Top Banner

of 23

2 a ZH EM I Coulombs Law E Field Intensity

Jun 02, 2018

Download

Documents

Ahmad Elhaj
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
  • 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