CHAPTER 2

transcript

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

FIELDS THEORY 3

Charles Agustin de Coulomb

FIELDS THEORY 4

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

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

t1t ˆ

NrR tt

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

4 rrrrE

FIELDS THEORY 8

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

0 ||)(

FIELDS THEORY 9

||4...

||4||4 N

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

FIELDS THEORY 14

FIELDS THEORY 15

mCddqq

L /lim0

Imagine charges distributed along a line

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

0/lim mC

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

We can find the charge density over the surface as:

qq qqq

And the total charge is

FIELDS THEORY 18

v vv dvQdvdQ

0/lim mC

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

+ ++ +

+ + ++++

RπεdSρ a E 204

Rπεdlρ a E 204

RRdv a

FIELDS THEORY 20

Surface charge

Volume charge

Line charge++ + + +ρL

+ ++ +

+ + ++++

FIELDS THEORY 21

E field due to line charge

E field due to surface charge

FIELDS THEORY 22

FIELDS THEORY 23

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.

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:

FIELDS THEORY 28

Recall for point charge,

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

FIELDS THEORY 29

The differential charge,

dzdldQ

The vector from source to test point P,

FIELDS THEORY 30

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

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

2222204 z

FIELDS THEORY 31

Since there is no component,

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:

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

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

ρdEdEρ

P(1, 2, 3)

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:

FIELDS THEORY 39

Here we use the general equation :

Where the total charge Q is

FIELDS THEORY 40

dzdldQ LL

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

(0, y’, 0) (0, y, 0)

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

(1, -2, 5)

P(1, 2, 3)

),,(),,(

linelineP zyxzyx

)2,4,0(

20)2(4

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

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:

/8.285.5720

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.

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

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

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.

Solution

FIELDS THEORY 59

Each term need to be determined:

The differential charge,

addldQ

Solution

FIELDS THEORY 60

The vector from source to test point,

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

Solution

FIELDS THEORY 61

The integral of continuous charge distribution becomes:

2222204 ha

ad zL aaE

ad aE2

Solution

FIELDS THEORY 62

Rearranging,

zL dha

Easily solved,

ah aE2

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

P(0,0,z’)

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

FIELDS THEORY 68

From figure ,

||,)(2/122

hRRhaaR

FIELDS THEORY 69

Replace in

Thus becomes

RRdSd a

2/3220

FIELDS THEORY 70

FIELDS THEORY 71

P(0,0,z’)

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

zddh aE 2/322

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 aaaE

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

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

nS aaE

SOLUTION

FIELDS THEORY 83

Step 3: solve the problem!

10854.821010

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

(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

P(0,0,z’)=P(z’,0,0)

dv at (r,θ’,Φ’)

The charge dQ is

The total charge in a sphere with radius a is

dvdQ v

FIELDS THEORY 91

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

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

FIELDS THEORY 100

EXAMPLE 6

Find the total charge over

the volume with volume

charge density,

FIELDS THEORY 101

SOLUTION

VdVQ The total charge, with volume:

dzdddV

10854.7

FIELDS THEORY 102

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

RdVdQ aR

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.

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

FIELDS THEORY 105

...)()()(

'1 RRRRRRE

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

CHAPTER 2

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