elektromagnetika teknik
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Elektromagnetika TeknikCHAPTER 2 COULOMB’S LAW AND ELECTRIC FIELD INTENSITY
Chapter 2Field Due to a Continuous Volume Charge
Distribution, Line Charge, and Sheet of Charge
Engineering Electromagnetics
Chapter 2 Coulomb’s Law and Electric Field Intensity
Field Due to a Continuous Volume Charge Distribution We denote the volume charge density by ρv, having the units of coulombs per cubic
meter (C/m3).
The small amount of charge ΔQ in a small volume Δv is
vQ v
We may define ρv mathematically by using a limit on the above equation:
0limvv
Q
v
The total charge within some finite volume is obtained by integrating throughout that volume:
vol
vQ dv
Chapter 2 Coulomb’s Law and Electric Field Intensity
ExampleFind the total charge inside the volume indicated by ρv = 4xyz2, 0 ≤ ρ ≤ 2, 0 ≤ Φ ≤ π/2, 0 ≤ z ≤ 3. All values are in SI units.
cosx
siny
24 sin cosv z
vol
vQ dv 23 2
2
0 0 0
(4 sin cos )( )z
z d d dz
23 2
3 2
0 0 0
4 sin cosz d d dz
23
2
0 0
16 sin cosz d dz
sin 2 2sin cos
3
2
0
8z dz 72 C
Field Due to a Continuous Volume Charge Distribution
Chapter 2 Coulomb’s Law and Electric Field Intensity
The incremental contribution to the electric field intensity at r produced by an incremental charge ΔQ at r’ is:
2
0
( )4
Q
r rE r
r rr r
The contributions of all the volume charge in a given region, let the volume element Δv approaches zero, is an integral in the form of:
2
0vol
( )1( )
4
v dv
r r rE r
r rr r
2
04
v v
r r
r rr r
Field Due to a Continuous Volume Charge Distribution
Field of a Line ChargeChapter 2 Coulomb’s Law and Electric Field Intensity
Now we consider a filamentlike distribution of volume charge density. It is convenient to treat the charge as a line charge of density ρL C/m.
Let us assume a straight-line charge extending along the z axis in a cylindrical coordinate system from –∞ to +∞.
We desire the electric field intensity E at any point resulting from a uniform line charge density ρL.
E z zd d dE E a a
Field of a Line ChargeChapter 2 Coulomb’s Law and Electric Field Intensity
The incremental field dE only has the components in aρ and az direction, and no aΦdirection.
• Why?
The component dEz is the result of symmetrical contributions of line segments above and below the observation point P.
Since the length is infinity, they are canceling each other ► dEz = 0.
The component dEρ exists, and from the Coulomb’s law we know that dEρ will be inversely proportional to the distance to the line charge, ρ.
E z zd d dE E a a
Field of a Line ChargeChapter 2 Coulomb’s Law and Electric Field Intensity
Take P(0,y,0),
E z zd d dE E a a
3
0
1 ( )
4
dQd
r rE
r r
2 2 3 2
0
( )1
4 ( )
L zdz z
z
a a
zz r a
yy r a a
2 2 3 2
0
1
4 ( )
L dz
z
a
2 2 3 2
0
1
4 ( )
L dzE
z
2 2 2 1 2
04 ( )
L z
z
02
LE
Field of a Line ChargeChapter 2 Coulomb’s Law and Electric Field Intensity
Now let us analyze the answer itself:
02
L
E a
The field falls off inversely with the distance to the charged line, as compared with the point charge, where the field decreased with the square of the distance.
Field of a Line ChargeChapter 2 Coulomb’s Law and Electric Field Intensity
Example D2.5.Infinite uniform line charges of 5 nC/m lie along the (positive and negative) x and yaxes in free space. Find E at: (a) PA(0,0,4); (b) PB(0,3,4).
PAPB
0 0
( )2 2
x
x y
L Ly
A
x y
P
E a a
9 9
0 0
5 10 5 10
2 (4) 2 (4)z z
a a
44.939 V mz a
0 0
( )2 2
x
x y
L Ly
B
x y
P
E a a
9 9
0 0
5 10 5 10(0.6 0.8 )
2 (5) 2 (4)y z z
a a a
10.785 36.850 V my z a a
• ρ is the shortest distance between an observation point and the
line charge
Field of a Sheet of ChargeChapter 2 Coulomb’s Law and Electric Field Intensity
Another basic charge configuration is the infinite sheet of charge having a uniform density of ρS C/m2.
The charge-distribution family is now complete: point (Q), line (ρL), surface (ρS), and volume (ρv).
Let us examine a sheet of charge above, which is placed in the yz plane.
The plane can be seen to be assembled from an infinite number of line charge, extending along the z axis, from –∞ to +∞.
Field of a Sheet of ChargeChapter 2 Coulomb’s Law and Electric Field Intensity
For a differential width strip dy’, the line charge density is given by ρL = ρSdy’.
The component dEz at P is zero, because the differential segments above and below the yaxis will cancel each other.
The component dEy at P is also zero, because the differential segments to the right and to theleft of z axis will cancel each other.
Only dEx is present, and this component is a function of x alone.
Field of a Sheet of ChargeChapter 2 Coulomb’s Law and Electric Field Intensity
The contribution of a strip to Ex at P is given by:
2 2
0
cos2
sx
dydE
x y
2 2
02
s xdy
x y
Adding the effects of all the strips,
2 2
02
sx
xdyE
x y
1
0
tan2
s y
x
02
s
Field of a Sheet of ChargeChapter 2 Coulomb’s Law and Electric Field Intensity
Fact: The electric field is always directed away from the positive charge, into the negative charge.
We now introduce a unit vector aN, which is normal to the sheet and directed away from it.
02
sN
E a
The field of a sheet of charge is constant in magnitude and direction. It is not a function of distance.
Chapter 2 Coulomb’s Law and Electric Field Intensity
Homework 2 D2.2.
D2.4.
D2.6. All homework problems from Hayt and Buck, 7th Edition.
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