Transcript charge - Erwin Sitompul
Engineering Electromagnetics
Lecture 2 Dr.-Ing. Erwin Sitompul President University President University
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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 compared to their size is
force
between two very small objects separated in vacuum or free space by a distance which is large
proportional
to the
charge
on each and
inversely proportional
to the
square of the distance
between them.
F
k Q Q
1 2
R
2 In SI Units, the quantities of charge
Q
coulombs (C), the separation
R
are measured in in meters (m), and the force
F
should be newtons (N).
This will be achieved if the constant of proportionality
k
written as: is
k
1 4
0 President University 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: 0
12
1 36
9
10 F m
The Coulomb’s law is now:
F
1 4
0
Q Q
1 2
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
a
12
R
12
R
12
R
12
R
12
r
2
r
2
r
1
r
1 In vector form, Coulomb’s law is written as:
F
2
1 4
0
Q Q
1 2 2
R
12
a
12
F
2 is the force on
Q
2 same sign, while
a
12 , for the case where line segment from
Q
1 is the unit vector in the direction of
R
12 , the to
Q
2 .
Q
1 and
Q
2 have the President University Erwin Sitompul EEM 2/5
Chapter 2 Coulomb’s Law and Electric Field Intensity
The Experimental Law of Coulomb
Example A charge
Q
1
Q
2 = –10 –4 the force exerted on
R
12
r
2 (2
a
x
r
1 = 3 10 –4 C at
y N
(2,0,5) are located in a vacuum. Determine
z Q
C at
M
(1,2,3) and a charge of 2
a
by
x
Q
1 .
y
a a
12 1
R
12
R
12 1
a
x
0
a
2
a
y
5
a
2
a
z
2
a
3 (1
a
x
2
a
y
2
a
z
)
F
2 1 4 0
Q Q
1 2 2
R
12 1
a
12 12 ) 4 3 2 4 3 (1
a
x
2
a
y
10
a
x
20
a
y
20
a
z
N
a
1 2 1 1 4 ?
0
Q Q
1 2 2
R
12
a
21 President University Erwin Sitompul EEM 2/6
Chapter 2 Coulomb’s Law and Electric Field Intensity
Electric Field Intensity
Let us consider one charge, say
Q
1 , fixed in position in space.
Now, imagine that we can introduce a second charge,
Q t
, 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:
F
t
1 4
0
Q Q
1
t R
1
t
2
a
1
t
Writing this force as a “force per unit charge” gives:
F
t Q t
1 4
0
Q
1
R
1
t
2
a
1
t
Vector Field, Electric Field Intensity
President University 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).
E = F
Q t t
4 1
0
Q R
1 2 1
t
a
1
t
The field of a single point charge can be written as:
1
Q
E =
4
0
R
2
a
R
a
R
is a unit vector in the direction point charge
Q
from
the point at which the 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:
1 4
0
Q
2
=
1 4
0
Q
( )
3
=
1 4
0
Q
(
x
x
)
a
x
(
x
x
)
2
(
y
(
y
y y
)
2
)
a
y z z z
)
2
z
)
a
z
3 2 President University Erwin Sitompul EEM 2/9
Chapter 2 Coulomb’s Law and Electric Field Intensity
Electric Field Intensity
The electric field intensity due to two point charges, say
Q
1 and
Q
2 at caused by
r
2 , is the sum of the electric field intensity on
Q
1 and
Q
2
Q t
acting alone (Superposition Principle).
at
r
1
1 4
0
1 4
0
Q
1
Q
2
2
a
1 2
a
2
President University Erwin Sitompul EEM 2/10
Chapter 2 Coulomb’s Law and Electric Field Intensity
Electric Field Intensity
Example A charge
Q
1 of 2
μ
C is located at at
P
1 (0,0,0) and a second charge of 3
μ
C is at
P
2 ( –1,2,3). Find
E
at
M
(3, –4,2).
r r
1 3
a
x
4
a
y
2
a
z
,
r r
1 29
r r
2 4
a
x
6
a
y
a
z
, 2 53 1 4 0
Q
1 1 2
a
1 1 4 0
Q
2 2 2
a
2
=
1 4 0
Q
1 ( 1 3 1 )
Q
2 ( 2 ) 2 3
=
1 4 0 6
a
x
4
a
29 3
y
a
623.7
a
x
879.92
a
y
160.17
a
z
V m 6
a
x
6
a
y
a
z
) 53 3 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
ρ v
, having the units of coulombs per cubic meter (C/m 3 ).
The small amount of charge Δ
Q Q
v v
in a small volume Δ
v
is We may define
ρ v
equation:
v
lim
v
0
Q v
mathematically by using a limit on the above The total charge within some finite volume is obtained by integrating throughout that volume:
Q
vol
v dv
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
ρ v
0 ≤
ρ
≤ 2, 0 ≤
Φ
≤
π
/2, 0 ≤
z
= 4
xyz
2 , ≤ 3. All values are in SI units.
x
cos
y
sin
v
sin cos
z
2
Q
vol
v dv
z
3 0 2
0 2 0 3 2
2 0 0 0 (4 sin 4 3 2
z
3
2 0 0 16
z
2 3 0
2 8
z dz
72 C cos
z
2 )(
d
dz
) sin 2 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 Δ
v
approaches zero, is an integral in the form of:
vol
4 1
0
v
r
dv
2
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
ρ L
C/m.
Let us assume a straight-line charge extending along the –∞ to +∞.
z
axis in a cylindrical coordinate system from We desire the electric field intensity
E
at any point resulting from a uniform line charge density
ρ L
.
President University Erwin Sitompul
d
E
d
E
dE z
a
z
EEM 2/15
Chapter 2 Coulomb’s Law and Electric Field Intensity
Field of a Line Charge
The incremental field direction, and no
a
Φ dE
only has the components in
a
ρ
direction.
•
Why?
and
a
z
The component
dE
observation point
P z
.
is the result of symmetrical contributions of line segments above and below the Since the length is infinity, they are canceling each other ►
dEz
= 0.
The component
dE ρ
exists, and from the Coulomb’s law we know that
dE ρ
will be inversely proportional to the distance to the line charge,
ρ
. President University Erwin Sitompul
d
E
d
E
dE z
a
z
EEM 2/16
Chapter 2 Coulomb’s Law and Electric Field Intensity Take
P
(0,
y
,0),
d
E
1 4 0
dQ
( 3 ) 1 4 0
L dz
( ( 2
a
z
2 3 2 ) 1
L
a
4 0 ( 2
dz
z
2 3 2 )
z
a
z
)
Field of a Line Charge
r
r
z
a
z y
a
y
a
E
E
4
L
4 0 1 0 ( 2 2
dz L
z
z
z
2 3 2 ) 2 1 2 ) 2
L
0
d
E
d
E
dE z
a
z
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:
E
2
L
0
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.
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
)
P A
(0,0,4); (
b
)
P B
(0,3,4).
E
(
P A
)
L x
2 0
x
a
x
Ly
2 0
y
a
y P A P B
9 9 2 0 (4)
a
z
2 0 (4)
a
z
44.939
a
z
V m
E
(
P B
)
L x
2 0
x
a
x
Ly
2 0
y
a
y
9 2 0 (5) 10.785
a
y
(0.6
a
y
36.850
a
z
a
z
V m 9 2 0 (4)
a
z
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
ρ S
C/m 2 .
The charge-distribution family is now complete: point (
Q
), line (
ρ L
), surface (
ρ S
), and volume (
ρ v
).
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
ρ L
=
ρ S dy
’.
The component
dE z
at above and below the
y P
is zero, because the differential segments axis will cancel each other.
The component zero, because the differential segments to the right and to the left of
z dE y
at
P
is also axis will cancel each other.
Only
dE x
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
E x
is given by:
dE x
2 0
s x
2
y
2 cos 2
s
0
x
2
xdy
y
2 at P Adding the effects of all the strips,
E x
2
s
0 2
s
0 2
s
0
x
2 tan 1
xdy
y x y
2 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
a
N
, which is normal to the sheet and directed away from it.
E
2
s
0
a
N
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