
>
>
>
=
=
<C1 =
<C1 =
V1
d1
V
d
Q
²
>
>
²2 S2
Q2
2S
:
>
:C2 =
=
=
C2 =
V2
d2
V
d
dl
Constant V
+ V ‐
Cp
² Parallel plate capacitors (Composite
dielectrics)
Mechanically supplied energy
Wm = –ʃ F1∙dl = ʃ F2∙dl
² It is made by postulating an infinitesimal
displacement, and then applying the
principle of conservation of energy. Here
² Two approaches (Fig.
CN
‐QN
+QN
² To calculate the electrostatic forces on the
conducting plates, we can use the concept
of the principle of virtual work, which is a
general and reliable method for calculating
forces.
¢Ws = ¢Wi
Increased = Increase in
supplied energy
the internal energy
+ V ‐
C1
+Q1 ‐Q1
C2
+Q2 ‐Q2
{ Parallel connection of capacitor
9. Electrostatic Forces
Constant Q
dz
z
FQ S
‐Q
z=0
dWb = VdQ
extra charge fed into the capacitor
by the battery
ϵ1
Electromagnetics
26
S
Ex. Force on the conductor of parallel-plate capacitor
² Calculate the force exerted on the conductor of parallel-plate
capacitor.
{ Case 1 :
With constant V
¡ The variation of the supplied energy by the battery
dWb = VdQ = V(VdC)
With the following expression of C(z)
C(z) =
²0 S
z
!
!
dC
¡²0 S
= 2
dz
z
¡²0 S
dC = 2 dz
z
¡ The resulting dWb can be found with E = V=z:
μ
¶
¡²0 S
2
2
dz
dWb = V dC = V
z2
= (¡²0 E2 ) S dz
Negative sign in dC and dWb implies that the energy flows
from the capacitor into the battery.
¡ On the other hand, dWe can be found with E = V=z:
μ 2¶
²0 V2 S
²0 E
We =
Sz =
2
2z
μ
¶
μ
¶
2
¡²0 V S
¡²0 E2
dz =
Sdz
dWe =
2z2
2
¡ Finally the force on the capacitor can be found as
Fdz + dWb = dWe
μ
¶
²0 E2
2
Fdz = ²0 E ¡
Sdz
2
μ 2¶
²0 E
F=
S
2
{ Case 2:
With constant Q (isolated capacitor, dWb = 0)
¡ The related differential energy become
dWm = dWe ;
dWm = Fdz
¡ In a similar way, dWe can be found with D = Q=S as
μ 2¶
D
Q2
z
Sz =
We =
2²0
2²0 S
Q2
E2
D2
dWe =
Sdz = Sdz
dz =
2²0 S
2²0
2
¡ The resulting F is found to be same as the previous result.
μ 2¶
²0 E
F=
S
2
Ex. Force on a capacitor with pulling
dielectric sheet
² For a dielectric-filled capacitor,
calculate the force exerted on the upper
conductor wheen a dielectric slab is
pull.
a
w
ϵ
d
x=0 x
{ With keeping constant Q (isolated capacitor), energy
conservation law is stated as
Fdx = dWe
{ It is found that We is a function of C, and C is a function
of x. Therefore dWe can be found as
8
² w
<C = C1 + C2 = 0 [x + ²r (a ¡ x)]
2
Q
d
We =
;
:dC = ¡²0 b (²r ¡ 1) dx
2C
d
¡Q2
Q2 ³ ²0 w ´
(²r ¡ 1) dx
dC = 2
dWe =
2C2
2C
d
{ Therefore F can be found with V = Q=C as
F=
Chung‐Ang University
with
Electromagnetics
Q2 ³ ²0 w ´
²0 E2
(²r ¡ 1)wd
¡
1)
=
(²
r
2C2 d
2
27
Pulling
10. Solutions of Electrostatic Problems
10.1 Uniqueness Theorem
² Up to now, we have studied the following methods for solving the
electrostatic problems:
² A solution of Poisson and Laplace eqs for
the specified B.C. is unique !!!
Q3
{ Direct calculation (vector operation)
Proof:
{ V ! E
{ Series Solutions
{ Let us assume that two different solutions
V1 and V2 with the same boundary condition
can exist.
¡½
¡½
r2 V2 =
;
r2 V1 =
²
²
(
V1 6= V1 ; inside the boundary
V1 = V2 ; on the boundary
{ Numerical Methods
With Vd = V1 ¡ V2
{ Gauss' law
² Much more geenral methods based on Poisson's and Laplace'
equations:
{ Image Method
V=0 on the boundary
Q2
V = ???
Q1
(
r2 Vd = 0
Vd = 0; on the boundary
¡ FDM (Finite Difference Method
¡ FEM (Finite Element Method)
¡ MOM (Method of Momentum)
{ Vector identity with f = Vd and A = rVd
² Poisson Equation
{ Two governing electrostatic equations in any medium
8
<r £ E = 0 ! E = ¡r V
:r ¢ D = ½ ! r ¢ E = ½
²
!
¡½
r ¢ (r V) =
²
!
¡½
r2 V =
²
r ¢ (f A) = f (r ¢ A) + (rf) ¢ A
:0
2
»»
r ¢ (Vd rVd ) = Vd »
(r»¢ »
rV
d ) + rVd ¢ rVd = jrVd j
{ With a volume integral and the divergence theorem
Z
Z
2
[r ¢ (Vd rVd )] dv = jrVd j dv
V
{ Lapalce' equation (with ½ = 0
r2 V =
¡½
²
½=0
¡¡¡¡¡¡!
S
r2 V = 0
2
jrVd j dv
V
{ Since Vd = 0 at the boundary,
Z
2
jrVd j dv = 0 ! rVd = 0
V
(
Vd = constant
Vd = 0on the boundary
!
Chung‐Ang University
V
(Vd rVd ) ¢ dS =
!
{ Electrostatic boundary-value problems are specified with
1) r2 V = 0
2) Boundary conditions
Z
I
Electromagnetics
V1 = V2
!
! Vd = 0 for all region
Contradiction !!!
28
10.2 Simple Boundary-Value Problems
Ex. Electric potential for a parallel-plate
capacitor
² Calculate a electric potential V for a
parallel-plate capacitor, where the plates
are assumed to be infinite and the applied
potentials are
μ
¶
@V
½
=0
@½
@V
A1
!
=
@½
½
1 @
½ @½
z
V0
z=d
z=0
V=0
x
1) V(z = 0) = 0
2) V(z = d) = V0
A1 ln a + A2 = V0
From the boundary conditions, it is known
that
1) V(½ = a) = V0
2) V(½ = b) = 0
{ It is known that V becomes a function of
½ only, and the Lapalce equation becomes
simplified as
8
>
<A1 =
!
ln
b
>
:
A2 = ¡A1 ln b
z
² Calculate a electric potential V for cylindrical wedge
having the following potentials:
1) V(Á = 0) = 0
2) V(Á = Á0 ) = V0
b
a
{ It is known that V becomes a function of Á only,
and the Lapalce equation becomes simpli¯ed as
ϵr
Insulating
gap

y
x
V=0
V0
1 @2
@2
=
0
!
=0
½2 @Á2
@Á2
! V = A1 Á + A2
V0
V=0
{ From the boundary conditions, we can determine
the coe±cients A2 and A2 , and the resulting V becomes
V = V0
Chung‐Ang University
V
³0a ´
Ex. Electric potential for cylindrical wedge
V0
A1 =
μ ¶ d
V0
V=
z
d
A2 = 0;
² Calculate a electric potential V for a
coaxial capacitor with the following
potentials:
V = A1 ln ½ + A2
³½´
V = A1 ln ½ ¡ A1 ln b = A1 ln
b
μ ¶
b
ln
½
= V0 μ ¶
b
ln
a
@2
V(z) = 0
@ z2
V(z) = A1 z + A2
Ex. Electric potential for a coaxial
capacitor
!
{ Therefore the potential becomes
r2 V = 0;
!
½
From the boundary conditions
(
A1 ln b + A2 = 0
{ The potential V becomes a function of z
only, and the related Lapalce equation
becomes
@V
= A1
@½
!
Electromagnetics
Á
Á0
29
{ B1 is given as
Ex. Electric potential V for spherical-shell capcitor
² Calculate a electric potential V for spherical-shell
capcitor having the following potentials:
1) V(r = a) = V0
2) V(r = b) = 0
{ It is known that V becomes a function of r only,
and the Lapalce equation becomes simpli¯ed as
μ
¶
1 @
2 @V
r
=0
r2 @ r
@r
{ The solution of V(r) can be expressed as
8A
1
>
< + A2 ; a < r < b1
r
V(r) =
>
B
: 1 +B ; b <r<b
2
1
2
r
and considering the given potentials,
¶
8 μ
1 1
>
>
<A1 r ¡ a + V0 ; a < r < b1
μ
¶
V(r) =
>
1
1
>
:B1
;
b1 < r < b2
¡
r b2
{ Therefore Er becomes
8A
1
>
< 2 ; a < r < b1
r
Er (r) = ¡rV(r) =
>
: B1 ; b < r < b
1
2
r2
{ From the boundary conditions at r = b1 are
(
V1 = V2
Dr1 = Dr2 ! ²1 Er1 = ²2 Er2
8
μ
¶
μ
¶
>
<V0 + A1 1 ¡ 1 = B1 1 ¡ 1
b1 a
b1 b2
!
>
:² A = ² B
1 1
2 1
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μ
b2
a
b1
ϵ1
ϵ2
B1 =
²1
²2
¶
A1
{ Therefore A1 can be found as
μ
¶
μ ¶μ
¶
1
1
1
²1
1
A1
¡
= A1
¡
¡ V0
b1 a
²2
b1 b2
V0
¶ μ
¶
A1 = μ ¶ μ
²1
1
1
1
1
¡
+
¡
²2
b1 b2
a b1
{ Now the potential V(r) can be determined.
Ex. Potential distribution of cone structure
² Calculate the potential distribution of cone
structure with the following potentials:
(
V(μ = ®) = V0
V(μ = 0) = 0
{ Considering the structure and assigned
potentials, the potential distribution
is known to be a function of μ only.
Therefore the Lapalace equation and its
solution become
μ
¶
A1
1 @
@ V(μ)
@ V(μ)
sin μ
=0 !
=
sin μ @μ
@μ
@μ
sin μ
μ
¶
μ
! V(μ) = A1 ln tan
+ A2
2
z
V0

P(r,,)

V=0
{ From the boudary conditions, the electric
potential V, having the axial symmetry,
satisfies the Laplace equation as
μ
¶
μ
ln tan
2
V(μ) = V0 ³
®´
ln tan
2
Electromagnetics
30
x
² From the boundary condition (4), the condition of k becomes
10.3 Series Solution
10.3.1 Lapalce Equation
k
² For a source-free region, let us consider how to ¯nd the
potential V if the boundary conditions are speci¯ed.
Ex. Two dimensional boundary-value problem
² One two-dimensional example is shown in Fig. XXX
with the following boundary conditions:
(
(
1) X(0) = V0
3) Y(0) = 0
;
2) X(1) = ¯nite
4) Y(b) = 0
x
x→
0V
² The Laplace equation for this problem becomes
r2 V = 0
!
@V
@V
+ 2 =0
2
@x
@y
x
m
² Multiplying sin(kn y) to both parts, and taking integrations them
from y = 0 to y = b
Z
V0
² The Lapalce equation can be simplified as
Ä
X
¡Ä
Y
=
= k2
X
Y
!
1
X
b
sin(kn y) dy =
0
V(x; y) = X(x) Y(y)
!
m = 1; 2; ¢:
Here the unkown coefficient Am can be found from the remaining
boundary condition (1):
X
Am sin(km y)
V0 =
² With the seperation of variables,
Ä
XY + XÄ
Y=0
m¼
;
b
m
0V
V0
0
km =
Now the general expression of V becomes
X
Am e¡km x sin(km y)
V(x; y) =
y
b
!
Ä
X Ä
Y
+ =0
X Y
(
Ä
X = k2 X
Ä
Y = ¡k2 Y
Am sin
m=1
³ n¼ ´
³ m¼ ´
y sin
y dy
b
b
Using the integral formula and the orthogonal relation
8
Z b
< 2b ; odd m
³ m¼ ´
y dy = m¼
sin
:
b
0
0;
even m
μ ¶
Z b
³ n¼ ´
³ m¼ ´
b
y sin
y dy = ±mn
sin
b
b
2
0
Now Am can be found as
where k(¸ 0) is a constant, and two dots over letter means
two times derivative with the related variable. The possible
solution sets are
(
(
e¡kx
cos(ky )
X»
;
Y»
kx
sin(ky)
e
8
< 4V0 ;
Am = m¼
:
0;
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even m
Therefore V is given as
X
and their linear combinations become the solution. From the
boundary conditions (2) and (3), V(x; y) can be written as
V(x; y) » e¡kx sin(ky)
odd m
V(x; y) =
4V0
exp
m¼
m=1;3;5;¢
μ
¶
³ m¼ ´
¡m¼
x sin
y
b
b
The above expression with only a few terms gives an accurate
result because of the decrasing term of 1=m as m increases.
Electromagnetics
31
10.3.2 Boundary Value Problem of Poisson's Equation
² AAA
Chung‐Ang University
Electromagnetics
32
{ In a matrix form
10.4 Numerical Solution
y
0V
10.4.1 Finite Di®erence Method (FDM)
b
AV = Vs
² Source free problem ! Laplace equation
2
¡4
V(x,y)
2
@ V @ V
r2 V = 2 + 2 = 0
@x
@y
A=
0
1
1
0
0
0
0
1
1
{ 2D difference approximation
V(x + ¢x=2) ¡ V(x ¡ ¢x=2)
@V
= V0 (x) !
@x
¢x
2
0
0
@ V
V (x + ¢x=2) ¡ V (x ¡ ¢x=2)
!
@ x2
¢x
V(x + ¢x; y) ¡ 2V(x; y) + V(x ¡ ¢x; y)
=
¢x2
2
V(x; y + ¢y) ¡ 2V(x; y) + V(x; y ¡ ¢y)
@ V
!
2
@y
¢y2
§
V(x,y+Δy)
V(x‐Δx,y) V(x,y)
0
Δy
Δx
V(x+Δx,y)
V(x,y‐Δy)
V1
.
.
.
V = V6 ;
..
.
V12
0
0
0
0
0
.
.
.
¡4
.
.
.
0
0
0
0
1
0
1
0
0
0
0
V0
1
0
0
1
3
4
6
7
8
9
10 11 12
x
V2
2
V5
V6
5
6
V7
7
h
V10
10
h
{ Volution is
V = A¡1 Vs
§
{ With ¢x = ¢y = h, x = x § h, y = y § h
{ Matlab expression of inverse matrix:
4V(x; y) ¡ V(x+ ; y) ¡ V(x¡ ; y) ¡ V(x; y+ ) ¡ V(x; y¡ ) = 0
! useful formula for programming
{ V(x; y) is found to be the average of 4 nearby potenials at (x; y).
V(x; y) =
>> V = inv(A) ¤ VB ;
² For the point of discrete charge with Qs = ½L
{ Poisson's equation
V(x+ ; y) + V(x¡ ; y) + V(x; y+ ) + V(x; y¡ )
4
Ex. Matlab programming of FDM (See Fig.XXX)
Equal grid spacing h = 1 cm, and a = 10 cm, b = 5 cm:
{ For the node 1
5 + V2 + 0 + V5 ¡ 4V1 = 0
! ¡4V1 + V2 + 0 + V5 = ¡5
r ¢ (rV) =
Ix = [Ex (x + ¢x=2) ¡ Ex (x ¡ ¢x=2)] ¢y¢z
£
¤
Iy = Ey (y + ¢y=2) ¡ Ey (y ¡ ¢y=2) ¢x¢z
V5 + V7 + V2 + V10 ¡ 4V6 = 0
! V2 + V5 ¡ 4V6 + V7 + V10 = 0
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¡½L
±(x)±(y)
²
{ Taking a volume integral and applying the divergence theorem (extending the region ¢z = 1 m
along z-axis with uniform ¯elds)
I
I
Q
D ¢ dS = Q !
E ¢ dS =
²
I
E ¢ dS = Ix + Iy
{ For the node 6
Q = ½v ¢v = ½L ¢x¢y
Electromagnetics
0V
a
0V
¡4
¡5
.
.
.
VB = 0
.
.
.
0
2
5
0
0
0
1
33
L
W
; i; m = 1 : M;
dy = ; j; l = 1 : N
M
N
Source points
Field points
xsm = (m ¡ 0:5) dx xi = (i ¡ 0:5) dx + ±x
ysn = (n ¡ 0:5) dy yj = (j ¡ 0:5) dy + ±y
§d
d
z=
z§
s =
2
2
V(x) ¡ V(x + ¢x)
¢x
V(x ¡ ¢x) ¡ V(x)
Ex (x ¡ ¢x=2) =
¢x
V(y) ¡ V(y + ¢y)
Ey (y + ¢y=2) =
¢y
V(y ¡ ¢y) ¡ V(y)
Ey (y ¡ ¢y=2) =
¢y
2V(x; y) ¡ V(x + ¢x; y) ¡ V(x ¡ ¢x; y)
Ix + Iy =
¢y
¢x
2V(x; y) ¡ V(x; y + ¢y) ¡ V(x; y ¡ ¢y)
¢x
+
¢y
½L
¢x¢y
Ix + Iy =
²
dx =
Ex (x + ¢x=2) =
{ The charge distribution is expressed by linear combination with
basis functions and unknown coefficients:
R
X
l = (m ¡ 1)N + j
qsl : point charge at the l-th index
4V(x; y) ¡ V(x+ ; y) ¡ V(x¡ ; y) ¡ V(x; y+ ) ¡ V(x; y¡ )
½L
=
h2
²
½L 2
4V(x; y) ¡ V(x+ ; y) ¡ V(x¡ ; y) ¡ V(x; y+ ) ¡ V(x; y¡ ) =
h
²
{ The charge distribution in the lower plate has the same but
different sign.
{ With an assumed potential difference V0 , the potential on the
upper plate becomes V+ = V0 =2.
{ Calculate the potential Vk at the reasonably chosen points on
the upper plate:
10.4.2 Method of Moments
² Here one numerical approach called method of
moments is studied. (Integral equation formula
z
y
k = (i ¡ 1)N + j;
x
1
2
M
μ
+Q
‐Q
L
N
2 W
1
l = (m ¡ 1)N + n
¶
1
1
¡ ¡
R+
Rkl
kl
R
1 X
al
4¼²0 l=1
q
R+
(xi ¡ xm )2 + (yj ¡ yn )2
kl =
q
R¡
(xi ¡ xm )2 + (yj ¡ yn )2 + d2
kl =
V+
k =
VR+£1 = FR£R QR£1
μ
¶
1
1
1
Fkl =
¡ +
4¼²0 R+
Rkl
kl
Ex. Capacitance of ¯nite parallel plate capacitor
{ The conducting plate is divided into small rectanglular section as
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on the upper plate
l=1
{ With ¢z = 1, ¢x = ¢y = h
² Up to now we have assumed a uniform charge distribution in conducting plates of a capacitor, and then
we can ¯nd electric ¯eld distribution and capacitance.
However, the size of a capacitor is ¯nite, and the real
distrbution is di®erent to the assumed. For this case,
some numerical approaches can give us a solution even
though it is not a closed form.
qsl ± (x ¡ xsm ; y ¡ ysn );
qs =
Q = F¡1 V+
X
Qsum =
Q !
Electromagnetics
C=
Qsum
V0
34
10.5 Image Theorem
ROI = Region of interest
² Same Poisson's equations for a region y ¸ 0
with the same B.C. at y = 0.
² Two problems are the same from the
uniqueness theorem.
ROI
z
E Q
h
V=0
² Equivalent with each other for y ¸ 0.
ρ
Planar
Conductor
ROI
z
E Q
h
V=0
ρ
h
‐Q
Image
y
ROI
d1
Q
d2
x
z
y
ROI
d1 Q
‐Q d1
d2
z
x
d2
‐Q
Q
ROI
Q
30∘ 60∘
x
z
‐Q
ROI
Q
Q
z
30∘ 60
∘
x
‐Q
‐Q
Q
Chung‐Ang University
Electromagnetics
35
Appendix
Ex. Non-uniform line-charge distribution
z
‐5a
Ex. Theory of operation of torsion balance
appratus
² One conducting sphere (A) connected to the
balance bar is charged with Qa . The other
conducting spahere (B) charged with Qb is
inseted carefully and placed (fixed) as shown
in Fig.
² Charge distribution
½L (z) =
1+
Torsion
spring
L/2
m
Z
m
ρ(z)
1
½L (z)dz
Q=2
0
With a change of variables
z = a tan μ ! dz = a sec2 μ dμ
Z ¼=2
¼
Q=2
½L (z)dz = 2a½0
2
0
= ¼ a½0
{ Then the rotation angle μ is measued, and
the torque is calculated as
L/2
L/2
spring constant
0
L/2
pendulum
² Now an electric force is exerted on the
conducting sphere A, resulting in rotation
of the balancing bar.
· :
a
Total charge
F
¿ = ·μ;
½0
³ z ´2
θ
Fixed Qb
F
Qa
R
{ The force exerted on the conducting sphere A
is calculated
2¿
L
! F=
¿ =F
2
L
{ The distance of two conducting spheres are also measured.
² E½ along ½-axis
² Charge distribution of Maxwellian-hat
function
‐5a
a
0
ρ(z)
‐a
½
½L (z) = p 2 0 2
a ¡z
{ Repeating this procedure with other conducting sphere B having
different amount of charge
{ The spring constant · can be measure as follows:
¤ A force is applied momentarily. Then the bar connected to
the torsion spring oscillates with its center.
¤ The oscillation period T becomes a function of spring
constant and length of bar as
r
μ ¶2
I
T
T = 2¼
! ·=I
·
2¼
where the Moment of inertia of the balance bar I is given as
μ ¶2
μ ¶2
m L2
L
L
I=m
+m
=
2
2
2
Chung‐Ang University
Electromagnetics
36
² Condition for zero electric field at the
point P ???
½L2
¡½L1
=
2
Ex. Chathode-Ray Tube (CRT)
= 60o
L1
² An electron ejected from the cathode in vacume tube (chathode-ray
tube (CRT)) travels to the right and passes through deflection
region, and finally arrivea on the screen. Calculate the
distance d0 .
P
L2
Ex. Circular disk with a uniform line charge
distribution ½L
=60o
² Electric field formed between two conducting plates and electric
force exerted on the electron
d
² Ez along the z-axis can be found as
2d
dq = ½S dS = ½S ½d½dÁ
¶
Z b Z 2¼ μ ^
R¢^
z
Ez = ke ½S
½d½dÁ
R2
0
0
The integrand is not a function of Á.
Z b
½z
Ez = ke ½S (2¼ )
d½
3
0 R
d½ = z sec2 ° d°
½ = z tan °;
¸
Z °0 · 3
z tan °
Ez = ke ½S (2¼ )
sec2 ° d°
3 sec3 °
z
Z0 °0
= ke ½S (2¼ )
sin ° d°
0
Circular disk charge
(0,0,z)

s
0
y
t1 =
x
Vp
dp
Fe = ¡e E0 = ^
zF0 ;
F0 =
e Vp
dp
w
;
v0
a=
e Vp
F0
=
me
me dp
ms = 9:31 £ 10¡31 kg (mass of electron)
Therefore d1 becomes
Infinite planar charge
(0,0,z)
Ez
d1 =
t2 =
Circular ring disk charge
s
b1
vx = v0 ;
vz = at1
Vp e w(L ¡ w)
dp me v20
Vp e w(L ¡ w=2)
d0 = d1 + d2 =
dp me v20
d2 = vz t2 =
2
0
L¡w
;
v0
Therefore d2 and d0 can be found as
Ez
(0,0,z)
1
Vp e w2
1 2
a t1 =
2
2dp me v20
² After the region of two conducting plates, the electrons moves to
the right and upward for a travel time t2 with uniform velocities
y
0
x
Ez = ke ½s (2¼ )
Ex. Circular ring disk with a uniform charge
distribution ½s
0
b
ρs
² For an infinite plane (b ! 1)
E0 =
² Within the region between two conducting plates, electrons move
to the right with a uniform velocity v0 and move up simultaneously
because of the force exerted on the electrons with acceleration a
for a time t1 :
Ez
dq
= ke ½S (2¼ ) (1 ¡ cos °0 )
z
cos °0 = p
b 2 + z2
Ex. Infinite plane with a uniform charge
distribution ½s
E = ¡^
z E0 ;
b2
y
z
Vp
Vc
v0
Cathode
Screen
Deflection
plates
d2
dp
d1
E
0
w
L
x
Chung‐Ang University
Electromagnetics
37
d0
x
Ex. Circular Line Charge (N-Polygon
Approximation)
E1
² Electric field along the z -axis for an
N-polygon uniform line charge
Ex. E along the z-axis for rectangular
surface charge
γ z
P(0,0,z)
For a single line section, Ez1 can be found
from the expression for a finite length line
cyrrent as
γ
R10

y
y

α

L
x
z
dE
¼
N
γ
(0,0,z)
γ
L
= p sin ®;
2
q = p cos ®
μ
¶μ
¶
L=2
z
Ez = N (2ke ½L )
R10
R2
μ
¶μ
¶ c0
CN
z
= ke ½L
R10
R2c0
CN = NL
dq = ½s dxdy = ½L dx ! ½L = ½s dy
p
½ ! Rc0 = z2 + y2
p
R10 = (L=2)2 + z2 + y2
μ
¶
sin ¯10
Ez1 = 2ke ½L
cos °
Rc0
L=2
z
sin ¯10 =
;
cos ° =
R
Rc0
μ 10¶ μ
¶
L=2
z
Ez1 = 2ke ½L
R10
R2c0
L
ρL
Thus the total electric field can be found
as
®=

0
x
R
0

y
ϕ′ r′
x
ρL
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L/2
R
ρs
r’
‐W/2
y
dq
‐L/2
x
W/2
z
Ez
R10
L/2
Rc0
ρs
L
‐W/2
x
y
‐L/2
W/2
² Thus the total electric field can be found as
¶μ
¶
Z W= 2 μ
L=2
z
Ez = 2ke ½L
dy = 2ke ½L Lz I(z)
R10
R2c0
¡W=2
Z W=2
Z W=2
dy
dy
p
I(z) =
2 =
2
R
R
(L=2) + z2 + y2 (z2 + y2 )
10 c0
0
0
dq
From cos ° = z=Rc0
Rc0
ydy
! sec ° tan ° d° =
z
zRc0
μ ¶
dy
y
sin °
sin ° =
=!
d° = sin °
2
Rc0
cos °
z
z
dy =
cos2 °
sec ° =
(Total length of polygon)
² For a circular line charge distribution
p
C1 = 2¼ p; Rc0 = R10 = p2 + z2
"
#
μ
¶
C1 z
2¼ p z
Ez = ke ½L
= ke ½L
R310
(p2 + z2 )3=2
Ez
² E=^
z Ez due to the structural symmetry.
Rc0
Due to the axial symmetry, E = ^
zEz . For a
single line section, Ez1 can be found from
the expression for a finite length line
cyrrent as
p
p
½ ! Rc0 = q2 + z2 ;
R10 = p2 + z2
μ
¶
sin ¯10
Ez1 = 2ke ½L
cos °
Rc0
L=2
z
sin ¯10 =
;
cos ° =
R10
Rc0
μ
¶μ
¶
L =2
z
Ez1 = 2ke ½L
R10
R2c0
z
² For a circular line charge distribution
p
C1 = 2¼ p; Rc0 = R10 = p2 + z2
"
#
μ
¶
C1 z
2¼ p z
Ez = ke ½L
= ke ½L
3= 2
R310
(p2 + z2 )
Electromagnetics
38
Ex. Circular Disk Charge and Its Variants
{ Therefore dEz becomes
² Circular disk with a radius a carrying a uniformly
distributed charge with a surface charge density ½s .
² Due to the axial symmetry. it is su±cient to ¯nd Ex
and Ez on the xz-plane as before. With
r = (x; 0; z)
r0 = (½0 cos Á0 ; ½0 sin Á0 ; 0)
R = r ¡ r0
= (x ¡ ½0 cos Á0 ; ¡½0 sin Á0 ; z)
dEx and dEz for a di®erential surface dS = ½0 d½0 dÁ0
becomes
μ
¶
^
½s
R¢x
dS
dEx =
4¼²0
R3
μ
¶
½s
R¢^
z
dS
dEz =
4¼²0
R3
where
R3 = (x2 + z2 + ½02 ¡ 2x½0 cos Á0 )3=2
^ = x ¡ ½0 cos Á0
R¢x
R¢^
z=z
² At ¯rst, let us calculate the electric ¯eld on the z-axis.
Due to the axial symmetry. we know that the electric
¯eld has only the z -component, and dEz becomes
dEz =
½s dS z 0 0 0
½ d½ dÁ
4¼²0 R3
where
R3 (z2 + ½02 )3=2
dEz =
½s
½0
z 02
d½0 dÁ0
4¼²0 (½ + z2 )3=2
{ Considering the integral along Á0 becomes 2¼ , the
total Ez becomes
Z a
½s
½0
z
d½0
Ez =
02 + z2 )3=2
2²0
(
½
0
0
With a change
p of variables ½ = z tan ® and introduce R0 = z2 + a2 ,
μ
¶
½s
z
Ez =
1¡
2²0
R0
This is valid for z > 0, and Ez including the region
z < 0 becomes
μ
¶
jzj
½s
Ez = §
1¡
2²0
R0
(
+ for z > 0
¡ for z < 0
{ One special case : circular ring disk with inner and
outer radius a and b
·μ
¶¸
½s
jzj jzj
¡
Ez = §
2²0
Ra
Rb
p
Ra = z2 + a2
p
Rb = z2 + b2
{ Second one : in¯nitely extended surface charge distribution
Taking a limit on a to in¯nity, then
Ez = §
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Electromagnetics
½s
2²0
39
Ex. Conical surface
(like an empty ice-cream cone)
{ Using the following integral
Z
p
x dx
p
= x2 + a2
x 2 + a2
Z
³
´
p
dx
p
= ln x + x2 + a2
x 2 + a2
z
² The cone surace with a height h carrys a
uniform surface charge ½S . Find the potential
difference between points A (the vertex) and B
(the center of the top).
γ
B
h
h‐z′
z′
{ We can use the previous result with
b = z0
p
p
R = b2 + (z ¡ z0 )2 = z02 + (z ¡ z0 )2
h
Iz = Iz1 + Iz2
Z h¡z=2
p
Iz1 =
z′
ρs
A
z=0
¡z=2
{ VA and VB
μ
VA = V(0) =
V(z) =
¶Z
hμ 0¶
0
z
R
μ
dz0 =
½
ps
2²0
¶Z
0
h
"
z0
p
z02 + (z ¡ z0 )2
#
dz0
{ With ´ = z0 ¡ z=2, the argument of square root can be written as
·
³
³ z ´2 ¸
z ´2 z2
2z02 ¡ 2zz0 + z2 = 2 z0 ¡
+
= 2 ´2 +
2
2
2
d´ = dz0
μ
¶
½s
V(z) =
Iz
2²0
Z h¡z=2
´ + (z=2)
p
Iz =
d´
´ 2 + (z=2)2
¡z=2
Chung‐Ang University
½s h
2²0
¶
p ¸
·
h
1=2 + 1= 2
p
VB = V(h) =
ln
2
¡1=2 + 1= 2
μ
¶
μp
¶
½s h 1
2+1
p
=
ln
2²0 2
2¡1
μ
¶ μq
¶
p
½s h
=
ln
3+2 2
2²0
μ
¶ ³
p ´
½s h
=
ln 1 + 2
2²0
μ
{ The total potential on the z-axis
½
ps
2²0
d´
p
p
(h ¡ z=2)2 + (z=2)2 ¡ z2 =2
Z h¡z=2
z=2
p
Iz2 =
d´
2
´ + (z=2)2
¡z=2
"
#
p
h ¡ z=2 + (h ¡ z=2)2 + (z=2)2
z
p
= ln
2
¡z=2 + z2 =2
½L = ½s dr0
!
dr0
z0 = r0 cos μ ! dz0 = p
2
μ
¶
½L b 1
V(z) =
2²0 R
μ
¶
μ
¶ 0
½s z0 1 p
½s
z
0
2 dz = p
dz0
dV(z) =
2²0
R
2²0 R
μ
+ (z=2)2
=
{ In addition, the uniform line charge density ½L can be found as
dq = ½s dr0 bdÁ0 = ½L bdÁ0
´
´2
½s
2²0
¶
{ Resulting electric field has only the z -component.
μ
¶μ
¶
¡½s
@ Iz1
@ Iz1
¡@ V(z)
Ez =
=
+
@z
2²0
@z
@z
p
p
2
2
2
Iz1 = (h ¡ z=2) + (z=2) ¡ z =2
"
#
p
h ¡ z=2 + (h ¡ z=2)2 + (z=2)2
z
p
Iz2 = ln
2
¡z=2 + z2 =2
Electromagnetics
40
Ex. Hemispherical Charge
² Uniformly distributed spherical charge.
² Because of a spherical symmetry, the
resulting electric filed is known to have
the r-component only and it is a function of
r only. It is sufficient to calculate Ez
at (0; 0; z).
² Second way : The result for a circular
disck charge can be applied extended with
the following surface charge density ½s as
dq = ½v dv = ½s dS
² What is noteworthy is that this expression can be rewritten in
terms of the total charge of the sphere Q as
z
!
½s = ½v dz0
E
P(0,0,z)
Q
4¼²0 z2
4¼ a3 ½v
Q=
3
Ez =
a
ρv
0
y
a
x
² Electric field inside the sphere, that is z < a
½v
(I3 + I4 )
2²0
¶
Z zμ
z ¡ z0
1¡ p
dz0
I3 =
z2 + a2 ¡ 2zz0
¡a
¶
Z aμ
z0 ¡ z
I4 = ¡
1¡ p
dz0
2
2
0
z + a ¡ 2zz
z
Ez =
z
dE
P(0,0,z)
R0
b
ρs z′
a
Therefore dEz for z > a (outside the sphere)
becomes
μ
¶
½v dz0
z ¡ z0
1¡
dEz =
2²0
R0
p
R0 = z2 + a2 ¡ 2zz0
0
² Taking an integration from z0 = ¡a to a, Ez can be found as
¶
Z aμ
z ¡ z0
½v
½v
Ez =
1¡ p
dz0 =
(a + Iz1 + Iz2 )
2
2
0
2²0 0
2
²0
z + a ¡ 2zz
Z a
1
p
dz0
Iz1 = ¡z
2
z + a2 ¡ 2zz0
0
Z a
z0
p
Iz2 = Iz2 =
dz0
2
2
0
z + a ¡ 2zz
0
Using the integral formula with
Z
1
dx =
p
px + q
Z
x
dx =
p
px + q
y
x
p = ¡2z and q = z2 + a2 :
p
2 px + q
p
2(px ¡ 2q) p
px + q
3p2
Following the similar procedures and algebraic manipulations,
μ 3¶
½v
2z
Ez =
2²0 3z2
This expression can be rewritten in terms of the charge Qin ,
which is total charge inside the sphere with a radius z as
Qin
4¼²0 z2
4¼ z3 ½v
=
3
Ez =
Qin
² Conclusions:
{ Only the charge inside the sphere with a radius z can make a
contribution to E. This can be easily proved by the Gauss' law
which we will study later.
{ One more interesting problem is E from a uniformly distributed
spherical shell charge. This problem can be solved by
extending the above results as
and taking some algebraic manipulations
Ez =???
Chung‐Ang University
Electromagnetics
41
Qin
4¼²0 z2
8
0;
z·a
>
>
>
< 4¼(z3 ¡ a3 )½v
; a·z·b
=
3 3
>
> 4¼(b3 ¡
>
a
)
½
v
:
; z¸b
3
{ z-point for maximum Ez
Ez =
Qin
dEz
=0
dz
Ex. Circular Line Charge (O® the Axis)
² Due to an axial symmetry, EÁ = 0.
² On the xz-plane dEx and dEz become
μ
¶
½L
R¢^
x
dEx =
a dÁ0
4¼²0
R3
μ
¶
½L
R¢^
z
dEz =
a dÁ0
4¼²0
R3
8
r = (x; 0; z)
>
>
>
>
>
r0 = (a cos Á0 ; a sin Á0 ; 0)
>
>
>
<R = r ¡ r0 = (x ¡ a cos Á0 ; ¡a sin Á0 ; z)
>
R3 = (a2 + x2 + z2 ¡ 2ax cos Á0 )3=2
>
>
>
>
>
R¢^
x = x ¡ a cos Á0
>
>
:
R¢^
z=z
·
¸
½ a
x ¡ a cos Á0
dEx = L
dÁ0
4¼²0 (a2 + x2 + z2 ¡ 2ax cos Á0 )3=2
·
¸
½L a
z
dEz =
dÁ0
4¼²0 (a2 + x2 + z2 ¡ 2ax cos Á0 )3=2
z
dE
P
R
r
b
0
r’
x ρ
ϕ’
dq
y
ρL
·
¸
Z
½L a 2¼
z
Ez =
dÁ0
4¼²0 0
(a2 + z2 )3=2
μ
¶
½L a
z
=
2²0 (a2 + z2 )3=2
!
!
a2 ¡ 2z2 = 0
·
² Ex and Ez off the axis
·
¸
Z
½L a 2¼
x ¡ a cos Á0
½ a
Ex =
dÁ0 = L
(I1 + wI2 )
4¼²0 0
4¼²0 x
(a2 + x2 + z 2 ¡ 2ax cos Á0 )3=2
Z ¼
1
p
I1 =
dÁ0
2
2
2
0
a + x + z ¡ 2ax cos Á
0
Z ¼
1
I2 =
dÁ0
2 + x2 + z2 ¡ 2ax cos Á0 )3=2
(a
0
w = x2 ¡ a 2 ¡ z2
Z 2¼
½ a
z
Ez = L
dÁ0
2
2
2
4¼²0 0 (a + x + z ¡ 2ax cos Á0 )3=2
=
½L a
(z I2 )
2¼²0
Ex. Circular disk charge (Off the axis)
² Ex and Ez off the z-axis
dEx =
½s
4¼²0
½s
dEz =
4¼²0
² On the z-axis (x = 0),
Ex = 0
¸
2z2
=0
(a2 + z2 )5=2
a
! zmax = § p
2
1
3
¡
2
2
3
=
2
2
(a + z )
·
·
(x ¡ ½0 cos Á0 ) ½0 d½0 dÁ0
(x2 + z2 + ½02 ¡ 2x½0 cos Á0 )3=2
z ½0 d½0 dÁ0
(x2 + z2 + ½02 ¡ 2x½0 cos Á0 )3=2
¸
¸
Considering the integral along Á0 at first
Z aZ ¼
½s
(x ¡ ½0 cos Á0 ) dÁ0 ½0 d½0
Ex =
2¼²0 0 0 (x2 + z2 + ½02 ¡ 2x½0 cos Á0 )3=2
Z aZ ¼
½s
z dÁ0 ½0 d½0
Ez =
4¼²0 0 0 (x2 + z2 + ½02 ¡ 2x½0 cos Á0 )3=2
² Numerical calculation is needed ???
Chung‐Ang University
Electromagnetics
42
Ex. Find V and then E for a circular ring charge
² Due to an axial symmetry, we can choose a field point P(x; 0; z) on
the xz-plane (Á = 0).
² Differential potential dV
½
dV = L
4¼²0
8
r = (x; 0; z)
>
>
>
<r0 = (a cos Á0 ; a sin Á0 ; 0)
>R = r ¡ r0 = (x ¡ a cos Á0 ; ¡a sin Á0 ; z)
>
>
p
:
R = a2 + x2 + z2 ¡ 2ax cos Á0
μ ¶
1
a dÁ0
R
² Electric potential (2 times of the contribution from the half
circle)
Z
½L a ¼
dÁ0
p
V=
2¼²0 0
a2 + x2 + z2 ¡ 2ax cos Á0
² With a change of variables
¼ ¡ Á0
! dÁ0 = ¡2d®
2
cos Á0 = cos(¼ ¡ 2®) = 2 sin2 ® ¡ 1
Z
½L a ¼=2
d®
½ a
1
p
V=
= L p
IV
2
¼²0 0
¼²0 (a + x)2 + z2
(a + x)2 + z2 ¡ 4ax sin ®
Z ¼=2
1
p
IV =
d® = K(k)
0
1 ¡ k sin2 ®
4ax
k=
(< 1)
(a + x)2 + z2
: Complete elliptic integral
®=
² Generalization (x to ½)
V=
½L a
1
p
K(k)
¼²0 (a + ½)2 + z2
4a½
k=
(a + ½)2 + z2
Chung‐Ang University
² Induced charges on metallic conductors
{ When the point charge Q is moved to a
point off from the center of the spherical
conducting shell, it is difficult to find
the analytical solution of induced charge
distribution.
+
{ E inside the shell is now distorted to
ensure the normal component of E only on
the inner surface of the shell.
+
+ ‐ ‐ ‐ +
‐
+Q ‐
‐
+
‐
+
‐
+
+
{ In the inner surface of the shell, the induced (¡) charge is
=
½s , being
non-unifomly distrbuted, in accordance with Dn
denser at points of the surface closer to the charge Q, where
the electric field magnitude is higher.
{ On the other hand, the (+) charge layer on the outer surface of
the shell remains uniformly distributed, resulting in the same
magnitude at all points as if the exterior field originated
from a point charge located at the center of the shell.
Ex. Connection of two conducting spheres
Q1  q1
² Different charges Q1 and Q2 are put on two
conducting spheres with radii R1 and R2
(R1 < R2 ), respectively. These conducting
spheres are places a long distance D apart
as shown in Fig. After connecting these
conduction spheres, calculate the normal
component of E on the two spheres.
Q2  q2
R1
R2
D
{ Because D
À
R1 + R2 , each sphere can be treated as isolated.
potential on each sphere is then
V1 =
Q1
;
4¼²0 R1
V2 =
The
Q2
4¼²0 R2
{ If a wire is connected between the spheres, they are forced to be at
the same potential V0 with redistribution of charges q1 and q2 :
q1
q2
=
4¼²0 R1
4¼²0 R2
q1 + q2 = Q1 + Q2
V0 =
Electromagnetics
43
{ The lower equation comes from the charge
conservation. Now we can find new
charges q1 and q2 as
q1 =
R1 (Q1 + Q2 )
;
R1 + R2
q2 =
Ex. Coaxial and Spherical shell capacitors
² Coaxial capacitor
{ Parallel connection
R2 (Q1 + Q2 )
R1 + R2
C1 , C2 , and C can be found from C0 , the
capacitance of air-filled coaxial capacitor
of length L (already been studied):
{ The resulting system potential is found
to be
V0 =
C1 = ²r1
C0
;
2
C2 = ²r2
C0
2
2¼²0 L
μ ¶
b
ln
a
μ
¶
²r1 + ²r2
C=
C0
2
En = ½s =²0 at the surface implies that the
surface electric field is stronger as the
surface charge density is larger.
a
ϵ1
L
b2
b1
aϵ
1
ϵ2
C0 =
{ Therefore even though the smaller sphere
carries less total charge, it is found
that E1 > E2 .
L
{ Series connection
q1
V0
E1 (r = R1 ) =
=
4¼²0 R21
R1
q2
V0
E2 (r = R2 ) =
=
4¼²0 R21
R2
1
1
1
=
+
C
C1
C2
2¼²1 L
C1 = μ ¶ ;
b1
ln
a
{ For this reason, E is always largest
near corners and edges of equipotential
surfaces, which is why sharp points must
be avoided in high-voltage equipment.
C2 =
2¼²2 L
μ ¶
b2
ln
b1
² Spherical shell capacitor
z
PB
{ Parallel connection
Circle
² When the electric field exceeds a critical
amount Eb , called the breakdown strength,
spark discharges occur as electrons are
pulled out of the surrounding medium.
Chung‐Ang University
b
C = C1 + C2
Q1 + Q2
4¼²0 (R1 + R2 )
² Air has a breakdown strength of Eb ' 3x106
V/m. If the two spheres had the same radius
of 1 cm, the breakdown strength is reached
when V0 ' 30; 000 V. This corresponds to a
total system charge of Q1 + Q2 = 6:7x10¡8
Coulombs.
ϵ2
C
PO
μ
C = C1 + C2 =
‐Q
PA
d
C0 = μ
s
Infinite
Conducting
plate
y
x
²r1 + ²r2
2
¶
b
ϵ1
{ Series connection
Electromagnetics
a
C0
4¼²0
¶
1 1
¡
a b
1
1
1
=
+
C
C1
C2
4¼²1
¶;
C1 = μ
1
1
¡
a b1
ϵ2
b2
a
C2 = μ
4¼²2
¶
1
1
¡
b1
b2
b1
ϵ1
44
ϵ2
Ex. Conducting cylinder in uniform E
² Initially
exists in
uncharged
placed in
x E0
a uniform electric field E0 = ^
an unbounded free space. If an
conducting cylinder of radius a is
this field, determine V and E.
{ Induced charge on the conducting cylinder
due to the uniform electric field, generates
scattered field Es . Therefore the total
electric field can be expressed as
^ E0Á
^ E0½ + Á
E0 = ½
Conducting
cylinder
E0½ = E0 cos Á;
E0Á = ¡E0 sin Á
Z x
V0 = ¡
E ¢ dl = ¡E0 x = ¡E0 ½ cos Á
a
0
y
{ The scattered field should be zero as ½ ! 1, therefore the ¡1
should be chosen for our solution as
Ei = E0
0
a
z
Etotal = E0 + Es
Vtotal = V0 + Vs
Vi (r = a) + Vs (r = a) = 0
C1
! ¡E0 a +
=0 !
a
Equipotential
surface
{ Furthermore ©(Á) is known to be 2¼ -periodic
and also even in Á, theerfore ©(Á) becomes
m=0
m 6= 0
{ Now the equation of R(½) becomes
μ
¶
@
@2
½2 2 + ½
¡ m2 R(½) = 0
@½
@½
(
ln ½;
m=0
R(½) =
m
¡m
½ ; ½ ; m 6= 0
{ We have to consider the expression of the original potential in
the cylindrical coordinates. With a zero reference potential
at x = 0,
Chung‐Ang University
cos Á
½
{ Applying the boundary condition V(r = a) = 0
Electric field
{ Considering structure, Vs = R(½)©(Á), and the
Laplace equation of scattered potential Vs
in cylindrical coordinates, becomes
·
μ
¶
¸
@
1 @
1 @2
½
+ 2 2 R(½)©(Á) = 0
½ @½
@½
½ @Á
μ
¶
Ä
©(Á)
½ d
@
½
R(½) +
=0
R(½) d½
@½
©(Á)
Ä
©(Á) = ¡m2 ©(Á)
(
Á, or constant;
©(Á) =
cos(mÁ);
Vs (½; Á) = C1
x
C1 = E0 a2
{ Therefore Vs and the total V are found as
E0 a2
cos Á
½
μ
¶
½ a
V = ¡E0 a
cos Á
¡
a ½
Vs =
{ The electric fields can be found as
μ
¶
@V ^ @V
+Á
E = ¡rV = ¡ ½^
@½
½@Á
μ
¶
μ
¶
2
a
a2
^
= ½^E0 1 + 2 cos Á ¡ ÁE0 1 ¡ 2 sin Á
½
½
As shown in Fig. XXX displaying the electric field and
potential distributions, field values increases as Á moves
to 0 and ¼ , and becomes maximum as (2E0 ) at ½ = a, two times of
original field.
{ Now the surface induced charge on the cylindrical conductor can
be found from the Gauss' law as
½s = D½ = ²0 E½ = 2²0 E0 cos Á
Therefore the total charge induced can be found by surface
integration, and is found to be zero.
Electromagnetics
45
Ex. Conducting sphere in uniform E
² Initially
exists in
uncharged
placed in
z E0
a uniform electric field E0 = ^
an unbounded free space. If an
conducting sphere of radius a is
this field, determine V and E.
{ Consodering the scattered electric filed
due to the induced charge on the conducting
sphere, the total electric field and
potential can be expressed as
Conducting
sphere
{ Two independent differential equations with a constant k
·
¸
@ R(r)
1 @
r2
= k2
R(r) @ r
@r
·
¸
@ £(μ )
1
1 @
sin μ
= ¡k2
£(μ ) sin μ @μ
@μ
a
x
Ei = E0
{ R(r), satisfying the following differential equation, becomes
0
dR(r)
d2 R(r)
+ 2r
¡ k2 R(r) = 0
2
dr
dr
n(n ¡ 1) = k2
R(r) » r¡(n+1) ;
n = 0; 1; 2; ¢ ¢ ¢ :
z
r2
Etotal = E0 + Es
Vtotal = V0 + Vs
{ The expressions of the original electric
field and potential in the spherical
coordinates, which are needed later, are
found as
(
E0r = E0 cos μ
^
E0 = ^r E0r + μ E0μ ;
E0μ = ¡E0 sin μ
Z z
V0 = ¡
E0 ¢ dl = ¡E0 z = ¡E0 r cos μ
Electric field
Equipotential
surface
x
0
z
{ £(μ) is known to satisfy the Legendre differential equation,
·
¸
d
d£(μ)
sin μ
+ n(n + 1)£(μ) sin μ = 0
dμ
dμ
k2 = n(n + 1)
and the solution of candidate, which is called the Lengendre
function, is
0
n
£(μ ) = Pn (cos μ)
0
1
1
cos μ
2
(3 cos2 μ ¡ 1)=2
3 (5 cos3 μ ¡ 3 cos μ)=2
Here, the potential at z = 0 is chosen as a
reference zero potential.
{ Considering the axial symmetry of Vs , the related Laplace
equation in the spherical coordinaes becomes
μ
¶
μ
¶
@
1 @
@ Vs
2 @ Vs
r
+
sin μ
=0
@r
@r
sin μ @μ
@μ
{ Applying the separation of variables
Vs (r; μ) = R(r) £(μ)
·
¸
·
¸
@ £(μ)
@
1 @
2 @ R(r)
r
£(μ) +
sin μ
R(r) = 0
@r
@r
sin μ @μ
@μ
·
¸
·
¸
1
1 @
@ £(μ)
1 @
2 @ R(r)
r
+
sin μ
=0
R(r) @ r
@r
£(μ) sin μ @μ
@μ
Chung‐Ang University
{ If n is a valid index, then a new n0 = ¡(n + 1) is alos valid as
n0 (n0 + 1) = ¡(n + 1)(¡n) = n(n + 1) = k2
This implies that functions with index n and ¡(n
valid. Therefore
+
1) become
Vs (r; μ ) = R(r) £(μ)
1 h
i
X
=
An r¡(n+1) + Bn rn Pn (cos μ )
n=0
Electromagnetics
46
{ The scattered field, formed according to
original, is known to be dependent of only
the cos μ (orthogonality), that is, n = 1.
Therefore the scattered potential and
field components are
£
¤
Vs = A1 r + B1 r¡2 cos μ ! Vs = B1 r¡2 cos μ
¡@ Vs
Esr =
= 2B1 r¡3 cos μ
@r
¡@ V s
Esμ =
= B1 r¡3 sin μ
r@μ
{ From the boundary condition V(r = a) = 0,
V0 (r = a) + Vs (r = a) = 0
¡2
E0 a cos μ = B1 a
cos μ
!
B1 = E0 a
3
{ Therefore total electric potential and
field components are
·
³ a ´3 ¸
V = V0 + Vs = ¡E0 r 1 ¡
cos μ
r
·
³ a ´3 ¸
Er = E0r + Esr = E0 1 + 2
cos μ
r
·
³ a ´3 ¸
Eμ = E0μ + Esμ = ¡E0 1 ¡
sin μ
r
{ Surface charge density induced on the
conductor surface
½s = Dr (r = a) = ²0 Er (r = a) = 3²0 E0 cos μ
{ Total induced charge is obviously zero.
ZZ
Q=
½s a2 sin μ dμdÁ
S
Z ¼
2
= 6¼²0 a E0
sin μ cos μ dμ
0
Z ¼
= 3¼²0 a2 E0
sin(2μ) dμ0 = 0
0
Chung‐Ang University
Ex. Dielectric sphere in uniform E
Dielectric
sphere
² A dielectric sphere of radius a is placed in
an initially formed uniform electric field
E0 = ^
z E0 . Determine V and E.
{ Similar to the case of conducting sphere,
the scattered electric potential and field
components are
(
Vso (r; μ) = B1 r¡2 cos μ
Vsi (r; μ) = C1 r cos μ
(
(Eso )r = 2B1 r¡3 cos μ
(Esi )r = ¡C1 cos μ
(
(Eso )μ = B1 r¡3 sin μ
(Esi )μ = C1 sin μ
clear all, close all, clc
a=1;
E0=1;
dx=0.05;
dz=0.05;
x=‐2.4:dx:2.4;
z=‐2:dz:2;
[xx,zz]= meshgrid(x,z);
rr=sqrt(xx.^2+zz.^2);
th=acos(zz./rr);
with the boundary conditions at r = a:
a
x
Ei = E0
0
z
Electric field
Equipotential
surface
x
???
z
0
1) Vo = Vi
2) Do r = (Di )r
V=‐E0*(1‐(a./rr).^3).*rr.*cos(th);
[Ax,Az]=gradient(V,dx,dz);
Ex=‐Ax;
Ez=‐Az;
(Eo )r = ²r (Ei )r
!
{ Therefore
ii0=find(rr<=1);
V(ii0)=0;
Ex(ii0)=0;
Ez(ii0)=0;
1)
B1
= C1 a
a2
!
C1 =
B1
a3
8
¡3
>
<E0 + 2B1 a = ²r (E0 ¡ C1 )
μ
¶
2) (E0 )r + (Eso )r = ²r [(E0 )r + (Esi )r ] !
2
²r
>
+ 3
:! E0 (²r ¡ 1) = B1
a3
a
· 3
¸
8
a (²r ¡ 1)
>
>
<B1 = E0
2 + ²r
μ
¶
!
>
² ¡1
>
:C1 = E0 r
2 + ²r
xth=(0:1:360)*pi/180;
figure(1)
pcolor(zz,xx,V), axis equal,
shading interp, colorbar, hold on
plot(cos(xth),sin(xth))
figure(2)
quiver(zz,xx,Ez,Ex)
axis equal, hold on
plot(cos(xth),sin(xth))
{ Final expression of electric potential is
·
μ ¶¸
8
²r ¡ 1 a3
>
>
r cos μ
<Vo = ¡E0 1 ¡ 2 + ²
r3
r
μ
¶
>
3
>
:Vi = ¡E0
r cos μ
2 + ²r
Electromagnetics
47
{ Electric field can be found from the obtained coeffcients and
combing the components.
{ However, it can also be found as
E = ¡rV
Applications of Image Theory
y
² Capacitance of two conducting cylinders with
(
a = radius
d = center to center distance
‐V0
d
a
0
x
V0
Outside the sphere
μ
¶
@ Vo ^ @ Vo
+μ
Eo = ¡ ^r
@r
r@μ
·
μ ¶¸
2(²r ¡ 1) a3
rE0 1 +
=^
cos μ
2 + ²r
r3
·
μ ¶¸
² r ¡ 1 a3
^
+ μ E0 1 ¡
sin μ
2 + ² r r3
Inside the sphere
μ
¶
@ Vi ^ @ Vi
Ei = ¡ ^r
+μ
@r
r@μ
μ
¶
3
= ^rE0
cos μ
2 + ²r
μ
¶
3
^ E0
+μ
sin μ
2 + ²r
Here we can find the Ei has only the z-component and uniform
as
μ
¶
3
Ei = ^
z E0
2 + ²r
{ Since this is equivalent to a conducting
cylinder above a grounded conducting
surface. Let V0 be the potential on the
conductinf cylinder
y d/2
V0
{ Our concern is if there is an equivalent
line charge. If it exists, then the
problem is solved by considering two
oposite line charges in free space (having
line charge densities ½L and ¡½L ).
y
E½ =
½L
2¼²0 ½
V1 (½) = ¡
x0
ρL
V0
{ E and V1 of a single line charge at ½ = 0
½L
ln(½) + Vref
2¼²0
‐ρL
x0
0
ρL
V(x = 0) = 0
(* ½1 = ½2 )
{ Now let us check if these line charges can generate the
potential V0 on the conducting cylinder surface. This can be
written with introducing a variable R as
s
½2
(x + x0 )2 + y2
=
(> 1 )
R=
½1
(x ¡ x0 )2 + y2
μ
¶
½L
2¼²0 V0
ln R = V0 ! R = exp
2¼²0
½L
Chung‐Ang University
Electromagnetics
x
y
x0
{ Potential for two opposite line charges at x = x0 and ¡x0
μ ¶
½L
½2
V(½) =
ln
2¼²0
½1
(
p
½1 = (x ¡ x0 )2 + y2
p
½2 = (x + x0 )2 + y2
This can be verified from the electric field and potential
distributions in Fig. XXX.
x
48
x
² Capacitance of two conducting spheres Two
identical conducting spheres havong radius
a are separated with a center to center
spacing 2c. Calculate the capacitance.
{ Therefore
μ
2
R +1
R2 ¡ 1
x2 ¡ 2
¶
x0 x + +y2 + x20 = 0
2
(x ¡ xc ) + y2 = a2c : equation of circle
μ 2
¶
8
R
+
1
>
>
x0
<xc =
2
μR ¡ 1¶
2R
>
>
:ac =
x0
2
R ¡1
8
2
2
2
>
<xc ¡ ac = x0 ¢ ¢ ¢ (1)
!
x
R2 + 1
>
: c =
¢ ¢ ¢ (2)
ac
2R
¡ From this, we can determine
8
d
>
>
<xc = 2
ac = a
>
>
:
V0
$
C=
:
(
x0
½L
Charge per unit length
Q
½
¼²0
= L =
=
2V0
2V0
ln R
¼²0
μ
cosh¡1
d
2a
¶
s
R=
² At first, let us consider a positive point
charge Q located at x = d and a conducting
sphere of a radius a centered at the origin
(a < d) having a potential ¡V0 .
{ The image charge Qi (negative point
charge) is expected to exist inside the
sphere at a distance x = di with:
{ Potential on the surface of the sphere
caused by these two chages
μ
¶
1
Q
Qi
+
=0
V=
4¼²0 ro
ri
!
sμ ¶
¶
2
d
d
¡1
¡ 1 ; x0 = a
x=a
2a
2a
sμ ¶
2
d
d
R=
+
¡1
2a
2a
μ
Chung‐Ang University
the conductor
z
0
a
Q
d z
c
{ Thetrefore there are two unknowns di and
Qi (This is different to the case of two
opposite line charges in cylindrical
coordinates).
With
(En )on
‐V0
ro
a ri
Qi
‐z0
di
d
a
ro
‐V0
Qi
0 Qa
Q
z
Q
z
d
Qi
ri
= ¡ = R (constant); 0 < R < 1
ro
Q
(
p
ro = x2 + y2 + (z ¡ d)2
p
ri = x2 + y2 + (z ¡ di )2
{ Equation of zero-potential surface
s
x2 + y2 + (z ¡ di )2
ri
=
=R
r0
x2 + y2 + (z ¡ d)2
=
Electromagnetics
z
c
V0
c
Qi 6= ¡Q
(x + x0 )2
(x ¡ x0 )2
¡ Charge density on the conducting cylinder
a
V0
a
a
c 0
because ¡Q and the original Q can not
make the spherical surface with a radius
a of a zero-potential (but a planar
surface).
¡ Capacitance per unit length of two conducting cylinders
Q = ½L
‐V0
49
{ Step 3: In order to restore VS
charge of ¡Q0 , is put at z = z1 .
x2 + y2 + (z ¡ z0 )2 = a2 : equation of sphere
μ
¶2
di ¡ R2 d
R(d ¡ di )
2
z0 =
;
a =
1 ¡ R2
1 ¡ R2
di
z0 = 0 ! R2 =
a 2 = di d
!
d
Q1 = Q0 °;
z1 = c ¡ di1 ;
di1 =
{ Now the image charge can be determined as
¡Qi
= R ! Qi = ¡RQ
Q
di
di d
a2
R2 =
= 2 = 2 !
d
d
d
³a´
!
Qi == ¡
Q
d
VS = V0 ;
{ Step 4:
a
R=
d
at z = 0
z2 = c ¡ di2 ;
Electromagnetic Fields
and Waves, Lorraine, Cordon, 2nd ed.
di2
=
d0
{ Step 1: A charge Q0 is set at z = z0 = c (center of the sphere) to provide a
potential V0 on the surface S.
Q0 = 4¼²0 a V0 ;
VS = V0 ;
@ z = z0 = c
but non-euipotential VS on S
VS = V0 ;
{ Step 6:
VSg 6= 0
μ
but
¶2
z2
d1
= 1 ¡ °2
d0
di2 =
a
d0
=
di1
= °2
d0
d1
di1
=1¡
!
d0
d0
μ
¶
1
2
Q 2 = Q0 °
1 ¡ °2
a2
d
² Next a Conducting Sphere above a Ground Plane is considered.
VSg 6= 0
d1 = d0 ¡ di1 ;
at z = d
at di =
di1
= °2
d0
{ Step 5: Putting a charge Q2 (image charge of ¡Q1 ) at z
restores the equipotemtial on S.
μ ¶
μ ¶
a
a
d0
d0
= Q1
= Q0 ° 2
Q2 = Q1
d1
d0 d1
d1
² Is it possible to make non-zero potential ¡V0 on the conducting sphere ? Sure, it
can be made by putting an additional point charge Qa at the center of conducting
sphere.
8
Q
>
>
>
³a´
<
Qi = ¡Q
d
>
>
>
:
Qa = ¡4¼²0 a2 V0
a2
;
d0
d0 = 2c
Putting a charge ¡Q1 at z = ¡z1 yields
VSg = 0;
{ Now the V and E external to the grounded sphere can be calculated by the
two point charges Q and ¡(a=d) Q in free space.
but
a
< 1;
d0
°=
where
V0 , a charge Q1 , the image
=
d0
d1
a2
d1
!
di2
°2
=
d0
1 ¡ °2
QSg 6= 0
Putting a charge ¡Q2 at z = ¡z2
VSg = 0;
but non-equipotention VS on S
{ Step 2: Putting an image charge ¡Q0 at z = ¡z0 yields
VSg = 0;
but non-euipotential VS on S
Chung‐Ang University
Electromagnetics
50
{ Step 7:
Putting a charge Q3 (image charge of ¡Q2 ) at z = z3
μ ¶
μ ¶μ ¶
μ
¶μ ¶
a
a
d0
°3
d0
Q3 = Q2
= Q2
= Q0
2
d2
d0
d2
1¡°
d2
d2 = d0 ¡ di2
‐V0
di4
=
d0
VS = V0 ;
{ Step 8:
but
{ Step 9:
0
z
c
¶
‐V0
S
‐Q0 ‐Q1
QSg 6= 0
Sg
0
di1
Q2Q1Q0
z
z2 z1
di2
c
μ
Q5 = Q4
di3
d3
1 ¡ °2
=1¡
= 1 ¡ °2
d0
d0
1 ¡ 2° 2
1 ¡ 3° 2 + ° 4
d3
!
=
d
1 ¡ 2° 2
μ0
¶
1
4
Q4 = Q0 °
1 ¡ 3° 2 + ° 4
a2
z4 = c ¡ di4 ;
di4 =
d3
a
d4
¶
μ
= Q4
a
d0
¶μ
d0
d4
¶
μ
= Q0 ° 5
1
1 ¡ 3° 2 + ° 4
¶μ
d0
d4
¶
d4 = d0 ¡ di4
Putting a charge Q4 (image charge of ¡Q3 ) at z = z4
μ ¶
μ ¶μ ¶
μ
¶μ ¶
a
a
d0
1
d0
Q4 = Q3
= Q3
= Q0 ° 4
d3
d0
d3
1 ¡ 2° 2
d3
μ
but non-equipotention VS on S
{ Step 11: Putting a charge Q5 (image charge of ¡Q4 ) at z = z5
μ
¶
di4
d4
1 ¡ 2° 2
=1¡
= 1 ¡ °2
d0
d0
1 ¡ 3° 2 + ° 4
1 ¡ 4° 2 + 3° 4
d4
!
=
d0
1 ¡ 3° 2 + ° 4
μ
¶
1
5
Q5 = Q0 °
1 ¡ 4° 2 + 3° 4
but non-equipotention VS on S
Chung‐Ang University
c
1 ¡ 2° 2
1 ¡ 3° 2 + ° 4
VSg = 0;
QSg 6= 0
d3 = d0 ¡ di3
a V0
a
{ Step 10: Putting a charge ¡ Q4 at z = ¡z4
Putting a charge ¡Q3 at z = ¡z3
VSg = 0;
but
d0
d3
μ
¶
c
a2
z3 = c ¡ di3 ;
di3 =
d2
μ ¶2 μ ¶
di3
a
d0
=
d0
d0
d2
μ
¶
di3
1 ¡ °2
2
=°
!
d0
1 ¡ 2° 2
VS = V0 ;
a
d0
¶2 μ
di4
= °2
d0
!
di2
d2
°2
= 1¡
=1¡
d0
d0
1 ¡ °2
1 ¡ 2° 2
d2
!
=
d0
1 ¡ °2
μ
¶
1
3
Q3 = Q0 °
1 ¡ 2° 2
μ
² Repeating this process to get the limit of potential on S and Sg .
² Total induced charge on S = the total point charges
Q = Q0 + Q1 + Q2 + Q3 + Q4 + ¢ ¢ ¢ = Q0 G
·
¸
°2
°3
°4
+
+
+ ¢¢¢
G= 1+°+
1 ¡ °2
1 ¡ 2° 2
1 ¡ 3° 2 + ° 4
¶
² Capacitances of two cases
Q
Q0 G
=
S = 4¼²0 aG
V0
V0
Q
C0
=
C=
2V0
2
Conducting sphere and ground plane: C0 =
Two conducting spheres:
Electromagnetics
51
Convolution Integral in Time Domain
² Impulse function (or Dirac delta function)
(
0;
for t =
6 0
± (t) =
1; for t = 0
Z 1
± (t) dt = 1
Convolution Integral in Space Domain
(t)
h((t)
(t)
Integral = 1
² Impulse function in the space domain
1
± (r) = ± (x) ± (y) ± (z)
Z
± (r) dv = 1
h(t)
0
V
t
¡1
² Approximate impulse function
(
0;
for jtj > ¢=2
±¢ (t) =
1=¢; for jtj < ¢=2
(t)
1/
Area = 1
(Integral = 1)
‐/2 /2
² Output y(t) for an arbitrary input x(t) in an
LTI (Linear Time Invariant) ststem
Input ± (t)
! Output h(t)
Input x(t)
! Output y (t) = ???
where h(t) is called the impulse response.
Then y(t) becomes a convolution:
Z 1
x(t0 ) h(t ¡ t0 ) dt0
y(t) = x(t) ¤ h(t) =
¡1
(t)
h(t)
1/
t
² If we introduce Gv (r) to denote an impulse
response of electyric potential (Green's
function) then total electric potential
for the source charge distribution can
be expressed by Space-domain convolution
representation as
Z
V(r) = Gv (r) ¤ ½(r) =
Gv (jr ¡ r0 j) ½(r0 ) dv0
v0
1
Gv (r) =
4¼²0 jrj
h(t)
‐/2 /2
x’=0
|X|=|x‐x’|=|x|
E0
qs
0
Source
point
x
Field
point
x’≠0
|X|=|x‐x′|
qs
0
x
Field
point
x’
Source
point
Discrete charge distribution
|Xn|=|x‐xn’|
qs1 qs2
t
E1
² Appropriate expression of a point source qs
at the origin
qsn
0
En
qsN
x
xn’
qs ! ½(r) = qs ± (r)
x(t)
x(t)
x(n) (t‐n) 
Continuous charge distribution
ρL(x)
|X|=|x‐x’|
Proof
dE
{ Approximate expression of x(t) and y(t)
X
x¢ (t) =
x(n¢) ±¢ (t ¡ n¢) ¢
0
0
n t
x’
Source
point
x
Field
point
n
X
y¢ (t) =
x(n¢) h¢ (t ¡ n¢) ¢
n
{ Taking a limit with ¢ ! 0 with the
corresponding variable parameters:
n¢ = t0 ;
¢ = (n + 1)¢ ¡ n¢ ! dt0
Z
y(t) = lim y¢ (t) = x(t0 )h(t ¡ t0 )dt0
Time domain
Input
Output
h(t)
δ(t)
LTI system
h(t)
x(t)
y(t)
y(t)=x(t)∗h(t)
Space domain
Input
δ(r)
ρ(r)
LSI system
Gv(r)
Output
GV(r)
V(r)
V(r)=GV(r)∗ρ(r)
¢!0
Chung‐Ang University
Electromagnetics
52
Integral of sec(x)
Expressions of Some Derivatives
² For a sec x
² For ln P with
Z
Z
1
dx
cos x
Z
Z
cos x
cos x
dx
=
dx =
2
cos x
1 ¡ sin2 x
Z
cos x
=
dx
(1 ¡ sin x)(1 + sin x)
N
D
N = R1 + z + L=2;
P=
sec x dx =
I=
Derivatives of ln P with respect to ½ and z
N0½
D0½
@ ln P
1 @P
=
=
¡
@½
P @½
N
D
0
@ ln P
1 @P
N
D0
=
= z¡ z
@z
P @z
N
D
With u = sin x,
du = cos x dx
Z
1
I=
du
(1 + u)(1 ¡ u)
¶
Z μ
1
1
1
=
+
du
2
1+u 1¡u
1
= [ln(1 + u) ¡ ln(1 ¡ u)] + C
2 · μ
¶¸
1
1+u
=
ln
+C
2
1¡u
where the subscripts ½ and z stand for
taking a derivative with respect to ½ and
z, respectively. The related derivatives
are
½
½
;
D0½ =
R1
R2
R
+
z
+
L
=
2
R
1
2 + z ¡ L=2
N0z =
; D0z =
R1
R2
·
@ ln Q
1
=½
@½
R1 (z + L=2 + R1 )
¸
1
¡
R2 (z ¡ L=2 + R2 )
½
1
½2
=
½ R21 [1 + (z + L=2)=R1 ]
¾
½2
¡ 2
R [1 + (z ¡ L=2)=R2 ]
μ2
¶
1
cos2 ¯1
cos2 ¯2
=
¡
½ 1 + sin ¯1
1 + sin ¯2
1
= (sin ¯2 ¡ sin ¯1 ))
½
1
@ ln Q
1
=
¡
@z
R1
R2
N0½ =
Replacing u with tt sin x
(1 + sin x)2
1+u
1 + sin x
=
=
1¡u
1 ¡ sin x
(1 ¡ sin x)(1 + sin x)
μ
¶2
1 + sin x
2
=
= (sec x + tan x)
cos x
Therefore
1
2
ln (sec x + tan x) + C
2
= ln (sec x + tan x) + C
I=
Summarizing
Z
sec x dx = ln (sec x + tan x) + C
Chung‐Ang University
D = R2 + z ¡ L=2
Electromagnetics
53