CHAPTER 2

Electrostatic Fields:Electric Field Intensity

ELECTROMAGNETIC FIELDS THEORY

Electrostatic Fields -Charge, charge density-Coulomb's law-Electric field intensity- Electric flux and electric flux density-Gauss's law-Divergence and Divergence Theorem-Energy exchange, Potential difference, gradient-Ohm's law-Conductor, resistance, dielectric and capacitance- Uniqueness theorem, solution of Laplace and Poisson equation

Chapter 2BEE 3113ELECTROMAGNETIC

FIELDS THEORY 2

Two like charges repel one another, whereas two charges of opposite polarity attract

The force acts along the line joining the charges

Its strength is proportional to the product of the magnitudes of the two charges and inversely propotional to the square distance between them

NrR tt

121

ˆ0

t1t 4

QQ F


FIELDS THEORY 3

Charles Agustin de Coulomb


FIELDS THEORY 4

Q1

Qt

F1t

F1t

F1t

This proof the existence of electric force field which radiated from Q1

Consider a charge fixed in a position, Q1.

Let’s say we have another charge, say Qt which is a test charge.

When Qt is moved slowly around Q1, there exist everywhere a force on this second charge.

Thus proof the existence of electric force field.


FIELDS THEORY 5

Electric field intensity, E is the vector force per unit charge when placed in the electric field


FIELDS THEORY 6

Qt1

Qt2

Qt3

Q1

Experience E due to Q1

Experience the most E due to Q1

Experience less E due to Q1

According to Coulomb’s law, force on Qt is

We can also write

Which is force per unit charge

NrR tt

1210

t1t ˆ

4QQ F

NrR tt

1210

1

t

t ˆ4

Q Q

F


FIELDS THEORY 7

If we write E as

Thus we can rewrite the Coulomb’s law as

Which gives the electric field intensity for Qt at r due to a point charge Q located at r

t

t

Q F

E

3

1

1

0

1

4 rrrrE

t

t

πεQ


FIELDS THEORY 8

For N points charges, the electric field intensity for at a point r is obtained by

Thus

N

k k

k

rrrr

13

0 ||)(

NQ4

1E


FIELDS THEORY 9

30

32

2

0

23

1

1

0

1

||4...

||4||4 N

nN

rrrr

πεQ

rrrr

πεQ

rrrr

πεQ

E


FIELDS THEORY 10

The behavior of the fields can be visualized using field lines:

Field vectors plotted within a regular grid in 2D space surrounding a point charge.

Field Lines


FIELDS THEORY 11

Some of these field vectors can easily be joined by field lines that emanate from the positive point charge.

The direction of the arrow indicates the direction of electric fields

The magnitude is given by density of the lines

Field Lines


FIELDS THEORY 12

The field lines terminated at

a negative point charge

The field lines for a pair of

opposite charges

Field Lines


FIELDS THEORY 13

RQ a

RE 2

04


FIELDS THEORY 14

q q

qq


FIELDS THEORY 15

q

qq

q

mCddqq

L /lim0

Imagine charges distributed along a line

q

We can find the charge density over the line as:

We can find the total charge by applying this equation:


FIELDS THEORY 16

L LL dlQdldQ

Charge density Length of the line


FIELDS THEORY 17

S Ss dSQdSdQ

2

0/lim mC

dsdq

sq

Ss

Similarly, for surface charge, we can imagine charges distributed over a surface

q

We can find the charge density over the surface as:

qq qqq

q q

q

And the total charge is


FIELDS THEORY 18

v vv dvQdvdQ

qq

q qq

qq

q qq

3

0/lim mC

dvdq

vq

vv

We can find the volume charge density as:

And the total charge is

for line charge distribution:

for surface charge distribution:

for volume charge distribution


FIELDS THEORY 19

S Ss dSQdSdQ

L LL dlQdldQ

v vv dvQdvdQ

++ + + +ρL

+ ++ +

ρS

+ + ++++

+

+

ρv

++++

RS

RπεdSρ a E 204

RL

Rπεdlρ a E 204

RRdv a

4 E

0

v 2


FIELDS THEORY 20

Surface charge

Volume charge

Line charge++ + + +ρL

+ ++ +

ρS

+ + ++++

+

+

ρv

++++


FIELDS THEORY 21

E field due to line charge

E field due to surface charge


FIELDS THEORY 22


FIELDS THEORY 23

qq

q qq

qq

q qq

qq

qq

q


FIELDS THEORY 24

Infinite Length of Line Charge:

To derive the electric field intensity at any point in

space resulting from an infinite length line of

charge placed conveniently along the z-axis

LINE CHARGE


FIELDS THEORY 25

Place an amount of charge in coulombs along the z axis.

The linear charge density is coulombs of charge per meter length,

Choose an arbitrary point P where we want to find the electric field intensity.

mCL

zP ,,

LINE CHARGE (Cont’d)


FIELDS THEORY 26


The electric field intensity is:

zzEEE aaaE But, the field is only vary with the radial distance from the line.

There is no segment of charge dQ anywhere on the z-axis that will give us . So,E

zzEE aaE


FIELDS THEORY 27

LINE CHARGE (Cont’d)Consider a dQ segment a distance z above radial axis, which will add the field components for the second charge element dQ.

The components cancel each other (by symmetry) , and the adds, will give:

zE

E

aE E


FIELDS THEORY 28


Recall for point charge,

RQ a

RE 2

04

For continuous charge distribution, the summation of vector field for each charges becomes an integral,

RdQ a

RE 2

04


FIELDS THEORY 29


The differential charge,

dzdldQ

L

L

The vector from source to test point P,

z

R

zR

aa aR


FIELDS THEORY 30


Which has magnitude, and a unit vector, 22 z R

22 z

z zR

aaa

So, the equation for integral of continuous charge distribution becomes:

2222204 z

z

z

dz zL

aaE


FIELDS THEORY 31


Since there is no component,

a

aE

23

220

23

220

4

4

z

dz

z

dz

L

L

za


FIELDS THEORY 32

IMPORTANT!!


FIELDS THEORY 33

IMPORTANT!!


FIELDS THEORY 34


FIELDS THEORY 35

Hence, the electric field intensity at any point ρ away from an infinite length is:

aE

02L

For any finite length, use the limits on the integral.


ρ is the perpendicular distance from the line to the point of interest

aρ is a unit vector along the distance directed from the line charge to the field point


FIELDS THEORY 36

aE

02L

A uniform line charge of 2 µC/m is located on the z axis. Find E in cartesian coordinates at P(1, 2, 3) if the charge extends from

a) −∞ < z < ∞: b) −4 ≤ z ≤ 4:


FIELDS THEORY 37


FIELDS THEORY 38

dEz

4

-4

dl

r

ρdEdEρ

z

x

y

P(1, 2, 3)

z

r’

With the infinite line, we know that the field will have only a radial component in cylindrical coordinates (or x and y components in cartesian).

The field from an infinite line on the z axis is generally

Therefore, at point P:

aE

02L


FIELDS THEORY 39

Here we use the general equation :

Where the total charge Q is

RdQ a

RE 2

04


FIELDS THEORY 40

dzdldQ LL

B

A

z

z LdzQ

Or, we can also write E as:

Whereand


FIELDS THEORY 41

So E would be:

Using integral tables, we obtain:


FIELDS THEORY 42


FIELDS THEORY 43


FIELDS THEORY 44

dl

R ρ

(0, y’, 0) (0, y, 0)

z

x

y

P(x, y, z)


FIELDS THEORY 45

A uniform line charge of 16 nC/m is located along the line defined by y = −2, z = 5. If ε= ε0. Find E at P(1, 2, 3)


FIELDS THEORY 46

1. Draw the line y = −2, z = 5.2. Find aρ 3. Find E


FIELDS THEORY 47


FIELDS THEORY 48

ρ

(1, -2, 5)

z

x

y

P(1, 2, 3)

-2

5

2

),,(),,(

P

P

linelineP zyxzyx

RRa

R

zy

P

P

P

P

aaRRa

R

R

2420

)2,4,0(

20)2(4

)2,4,0()5,2,1()3,2,1(

2

2222


FIELDS THEORY 49

To find aρ , we must draw a “projection” line from point P to our line. Then find out what is the x-axis value at that projection point

Then aρ can be calculated as:

aE

02L

mVzy

zy

/8.285.5720

242

1016

0

9

aa

aaE


FIELDS THEORY 50


FIELDS THEORY 51


FIELDS THEORY 55


FIELDS THEORY 56

EXAMPLE

Use Coulomb’s Law to find

electric field intensity at

(0,0,h) for the ring of

charge, of charge density,

centered at the origin in

the x-y plane.

L

Step 1: derive what are dl and R, R2 and aR

Step 2: Replace dl, R2 and aR into line charge equation:R

L

Rπεdlρ a E 204


FIELDS THEORY 57


FIELDS THEORY 58

By inspection, the ring

charges delivers only

and contribution to

the field.

component will be

cancelled by symmetry.

zdEEd

Ed

Solution


FIELDS THEORY 59

RdQ a

RE 2

04

Each term need to be determined:

The differential charge,

addldQ

L

L

Solution


FIELDS THEORY 60

The vector from source to test point,

z

R

haR

aa aR

Which has magnitude, and a unit vector, 22 ha R

22 ha

ha zR

aaa

Solution


FIELDS THEORY 61

The integral of continuous charge distribution becomes:

2222204 ha

ha

ha

ad zL aaE

zL h

ha

ad aE2

322

04

Solution


FIELDS THEORY 62

Rearranging,

2

023

2204

zL dha

ah aE

Easily solved,

zL

ha

ah aE2

322

02

Solution


FIELDS THEORY 63


FIELDS THEORY 64

Static charges resides on conductor surfaces and not in their interior, thus the charges on

the surface are known as surface charge density or ρs

Consider a sheet of charge in the xy plane shown below


FIELDS THEORY 65


FIELDS THEORY 66

Φ

z’

y

x

z

R

ρ

P(0,0,z’)

1

2

The field does not vary with x and y Only Ez is present The charge associated with an elemental

area dS is

dddQ s


FIELDS THEORY 67

dSdQ S

The total charge is

From figure,

RRdQd aE 2

04

dSQ S


FIELDS THEORY 68

From figure ,

,

||,)(2/122

RRa

hRRhaaR

R

z


FIELDS THEORY 69

h

R

-ρ

Replace in

Thus becomes

RRdSd a

4 E

0

S2

2/3220

'

][4][

hhdd

d zS

aaE


FIELDS THEORY 70


FIELDS THEORY 71

Φ

z’

y

x

z

R

ρ

P(0,0,z’)

1

2

For an infinite surface, due to symmetry of charge distribution, for every element 1, there is a corresponding element 2 which cancels element 1.

Thus the total of Eρ equals to zero. E only has z-component.

For infinite surface, 0 < ρ < ∞

zzdE aE

zS

zddh aE 2/322

0

2

00 ]'[4


FIELDS THEORY 72

Infinite Surface charge

Thus in general, for an infinite sheet of charge,

Where an is a vector normal to the sheet.E is normal to the sheet and independent of

the distance between the sheet and the point of observation, P .

nS aE02


FIELDS THEORY 74

In parallel plate capacitor, the electric field existing between the two plates having equal and opposite charges is given by

nS

nS

nS aaaE

000

)(22


FIELDS THEORY 75


FIELDS THEORY 80

EXAMPLE

An infinite extent sheet of charge

exists at the plane y = -2m. Find the electric field

intensity at point P (0, 2m, 1m).

210mnC

S


FIELDS THEORY 81

SOLUTIONStep 1: Sketch the figure


FIELDS THEORY 82

Step 2: find an.

The unit vector directed away from the sheet and toward the point P is

Thus

ya

yS

nS aaE

00 22

SOLUTION


FIELDS THEORY 83

Step 3: solve the problem!

mV

y

y

yS

a

a

aE

565

10854.821010

2

12

9

0

SOLUTION

Given the surface charge density, ρs = 2µC/m2, in the region ρ < 0.2 m, z = 0, and is zero elsewhere, find E at PA(ρ = 0, z = 0.5):


FIELDS THEORY 84

Sketch the figure:


FIELDS THEORY 85

2

2

(0,0,0.5)

First, we recognize from symmetry that only a z component of E will be present. Considering a general point z on the z axis, we have r = zaz. Then, with r = ρaρ, we obtain r − r = zaz − ρaρ. The superposition integral for the z component of E will be:


FIELDS THEORY 86

With z = 0.5 m,


FIELDS THEORY 87

Planes x = 2, and y = -3, respectively, carry charges 10nC/2 and 15nC/m2. if the line x = 0, z = 2 carries charge 10π nC/m, calculate E at (1, 1, -1) due to three charge distributions.


FIELDS THEORY 88


FIELDS THEORY 89

Consider the volume charge distribution with uniform charge density, in the figure below


FIELDS THEORY 90

Φ’

θ’

y

x

Rρv

P(0,0,z’)=P(z’,0,0)

r’

dv at (r,θ’,Φ’)

Z’

α

The charge dQ is

The total charge in a sphere with radius a is

34 3a

dv

dvQ

v

v

v

dvdQ v


FIELDS THEORY 91

The electric field dE at point P(0,0,z) due to the elementary volume charge is

Where

Rv

Rdvd aE 204


FIELDS THEORY 92

aaa sincos zR

For a point P2 at (r,θ,Φ) E is

Which is identical to the electric field at the same point due to a point charge Q at the

origin .


FIELDS THEORY 99

rrQ aE 2

04


FIELDS THEORY 100

EXAMPLE 6

Find the total charge over

the volume with volume

charge density,

3105

5m

Ce zV


FIELDS THEORY 101

SOLUTION

V

VdVQ The total charge, with volume:

dzdddV

Thus,

C

dzdde

dVQ

z

z

VV

14

01.0

0

2

0

04.0

02.0

10

10854.7

55


FIELDS THEORY 102

To find the electric field intensity resulting from a volume charge, we use:

RV

RdVdQ aR

aR

E 20

20 44

Since the vector R and will vary over the volume, this triple integral can be difficult. It can be much simpler to determine E using Gauss’s Law.

V

SOLUTION

Total E field at a point is the vector sum of the fields caused by all the individual

charges.

Consider the figure below. Positions of the charges q1, q2, q3, …… qn, (source points )

be denoted by position vectors R1’, R2’, R3’, R4’, ….. Rn’. The position of the field

point at which the electric field intensity is to be calculated is denoted by R


FIELDS THEORY 103


FIELDS THEORY 104

Thus


FIELDS THEORY 105

...)()()(

41 3'

3

33'

2

23'

1

1

0 RR

q

RR

q

RR

q '3

'2

'1 RRRRRRE

n

k k

k

RR

q

1

3'0

)(4

1 'kRRE

Three infinite lines of charge, all parallel to the z-axis, are located at the three corners of the kite-shaped arrangement

shown in the figure below. If the two right triangles are symmetrical and of equal corresponding sides, show that the

electric field is zero at the origin .


FIELDS THEORY 106

Date post:	04-Feb-2016
Category:	Documents
Upload:	zia
View:	61 times
Download:	0 times

CHAPTER 2

Documents