8/10/2019 2 a ZH EM I Coulombs Law E Field Intensity
1/23
Dr. Zuhair M. Hejaz Set 2 a Coulombs Law & E-Field Intensity 1
Coulombs Law and
E-Filed Intensity
Set 2 a
8/10/2019 2 a ZH EM I Coulombs Law E Field Intensity
2/23
Coulomb's Law
Dr. Zuhair M. Hejaz Set 2 a Coulombs Law & E-Field Intensity 2
8/10/2019 2 a ZH EM I Coulombs Law E Field Intensity
3/23
Coulomb's Law Coulomb's law: The force between two
electrical charges and , separated in
vacuum or free space is:
1. Along the line joining them.
2. Directly proportional to the product of the
charges.
3. Inversely proportional to the square of the
distance between them.
Mathematically
Dr. Zuhair M. Hejaz Set 2 a Coulombs Law & E-Field Intensity 3
8/10/2019 2 a ZH EM I Coulombs Law E Field Intensity
4/23
Coulomb's Law
is proportionality constant
is The constant known as thepermittivity
of free space and has the value
In SI units, charges and are measuredin Coulombs (C), the distance is in meters
(m), and the force is in Newtons (N).
Dr. Zuhair M. Hejaz Set 2 a Coulombs Law & E-Field Intensity 4
8/10/2019 2 a ZH EM I Coulombs Law E Field Intensity
5/23
Coulomb's Law Note that 1 C is approximately equivalent to
electrons. It is a very large unit of
charge because one electron charge is:
Coulomb's law is now
If point charges and are located at
points having position vectors and as
shown in the fig. to follow Dr. Zuhair M. Hejaz Set 2 a Coulombs Law & E-Field Intensity 5
8/10/2019 2 a ZH EM I Coulombs Law E Field Intensity
6/23
Coulomb's Law
then the force on caused by
is given by: Vector form
where ; and Dr. Zuhair M. Hejaz Set 2 a Coulombs Law & E-Field Intensity 6
F1
8/10/2019 2 a ZH EM I Coulombs Law E Field Intensity
7/23
Coulomb's Law
Example 1: A charge is located atpoint and a charge is
located at point in a vacuum, find the
force exerted on . Solution:
Dr. Zuhair M. Hejaz Set 2 a Coulombs Law & E-Field Intensity 7
8/10/2019 2 a ZH EM I Coulombs Law E Field Intensity
8/23
Coulomb's Law- Example 1
check!
The unit vector is then:
Thus, the force is found using:
Dr. Zuhair M. Hejaz Set 2 a Coulombs Law & E-Field Intensity 8
8/10/2019 2 a ZH EM I Coulombs Law E Field Intensity
9/23
8/10/2019 2 a ZH EM I Coulombs Law E Field Intensity
10/23
Electric FieldIntensity
Dr. Zuhair M. Hejaz Set 2 a Coulombs Law & E-Field Intensity 10
8/10/2019 2 a ZH EM I Coulombs Law E Field Intensity
11/23
Electric Field Intensity The Electric Field Intensity (or electric field
strength) is the vector force on a unitpositive test charge when placed in the
electric field.
The force exerted on the test charge isexpressed as:
As a force per unit test charge, the eqn.becomes:
Dr. Zuhair M. Hejaz Set 2 a Coulombs Law & E-Field Intensity 11
8/10/2019 2 a ZH EM I Coulombs Law E Field Intensity
12/23
Electric Field Intensity The term on the right and the directed line
from to the position of test charge
describe a vector force called Electric Field
Intensity and expressed as:
so,
and measured by Volts per meter (V/m) .
Note that: the last eqn. is the Electric Field
Intensity caused by a single point charge
Dr. Zuhair M. Hejaz Set 2 a Coulombs Law & E-Field Intensity 12
8/10/2019 2 a ZH EM I Coulombs Law E Field Intensity
13/23
Electric Field Intensity For a point charge located at the origin of a
Cartesian coordinates:
We have:
;
and
So,
This shows the field complexity in 3D coordinates.
Dr. Zuhair M. Hejaz Set 2 a Coulombs Law & E-Field Intensity 13
8/10/2019 2 a ZH EM I Coulombs Law E Field Intensity
14/23
Electric Field Intensity
Now, consider a point charge, not located at the
origin of our system, but at , so
We find the electric field
intensity at point due to
the point charge, where
Expressing , we finally find the field
Dr. Zuhair M. Hejaz Set 2 a Coulombs Law & E-Field Intensity 14
8/10/2019 2 a ZH EM I Coulombs Law E Field Intensity
15/23
Electric Field Intensity
Now for 2 point charges
the electric field intensity on test charge is
obtained as the sum:
Recall that Coulomb forces are linear.Dr. Zuhair M. Hejaz Set 2 a Coulombs Law & E-Field Intensity 15
8/10/2019 2 a ZH EM I Coulombs Law E Field Intensity
16/23
8/10/2019 2 a ZH EM I Coulombs Law E Field Intensity
17/23
Electric Field Intensity-Example 2
Example 2:
Find the
Electric
FieldIntensity
at point
P(1,1,1) caused by the 4 identical charges locatedat P1(1,1,0), P2(-1,1,0), P3(-1,-1,0) and P4(1,-1,0)
(each has 3 nC charge). Solution
Dr. Zuhair M. Hejaz Set 2 a Coulombs Law & E-Field Intensity 17
8/10/2019 2 a ZH EM I Coulombs Law E Field Intensity
18/23
Electric Field Intensity-Example 2
Solution:
r at P(1,1,1): and
Then and the magnitudes are:
Using the expression for the field , we find
As , we find the term:
Dr. Zuhair M. Hejaz Set 2 a Coulombs Law & E-Field Intensity 18
8/10/2019 2 a ZH EM I Coulombs Law E Field Intensity
19/23
Electric Field Intensity-Example 2 Recall the unit vectors:
;
Now substitute the other terms in:
Simplify to
Dr. Zuhair M. Hejaz Set 2 a Coulombs Law & E-Field Intensity 19
8/10/2019 2 a ZH EM I Coulombs Law E Field Intensity
20/23
Electric Field Intensity Note that if the field intensity is required NOT
just at a point without charge, BUT on anotherpoint charge having a charge, then the eqn.:
Becomes:
Dr. Zuhair M. Hejaz Set 2 a Coulombs Law & E-Field Intensity 20
8/10/2019 2 a ZH EM I Coulombs Law E Field Intensity
21/23
Practice Exercise
Problem:
Point charges with (1 mC) and (-2 mC) are
located at P1(3, 2, -1 ) and P2(-1, -1, 4),
respectively. Calculate the electric force on a
(10 nC) charge located at P(0, 3, 1) and theelectric field intensity at that point.
Answers:
At the point P(0, 3, 1)
Dr. Zuhair M. Hejaz Set 2 a Coulombs Law & E-Field Intensity 21
8/10/2019 2 a ZH EM I Coulombs Law E Field Intensity
22/23
You have learned about
Coulomb's law for the force between two
static point charges.
The vector force of electric field intensity
caused by 1, 2 and n point charges at a point.
The vector force of electric field intensitycaused by n point charges on another point
charge.
Dr. Zuhair M. Hejaz Set 2 a Coulombs Law & E-Field Intensity 22
8/10/2019 2 a ZH EM I Coulombs Law E Field Intensity
23/23
End of Set 2 a
Thank You for YourAttention
Dr. Zuhair M. Hejaz Set 2 a Coulombs Law & E-Field Intensity 23