Engineering Electromagnetics; William Hayt & John...

Dr. Naser Abu-Zaid; Lecture notes on Electromagnetic Theory(1); Ref:Engineering

Electromagnetics; William Hayt& John Buck, 7th & 8th editions; 2012

Preliminary material

(mathematical

requirements)

Vector:A quantity with both magnitude and direction. (Force N10F to the

east).

Scalar:A quantity that does not posses direction, Real or complex. (Temperature oT 20 .

Vector addition:

1) Parallelogram:

2) Head to Tail:

A

B

A

B

BA

A

B

A

B

BA

Vector Analysis

Vector algebra:

Addition; Subtraction;

Multiplication

Vector Calculus:

Differentiation; Integration



Vector Subtraction:

Multiplication by scalar: AB k

AB 2

AB 50.

AB 3

Commulative law: ABBA

Associative law: CBACBA

Equal vectors: BA if 0BA (Both have same length and direction) Add or subtract vector fields which are defined at the same point. If non vector fields are considered then vectors are added or subtracted

which are not defined at same point (By shifting them)

A

A3

A50.

A

A A2

A

B

A

B

B

A

B

B

BA



THE RECTANGULAR COORDINATE SYSTEM

z,y,x are coordinate

variables (axis) which are

mutually perpendicular.

A point is located by its y,x

and z coordinates, or as the

intersection of three constant

surfaces (planes in this case)

x

y

z

321 ,,P

1x

2y

3z

x

y

z

Right Handed System

Out of page



x

y

z

321 ,,P

1x

Surface

(plane)

2y

surface

(plane)

3z

surface

(plane)

Three mutually perpendicular surfaces intersect at a common point



Increasing each coordinate variable by a differential amount dx , dy ,

and dz , one obtains a parallelepiped.

Differential volume: dxdydzdv

x

y

z

z,y,xP

dz

dx dy

dzz,dyy,dxx'P



Differential Surfaces: Six planes with dierential areas dxdyds ; dzdyds ;

dxdzds

Differential length: from P to P’ 222dzdydxdl

VECTOR COMPONENTS AND UNIT VECTORS

A general vector r may be written as the sum of three vectors;

,,BA and C arevector components with constant directions.

Unit vectors xa , ya , and za directed along x, y, and z respectively with unity

length and no dimensions.

x

y

z

321 ,,P

1x

2y

3z

r

A B

C

Projection of

r into x-axis

Projection of

r into z-axis

Projection of

r into y-axis

CBAr



So, the vector CBAr may be written in terms of unit vectors as:

componentsscalarcomponentsvector

zyx

CBA

CBA

,,,,

ˆˆˆ

CBA

aaaCBAr

Where:

A is the directed length or signed magnitude of A .

B is the directed length or signed magnitude of B .

C is the directed length or signed magnitude of C .

As a simple exercise, let pr (Position vector) point from origin (0,0,0) to P(1,2,3),

then

zyxP aaar ˆ3ˆ2ˆ1

Scalar components of Pr are:

1 ArPx ; 2 BrPy ; 3CrPz .

Vector components of Pr are:

xPx aAr ˆ1 ; yPy aBr ˆ2 ; zPz aCr ˆ3 .

x

y

z

xa

xa

xa

xa

ya

ya

ya

ya za

za

za



If Q(2,-2,1) then

zyxQ aaar ˆˆ2ˆ2

And the vector directed from P to Q, PQPQ rrr (displacement vector)

which is given by

zyxzyxPQ ˆˆˆˆˆˆ aaaaaar 24312212

The vector Pr is termed position vector which is directed from the origin toward

the point in quesion.

Other types of vectors (vector fields such as Force vector) are denoted:

x

y

z

Qr

Pr

POr

Pr



zzyyxx ˆFˆFˆF aaaF

Where ,F,F yx zF are scalar components, and zzyyxx ˆF,ˆF,ˆF aaa are the

vector components.

The magnitude of a vector zzyyxx ˆBˆBˆB aaaB is;

222

zyx BBBB B

A unit vector in the direction of B is;

222

zyx

zzyyxxB

BBB

ˆBˆBˆBˆ

aaa

B

Ba

Let zzyyxx ˆBˆBˆB aaaB and zzyyxx ÂÂÂ aaaA , then

zzzyyyxxx ˆBAˆBAˆBA aaaBA

zzzyyyxxx ˆBAˆBAˆBA aaaBA

Ex:Specify the unit vector extending from the origin toward the point G(2,-2,-1).

Ex: Given M(-1,2,1), N(3,-3,0) and P(-2,-3,-4) Find:

(a) MNR

(b) MPMN RR

(c) Mr

(d) MPa

(e) NP rr 32

B

Ba



THE VECTOR FIELD AND SCALAR FIELD

Vector Field: vector function of a position vector r . It has a magnitude and

direction at each point in space.

zzyyxx

zzyyxx

ˆz,y,xvˆz,y,xvˆz,y,xv

ˆvˆvˆv

aaa

arararrv

Scalar field: A scalar function of a position vector r . Temperature is an

example z,y,xTT r which has a scalar value at each point in space.

Velocity or air flow in a pipe



Ex:A vector field is expressed as

zyx ˆzˆyˆx

zyxaaaS 121

121

125222

(a) Is this a scalar or vector field?

(b) Evaluate S @ 342 ,,P .

(c) Determine a unit vector that gives the direction of S @ 342 ,,P .

(d) Specify the surface z,y,xf on which 1S .

1T

2T

3T



THE DOT PRODUCT

ABcos BABA

Which results in a scalar value, and AB is the smaller angle between A and B .

ABBA since ABAB coscos ABBA

11110cosˆˆˆˆ xxxx aaaa

11110cosˆˆˆˆ yyyy aaaa

11110cosˆˆˆˆ zzzz aaaa

xy

o

yxyx aaaaaa ˆˆ001190cosˆˆˆˆ

xz

o

zxzx aaaaaa ˆˆ001190cosˆˆˆˆ

yz

o

zyzy aaaaaa ˆˆ001190cosˆˆˆˆ

A

B

Projection of A into B

ABcos A

Projection of B into A

ABcos B

AB



Let zzyyxx BBB aaaB ˆˆˆ and zzyyxx AAA aaaA ˆˆˆ , then

zzyyxx BABABA BA

AAAAA AAAAA zyx22222

The scalar component of B in the direction of an arbitrary unit vector a is given

by aB ˆ

The vector component of B in the direction of an arbitrary unit vector a is given

by aaB ˆˆ .

B

Scalar Projection of B into a

cosB aB ˆ

cos

cosˆˆ

B

aBaB

a



Distributive property: CABACBA

Ex: Given zyx ˆˆx.ˆy aaaE 352 and Q(4,5,2) Find:

(a) E @ Q.

(b) The scalar component of E @ Q in the direction of

zyxnˆˆˆˆ aaaa 22

3

1 .

(c) The vector component of E @ Q in the direction of

zyxnˆˆˆˆ aaaa 22

3

1 .

(d) The angle Ea between QrE and na .

B

Vector Projection of B into a

aaB ˆˆ

a



THE CROSS PRODUCT

nAB aBABA ˆsin results in a vector

ABsinBABA

nofDirection aBA ˆ

na is a unit vector normal to the plane containing A and B . Since there are two

possible s'ˆ na , we use the Right Hand Rule (RHR) to determine the direction of

BA .

Cross product clearly results in a vector, and AB is the smaller angle between A

and B .

Properties:

ABBA

CABACBA

A

B

ABsin B

which is the height

Of the parallelogram

AB

BA

ABBA

ABsin BA

Is the area of the

parallelogram

na



CBACBA

00 nxxxx ˆsinˆˆˆˆ aaaaa

00 nyyyy ˆsinˆˆˆˆ aaaaa

00 nzzzz ˆsinˆˆˆˆ aaaaa

znno

yxyx ˆˆˆsinˆˆˆˆ aaaaaaa 11190

ynno

zxzx ˆˆˆsinˆˆˆˆ aaaaaaa 11190

znno

xyxy ˆˆˆsinˆˆˆˆ aaaaaaa 11190

xnno

zyzy ˆˆˆsinˆˆˆˆ aaaaaaa 11190

ynno

xzxz ˆˆˆsinˆˆˆˆ aaaaaaa 11190

xnno

yzyz ˆˆˆsinˆˆˆˆ aaaaaaa 11190

Let zzyyxx ˆBˆBˆB aaaB and zzyyxx ÂÂÂ aaaA , then

zxyyxyxzzxxyzzy

zyx

zyx

zyx

nAB

ˆBABAˆBABAˆBABA

BBB

AAA

ˆˆˆ

ˆsin

aaa

aaa

aBABA

x

y

z

RHR

Out of page



CIRCULAR CYLINDRICAL COORDINATES

z,, are coordinate variables

which are mutually perpendicular.

Remember polar coordinates (The

2D version)

cosx

siny

is measured from x-axis toward y-

axis.

Including the z-coordinate, we obtain the cylindrical coordinates (3D version)

A point is

located by its

, and z

coordinates. Or

as the

intersection of

three mutually

orthogonal

surfaces.

x

y

z

1x

2y

3z )z,.,(P

or)z,y,x(Po 34635

321

x

y

z x

y ),(P

or)y,x(P



1) Infinitely long cylinder of radius 1 .

2) Semi-infinite plane of constant angle 1 .

3) Infinite plane of constant elevation 1zz .

x

y

z

111 z,,P

1

Surface

(plane) 1

surface

(cylinder)

1zz

surface

(plane)

1

1

1z

a

a

za

http://upload.wikimedia.org/wikipedia/commons/b/b2/Cylindrical_coordinate_surfaces.png



The three unit vectors za , a , and

a are in the direction of increasing variables

and are perpendicular to the surface at which the coordinate variable is

constant.

x

y

z x

y

a

a

za



Note that in Cartesian coordinates, unit vectors are not functions of coordinate

variables. But in cylindrical coordinates a , and

a are functions of .

The cylindrical coordinate system is Right Handed: zˆˆˆ aaa .

Increasing each coordinate variable by a differential amount d , d , and dz ,

one obtains:

x

y

z

1

1 2

2 xa

ya ya

xa

x

y

z

1

1

1a

1a

2 2

2a

2a



Note that and z are lengths, but is an angle which requires a metric

coefficient to convert it to length.

d

tcoefficienmetric

lengtharc

x

y

z

d

d

lengtharcd

areadd



Differential volume: dzdddv

Differential Surfaces: Six planes with differential areas shown in the figure above.

(Try it!)

x

y

z

d d

dz

d

dzdds

dzdds

ddds



Transformations between Cylindrical and Cartesian Coordinates

From cylindrical to cart:

cosx

siny

zz

From cart. To cyl.:

22 yx

x

ytan 1

zz

x

y

z

1xx

1yy

1zz

1 1



Consider a vector in rectangular coordinates;

zzyyxx EEE aaaE ˆˆˆ

Wishing to write E in cylindrical coordinates:

zzEEE aaaE ˆˆˆ

From the dot product:

aE ˆE aE ˆE zzE aE ˆ

aaaa ˆˆˆˆ zzyyxx EEEE

?

ˆˆ

?

ˆˆ

?

ˆˆ aaaaaa zzyyxx EEE


?

ˆˆ

?

ˆˆ

?


zzzyyxxz EEEE aaaa ˆˆˆˆ

?

ˆˆ

?

ˆˆ

?

ˆˆzzzzyyzxx EEE aaaaaa



Clearly:

coscosˆˆˆˆ aaaa xx

sin90cosˆˆˆˆ o

yy aaaa

sin90cosˆˆˆˆ o

xx aaaa

coscosˆˆˆˆ aaaa yy

090cosˆˆˆˆ o

zz aaaa

090cosˆˆˆˆ o

zz aaaa

So:

sincos yx EEE

cossin yx EEE

Or in matrix form

y

x

E

E

E

E

cossin

sincos

x

y

z x

y

a

a

za

xa

ya

090



And, the inverse relation is:

E

E

E

E

y

x

cossin

sincos

Note that the story is not finished here, after transforming the components; you

should also transform the coordinate variables.



THE SPHERICAL COORDINATE SYSTEM

,,r are

coordinate

variables.

is measured

from x-axis

toward y-axis,

and is

measured from

the z-axis

toward the xy

plane.

A genral point is loacated by its coordinate variables ,,r , or as the

intersection of three mutually perpendicular surfaces.

x

y

z

1x

2y

3z

oo .,.,rP

orz,y,xP

463693514

321

r

http://upload.wikimedia.org/wikipedia/commons/a/ad/Spherical_coordinate_surfaces.png



1) Sphere of radius 1rr , centered at the origin.

2) Semi-infinite plane of constant angle 1 with it’s axis aligned with z-

axis. 3) Right angular cone with its apex centered at the origin, and it axis aligned

with z-axis, and a cone angle 1 .

The three unit vectors ra , a , and a are in the direction of increasing

variables and are perpendicular to the surface at which the coordinate variable

is constant.

x

y

z

111 ,, rP

1

Surface

(plane)

1rr

surface

(sphere)

1

surface

(cone)

ra

a

a



x

y

z 111 ,,rP

1

ra

a

1

1r

a



Note that in spherical coordinates, unit vectors are functions of coordinate

variables. a , a and ra are functions of and .

x

y

z x

y

a

za

planexy

z

1

r

1ˆa

1ˆ

ra 2

ˆra

2ˆa r

2

x

y

z x

y

a

za

planexy

z

z

r

a

ra



The spherical coordinate system is Right Handed:

aaa ˆˆˆ r .

Increasing each coordinate variable by a differential amount dr , d , and d ,

one obtains:

x

y

z

1

1 2

2 xa

ya ya

xa

x

y

z

1

1

1a

1a

2 2

2a

2a



Note that r is length, but and are angles which requires a metric coefficient

to convert them to lengths.

dr

tcoefficienmetric

lengtharc

dr

tcoefficienmetric

sinlengtharc

Differential volume: dddrrdv sin2

Differential Surfaces: Six surfaces with differential areas shown in the figure. (Try

itttttttttttt!)

Transformations between Spherical and Cylindrical Coordinates

From spherical to cart:

cossinrx

sinsinry

cosrz

From cart. To spherical:

222 zyxr ;

;

x

y1tan

x

y

z

1xx

1yy

1zz

1

1

1

1r



Consider a vector in rectangular coordinates;

zzyyxx EEE aaaE ˆˆˆ

Wishing to write E in spherical coordinates:

aaaE ˆˆˆ EEE rr

From the dot product:

rrE aE ˆ aE ˆE aE ˆE

rzzyyxxr EEEE aaaa ˆˆˆˆ

?

ˆˆ

?

ˆˆ

?

ˆˆrzzryyrxx EEE aaaaaa


?

ˆˆ

?

ˆˆ

?



?

ˆˆ

?

ˆˆ

?




From figure

coscosˆˆˆˆ rzrz aaaa

sin90cosˆˆˆˆ o

rr aaaa

sin90cosˆˆˆˆ o

zz aaaa

And the rest is left to you as an exercise!

So:

cossinsincossin zyxr EEEE

Note that, after transforming the components; you should also transform the

coordinate variables.

planexy

z

z r

a

ra za

a

090



The Study of Electromagnetics

(a) Electrostatics: Stationary charges. (b) Magnetostatics: DC (steady currents), charges motion with constant

velocity. (c) Electromagnetics (Electrodynamics): Time varying fields.

Brief History:

Plato & Socrates: load stone. John Mitchell 1750: Inverse square law. Charles de Coulomb 1785: Confirmed inverse law. Karl Frederick Gauss 1800: Gauss law for electrostatics. Alessandro Volta 1800: Invention of a battery. Andre’ Ampere 1820 : Two current carrying wires create forces on each

other. Jean-Paptiste Biot & Felix Savart 1820: Forces between two current

elements. Michael Faraday 1831: Time-changing magnetic field produces electric

field. James Clerk Maxwell 1873: Mathematical prediction of electromagnetic

(EM) waves. Heinrich hertz 1887: Confirmed existence of EM waves.

SI (International System of Units):

Name Symbol Unit Abbreviation

Mass m Kilogram Kg

Length ....,d,x,l Meter m

Time t Seconds s

Electric Current I,i Ampere A

Powers of ten:

910 610

310 310

610

910

1210

Giga Mega Kilo Milli micro nano pico

G M K m n p



Charge Distributions

Electric Charge is a fundamental property of matter, and the charge on an

electron is e , where: C.e 191061 , and C is the unit of charge

(Coulomb).

Point charge distributionQ : assumed to

exist (concentrated) at isolated points in

space.

Line charge distribution is the distribution

of charge over a line. The line charge

density is the amount of electric charge

per unit length in a line of vanishingly

small radius.

Surface charge distribution is the

distribution of charge over a surface.

The surface charge density is the

amount of electric charge per unit area

in a surface of vanishingly small

thickness.

x

y

z

+ +

+ +

+ + +

- - -

- - -

2m

Cz,y,xs

x

y

z

+ +

+ + + + +

- - - -

- -

m

Cz,y,xl

x

y

z

1Q

2Q 3Q

4Q



Volume charge distribution is the

distribution of charge over a volume.

The volume charge density is the

amount of electric charge per unit

volume in a volume.

Charge density in general may depend on position (nonuniform). But if it is

constant over the respective region (independent of x, y, and z), it is said to be

uniform charge density.

Perhaps the most frequently used method for solving electrostatic field

problems is the differential element of charge, with this method, the charge

distribution is reduced to a differential element of charge, dQ , which is treated as

point charges.

Let v be the volume charge density measured in 3mC , then:

3

3m

m

CC

vQ v

As 0v (differential volume dv ) then dQQ (a point charge), so:

dvdQ v

v

v

v

v dvdvQ

Infinitely large number of

vanishingly small

charges spaced by

differentialy small

distances.

Volume v containing

total charge Q

Considering a volume element

v containing elemental

amount of charge Q

x

y

z

+ +

+ + +

+ +

- - - -

- -

3m

Cz,y,xv



zv e

5106105

20

10

42

cmcm

cmzcm

A charge distribution of vanishingly small thickness is referred to as

surface charge. Like the charge on the surface of a perfect conductor.

Let s be the surface charge density measured in 2mC , then:

x

y

3m

Cz,y,xv

z



2

2m

m

CC

sQ s

As 0s (differential surface ds ) then dQQ (a point charge), so:

dsdQ s

s

s

s

s dsdsQ

A charge distribution of filament shape (vanishingly small radius) is

referred to as line charge. Like the charge on cylinder of a very small radius

(approaching zero).

Let l be the line charge density measured in m

C , then:

mm

CC

lQ l


vanishingly small

charges spaced by

differentialy small

distances.

Surface s containing

total charge Q

Considering a surface element

s containing elemental

amount of charge Q



As 0l (differential line dl ) then dQQ (a point charge), so:

dldQ l

l

l dlQ

Note that

Coulomb’ Law

The force between two very small objects separated in a 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.

x

y

z

1r

2r

21R 1Q 2Q

1F

2F 21a

21a

Assuming both charges have

the same polarity


vanishingly small

charges spaced by

differentialy small

distances.

Line l containing total

charge Q

Considering a line element l

containing elemental amount

of charge Q



The electric force exerted on 2Q by 1Q is:

NQQ

R

QQQQ

ooo

122

21

21122

12

21122

12

212

ˆ4

ˆ4

ˆ4

arr

aaR

F

Where:

2

212

9

.10854.8

36

10

mN

C

m

Fo

is the permittivity of free space.

12a is a unit vector directed from 1Q toward 2Q .

Note that:

NQQ

o

2212

21

211

ˆ4

FaR

F

2

12

2121

1,,

RandQQFF



The Electric Field Intensity E

Assume a test charge tQ is moving around (still stationary) a fixed charge

Q as shown:

The Electric field intensity E is the vector electric force on a unit positive test

charge. Since

NQQ

Qt

Qto

tt a

RF ˆ

42

So;

m

V

C

NQ

QQt

Qtot

t aR

FE ˆ

42

Q

FE

x

y

z

r tr

QtR Q

tQ tF

Qta

tQ

QtR

Qta

tr

tF

tF

tr

tQ

QtR



Note that E exists regardless tQ exists or not, so E is measured at any

point in space called the observation point (field point or test point) which is a

distance R from the source point Q . To remove all the subscripts, we use

primed coordinates to indicate source point(s) and unprimed coordinates to

indicate observation point as illustrated in the following figure.

m

V

C

NQQQ

o

R

o

R

o

322'4

ˆ'4

ˆ4 rr

Ra

rra

RE

Where:

'rrR ; R

Ra Rˆ

r : Position vector locating the observation point zyx zyx aaar ˆˆˆ

'r : Position vector locating the source point zyx zyx aaar ˆ'ˆ'ˆ''

zyx zzyyxx aaaR ˆ'ˆ'ˆ'

x

y

'r

r

R Q

z,y,xP

Ra

Assuming Q is positive

E

'z,'y,'xP

z



E Due to n discrete point charges

In this case superposition applies.

n

i

Ri

io

i

Rn

no

nR

o

R

o

n

Q

QQQ

12

222

2

212

1

1

321

ˆ4

ˆ4

...ˆ4

ˆ4

...

aR

aR

aR

aR

EEEEE

Where;

'

ii rrR and i

iRi

R

Ra ˆ

x

y

z

'2r

'3r

2R

2Q 3Q

1Q 1R

'1r

'nr

nQ

nR

r

3R

x,y,xP



THE SUPERPOSITION INTEGRALS

E Due to volume charge density v

Let v be the volume charge density measured in 3mC . How do we

obtain E for such a charge distribution?

Partition the charge Q contained in volume 'v into small charges dQ

(approaching a point charge 'dvdQ v ). Then each dQ produces an

incremental field Ed given by:

m

V

C

NdvdQdQdQd

o

v

o

R

o

R

o

3322'4

''

'4ˆ

'4ˆ

4 rr

rr

rr

Ra

rra

RE

Where:

'rrR ; R

Ra Rˆ



zyx zzyyxx aaaR ˆ'ˆ'ˆ''

Integrating over the volume containing the source, then:

m

V

C

N

R

dvdv

v o

vR

v o

v

'

2

'

34

'ˆ

'4

''

a

rr

rrE

x

y

z

'r

R

Observation point

z,y,xP

Source point 'dv containing dQ with coordinates 'z,'y,'x'P

Volume 'v containing

total charge Q

r



E Due to line charge density l

Let l be the line charge density measured in m

C . How do we obtain E

for such a charge distribution? Partition the charge Q contained on line 'l into

small charges dQ (approaching a point charge 'dldQ l ). Then each

elemental line 'dl containing vanishingly small charge dQ produces an

incremental field Ed given by:

m

V

C

NdldQdQdQd

o

l

o

R

o

R

o

3322'4

''

'4ˆ

'4ˆ

4 rr

rr

rr

Ra

rra

RE

Where:

'rrR ; R

Ra Rˆ



zyx zzyyxx aaaR ˆ'ˆ'ˆ'

Integrating over the line containing the source, then:

m

V

C

N

R

dldl

s o

lR

l o

l

'

2

'

34

'ˆ

'4

''

a

rr

rrE

x

y

z

'r

R

Observation point

z,y,xP

Source point 'dl containing dQ with coordinates 'z,'y,'x'P

Line 'l containing total

charge Q

r



Example: An infinite filament of uniform charge density



What if the filament is not aligned with the z-axis?



Observations:

aE ˆ

2 o

l For an infinite length line.

E is perpendicular to the line charge, aE ˆ?E .

E does not vary with or z , aE ˆE .

R

o

l

RaE ˆ

2

For an infinite length line, that is arbitrarily aligned.

E Due to surface charge density s

Let s be the surface charge density measured in 2mC . How do we

obtain E for such a charge distribution? Partition the charge Q contained on

surface 's into small charges dQ (approaching a point charge 'dsdQ s ).

Then each elemental surface 'ds containing vanishingly small charge dQ

produces an incremental field Ed given by:

x

y

z

'r

R

Observation point

z,y,xP

Source point 'ds containing dQ with coordinates 'z,'y,'x'P

Surface 's containing

total charge Q

r



m

V

C

NdsdQdQdQd

o

s

o

R

o

R

o

3322'4

''

'4ˆ

'4ˆ

4 rr

rr

rr

Ra

rra

RE

Where:

'rrR ; R

Ra Rˆ



zyx zzyyxx aaaR ˆ'ˆ'ˆ''

Integrating over the surface containing the source, then:

m

V

C

N

R

dsds

s o

sR

s o

s

'

2

'

34

'ˆ

'4

''

a

rr

rrE

Example: Consider an infinite sheet lying in the yz-plane, having a uniform

charge distribution of

2m

Cs . Determine an expression for E at an arbitrary

point lying on the x-axis.



Observations:

z

o

s aE ˆ2

E is constant in magnitude and direction. E is always normal to the sheet, away from positively charged sheet, and

into negatively charged one.

In general n

o

s aE ˆ2

, with na being the outward unit normal to the sheet

and s is signed.

For two oppositely charged parallel infinite sheets, the electric field outside

the sheets is zero. And between the sheets the field is n

o

s aE

2 with na being

the unit normal directed from positively charged sheet to the negatively charged

one.



1) :x 0 x

o

s aE

2

and x

o

s aE

2 0 EEE .

2) :x 0 x

o

s aE

2 and x

o

s aE

2

0 EEE .

3) :ax 0 x

o

s aE

2 and x

o

s aE

2 x

o

s aEEE

.

x

+ +

+

+

+

- -

-

-

-

E E

E E

E

E

+ - 0x ax

E



Stream Lines (Field lines or Flux lines)

Lines start at positive charge and terminate on negative charge. Lines start at positive charge and terminate at infinity . Lines start at infinity and terminate on negative charge. Spacing between lines is inversely proportional to the strength of the field. E is tangent to field lines at any point on the line.

Positive point

charge

Negative point

charge

Negative line

charge l

Positively charged

surface Two oppositely charged

point charges

All drawings above are just approximations

Engineering Electromagnetics; William Hayt & John...

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