chapter 2 coulomb’s law and electric field intensity

President University Erwin Sitompul EEM 2/1

Lecture 2

Engineering Electromagnetics

Dr.-Ing. Erwin SitompulPresident University

http://zitompul.wordpress.com

2 0 1 3


Chapter 2Coulomb’s Law and Electric Field Intensity

Engineering Electromagnetics


Chapter 2 Coulomb’s Law and Electric Field Intensity

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.

The Experimental Law of Coulomb


The Experimental Law of CoulombChapter 2 Coulomb’s Law and Electric Field Intensity

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.

1 22

QQF k

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:

0

1

4k


The Experimental Law of CoulombThe permittivity of free space ε is measured in farads per meter

(F/m), and has the magnitude of: 12 9

0

18.854 10 10 F m

36

The Coulomb’s law is now:1 2

20

1

4

QQF

R

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.



The Experimental Law of Coulomb

1 22 122

0 12

1

4

QQ

RF a

In vector form, Coulomb’s law is written as:


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.

1212

12

R

aR

12

12RR 2 1

2 1

r r

r r


1 ?F =1 2

1 2 2120 12

1

4

QQ

R F = F a

The Experimental Law of CoulombChapter 2 Coulomb’s Law and Electric Field Intensity

ExampleA charge Q1 = 310–4 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.

12 2 1 R r r

(2 0 5 ) (1 2 3 )x y z x y z a a a a a a

1 2 2x y z a a a

1212

12

R

aR

1(1 2 2 )

3 x y z a a a

1 22 122

0 12

1

4

QQ

RF a

4 4

12 2

1 (3 10 )( 10 ) 1(1 2 2 )

4 (8.854 10 ) 3 3 x y z

a a a

10 20 20 Nx y z a a a


Electric Field IntensityLet 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.


112

0 1

1

4t

t tt

QQ

RF a

The force on it is given by Coulomb’s law as:

Writing this force as a “force per unit charge” gives:

112

0 1

1

4t

tt t

Q

Q R

Fa Vector Field,

Electric Field Intensity


Electric Field IntensityChapter 2 Coulomb’s Law and 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).

112

0 1

1

4t

tt t

Q

Q R

FE = a

The field of a single point charge can be written as:

20

1

4 R

Q

RE = a

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.



For a charge which is not at the origin of the coordinate, the electric field intensity is:

20

1( )

4

Q

r rE r =

r rr r

30

1 ( )

4

Q

r r=

r r

3 22 2 20

( ) ( ) ( )1

4 ( ) ( ) ( )

x y zQ x x y y z z

x x y y z z

a a a=



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).

120 1

220 2

1( )

4

1

4

Q

Q

E r = ar r

ar r



ExampleA 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).

1 3 4 2 ,x y z r r a a a

2 4 6 ,x y z r r a a a 2 53 r r

1 29 r r

1 21 22 2

0 01 2

1 1( )

4 4

Q Q

E r = a a

r r r r

1 1 2 23 3

0 1 2

( ) ( )1

4

Q Q

r r r r=

r r r r

6 6

3 30

2 10 (3 4 2 ) 3 10 (4 6 )1

4 29 53

x y z x y z

a a a a a a=

623.7 879.92 160.17 V mx y z a a a



Field Due to a Continuous Volume Charge DistributionWe 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

vQ v

We may define ρv mathematically by using a limit on the above equation:

0limvv

Q

v

The total charge within some finite volume is obtained by integrating throughout that volume:

vol

vQ dv



Field Due to a Continuous Volume Charge DistributionExample

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.

cosx

siny 24 sin cosv z

vol

vQ dv23 2

2

0 0 0

(4 sin cos )( )z

z d d dz

23 2

3 2

0 0 0

4 sin cosz d d dz

23

2

0 0

16 sin cosz d dz

sin 2 2sin cos

32

0

8z dz 72 C



Field Due to a Continuous Volume Charge DistributionThe incremental contribution to the electric field intensity at r

produced by an incremental charge ΔQ at r’ is:

2

0

( )4

Q

r rE r

r rr r

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:

20vol

( )1( )

4v dv

r r rE r

r rr r

2

04v v

r r

r rr r


Field of a Line ChargeChapter 2 Coulomb’s Law and Electric Field Intensity

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.

E z zd d dE E a a



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.

E z zd d dE E a a



Take P(0,y,0),

E z zd d dE E a a

30

1 ( )

4

dQd

r rE

r r

2 2 3 20

( )1

4 ( )L zdz z

z

a a

zz r a

yy r a a

2 2 3 20

1

4 ( )L dz

z

a

2 2 3 20

1

4 ( )L dzEz

2 2 2 1 204 ( )

L z

z

02LE



Now let us analyze the answer itself:

02L

E 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.



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).

PA

PB

0 0

( )2 2

x

x y

L LyA

x y

P

E a a

9 9

0 0

5 10 5 10

2 (4) 2 (4)z z

a a

44.939 V mz a

0 0

( )2 2

x

x y

L LyB

x y

P

E a a

9 9

0 0

5 10 5 10(0.6 0.8 )

2 (5) 2 (4)y z z

a a a

10.785 36.850 V my z a a

• ρ is the shortest distance between an observation point and the line charge


Field of a Sheet of ChargeChapter 2 Coulomb’s Law and Electric Field Intensity

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 +∞.



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 theleft of z axis will cancel each other.

Only dEx is present, and this component is a function of x alone.



The contribution of a strip to Ex at P is given by:

2 20

cos2

sx

dydE

x y

2 202s xdy

x y

Adding the effects of all the strips,

2 202s

x

xdyE

x y

1

0

tan2

s y

x

02s



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.

02s

N

E a

The field of a sheet of charge is constant in magnitude and direction. It is not a function of distance.


Chapter 1 Vector Analysis

Homework 2D2.2. D2.4. D2.6. All homework problems from Hayt and Buck, 7th Edition.

Due: Monday, 22 April 2013.

chapter 2 coulomb’s law and electric field intensity

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