Chapter 2 Coulomb’s Law and Electric Field Intensity
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Transcript of Chapter 2 Coulomb’s Law and Electric Field Intensity
President University Erwin Sitompul EEM 2/1
Lecture 2
Engineering Electromagnetics
Dr.-Ing. Erwin SitompulPresident University
http://zitompul.wordpress.com
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Chapter 2Coulomb’s Law and Electric Field Intensity
Engineering Electromagnetics
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Chapter 2 Coulomb’s Law and Electric Field Intensity
In 1600, Dr. Gilbert, a physician from England, published the first major classification of electric and non-electric materials.
He stated that glass, sulfur, amber, and some other materials “not only draw to themselves straw, and chaff, but all metals, wood, leaves, stone, earths, even water and oil.”
In the following century, a French Army Engineer, Colonel Charles Coulomb, performed an elaborate series of experiments using devices invented by himself.
Coulomb could determine quantitatively the force exerted between two objects, each having a static charge of electricity.
He wrote seven important treatises on electric and magnetism, developed a theory of attraction and repulsion between bodies of the opposite and the same electrical charge.
The Experimental Law of Coulomb
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The Experimental Law of CoulombChapter 2 Coulomb’s Law and Electric Field Intensity
Coulomb stated that the force between two very small objects separated in vacuum or free space by a distance which is large compared to their size is proportional to the charge on each and inversely proportional to the square of the distance between them.
1 22
QQF kR
In SI Units, the quantities of charge Q are measured in coulombs (C), the separation R in meters (m), and the force F should be newtons (N).
This will be achieved if the constant of proportionality k is written as:
0
14
k
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The Experimental Law of CoulombThe permittivity of free space ε is measured in farads per meter
(F/m), and has the magnitude of: 12 9
018.854 10 10 F m
36
The Coulomb’s law is now:1 2
20
14
QQFR
The force F acts along the line joining the two charges. It is repulsive if the charges are alike in sign and attractive if the are of opposite sign.
Chapter 2 Coulomb’s Law and Electric Field Intensity
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The Experimental Law of Coulomb
1 22 122
0 12
14
QQR
F a
In vector form, Coulomb’s law is written as:
Chapter 2 Coulomb’s Law and Electric Field Intensity
F2 is the force on Q2, for the case where Q1 and Q2 have the same sign, while a12 is the unit vector in the direction of R12, the line segment from Q1 to Q2.
1212
12
RaR
12
12RR 2 1
2 1
r rr r
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1 ?F =1 2
1 2 2120 12
14
QQR
F = F a
The Experimental Law of CoulombChapter 2 Coulomb’s Law and Electric Field Intensity
ExampleA charge Q1 = 310–4 C at M(1,2,3) and a charge of Q2 = –10–4 C at N(2,0,5) are located in a vacuum. Determine the force exerted on Q2 by Q1.
12 2 1 R r r
(2 0 5 ) (1 2 3 )x y z x y z a a a a a a
1 2 2x y z a a a
1212
12
RaR1 (1 2 2 )3 x y z a a a
1 22 122
0 12
14
QQR
F a4 4
12 2
1 (3 10 )( 10 ) 1 (1 2 2 )4 (8.854 10 ) 3 3 x y z
a a a
10 20 20 Nx y z a a a
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Electric Field IntensityLet us consider one charge, say Q1, fixed in position in space.Now, imagine that we can introduce a second charge, Qt, as a
“unit test charge”, that we can move around.We know that there exists everywhere a force on this second
charge ► This second charge is displaying the existence of a force field.
Chapter 2 Coulomb’s Law and Electric Field Intensity
112
0 1
14
tt t
t
QQR
F a
The force on it is given by Coulomb’s law as:
Writing this force as a “force per unit charge” gives:
112
0 1
14
tt
t t
QQ R
F a Vector Field,
Electric Field Intensity
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Electric Field IntensityChapter 2 Coulomb’s Law and Electric Field Intensity
We define the electric field intensity as the vector of force on a unit positive test charge.
Electric field intensity, E, is measured by the unit newtons per coulomb (N/C) or volts per meter (V/m).
112
0 1
14
tt
t t
QQ R
FE = a
The field of a single point charge can be written as:
20
14 R
QR
E = a
aR is a unit vector in the direction from the point at which the point charge Q is located, to the point at which E is desired/measured.
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Electric Field IntensityChapter 2 Coulomb’s Law and Electric Field Intensity
For a charge which is not at the origin of the coordinate, the electric field intensity is:
20
1( )4
Q
r rE r =r rr r
30
1 ( )4
Q
r r=r r
3 22 2 20
( ) ( ) ( )14 ( ) ( ) ( )
x y zQ x x y y z z
x x y y z z
a a a=
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Electric Field IntensityChapter 2 Coulomb’s Law and Electric Field Intensity
The electric field intensity due to two point charges, say Q1 at r1 and Q2 at r2, is the sum of the electric field intensity on Qt caused by Q1 and Q2 acting alone (Superposition Principle).
120 1
220 2
1( )4
1 4
Q
Q
E r = ar r
ar r
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Electric Field IntensityChapter 2 Coulomb’s Law and Electric Field Intensity
ExampleA charge Q1 of 2 μC is located at at P1(0,0,0) and a second charge of 3 μC is at P2(–1,2,3). Find E at M(3,–4,2).
1 3 4 2 ,x y z r r a a a
2 4 6 ,x y z r r a a a 2 53 r r1 29 r r
1 21 22 2
0 01 2
1 1( )4 4
Q Q
E r = a ar r r r
1 1 2 23 3
0 1 2
( ) ( )14
Q Q
r r r r=r r r r
6 6
3 30
2 10 (3 4 2 ) 3 10 (4 6 )14 29 53
x y z x y z
a a a a a a=
623.7 879.92 160.17 V mx y z a a a
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Chapter 2 Coulomb’s Law and Electric Field Intensity
Field Due to a Continuous Volume Charge DistributionWe 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:
0limv v
Qv
The total charge within some finite volume is obtained by integrating throughout that volume:
volvQ dv
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Chapter 2 Coulomb’s Law and Electric Field Intensity
Field Due to a Continuous Volume Charge DistributionExample
Find 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
volvQ dv
23 22
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
32
0
8z dz 72 C
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Chapter 2 Coulomb’s Law and Electric Field Intensity
Field Due to a Continuous Volume Charge DistributionThe incremental contribution to the electric field intensity at r
produced by an incremental charge ΔQ at r’ is:
20
( )4
Q
r rE rr 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:
20vol
( )1( )4
v dv
r r rE r
r rr r
204v v
r rr rr r
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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
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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 aΦ direction. •
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
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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
30
1 ( )4
dQd
r rE
r r
2 2 3 20
( )14 ( )
L zdz zz
a a
zz r ayy r a a
2 2 3 20
14 ( )
L dzz
a
2 2 3 20
14 ( )
L dzEz
2 2 2 1 204 ( )
L zz
02LE
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Field of a Line ChargeChapter 2 Coulomb’s Law and Electric Field Intensity
Now let us analyze the answer itself:
02L
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.
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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 y axes in free space. Find E at: (a) PA(0,0,4); (b) PB(0,3,4).
PA
PB
0 0
( )2 2
x
x y
L LyA
x y
P
E a a
9 9
0 0
5 10 5 102 (4) 2 (4)z z
a a
44.939 V mz a
0 0
( )2 2
x
x y
L LyB
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
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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 +∞.
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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 y axis 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.
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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 20
cos2
sx
dydEx y
2 202s xdyx y
Adding the effects of all the strips,
2 202s
xxdyEx y
1
0
tan2
s yx
02s
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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.
02s
N
E a
The field of a sheet of charge is constant in magnitude and direction. It is not a function of distance.
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Chapter 1 Vector Analysis
Homework 2D2.2. D2.4. D2.6. All homework problems from Hayt and Buck, 7th Edition.
Due: Monday, 22 April 2013.