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CHAPTER 2

ELECTROMAGNETIC FIELDS THEORY

- Electrostatic Fields
- Electric Field Intensity

In this chapter you will learn..

- 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

- 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

Charles Agustin de Coulomb

Electric Field Intensity

This proof the existence of electric force field

which radiated from Q1

Electric Field Intensity

- Consider a charge fixed in a position, Q1.
- Lets 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.

Electric Field Intensity

Experience E due to Q1

- Electric field intensity, E is the vector force

per unit charge when placed in the electric field

Q1

Experience the most E due to Q1

Experience less E due to Q1

Electric Field Intensity

- According to Coulombs law, force on Qt is
- We can also write
- Which is force per unit charge

Electric Field Intensity

- If we write E as
- Thus we can rewrite the Coulombs law as
- Which gives the electric field intensity for Qt

at r due to a point charge Q located at r

Electric Field Intensity

- For N points charges, the electric field

intensity for at a point r is obtained by - Thus

Field Lines

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

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

The field lines terminated at a negative point

charge

The field lines for a pair of opposite charges

Example 4.1

- Point charges of 1 mC and -2mC are located at (3,

2, -1), - B(-1, -1, 4) respectively in free space.

Calculate the electric force on a 10 nC charge

located at (0, 3, 1) and the electric field

intensity at that point.

Solution

Line charge

Imagine charges distributed along a line

We can find the charge density over the line as

Charge density and charge distribution

- We can find the total charge by applying this

equation

Charge density

Length of the line

Surface Charge

Similarly, for surface charge, we can imagine

charges distributed over a surface

We can find the charge density over the surface

as

And the total charge is

Volume Charge

We can find the volume charge density as

And the total charge is

Continuous Charge Distribution

- for line charge distribution
- for surface charge distribution
- for volume charge distribution

E due to Continuous Charge Distributions

Line charge

?S

Surface charge

?v

Volume charge

What About the Field Lines?

E field due to line charge

E field due to surface charge

E field due to line charge

What about volume charge?

LINE CHARGE

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 (Contd)

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 (Contd)

The electric field intensity is

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,

LINE CHARGE (Contd)

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

LINE CHARGE (Contd)

Recall for point charge,

For continuous charge distribution, the summation

of vector field for each charges becomes an

integral,

LINE CHARGE (Contd)

The differential charge,

The vector from source to test point P,

LINE CHARGE (Contd)

Which has magnitude, and

a unit vector,

So, the equation for integral of continuous

charge distribution becomes

LINE CHARGE (Contd)

Since there is no component,

IMPORTANT!!

IMPORTANT!!

LINE CHARGE (Contd)

Hence, the electric field intensity at any point

? away from an infinite length is

For any finite length, use the limits on the

integral.

? is the perpendicular distance from the line to

the point of interest a? is a unit vector along

the distance directed from the line charge to the

field point

Example of Line Charge

- 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) -8 lt z lt 8
- b) -4 z 4

Example of Line Charge

z

dEz

4

dE

?

dE?

P(1, 2, 3)

r

dl

z

r

y

x

-4

Solution (a)

- 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

Solution (b)

- Here we use the general equation
- Where the total charge Q is

Solution (b)

- Or, we can also write E as
- Where
- and

Solution (b)

- So E would be
- Using integral tables, we obtain

What if the line charge is not along the z-axis??

Line Charge

z

P(x, y, z)

?

R

y

dl

(0, y, 0)

(0, y, 0)

x

What if the line is not along any axis??

Example

- A uniform line charge of 16 nC/m is located along

the line defined by - y -2, z 5. If e e0.
- Find E at P(1, 2, 3)

Solution

- Draw the line y -2, z 5.
- Find a?
- Find E

Solution

z

(1, -2, 5)

5

?

P(1, 2, 3)

-2

y

x

Solution

- To find a? , 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 a? can be calculated as

Then finally, calculate E

What if it is a circular line of charge??

EXAMPLE

Use Coulombs 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.

Solution

- Step 1 derive what are dl and R, R2 and aR
- Step 2 Replace dl, R2 and aR into line charge

equation

Solution

By inspection, the ring charges delivers only

and contribution to the field.

component will be cancelled by symmetry.

Solution

Each term need to be determined

The differential charge,

Solution

The vector from source to test point,

Which has magnitude,

and a unit vector,

Solution

The integral of continuous charge distribution

becomes

Solution

Rearranging,

Easily solved,

Infinite Surface charge

- Static charges resides on conductor surfaces and

not in their interior, thus the charges on the

surface are known as surface charge density or ?s

- Consider a sheet of charge in the xy plane

shown below

Infinite Surface charge

Infinite Surface charge

Infinite Surface charge

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

is

Infinite Surface charge

- The total charge is
- From figure,

Infinite Surface charge

- From figure,

Infinite Surface charge

- Replace in
- Thus becomes

Infinite Surface charge

For an infinite surface, due to symmetry of

charge distribution, for every element 1, there

is a corresponding element 2 which cancels

element 1.

Infinite Surface charge

- Thus the total of E? equals to zero. E only has

z-component. - For infinite surface, 0 lt ? lt 8

Solution (cont)

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.

Infinite Surface charge

- In parallel plate capacitor, the electric field

existing between the two plates having equal and

opposite charges is given by

EXAMPLE

An infinite extent sheet of charge

exists at the plane y -2m. Find the

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

SOLUTION

Step 1 Sketch the figure

SOLUTION

Step 2 find an. The unit vector directed away

from the sheet and toward the point P is Thus

SOLUTION

Step 3 solve the problem!

Example

- Given the surface charge density, ?s 2µC/m2, in

the region ? lt 0.2 m, z 0, and is zero

elsewhere, find E at PA(? 0, z 0.5)

Solution

- Sketch the figure

(0,0,0.5)

2

2

Solution

- 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 ?a?, we obtain r - r zaz - ?a?.

The superposition integral for the z component of

E will be

Solution

- With z 0.5 m,

Exercise

- Planes x 2, and y -3, respectively, carry

charges 10nC/2 and 15nC/m2. if the line x 0, z

2 carries charge 10p nC/m, calculate E at (1,

1, -1) due to three charge distributions.

Volume charge

Volume charge sphere

- Consider the volume charge distribution with

uniform charge density, in the figure below

P(0,0,z)P(z,0,0)

a

R

?v

dv at (r,?,F)

Z

r

?

y

F

x

Volume charge sphere

- The charge dQ is
- The total charge in a sphere with radius a is

Volume charge sphere

- The electric field dE at point P(0,0,z) due to

the elementary volume charge is - Where

Volume charge sphere

- Due to symmetry of charge distribution, Ex or Ey

add up top zero. - Thus

Volume charge sphere

- For spherical coordinate,
- Then by applying the cosine rule,

Volume charge sphere

- Or we can express it as
- Differentiate with respect to ?

Volume charge sphere

- Then dv now becomes
- And Ez becomes

Volume charge sphere

- And Ez becomes (cont)

(a)

Volume charge sphere

- But
- So
- Replace in equation (a). So E now becomes

Volume charge sphere

- For a point P2 at (r,?,F) E is
- Which is identical to the electric field at the

same point due to a point charge Q at the origin.

EXAMPLE 6

Find the total charge over the volume with volume

charge density,

SOLUTION

The total charge, with

volume

Thus,

SOLUTION

To find the electric field intensity resulting

from a volume charge, we use

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 Gausss

Law.

Principle of superposition

- 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

Principle of superposition

Principle of superposition

- Thus

Exercise

- 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.