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Chapter 2 Static Electric Fields 2.1 Two Fundamental...
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Lesson 6 Basic Electromagnetics, Dept. of Elec. Eng., The Chinese University of Hong Kong, Prof. K.-L. Wu / Prof. Th. Blu Chapter 2 Static Electric Fields 2.1 Two Fundamental Postulates of Electrostatics in Free Space where is the volume charge density of free charges (C/m 3 ), and is the permittivity of free space, universal constant. ! 0 ! 0 ! " = # $ E v 0 = ! " E v Fields are irrotational Field is not solenoidal unless 0 = ! These two postulates are concise, simple, and independent of any coordinate system; and they can be used to derive all other relations, laws and theorems in electrostatics! Their integral form can be obtained by using Stokes ʼ law and Gauss ʼ law as 0 = ! " C l d E v v 0 ! Q s d E S = " # v v 0 = ! " E v 0 ! " = # $ E v Fields form of KVL Fields form of KCL
Transcript of Chapter 2 Static Electric Fields 2.1 Two Fundamental...
Lesson 6Basic Electromagnetics, Dept. of Elec. Eng., The Chinese University of Hong Kong, Prof. K.-L. Wu / Prof. Th. Blu
Chapter 2 Static Electric Fields
2.1 Two Fundamental Postulates of Electrostatics in Free Space
where is the volume charge density of free charges (C/m3), andis the permittivity of free space, universal constant.
!0!
0!
"=#$ E
v
0=!" Ev
Fields are irrotational Field is not solenoidal unless 0=!
These two postulates are concise, simple, and independent of anycoordinate system; and they can be used to derive all other relations,laws and theorems in electrostatics!
Their integral form can be obtained by using Stokesʼ law and Gaussʼlaw as
0=!"C ldEvv
0!
QsdE
S="#
vv
0=!" Ev
0!
"=#$ E
v
Fields form of KVL Fields form of KCL
Lesson 6Basic Electromagnetics, Dept. of Elec. Eng., The Chinese University of Hong Kong, Prof. K.-L. Wu / Prof. Th. Blu
2.2 Coulombʼs Law
When a point charge Q2 is placed in the field of another point chargeQ1 at the origin, a force F12 is experienced by Q2 due to electric fieldintensity E12 of Q1 at Q2
A single point charge, Q, sits in origin of a boundless freespace, its E field must be everywhere radial and has thesame intensity at all points on the spherical surface,
0
ˆˆ!
QdsaEasdE
SRRR
S="=" ##
vv
0
2 )4(!
"Q
REdsE RS
R ==#or
)/( 4
ˆˆ2
0
mVR
QaEaE RRR
!"==
v
Therefore,
)( 4
ˆ2
0
2112212 N
R
QQaEQF R
!"==
vv
The permittivity of air is essentially the same as that of the free space:
)/(1 0
2
0 µ! c= )/(104 7
0 mH!
"= #µand
Coulombʼs Law
Lesson 6Basic Electromagnetics, Dept. of Elec. Eng., The Chinese University of Hong Kong, Prof. K.-L. Wu / Prof. Th. Blu
2.2 Coulombʼs Law (cont.)
If a single point charge, Q, sits in an arbitrary locationof a boundless free space, its E field would be
)/( ||||4
||4
ˆ
ˆ
2
0
2
0
mVRR
RR
RR
Q
RR
QaE
qPa
qPP
43421
vv
vv
vvvvv
!"
!"
!"=
!"=
#$#$
Suppose an electrostatic field is created by agroup of n discrete point charges Q1,Q2,…,Qn located at different positions, thetotal E field at a point is the vector sum of thefields caused by all the individual charges.
)/( ||
)(
4
1
13
0
mVRR
RRQE
n
k k
kk!= "#
"#= vv
vvv
$%
Lesson 6Basic Electromagnetics, Dept. of Elec. Eng., The Chinese University of Hong Kong, Prof. K.-L. Wu / Prof. Th. Blu
2.2 Coulombʼs Law (cont.)
3D case
The electric field caused by a continuous distribution ofcharge can be obtained by integrating (superposing) thecontribution of charge over the charge distribution.
)/( ˆ4
12
0
mVdv'R
aE
V
R!"
=#
$%
v
)/( 'ˆ4
12
0
mVdsR
aE
S
s
R!"
=#
$%
v
)/( 'ˆ4
12
0
mVdlR
aE
L
l
R!"
=#
$%
v
2D case
1D case
Lesson 6Basic Electromagnetics, Dept. of Elec. Eng., The Chinese University of Hong Kong, Prof. K.-L. Wu / Prof. Th. Blu
Solution Let us assume that the line charge lies along the zʼ-axis.(Weare perfectly free to do this because the field obviously does not dependon how we designate the line. It is an accepted convention to useprimed coordinates for source points and unprimed coordinates for fieldpoints when there is a possibility of confusion.)
FIGURE3-7 An infinitelylong,straight,line charge.
!
2.2 Coulombʼs Law (cont.)
The problem asks us to find the electric fieldintensity at a point P, which is at a distance rfrom the line.Since the problem has a cylindricalsymmetry (that is,the electric field is independentof the azimuth angle ),it would be mostconvenient to work with cylindrical coordinates.
Lesson 6Basic Electromagnetics, Dept. of Elec. Eng., The Chinese University of Hong Kong, Prof. K.-L. Wu / Prof. Th. Blu
2.2 Coulombʼs Law (cont.)
Lesson 6Basic Electromagnetics, Dept. of Elec. Eng., The Chinese University of Hong Kong, Prof. K.-L. Wu / Prof. Th. Blu
2222322
2
)( aaxa
x
ax
dx=
+=
+
+!
!"
!+
!"
#
2.2 Coulombʼs Law (cont.)
Lesson 6Basic Electromagnetics, Dept. of Elec. Eng., The Chinese University of Hong Kong, Prof. K.-L. Wu / Prof. Th. Blu
Gaussʼs Law
Gaussʼs law asserts that the total outward flux of the E-fieldover any closed surface in free space is equal to the totalcharge enclosed in the surface divided by .
Gaussʼs law is particularly useful in determining the E field ofcharge distributions with some symmetry conditions. On theother hand, when symmetry conditions do not exist, Gaussʼslaw would not be of much help.
Gaussʼ law follows directly from the divergencepostulate of electrostatics by the application ofdivergence theorem. 0
!
QsdE
S="#
vv
0!
2.3 Gaussʼs Law
