Electromagnetism New

ELECTROMAGNETISM-ECE 323-

ANGELITO A. SILVERIO, EcESir Mickey/ Sir Bangles

[email protected]

- SYLLABUS -

1. Electrostatics2. Magnetostatics3. Electrodynamics (Steady Electric Currents)4. Time Varying Electromagnetic Fields5. Electromagnetic Waves

- Electrostatics -• The Concept of Charges• Coulomb’s Law • Superposition Principle • Electric Field Intensity and

Electric Flux density due to charge distributions

• Gauss Law• Energy in an electrostatic field• Effect of electric field on

conductors and dielectrics• Boundary conditions and

boundary value problems • Electrostatic potential energy

and electrostatic potential

• Polarization• Field of a polarized object• Linear dielectrics and

capacitance• Boundary conditions and

boundary value problems applied to different capacitance geometries

• Special techniques in solving for electric fields and potentials: Laplace and Poisson Equation, Method of Mirrors

• Applications of electrostatic fields

- Magnetostatics -• Magnetic Fields, Field

Intensity and Flux Density • Gauss’s Law applied to

Magnetic Fields• Lorentz Force Law• Biot-Savart’s Law• Ampere’s Force Law• Magnetic Torque• Magnetic Flux and Gauss’s

Law for Magnetic Fields• Magnetic Vector Potential• Magnetic Field Intensity and

Ampere’s Circuital law• Magnetic Scalar Potential

• Energy in a magnetic field• Magnetic Materials:

diamagnetism, parramagnetism and ferromagnetism

• Magnetization • Field of a magnetized object • Boundary conditions for

magnetic fields• Magnetic circuits• Applications of magnetostatic

fields

- Electrodynamics -(Steady Electric Currents)

• Nature of current and current density: conduction current and convection current

• Conductor resistance• Equation of Continuity• Relaxation Time• Joule’s Law

• Boundary conditions for current density

• Analogy between electric flux density and current density

• Electromotive force

-Time-Varying Electromagnetic Fields -

• Motional electromotive Force

• Faraday’s Law of Induction• Self and mutual

inductance, • Magnetic energy

• Maxwell’s Equations:

Ampere’s law, Gauss law and Faraday’s law

• Applications of time-varying electromagnetic fields

-Electromagnetic Waves-

• General Wave Equations • Propagation Constant• Plane wave propagation in

perfect dielectrics free space, partially conducting medium and perfect conductors

• Skin depth• Interface conditions at normal

and oblique incidence• Brewster’s Angle

• Polarization types: parallel,

perpendicular, elliptical and circular

• Power and the Poynting’s Vector

• Radiation pressure• Applications of the

electromagnetic waves: waveguides, resonators, antennas and transmission lines

• Effects of electromagnetic waves to health

- REFERENCES -

• Guru, B.S. and Hiziroglu, H.R. (2004). Electromagnetic Field Theory Fundamentals, 2nd ed. Cambridge: Cambridge University Press.

• Wentworth, S.M., (2005). Fundamentals of Electromagnetics with Engineering Applications. New Jersey: John Wiley & Sons, Inc.

• Hayt Jr., W.H. and Buck J.A. (2005). Engineering Electromagnetics, 7th ed. New York: McGraw Hill.

• Fleisch, D. A Student’s Guide to Maxwell’s Equations (2008). Cambridge: Cambridge University Press.

• Young, H.D., Freedman, R.A. and Ford, L. (2009). University Physics with Modern Physics, 12th ed. Singapore: Pearson Education South Asia Pte. Ltd.

• Resnick, R., Halliday, D. and Krane, K.S. Physics, 5th ed. (2002). New York: John Wiley and Sons, Inc.

- COURSE POLICIES-“strict compliance”

• No make-up quiz will be given. If a student missed a quiz the first time, his score for that quiz will be his lowest quiz taken for the entire semester. A grade of zero will be given for succeeding missed quiz/quizzes. Only missed major exams can be given as special exams.

• A grade of Failure due to Absences (FA) will be given to students who incurred more than 11 hours of absence. The student is considered late if the elapsed time is 15 minutes or less. Otherwise he is considered absent. Succeeding 3 lates, is equivalent to 1 absent.

• A student caught cheating in a quiz or in preliminary examination will be given a score of zero for that quiz or preliminary examination. Outright failure will be given to a student caught cheating in the final examinations.

• Passing score is 50%. No remedial examination will be given.

- GRADING SYSTEM -

• 5 QUIZZES (50%) >> one per unit • 2 MAJOR EXAMS (40%)

>> PRELIMS (20%) & FINALS (20%)• CLASS STANDING (10%)

>> RECITATION Volunteer (3pts)

Valid Question (1pt) Maximum recitation (50pts adjustable) >> excess points added to the quiz

~ ELECTROSTATICS ~

1. The Concept of Charges• Ancient Greeks noticed when amber and fur were

rubbed together, they would become “attractive”

• The Greek work for amber is “elektron”• Today this attractive property is called “electric”• In the1750’s, Benjamin Franklin studied “static

electricity” he also denoted the

convention:

“positive” too little of electrons

“negative” too much of electrons

1. The Concept of Charges• An atomic viewpoint

nucleus

~10^-15m

~10^-10mProton:Mass: 1.673 x 10 ^-27kgCharge: 1.6 x 10^-19Neutron:Mass: 1.675 x 10 ^-27kg

Electron:Mass: 9.109 x 10^-31kgCharge: -1.6x10^-19

The Atom is ElectricallyNeutral!

Nucleus comprises most of the mass of the Atom (99%)

Strong Nuclear Force holdsthe protonstogether

1. The Concept of ChargesWhat is electric charge? Property of matter that creates electric and magnetic forces and interactions.A property that causes subatomic particles, protons and electrons, to attract or repel one another.“like charges repel, opposite charges

attract”Microscopic View of ChargeElectric charges exist within atoms.Charge is quantized:

“ The magnitude of charge of the electron or proton is a natural unit

of charge”

The Elementary Particles (from wikipedia)

1. The Concept of ChargesPrinciple of Charge Conservation

>> The algebraic sum of all the electric charges in a closed system is constant.

“ in any charging process, charge is not created nor

destroyed, it is only transferred from one body to another.”

1. The Concept of Charges 3 Ways to Charge an Object

a) Charging by Friction (Triboelectric Effect)• If two neutral objects are rubbed together, each can become

charged. For instance, when rubber and wool are rubbed together, electrons from the atoms

on the wool are transferred to the rubber. • The extra electrons on the rubber result in

a net negative charge. The electrons missing from the wool result in a net positive charge.

• The combined total charge remains the same. All that happens is that the positive and negative charges are separated through a transfer of electrons.

• For insulators, the transferred charges reside only at the contact area. They do not spread all through out the medium.

• For conductors, the transferred charges spread all through out the surface of the medium


b) Charging by Conduction

Charging a neutral body by touching it with a charged body Note! The resulting charge of the body is the

same as that of the charged body.

For CONDUCTORS

only!


c) Charging by Induction

Charging an object without touching itNote! The resulting charge of the body is opposite to that of the charged body.

Neutral

Highly Positive

Attracts electrons fromground

Transfer of charges

Neutral body becomes negatively charged

1. The Concept of Charges “3 Ways to Charge an Object”

c) Charging by Induction

1. The Concept of Charges Charge PolarizationSeparation of charges by the electricalalignment of molecules

The electron cloud surrounding the nucleus can become distorted in response to a nearby object. Depending upon the charge of the object, the electron cloud can be repelled or attracted towards the object.

+ -+ -+ -+ -+ -+ -

+ -+ -+ -+ -+ -+ -

Charge re-alignment

Electron cloud

POLAR COVALENT BONDSproduced by the unequal sharing of electrons

Water getsDeflected!

1. The Concept of Charges How to determine the type of charge? ~ tHE electroscope ~

2. Coulomb’s Law

212

2112 r

qqkF

))((

12212

2112 r

rqqk

F ˆ))((

Scalar Form Vector Form

• Charge (q) measured in coulombs (C)• Distance (r) measured in meters• Force (F) in newtons• Electrostatic constant (K) = 9.0 x 109 Nm2/C2 (1/4πЄ0)• One coulomb is the charge of 6.24 x 1018 electrons or protons.• The charge on a single electron is 1.60 x 10 −19 C.• The magnitude of the charge of an electron is called the elementary charge.

2112

2112

q to q direction the in vector unit a is r̂

q on q by exerted force the is F where

Coulomb’s Apparatus

Similar to:

In 1785, Charles Coulombdetermined the relationshipbetween force, charge anddistance

2. Coulomb’s Law

• Force acting on one charge due to another charge is independent on whether or not other charges are present.

• We can calculate the forces separately

• The net force acting on any charge due to a collection of charges is the vector sum of these forces

Superposition Principle for a collection of point charges

q1 q2

q3

23133 FFFq

13F

3qF

23F

2. Coulomb’s LawSample Problems

1. Compare the force of the electric repulsion between two alpha particles with the force of gravitational attraction between them. The alpha particle has a mass of 6.64 x 10^-27kg and a charge of +2e. G = 6.67 x 10-11N-m^2/kg^2.

2. Three equal charges of 200nC are placed in free space at (0,0,0), (2,0,0) and (0,2,0). Determine the force acting on a charge of 500nC at (2,2,0).


3. Three charged particles lie on a straight line and are separated by a distance d.

q1 and q2 charges are held fixed. Charge q3 which is free to move is found to be in equilibrium under the action of the electric forces. Find q1 in terms of q2.

q1 q2 q3

d d


4. Find the force experienced by q1 . Assume q= 2C and a= 2cm.

q1 = +q q2 = -qd

d

q3 = +2q q4 = -2qd

d

2. Coulomb’s LawSample Problem

2

2262.0

a

qF

o

3. A cube of edge a carries a point charge q at each corner. Show that the resultant force on any one of the charges is:

3. Electric Field

3. Electric Field

AB

FABFBA

AEo = Fo/qoP

qo = +1C

“The force that a unit positive test charge (1C) will experience when placed at that point .”

3. Electric FieldElectric Field Lines An imaginary line or curve drawn through a region of space so that its tangent at any point is in the direction of the electric field at that point. This was introduced by Michael Faraday

• They are drawn from positive charge to negative charge.• They never cross.• The density of field lines on a diagram is indicative of the

strength of the field.• Neutral point : no field exists (where?)

3. Electric FieldElectric Field of a point charge q

+ -24

1

rq

qF

Eoo

Electric Field produced by a collection of charges

++

+ ++ P

i

ioo

ir

rq

qF

E ˆ24

1

==

∑1

n

i ipEE

==

rr

qqF

Eoo

ˆ24

1

==

3. Electric Field

1. Two point charges of 20nC and -20nC are situated at (1,0,0) and (0,1,0) in free space. Determine the electric field intensity at (0,0,1).

2. Three negative point charges lie along a line as shown. Find the magnitude and direction of the electric field this combination of charges produces at point P.

Sample Problem

P

-5uC

8cm

8cm

-2uC

-5uC

6cm

3. Electric Field

3. Point charges q1 and q2 of +12nC and -12nC, respectively are placed 0.10m apart (refer to figure). Compute the electric field caused by q1, the field caused by q2, and the total field (a) at point a; (b) at point b; and (c) at point c.

Sample Problem (Assignment)

q1 q1

x

y

4cm6cm4cm

(a) (b)

(c) Electric dipole!>> each molecule of a neutralinsulator is an electric dipole 13cm

13cm

3. Electric FieldMotion of Charged Particle in an Electric Field

• When a charged particle is placed in an electric field, it experiences an electrical force

• If this is the only force on the particle, it must be the net force• The net force will cause the particle to accelerate according to

Newton’s second law.

Coulomb’s LawF = qE

Newton’s 2nd LawF = maay= - |q/m|Eax = 0 (constant VX)

3. Electric FieldMotion of Charged Particle in an Electric Field

• What is the final velocity?

VX

VY

2

2

22

Etmq

VV

VVV

VV

Etmq

atV

FXF

FYFXF

INITIALXFX

FY

)(

V

Pythagorean Theorem!

5. A 50v battery is connected to two plates separated by a distance 1.0cm. The upper plate is connected to the negative voltage whereas the lower plate is connected to the positive. If an electron is released from the upper plate what is its acceleration? At what speed and kinetic energy does the electron acquire while traveling 1.0cm to the lower plate? If the electron is released to a point parallel to the two plates at initial velocity 1x10^6m/s. What is its kinetic and final velocity, given that the plates have lengths of 5cm?

Sample Problem

3. Electric Field

Note: Electric Field = Voltage/length (V/m)

4. Continuous Charge Distributions

rrq

kE ˆ∫ 2

d=

rrq

kqFo

ˆ∫ 2

d=

• Charges are distributed over a line, surface or volume.

• The net force of those charges can be obtained by integration

• This involves the concept of charge density:λ = q/L : linear charge density

σ = q/A: surface charge densityρ = q/V: volume charge density


• The total net charge q is subdivided into infinitesimally small charges dq.

dq = λdx dq = σdA

dq = ρdV

20 ||||

r

qdqkdF 2

||

r

dqkdE


• The net force and the electric field is can be decomposed into components depending on symmetry.

dF = dFx + dFy + dFz

dE = dEx + dEy + dEz


• Uniform Line of Charge


yL

qk2E

qqk2F

y

qkE

y

qqkF

4Lyy

qkE

4Lyy

qqkF

zy

dzkyλE

zy

dzλykqF

0F 0F Lq

y0

y

2y20

y

22y22

0y

L/2

L/23/222y

L/2

L/23/2220y

zx

yL

When y>>L, Equation for point charges!

When L>>y, Infinite Line of Charge!

Due to symmetry

4. Continuous Charge Distributions• Uniform Line of Charge

1. A very long, straight wire has charge per unit length of 1.50e-10 C/m. At what distance from the wire is the electric field equal to 2.50N/C?2. Find the force on a point charge +q located a distance c from the end of a rod of length L with uniformly distributed positive charge Q.

Sample Problem


+ + + + + ++qc

3. Obtain an expression for the electric force exerted on charge q1 (positron)

Sample Problem


+ +

q1

++++++---------

L

L

L

L

Q

-Q

3. Two 1.20m non-conducting wires meet at a right angle. One segment carries +2.50uC of charge distributed uniformly along its length, and the other charge carries -2.50uC distributed uniformly along it. (a) Find the magnitude and direction of the electric field these wires produce at point P, which is 6.0cm from each wire. If an electron is released at point P, what are the magnitude and direction of the net force that the wires exert on it?


Sample Problem (Assignment)

+

4. Continuous Charge Distributions• Uniform Ring of Charge

( ) ( )

( ) ( )

3z30

z

2z20

z

2z20

z

2π

2z

2π

2

0z

yx

qzk =E

qzqk=F

z

qk=E

z

qqk=F

+z

qzk=E

+z

qzq2k=F

+z

λRzk =E

+z

λRzqk=F

0=F 0=F L=q

RR

RR

dR

dR

232232

023

2023

2

2

∫∫

//

Due to symmetry

When z>>R, Equation for point charges!

When R>>z, Infinite Ring of Charge!

4. Continuous Charge Distributions• Uniform Ring of Charge

4. A ring shaped conductor with radius r = 2.50cm has a total positive charge Q = +0.125nC uniformly distributed around it. The center of the ring is at the origin. What is the electric field (magnitude and direction) at point P which is at z =40cm? A point charge q=-2.50uC is placed at the same point P. What is the magnitude and direction of the force exerted by the ring on charge q, and that of q on the ring.

4. Continuous Charge Distributions• Uniform Ring of ChargeSample Problem

5. Obtain an equation for the electric field (vector) at a point in the z-axis given the following uniform ring of charge (oriented in the xy plane). The + and – sections have equal charge densities.

4. Continuous Charge Distributions• Uniform Ring of ChargeSample Problem

Z

R

r

6. Obtain an equation for the electric field (vector) at a point P for the system of charges.

4. Continuous Charge Distributions• Uniform Ring of ChargeSample Problem (Assignment)

R

4. Continuous Charge Distributions• Uniform Disk of Charge (Uniform Sheet of

Charge)

k2π E 2πkqF

)/(1

1-12πkE

)/(1

1-12πkqF

z

1-

z

1zk2πE

z

1-

z

1z2πkqF

)w(z

wdwzk2π E

)w(z

wdwz2πkq F

wz

z

wz

σ2πwdwqk cosθ

wz

dqqkdF

0F 0F σ2πwdw dq σπw σA q

z0z

22z220z

22z220z

R

0 2322

z

R

0 2322

0z

22220

220

z

yx2

zRzR

RR

When R>>z, Infinite Sheet of Charge!

Due to symmetry

When z>>R, becomes a point- charge system

4. Continuous Charge Distributions• Uniform Disk of Charge

5. A negative charge, -Q, is uniformly distributed over the circular sector shown. The sector is situated in the xy plane whose center of curvature is located at the origin. Obtain the x and y components of the electric field at the origin.

a

b

- - - -- - -- ---

------ a

- - - -- - -- ---

------

+ ++ +++++

++++

+

b

6. Given the following system of charges. Find the electric field at the center. The two sectors have uniformly distributed charges +Q and –Q respectively

4. Continuous Charge Distributions• Uniform Disk of ChargeSample Problem

7. A charged sheet with σ1 = 100nC/m^2 occupies the z = -3 plane, and a second charged sheet with σ2 = -100nC/m^2 occupies the z = +3 plane. Find the electric field at (a) the origin, (b) P(0,0,6), (c) P(0,0,-6).

8. Charge lies in the z=-3 plane in the form of a square sheet defined by -2≤x≤2, -2≤y≤2 with charge density σ = 2(x^2 + y^2 + 9)^(3/2) nC/m^2. Find E at the origin.

4. Continuous Charge Distributions• Uniform Sheet of ChargeSample Problem

4. Continuous Charge Distributions• Volume Charge

rdVRk

F

rdVRk

E

dVdQ

rR

kdQEd

v

v

ˆ

ˆ

ˆ

∫

∫

2

2

2

=

=

=

=

Rdq dE

9. Find the total amount of charge contained within the region bounded by: 2<r<4, 0<Φ<π/2, -2<z<2. Find also the electric field on the origin generated by this charged region. ρ = +100nC/m^3. 10. Find the electric force experienced by the positron due to a charged solid sphere with radius = 0.1m and ρ = 100uC/m^3. The positron is 100m away from the sphere.

4. Continuous Charge Distributions• Volume ChargesSample Problem

4. Gauss Law

Karl Friedrich Gauss

• 1777 – 1855• Made contributions in

– Electromagnetism– Number theory– Statistics– Non-Euclidean

geometry– Cometary orbital

mechanics– A founder of the

German Magnetic Union

Oh no daddy, you are so

wrong!

It has been said that at the age of 3 Gauss corrected his father's computations.

TRIVIA One of the problems Gauss’ math teacher gave the class was "add all the whole numbers from 1 to 100". His teacher Master Büttner was amazed that Gauss could add all the whole numbers 1 to 100 in his head. Master Büttner didn’t believe Gauss could do it, so he made him show the class how he did it. Gauss showed Master Büttner how to do it and Master Büttner was amazed at what Gauss just did. The system of how he did it is add 1+100, 2+99, 3+98…49+52 and he had 50 pairs of 101 and he multiplied 101x50 to get 5050, which is the answer.

Electric Flux (Φ) • a measure of the number of

field lines which pass through a surface.

• the electric field times the component of the area perpendicular to the field.

Preliminary Concepts

3. Gauss Law

+Gaussian Surface

“an imaginary surface”

area A

E n̂

outward.

directed and area the to

normal vectorunit a is n

AnA where

A•E=

ˆ

ˆ≡

3. Gauss LawSample Problem

1.Calculate the flux of the electric field E, through the surface A, in each of the three cases shown:a) =b) =c) =

3. Gauss LawSample Problem2. What is the flux Φ of the electric field

through this closed surface?

3. Gauss LawSample Problem3. What is the electric flux through a spherical shell

with a point charge at the center?

3. Gauss LawSample Problem4. What is the electric flux through each of the

6 faces of the cube?

jixE ˆˆ 34 +=

• Some more concepts

3. Gauss Law

• The flux is positive for field lines that leave the enclosed volume

• The flux is negative for field lines that enter the enclosed volume

• If a charge is outside a closed surface, the net flux is zero. As many lines leave the surface, as lines enter it.


5. For which of these closed surfaces (a, b, c, d) the flux of the electric field, produced by the charge +2q, is zero?

Gaussian Surfaces

• Finally, the statement of Gauss’ Law:

• or

where D = ЄE, D = Electric Flux Density

3. Gauss Law

o

enclosedQ=•=

AE

The total flux within a closed surface …

… is proportional tothe enclosed charge.

enclosedQ=•= AD


6. Calculate the flux of the electric field and the flux density (D)for each of the closed surfaces a, b, c, and d

Surface a, a =Surface b, b =Surface c, c =Surface d, d =


7.Find the electric field of an infinite sheet of charge. (Note: “σ “ is surface charge density)

Note: charge density is the amount of charges on either a line, a surface or a volume.

λ = q/L : linear charge densityσ = q/A: surface charge densityρ = q/V: volume charge density


8. Obtain an equation for the electric field of a uniformly charged solid sphere (of total charge “Q”)of radius R at r<R and r>R.

RE

E


8. Solution


9. Find the electric field inside and outside at a distance x from the center of an infinite plane of thickness s and uniform surface charge density .

R


9. Solution

R

E

xs/2

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