Transcript charge - Erwin Sitompul

Engineering Electromagnetics
Lecture 2
Dr.-Ing. Erwin Sitompul
President University
http://zitompul.wordpress.com
President University
Erwin Sitompul
EEM 2/1
Chapter 2
Coulomb’s Law and Electric Field Intensity
The Experimental Law of Coulomb
 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.
President University
Erwin Sitompul
EEM 2/2
Chapter 2
Coulomb’s Law and Electric Field Intensity
The Experimental Law of Coulomb
 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.
Q1Q2
F k 2
R
 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:
k
1
4 0
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Erwin Sitompul
EEM 2/3
Chapter 2
Coulomb’s Law and Electric Field Intensity
The Experimental Law of Coulomb
 The permittivity of free space ε is measured in farads per meter
(F/m), and has the magnitude of:
1
12
 0  8.854 10
109 F m
36
 The Coulomb’s law is now:
1 Q1Q2
F
4 0 R 2
 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.
President University
Erwin Sitompul
EEM 2/4
Chapter 2
Coulomb’s Law and Electric Field Intensity
The Experimental Law of Coulomb
a12 
R12
r r
R
 12  2 1
r2  r1
R12
R12
 In vector form, Coulomb’s law is written as:
1 Q1Q2
F2 
a12
2
4 0 R12
 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.
President University
Erwin Sitompul
EEM 2/5
Chapter 2
Coulomb’s Law and Electric Field Intensity
The Experimental Law of Coulomb
 Example
A charge Q1 = 310–4 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.
R12  r2  r1
 (2ax  0a y  5az )  (1ax  2a y  3az )
 1ax  2a y  2az
R12
a12 
R12
1
 (1a x  2a y  2a z )
3
1 Q1Q2
F2 
a12
2
4 0 R12
1
(3 104 )(104 ) 1

(1a x  2a y  2a z )
12
2
4 (8.854 10 )
3
3
10a x  20a y  20a z N
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1 Q1Q2
F1 = F2 F =
a 21
2
14
0 R12
?
Erwin Sitompul
EEM 2/6
Chapter 2
Coulomb’s Law and Electric Field Intensity
Electric Field Intensity
 Let 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.
 The force on it is given by Coulomb’s law as:
1 Q1Qt
Ft 
a1t
2
4 0 R1t
 Writing this force as a “force per unit charge” gives:
Ft
1 Q1

a
2 1t
Qt 4 0 R1t
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Vector Field,
Electric Field Intensity
Erwin Sitompul
EEM 2/7
Chapter 2
Coulomb’s Law and Electric Field Intensity
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).
Ft
1 Q1
E=

a
2 1t
Qt 4 0 R1t
 The field of a single point charge can be written as:
1
Q
E=
a
2 R
4 0 R
 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.
President University
Erwin Sitompul
EEM 2/8
Chapter 2
Coulomb’s Law and Electric Field Intensity
Electric Field Intensity
 For a charge which is not at the origin
of the coordinate, the electric field
intensity is:
r  r
E(r) =
4 0 r  r 2 r  r
1
Q
1 Q(r  r)
=
4 0 r  r 3
1 Q ( x  x)a x  ( y  y)a y  ( z  z)a z 
=
4 0 ( x  x)2  ( y  y)2  ( z  z)2 3 2


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Erwin Sitompul
EEM 2/9
Coulomb’s Law and Electric Field Intensity
Chapter 2
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).
E(r ) =
1
Q
a1 
4 0 r  r1
1
Q
a
2 2
4 0 r  r2
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Erwin Sitompul
EEM 2/10
Coulomb’s Law and Electric Field Intensity
Chapter 2
Electric Field Intensity
 Example
A 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).
r  r1  3ax  4a y  2a z , r  r1  29
r  r2  4ax  6a y  az ,
E(r) =
1
Q1
4 0 r  r1
a 
2 1
r  r2  53
1
Q2
4 0 r  r2
2
a2
1 
 Q1 (r  r1 ) Q2 (r  r2 ) 

=


3
3 
4 0 
r

r
r

r


1
2



6
6
1  2 10 (3a x  4a y  2a z ) 3 10 (4a x  6a y  a z ) 
=



3
3
4 0 

29
53


 623.7a x  879.92a y  160.17a z V m
President University
Erwin Sitompul
EEM 2/11
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
Q  v v
 We may define ρv mathematically by using a limit on the above
equation: Q
v  lim
v 0 v
 The total charge within some finite volume is obtained by
integrating throughout that volume:
Q    v dv
vol
President University
Erwin Sitompul
EEM 2/12
Chapter 2
Coulomb’s Law and Electric Field Intensity
Field Due to a Continuous Volume Charge Distribution
 Example
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.
x   cos 
v  4   sin    cos  z 2
y   sin 
Q

vol
3  2 2
  
 v dv 
(4   sin    cos   z 2 )(d    d  dz )
z 0  0  0
3 22


3 2
4

 z sin  cos  d  d dz
0 0 0
3 2

2
16
z
 sin  cos  d dz
sin 2  2sin  cos 
0 0
3
  8z 2 dz
0
 72 C
President University
Erwin Sitompul
EEM 2/13
Chapter 2
Coulomb’s Law and Electric Field Intensity
Field Due to a Continuous Volume Charge Distribution
 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:
1 v (r)dv r  r
E(r)  
2
4

r  r r  r
0
vol
President University
Erwin Sitompul
EEM 2/14
Chapter 2
Coulomb’s Law and Electric Field Intensity
Field of a Line Charge
 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.
dE  dE a  dEzaz
President University
Erwin Sitompul
EEM 2/15
Chapter 2
Coulomb’s Law and Electric Field Intensity
Field of a Line Charge
 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.
dE  dE a  dEzaz
President University
Erwin Sitompul
EEM 2/16
Chapter 2
Coulomb’s Law and Electric Field Intensity
Field of a Line Charge
 Take P(0,y,0),
r  za z
r  ya y  a
1 dQ(r  r)
dE 
4 0 r  r 3
1 L dz( a   za z )

4 0 (  2  z2 )3 2
1 L a  dz

4 0 (  2  z2 )3 2

E 
1
 4
L  dz
2
2 32

(


z
)
0



L 
z

4 0  2 (  2  z2 )1 2  
E 
dE  dE a  dEzaz
L
2 0 
President University
Erwin Sitompul
EEM 2/17
Chapter 2
Coulomb’s Law and Electric Field Intensity
Field of a Line Charge
 Now let us analyze the answer itself:
L
E
a
2 0 
 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.
President University
Erwin Sitompul
EEM 2/18
Chapter 2
Coulomb’s Law and Electric Field Intensity
Field of a Line Charge
 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).
L
Ly
PB
E( PA ) 
a 
a
PA
2 0  x
2 0  y
x
x
y
5 109
5 109

az 
az
2 0 (4)
2 0 (4)
 44.939a z V m
L
Ly
E( PB ) 
a 
a
2 0  x
2 0  y
x
x
y
5 109
5 109

(0.6a y  0.8a z ) 
az
2 0 (5)
2 0 (4)
 10.785a y  36.850a z V m
President University
Erwin Sitompul
• ρ is the shortest
distance between an
observation point and
the line charge
EEM 2/19
Chapter 2
Coulomb’s Law and Electric Field Intensity
Field of a Sheet of Charge
 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
+∞.
President University
Erwin Sitompul
EEM 2/20
Chapter 2
Coulomb’s Law and Electric Field Intensity
Field of a Sheet of Charge
 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 the
left of z axis will cancel each
other.
 Only dEx is present, and this
component is a function of x alone.
President University
Erwin Sitompul
EEM 2/21
Chapter 2
Coulomb’s Law and Electric Field Intensity
Field of a Sheet of Charge
 The contribution of a strip to Ex at P
is given by:
s dy
dEx 
cos 
2
2
2 0 x  y

xdy
 s 2
2 0 x  y2
 Adding the effects of all the strips,
s  xdy
Ex 
 x 2  y 2
2 0 


s
y


tan 1 
2 0
x  

 s
2 0
President University
Erwin Sitompul
EEM 2/22
Chapter 2
Coulomb’s Law and Electric Field Intensity
Field of a Sheet of Charge
 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.
s
E
aN
2 0
 The field of a sheet of charge is constant in magnitude and
direction. It is not a function of distance.
President University
Erwin Sitompul
EEM 2/23
Chapter 1
Vector Analysis
Homework 2
 D2.2.
 D2.4.
 D2.6. All homework problems from Hayt and Buck, 7th Edition.
 Deadline: 25 January 2011, at 07:30.
President University
Erwin Sitompul
EEM 2/24