electromagnetics

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

2013 - 2014

Electromagnetic Fields

Coulomb's Law and Electric Field Intensity

University of Missan College of Engineering Electrical Engineering Dept.

Lecture No.4

1 st Semester:2013-2014

Lesson Year Stage Subject Lecture No. Lecturer

2 nd Year Charge Density 4 Dr. Ahmed Thamer

3- Various Charge Distributions + + + ++

+ + + + + + + + +

L

Q

+ + +

++++++++

point Charge

Line Charge

C

+++

Surface Char e



Line charge density (



Surface charge density (



Volume charge density (

m

S

+ + + + +

V

Volume Charge

Q = ∫ ρ L d L

)

R

C m2 C m3

) )

Q = ∬ ρ S d S R

Q = ∭ ρ V d V R

Notes that: •

•

•

ρL = QL where (L) is any given line length or circumference ρS = QA where (A) is any given area such as area of circle or sphere ρV = QV where (V) is any given volume such as volume of sphere

Since, E = K

Q R2

a  R R

So by replacing Q in above equations with line charge density, surface charge density, and volume charge density, we get:   

E = ∫ K ρ L2 L a R R K ρ sds E = ∬ 2 a R R K ρ vdv  = ∭ 2 a R E R

R

R

R

4- Field Due to Continuous Volume Charge Distribution

Volume charge density is measured in Coulomb per cubic meter (

C m3

). The total charge

within some finite volume is obtained by integrating throughout that volume as: Q = ∫Vol Q = ∫Vol

ρV d v Coulomb's Law and Electric Field Intensity

Page 1



Lecture No.4

Example 4.1: Find the total charge inside each of the volumes indicated as: -0.1x (a) ρ V = 10 ze siny; -1≤ x ≤2; 0≤ y ≤1; 3≤ z ≤3.6 (b) ρ V = 4xyz; 0≤ ρ ≤2; 0≤ Ф ≤  /2; 0 ≤ z ≤3 (c) ρ V = 3  sinθ cos2Ф/[2r 2(r 2+1)]; universe R

R

R

Solution: (a) Cartesian coordinate:

Q = ∭ ρV d xd yd z Q = ∭ 10 ze−0.1x siny d xd yd z 3.6

1

2

Q = 10 [ ∫z=3 z z ∫y=0 siny y ∫x= −1 e−0.1x x ] Q = 10

z 2 3.6 [ ]3 2

− y 1 e −0.1x 2 [ ]0 [ ] = 36.1 C  −0.1 −1

(b) Cylindrical coordinate: Q = ∭ ρV ρ d ∅d z 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 d ∅d z = ∭ 4 z ρ3 cosФ sinФ d ∅d z R

R

2

π

3  ∫Ф2=0 cosФ sinФ ∅ ∫z=0 z z]

Q = 4[∫ρ =0 ρ Q=4[

ρ4 4

3

]20 [

sin 2 Ф 2

π

z 2

]02 [ ]30 = 36 C 2

(c) Spherical coordinate: Q = ∭ 3sinθ cos 2 Ф/[2r 2 (r 2 + 1)] r sinθrθd Ф ∞ 1  2 3π Q = [∫r=0 2 r ∫θ =0 sin2 θ θ ∫Ф=0 cos 2 Ф Ф] 2 r +1 3π 1 sin 2θ π 1 sin 2 θ 2 π Q= * [θ − [ tan−1 (r)]∞ ]0 * [θ + ]0 0 2 2 2 2 2 3π π 1 1 Q = * * * π * * 2π 2 2 2 2 3( π )4 = 36.5 C Q = 2

8

Hint:  



1

x

∫ x 2 + a2 = a tan−1 (a) + c  −1 ∫ (x 2 + a2 )3 = (x 2 + a2 )1/2 + c 





∫ (x 2 + a2 )3/2



sin 2x = (1-cos2x)

=

a 2 (x 2 + a 2 )1/2

+c

1

P

2

2 1



cos x = (1+cos2x)



sin2x = 2 sinx cosx

2


Page 2



Lecture No.4

5- Field of a Line Charge Let us assume a straight line charge extending a long z-axis in cylindrical coordinate system from - ∞ to ∞ as shown in Fig. 4.1. We desire the electric field intensity E at any 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:

E =

Q

 R a R

4πϵ o R 2

ρ

d Q = L dL =



d E = k

Q

ρL

dz'

 R = k a

ρ L ′

R

R2

 R a R

R2

 y = ρa ρ r⃗= ya z r′⃗ = z' a R

R


Page 3



Lecture No.4

R=r⃗- r′⃗ = ρa ρ - z' a z �R � = √ ρ 2 + z'2 ρ a − z′ a z  R = √ ρ ρ2 + z′ 2 a R

P

R

ρ L ′ d E = k

 ρ − z ′ a z) (ρ a

( ρ 2 + z′ 2)

3/2

Due to symmetry, d E z =0, then, d E = k

ρ L  ′ ρ a ρ

R

( ρ 2 + z′ 2)

3/2

Since only the Eρ component is present, we may simplify: d E = k

ρ L ρ ′

R

( ρ 2 + z′ 2)

3/2

−∞ ≤ L ≤ ∞);

For infinite line charge (i.e

∞ ρL ρ ′ ρL ρ ′ ∞ [ ] = ∫ −∞ −∞ 3/2 4πϵ o 4πϵ o ( ρ 2 + z′ 2) ρ 2( ρ 2 + z′ 2)1/2 ρ ρ E = L * 2 4πϵo ρ2 ρL E = 2πϵ o ρ ρL Or, E = a 2πϵ o ρ ρ  . We might have used the angle θ as our There are many other ways of obtaining E variable of integration, from Fig.4.1 z'= ρ cot θ and d z' = - ρ csc2 θ d θ . Since R= ρ c scθ , our integral becomes, simply,

E = R

R

R

ρ L ′ ρ L sin θ θ sin = θ 4πϵo ρ 4πϵo R 2 0 ρ ρL 0 E = - L = θ θ θ sin [cos ] ∫ π π 4 πϵ o ρ 4πϵ o ρ ρ E = L 2πϵ o ρ ρL Or, E = a 2πϵ o ρ ρ As an example, let us consider an infinite line charge parallel to the z-axis at x=6, y=8,



d E = R

R

R

Fig.4.2. We wish to find E at the general field point P(x, y, z). We replace ρ in above equation 2 2 by the radial distance between the line charge and point P, R=√(x -6) + (y-8) , and let a  ρ be a R . Thus,

E =

ρL 2πϵ o √ (x −6)2 + (y −8)2

R a

(x −6)a  x + (−8)a y

Where a  R=

E =

√ (x −6)2 + (y −8)2

(x −6)a  x + (−8)a y ρL ( ) 2πϵ o (x −6)2 + (y −8)2

Figure 4.2 Coulomb's Law and Electric Field Intensity

Page 4



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 o (c) Q (ρ =4, Ф=60 , z=2) Solution: ρL (a) ρ L in the direction of y, by replace ρ and aρ in E = aρ by R and a R , 2πϵ o ρ respectively, then: R

R

R

Z

E = ρ L a R 2πϵ o R  =3a x − 4a z R �R � =√ 32 + (-4)2=√ 25=5  R =3a x −5 4a z a z 25 ∗10 −9 3a x −4a E = ( ) 2∗3.14∗8.854 ∗10 −12 ∗5 5 E= 53.9a x − 71.9a z (V/m)

ρL

(-3, 0, 4) -X

 Y

(0, 0, 0)

X

Z

(b) P(2, 15, 3)

ρL

R=5a x − a z �R � =√ 52 + (-1)2=√ 26  R =5a√ x2−6a z a z 25 ∗10 −9 5a x −a E = ( ) √ 26 2∗3.14∗8.854 ∗10 −12 ∗√ 26 E= 86.4a x − 17.3a z (V/m)

(-3,15, 4)

(2,15, 3) 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

R=5a x − 2az �R � =√ 52 + (-2)2=√ 29 z x −2a a  R =5a √ 29 25 ∗10 −9  = E −12 2∗3.14∗8.854 ∗10

∗√ 29

(

5a x −2a z

√ 29

) = 77.5a  x − 31a z (V/m)


Page 5



Lecture No.4

6- Field of a Sheet Charge Another basic charge configuration is the infinite sheet of charge having a uniform 2 density of ρS C/m . 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 d y', and the distance from this line charge to our general point P on the x-axis is 2 2 R=√ x + y' .

Figure 4.3

The contribution to E x at point P from this differential-width strip is: ρL We have E =  (line charge), then; a 2πϵ o R R ρ S  y′ ρ S  y′  ρ S x  y′ cosθ = .( ) = d E x = 2πϵ o √ x2 + y′ 2 2πϵo √ x2 + y′ 2 √ x2 + y′ 2 2πϵo (x2 + y ′ 2) Adding the effects of all the strips, ∞ ∞ x  y′ ρ S x  y′ ρS ρS −1 (y′ )]∞ E x =∫−∞ = = [tan ∫ 2πϵ o (x2 + y ′ 2) 2πϵo −∞ (x2 + y ′ 2) 2πϵ o x −∞ ρ E x = S * π 2πϵ o ρ Ex = S 2 ϵo If the point P were chosen on the negative x-axis, then,

Ex


Page 6



Lecture No.4

Example 4.3: Three uniform sheets of charge are located in free space as follow:

2C/m at x=-3, -5 C/m at x=1, and 4 C/m at x=5. Determine �E at the points: (a) (0, 0, 0) (b) (2.5, -1.6, 4.7) (c) (8, -2, -5) (d) (-3.1, 0, 3.1) 2

2

2

S1= 2C/m2

Z

-X

−�

X= - 3.1 X= - 3

�

S2= - 5/

−�

Y X= 0 X= 1

S3= /

�

−� X= 2.5 X= 5

�

X= 8

X

Solution: (a) Since the position of plates in x-axis,

∴ we determine �E depended on (x) in each point.

Point (0, 0, 0), and normal on plates is ( ±⃗ ) X

�E T = �E S1 + �E S2 + �E S3 = R

R

�E T = R

R

R

2∗10−6 2∗8.85 ∗10 −12

�E T = 169.4 � R

� +

ρS 1

2ϵo −5∗10 −6

� +

2∗8.85 ∗10 −12

ρS 2

(- ⃗ ) +

2ϵo

(- ⃗ ) +

4 ∗10 −6

ρS 3 2ϵo

2∗8.85 ∗10 −12

(- ⃗ ) (- ⃗ )

(KV/m)

(b) Point (2.5, -1.6, 4.7) X

�T= E R

2∗10 −6 2∗8.85 ∗10 −12

�E T = - 395 � R

� +

−5∗10−6 2∗8.85 ∗10 −12

� +

4∗10 −6 2∗8.85 ∗10 −12

(- ⃗ )

(KV/m)


Page 7



Lecture No.4

(c) Point (8, -2, -5) X

2∗10 −6

� T = E

2∗8.85 ∗10 −12

R

�E T = 56.5 � R

� +

−5∗10−6 2∗8.85 ∗10 −12

� +

4∗10 −6 2∗8.85 ∗10 −12

�

(KV/m)

(d) Point (-3.1, 0, 3.1) X

2∗10 −6

�E T =

2∗8.85 ∗10 −12

R

�E T = - 56.5 � R

(- �) +

−5∗10−6 2∗8.85∗10 −12

(- ⃗ ) +

4∗10 −6 2∗8.85 ∗10 −12

(- ⃗ )

(KV/m)

Example 4.4: Two infinite uniform sheets of charge, each with charge density  S , are located at R

x= ±1 as shown. Determine �E everywhere. Solution: a) x<-1

� =E � 1+E � 2 = E R

�E =

−ρ S ϵo

R

ρS 2ϵo

(- ⃗ ) +

ρS 2ϵ o

Z

(- ⃗ )

S

-X

 ⃗ (V/m)

−�

b) -1
X= - 1

� =E � 1 + E � 2 = E R

R

ρS 2ϵ o

(- ⃗ ) +

ρS 2ϵ o

� =

0

S

�

−�

Y

c) x>1 X= 1

�E = �E 1 + �E 2 = R

ρS

� = E

ϵo

R

ρS 2 ϵo

 ⃗ +

ρS 2ϵo

� �

 ⃗ (V/m)

X

4- Streamlines and Sketches of Fields � is represented by lines from the charge which are everywhere tangent The direction of E � . These lines are usually called streamlines, although other terms such as flux lines to E 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

=

y x


Page 8


�E =

Let

Coulomb's Law and Electric Field Intensity x

x2

+ y2

� x + a

y x2

+ y2

�y a

Thus we form the differential equation, Therefore, lny = lnx + C 1

Lecture No.4

or,

y x

Ey

=

Ex

y

= x

or,

y y

x

=

x

lny = lnx + lnC

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:



Cartesian coordinates:

Eρ EФ Er Eθ Ey Ex

=

=

=

ρ ρ Ф r rθ y

x

Home Work: Q 4.1: Calculate the total charge within each of the indicated volumes:

(a) 0.1 ≤ |x| , |y| , |z| ≤ 0.2 ; ρ V = R

1 x3

y3 z3 2

2

(b) 0 ≤ ρ ≤ 0.1 , 0 ≤ Ф ≤ π , 2 ≤  ≤ 4 ; ρ V = ρ z sin0.6Ф R

(c) Universe; ρ V = R

P

e −2r r2

Q 4.2: Infinite uniform line charges of 5nC/m lie along the (positive and negative) x and y axes

� at: (a) P A (0, 0, 4); in free space. Find E

(b) P B (0, 3, 4) 2

Q 4.3: Three infinite uniform sheets of charge are located in free space as follows: 3nC/m at z = - 4, 6nC/m2 at z = 1, and -8nC/m2 at z = 4. Find �E at the points: (a) P A (2, 5, -5); (b) P B (4, 2, -3); (c) P C (-1, -5, 2); (d) P D (-2, 4, 5) Q 4.4: Find the equation of that streamline that passes through the point P(1, 4, -2) in the field: 2

−8x 4x � x + 2 a� y (a) �E = a y y

� x + x a� y ] (b) �E = 2e [y(5x+1) a 5x

o

Q 4.5: The region in which 4 ≤ r ≤ 5, 0 ≤ θ ≤ 25 , 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 R

2

R

within the region. Coulomb's Law and Electric Field Intensity

Page 9

electromagnetics

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