University of Missan
College of Engineering
Electrical Engineering
Department
1st Semester Year
2013-2014
2nd Lesson Stage
Engineering Electromagnetic Fields
Subject: Coulomb's Law and Electric
Field Intensity
Lecture No. 4
Dr. Ahmed Thamer Radhi
2013 - 2014
Electromagnetic Fields
Coulomb's Law and Electric Field Intensity
Lecture No.4
Lesson Year 1 st Semester:2013-2014
Stage
2 nd Year
Subject
Charge Density
Lecture No.
4
Lecturer
Dr. Ahmed Thamer
University of Missan
College of Engineering
Electrical Engineering Dept.
Engineering
Electromagnetics Fields
3- Various Charge Distributions
++ +++
+++++++++
+ + + 𝛒𝛒S + + +
+++++++
Surface Charge
𝛒𝛒L
++++++++
Line Charge
Q
point Charge
 Line charge density (
C
m
)
 Surface charge density (
 Volume charge density (
Q = ∫ ρ L dL
+ +++
+ 𝛒𝛒V
Volume Charge
R
C
m2
C
m3
)
)
Q = ∬ ρ S dS
R
Q = ∭ ρ V dV
R
Notes that:
Q
• ρL = where (L) is any given line length or circumference
• ρS =
• ρV =
L
Q
A
Q
V
where (A) is any given area such as area of circle or sphere
where (V) is any given volume such as volume of sphere
Q
Since, �E⃗ = K
R2
a�⃗ R
R
So by replacing Q in above equations with line charge density, surface charge density, and
volume charge density, we get:
�⃗ = ∫ K ρL𝑑𝑑L
E
a�⃗ R
2
�⃗ = ∬
E
R
K ρsds
 �E⃗ = ∭
R
a�⃗ R
R2
K ρvdv
R2
R
a�⃗ R
R
4- Field Due to Continuous Volume Charge Distribution
C
Volume charge density is measured in Coulomb per cubic meter ( 3 ). The total charge
m
within some finite volume is obtained by integrating throughout that volume as:
Dr. Ahmed Thamer
Q = ∫Vol 𝑑𝑑Q = ∫Vol ρV dv
Coulomb's Law and Electric Field Intensity
Page 1
Electromagnetic Fields
Coulomb's Law and Electric Field Intensity
Lecture No.4
Example 4.1: Find the total charge inside each of the volumes indicated as:
(a) ρ V = 10 ze-0.1x sin𝜋𝜋y; -1≤ x ≤2; 0≤ y ≤1; 3≤ z ≤3.6
(b) ρ V = 4xyz; 0≤ ρ ≤2; 0≤ Ф ≤ 𝜋𝜋/2; 0 ≤ z ≤3
(c) ρ V = 3 𝜋𝜋 sinθ cos2Ф/[2r2(r2+1)]; universe
R
R
R
Solution: (a) Cartesian coordinate:
Q = ∭ ρV dxdydz
Q = ∭ 10 ze−0.1x sin𝜋𝜋y dxdydz
3.6
1
2
Q = 10 [ ∫z=3 z 𝑑𝑑z ∫y=0 sin𝜋𝜋y 𝑑𝑑y ∫x=−1 e−0.1x 𝑑𝑑x ]
z2
Q = 10 [ ]3.6
3 [
2
−𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 y
e −0.1x
]10 [
𝜋𝜋
−0.1
]2−1 = 36.1 C
(b) Cylindrical coordinate:
Q = ∭ ρV ρ 𝑑𝑑𝑑𝑑d∅dz
Since, ρ V = 4xyz, we have x= ρ cosФ; y= ρ sinФ, and z=z; then,
ρ V = 4 z ρ2 cosФ sinФ
Hence, Q = ∭ 4 z ρ2 cosФ sinФ ρ 𝑑𝑑𝑑𝑑d∅dz = ∭ 4 z ρ3 cosФ sinФ 𝑑𝑑𝑑𝑑d∅dz
R
R
Q=
2
4[∫ρ=0 ρ3
Q=4
ρ4
[ ]20
4
[
π
2
3
𝑑𝑑𝑑𝑑 ∫Ф=0 cosФ sinФ 𝑑𝑑∅ ∫z=0 z 𝑑𝑑z]
sin 2 Ф
2
(c) Spherical coordinate:
π
2
z2
]0 [ ]30 = 36 C
2
Q = ∭ 3𝜋𝜋sinθ cos 2 Ф/[2r 2 (r 2 + 1)] r2 sinθ𝑑𝑑r𝑑𝑑θdФ
Q=
Q=
Q=
Q=
Hint:
 ∫
𝑑𝑑𝑑𝑑
x2+ a2
=
𝑥𝑥𝑥𝑥𝑥𝑥
1
a
3π
2
3π
2
3π
∞
2𝜋𝜋
𝑑𝑑r ∫θ=0 sin2 θ 𝑑𝑑θ ∫Ф=0 cos2 Ф 𝑑𝑑Ф]
r 2 +1
1
[ tan−1 (r)]∞
0 * [θ −
π
1
1
2
* * * π * * 2π
2
2
3(π)4
8
𝜋𝜋
1
[∫r=0
2
= 36.5 C
2
sin 2θ π
]0
2
1
* [θ +
2
sin 2θ 2π
]0
2
x
tan−1 ( ) + c
−1
a
 ∫ 2 2 3 = 2 2 1/2 + c
(x + a )
(x + a )
𝑑𝑑𝑑𝑑
𝑥𝑥
 ∫ 2 2 3/2 = 2 2 2 1/2 + c
(x + a )
a (x + a )
1
 sin 2x = (1-cos2x)
P
2
2
1
 cos x = (1+cos2x)
2
 sin2x = 2 sinx cosx
Dr. Ahmed Thamer
Coulomb's Law and Electric Field Intensity
Page 2
Electromagnetic Fields
Coulomb's Law and Electric Field Intensity
Lecture No.4
5- Field of a Line Charge
Let us assume a straight line charge extending a long z-axis in cylindrical coordinate
�⃗ at any
system from - ∞ to ∞ as shown in Fig. 4.1. We desire the electric field intensity E
and every point resulting from a uniform line charge density ρL .
Figure 4.1
Symmetry should be always considered first in order to determine two specific factors:
 With which coordinates the field does not vary.
 Which components of the field are not present.
Now, which components are present? Each incremental length of line charge acts as a point
charge and produces an incremental contribution to the electric field intensity which is directed
away from the bit of charge.
No element of charge produce as Ф component of electric intensity, (EФ is zero). However, each
element does produce an Er and Ez component, but the contribution to Ez by elements of charge
which are equal distance above and below the point at which we are determining the field will
cancel. We therefore have found that we have only an (Eρ )component and it is varies only with
(r), now to find this component:
We choose a point P(0, y,0) on the y-axis at which to determine the field. This is a perfectly
general point in view of the lack of variation of the field with Ф and z we have:
�E⃗ =
Q
a�⃗ R
4πϵo R 2
R
dQ = ρL dL = ρL dz'
�⃗ = k 𝑑𝑑Q
dE
2
R
a�⃗ R = k
r⃗= ya�⃗ y = ρa�⃗ρ
r⃗′ = z' �a⃗ z
R
R
ρ L 𝑑𝑑𝑑𝑑 ′
R2
a�⃗ R
R
R
Dr. Ahmed Thamer
Coulomb's Law and Electric Field Intensity
Page 3
Electromagnetic Fields
Coulomb's Law and Electric Field Intensity
Lecture No.4
�⃗=r⃗- r⃗′= ρa�⃗ρ - z' a�⃗ z
R
�⃗�= √ ρ 2 + z'2
�R
R
P
ρa�⃗ρ − z′ a�⃗z
�a⃗ R =
√ ρ2 + z′ 2
ρ L 𝑑𝑑𝑑𝑑 ′ (ρa�⃗ρ − z ′ a�⃗z)
R
�⃗ = k
dE
3/2
( ρ2 + z′ 2)
�⃗ z =0, then, dE
�⃗ = k
Due to symmetry, dE
ρ L 𝑑𝑑𝑑𝑑 ′ ρa�⃗ρ
R
3/2
( ρ2 + z′ 2)
Since only the Eρ component is present, we may simplify: dE𝜌𝜌 = k
For infinite line charge (i.e −∞ ≤ L ≤ ∞);
ρL ρ
E𝜌𝜌 =
4πϵo
R
E𝜌𝜌 =
R
E𝜌𝜌 =
R
Or, �E⃗ =
ρL ρ
4πϵo
ρL
2πϵo ρ
ρL
2πϵo ρ
∞
∫−∞
2
*
ρ2
𝑑𝑑𝑑𝑑 ′
3/2
( ρ2 + z′ 2)
=
ρL ρ
4πϵo
[
R
𝑧𝑧′
1/2
ρ2( ρ2 + z′ 2)
ρ L ρ 𝑑𝑑𝑑𝑑 ′
3/2
( ρ2 + z′ 2)
]∞
−∞
a�⃗ρ
�⃗ . We might have used the angle θ as our
There are many other ways of obtainingE
variable of integration, from Fig.4.1 z'= ρcotθ and dz' = - ρcsc2 θ dθ. Since R= ρcscθ, our
integral becomes, simply,
ρ L 𝑑𝑑𝑑𝑑 ′
dE𝜌𝜌 =
R
E𝜌𝜌 =
R
E𝜌𝜌 =
R
�⃗ =
Or, E
sinθ= -
4πϵo R 2
0
ρ
- L
sinθ
∫
π
4πϵo ρ
ρL
2πϵo ρ
ρL
2πϵo ρ
ρ L sin θ 𝑑𝑑θ
4πϵo ρ
𝑑𝑑θ= -
ρL
4πϵo ρ
[cosθ]0π
a�⃗ρ
As an example, let us consider an infinite line charge parallel to the z-axis at x=6, y=8,
�⃗ at the general field point P(x, y, z). We replace ρ in above equation
Fig.4.2. We wish to find E
by the radial distance between the line charge and point P, R=√(x-6)2 + (y-8)2, and let a�⃗ρ be a�⃗R .
Thus,
�E⃗ =
ρL
2πϵo √(x−6)2 + (y−8)2
Where �a⃗R =
�E⃗ =
Dr. Ahmed Thamer
(x−6)a�⃗x + (𝑦𝑦−8)a�⃗y
a�⃗R
√(x−6)2 + (y−8)2
ρL
2πϵo
(
(x−6)a�⃗x + (𝑦𝑦−8)a�⃗y
(x−6)2 + (y−8)2
)
Figure 4.2
Coulomb's Law and Electric Field Intensity
Page 4
Electromagnetic Fields
Coulomb's Law and Electric Field Intensity
Lecture No.4
Example 4.2: A uniform line charge, ρ L = 25nC/m, lies on the line x=-3, z=4, in free space.
�⃗ in Cartesian components at: (a) the origin (b) point P (2, 15, 3)
Find E
(c) Q (ρ =4, Ф=60o, z=2)
Solution:
ρL
(a) ρ L in the direction of y, by replace ρ and a�⃗ρ in �E⃗ =
a�⃗ρ by R and a�⃗R ,
R
R
R
respectively, then:
ρL
�⃗ =
E
Z
ρL
a�⃗R
2πϵo R
�R⃗=3a�⃗x − 4a�⃗z
�⃗�=√32 + (-4)2=√25=5
�R
�a⃗R =
3a�⃗x −4a�⃗z
5
X
Z
ρL
�R⃗=5a�⃗x − a�⃗z
�⃗�=√52 + (-1)2=√26
�R
√26
-X
�⃗
R
(0, 0, 0)
(b) P(2, 15, 3)
5a�⃗x −a�⃗z
(-3, 0, 4)
Y
−9
25∗10
3a�⃗ −4a�⃗
�⃗ =
E
( x z)
−12
∗5
5
2∗3.14∗8.854∗10
�⃗= 53.9a�⃗x − 71.9a�⃗z (V/m)
E
�a⃗R =
2πϵo ρ
(-3,15, 4)
(2,15, 3)
−9
25∗10
5a�⃗ −a�⃗
�E⃗ =
( x z)
−12
∗√26
√26
2∗3.14∗8.854∗10
�E⃗= 86.4a�⃗x − 17.3a�⃗z (V/m)
�
Y
X
(c) x= ρ cosФ=4cos60 =2
y= ρ sinФ=4cos60o =3.46
z=2
Hence, Q(2, 3.46, 2)
o
R
R
�R⃗=5a�⃗x − ����⃗
2az
2
�⃗�=√5 + (-2)2=√29
�R
�a⃗R =
�⃗ =
E
Dr. Ahmed Thamer
5a�⃗x −2a�⃗z
√29
25∗10 −9
2∗3.14∗8.854∗10 −12 ∗√29
(
����⃗z
5a�⃗x −2a
√29
) = 77.5a�⃗x − 31a�⃗z (V/m)
Coulomb's Law and Electric Field Intensity
Page 5
Electromagnetic Fields
Coulomb's Law and Electric Field Intensity
Lecture No.4
6- Field of a Sheet Charge
Another basic charge configuration is the infinite sheet of charge having a uniform
density of ρS C/m2. Let us place a sheet of charge in y-z plane and again consider
symmetry as shown in Fig. 4.3. We see first that the field cannot vary with y or with z.
Hence only E x is present, and this component is a function of(x) alone. Let us use the
field of the infinite line charge by dividing infinite sheet into differential-width strips.
Once such strip is shown in Fig. 4.3, the line charge density, or charge per unit length, is
( ρL = ρS dy', and the distance from this line charge to our general point P on the x-axis is
R=√x2 + y'2.
Figure 4.3
The contribution to E x at point P from this differential-width strip is:
�⃗ = ρ L a�⃗R (line charge), then;
We have E
2πϵo R
ρ S 𝑑𝑑y′
dE x =
2πϵo √x2 + y′ 2
cosθ =
Adding the effects of all the strips,
∞
E x =∫−∞
Ex =
Ex =
ρS
2πϵo
ρS
2ϵo
ρ S x 𝑑𝑑y′
2πϵo (x2 + y ′ 2)
*π
=
ρS
ρ S 𝑑𝑑y′
2πϵo √x2 + y′ 2
2πϵo
∞
x 𝑑𝑑y′
.(
𝑥𝑥
√x2 + y′ 2
ρ
)=
ρ S x 𝑑𝑑y′
2πϵo (x2 + y ′ 2)
y′
S
[tan−1 ( )]∞
∫−∞ (x2 + y ′ 2) = 2πϵ
x −∞
o
If the point P were chosen on the negative x-axis, then,
Ex
Dr. Ahmed Thamer
Coulomb's Law and Electric Field Intensity
Page 6
Electromagnetic Fields
Coulomb's Law and Electric Field Intensity
Lecture No.4
Example 4.3: Three uniform sheets of charge are located in free space as follow:
�⃗ at the points:
2𝜇𝜇C/m2 at x=-3, -5 𝜇𝜇C/m2 at x=1, and 4 𝜇𝜇C/m2 at x=5. Determine E
(a) (0, 0, 0)
(b) (2.5, -1.6, 4.7)
(c) (8, -2, -5)
(d) (-3.1, 0, 3.1)
𝛒𝛒S1= 2𝝁𝝁C/m2
Z
-X
−𝐚𝐚�⃗𝐱𝐱
X= - 3.1
X= - 3
𝐚𝐚�⃗𝐱𝐱
𝛒𝛒S2= - 5𝝁𝝁𝐂𝐂/𝐦𝐦𝟐𝟐
𝛒𝛒S3= 𝟒𝟒𝟒𝟒𝐂𝐂/𝐦𝐦𝟐𝟐
−𝐚𝐚�⃗𝐱𝐱
X= 1
𝐚𝐚�⃗𝐱𝐱
−𝐚𝐚�⃗𝐱𝐱
Y
X= 0
X= 2.5
X= 5
𝐚𝐚�⃗𝐱𝐱
X= 8
X
Solution: (a) Since the position of plates in x-axis,
�⃗ depended on (x) in each point.
∴ we determine E
Point (0, 0, 0), and normal on plates is (±𝐚𝐚⃗𝐱𝐱 )
X
�⃗𝛒𝛒 S1 + E
�⃗𝛒𝛒 S2 + E
�⃗𝛒𝛒 S3 =
�⃗ T = E
E
R
R
R
R
−6
ρS 1
2ϵo
−5∗10 −6
𝐚𝐚�⃗𝐱𝐱
+
ρS 2
2ϵo
(- 𝐚𝐚⃗𝐱𝐱 ) +
−6
ρS 3
2ϵo
(- 𝐚𝐚⃗𝐱𝐱 )
4∗10
�⃗ T = 2∗10 −12 𝐚𝐚�⃗𝐱𝐱 +
(- 𝐚𝐚⃗𝐱𝐱 ) +
(- 𝐚𝐚⃗𝐱𝐱 )
E
−12
2∗8.85∗10
2∗8.85∗10
2∗8.85∗10 −12
�E⃗ T = 169.4 𝐚𝐚�⃗𝐱𝐱 (KV/m)
R
R
(b) Point (2.5, -1.6, 4.7)
X
−6
−6
�E⃗ T = 2∗10 −12 𝐚𝐚�⃗𝐱𝐱 + −5∗10 −12
2∗8.85∗10
2∗8.85∗10
�⃗ T = - 395 𝐚𝐚�⃗𝐱𝐱 (KV/m)
E
R
R
Dr. Ahmed Thamer
𝐚𝐚�⃗𝐱𝐱
+
4∗10 −6
2∗8.85∗10 −12
(- 𝐚𝐚⃗𝐱𝐱 )
Coulomb's Law and Electric Field Intensity
Page 7
Electromagnetic Fields
Coulomb's Law and Electric Field Intensity
Lecture No.4
(c) Point (8, -2, -5)
X
−6
−6
�E⃗ T = 2∗10 −12 𝐚𝐚�⃗𝐱𝐱 + −5∗10 −12
2∗8.85∗10
2∗8.85∗10
�E⃗ T = 56.5 𝐚𝐚�⃗𝐱𝐱 (KV/m)
R
R
𝐚𝐚�⃗𝐱𝐱
+
4∗10 −6
2∗8.85∗10 −12
𝐚𝐚�⃗𝐱𝐱
(d) Point (-3.1, 0, 3.1)
X
−6
−6
−6
�⃗ T = 2∗10 −12 (- 𝐚𝐚�⃗𝐱𝐱) + −5∗10 −12 (- 𝐚𝐚⃗𝐱𝐱 ) + 4∗10 −12 (- 𝐚𝐚⃗𝐱𝐱 )
E
2∗8.85∗10
2∗8.85∗10
2∗8.85∗10
�E⃗ T = - 56.5 𝐚𝐚�⃗𝐱𝐱 (KV/m)
R
R
Example 4.4: Two infinite uniform sheets of charge, each with charge density 𝛒𝛒 S , are located at
�⃗ everywhere.
x= ±1 as shown. Determine E
R
Solution: a) x<-1
�E⃗ = �E⃗ 1 + �E⃗ 2 =
R
−ρ S
�E⃗ =
ϵo
b) -1<x<1
R
𝐚𝐚
⃗𝐱𝐱
R
�E⃗ = �E⃗ 1 + �E⃗ 2 =
ρS
�E⃗ =
ϵo
R
𝐚𝐚
⃗𝐱𝐱
(- 𝐚𝐚⃗𝐱𝐱 ) +
ρS
2ϵo
Z
(- 𝐚𝐚⃗𝐱𝐱 )
𝛒𝛒S
−𝐚𝐚�⃗𝐱𝐱
-X
X= - 1
c) x>1
R
2ϵo
(V/m)
�⃗ 1 + E
�⃗ 2 =
�⃗ = E
E
R
ρS
ρS
2ϵo
ρS
2ϵo
(V/m)
(- 𝐚𝐚⃗𝐱𝐱 ) +
𝐚𝐚
⃗𝐱𝐱
+
ρS
2ϵo
ρS
2ϵo
𝐚𝐚�⃗𝐱𝐱
𝐚𝐚�⃗𝐱𝐱
𝛒𝛒S
=0
𝐚𝐚�⃗𝐱𝐱
−𝐚𝐚�⃗𝐱𝐱
𝐚𝐚�⃗𝐱𝐱
Y
X= 1
X
4- Streamlines and Sketches of Fields
�⃗ is represented by lines from the charge which are everywhere tangent
The direction of E
to �E⃗. These lines are usually called streamlines, although other terms such as flux lines
and direction lines are also used. In the case of the two-dimensional fields in Cartesian
coordinates, the equation of the streamline is obtained by solving the differential
equation as:
Ey
Ex
Dr. Ahmed Thamer
=
𝑑𝑑 y
𝑑𝑑 x
Coulomb's Law and Electric Field Intensity
Page 8
Electromagnetic Fields
�⃗ =
E
Let
x
x2 + y2
a�⃗x +
y
Coulomb's Law and Electric Field Intensity
x2 + y2
a�⃗y
Thus we form the differential equation,
Therefore, lny = lnx + C 1
or,
𝑑𝑑 y
𝑑𝑑 x
=
Ey
Ex
=
y
x
or,
lny = lnx + lnC
𝑑𝑑 y
y
Lecture No.4
𝑑𝑑 x
=
x
From which the equations of the streamlines are obtained, y = Cx
If we want to find the equation of one particular streamline, say that one passing through
P(-2, 7, 10). Here, 7 = C(-2), and C = - 3.5, so that:
y = - 3.5 x
Each streamline is associated with a specific value of C.
The equations of streamlines may also be obtained directly in cylindrical or spherical
coordinates as:
 Cylindrical coordinates:
 Spherical coordinates:
Eρ
EФ
Er
Eθ
Ey
 Cartesian coordinates:
Ex
Home Work:
=
=
=
𝑑𝑑 ρ
ρ𝑑𝑑 Ф
𝑑𝑑 r
r𝑑𝑑 θ
𝑑𝑑 y
𝑑𝑑 x
Q 4.1: Calculate the total charge within each of the indicated volumes:
1
(a) 0.1 ≤ |x| , |y| , |z| ≤ 0.2 ; ρ V = 3 3 3
R
x y z
(b) 0 ≤ ρ ≤ 0.1 , 0 ≤ Ф ≤ π , 2 ≤ 𝑧𝑧 ≤ 4 ; ρ V = ρ 2 z2 sin0.6Ф
(c) Universe; ρ V =
R
e −2r
R
P
r2
Q 4.2: Infinite uniform line charges of 5nC/m lie along the (positive and negative) x and y axes
in free space. Find �E⃗ at: (a) P A (0, 0, 4); (b) P B (0, 3, 4)
Q 4.3: Three infinite uniform sheets of charge are located in free space as follows: 3nC/m2 at
�⃗ at the points: (a) P A (2, 5, -5);
z = - 4, 6nC/m2 at z = 1, and -8nC/m2 at z = 4. Find E
(c) P C (-1, -5, 2);
(d) P D (-2, 4, 5)
(b) P B (4, 2, -3);
Q 4.4: Find the equation of that streamline that passes through the point P(1, 4, -2) in the field:
−8x
4x 2
a�⃗x + 2 �a⃗y
(b) �E⃗ = 2e5x[y(5x+1) a�⃗x + x a�⃗y ]
(a) �E⃗ =
y
y
Q 4.5: The region in which 4 ≤ r ≤ 5, 0 ≤ θ ≤ 25o, and 0.9π ≤ Ф ≤ 1.1π, contains the volume
1
charge density ρ V = 10(r-4)(r-5) sinθsin Ф. Outside that region ρ V = 0. Find the charge
2
within the region.
R
Dr. Ahmed Thamer
R
Coulomb's Law and Electric Field Intensity
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