Electromagnetic Field Theory [Chapter[Chapter 4 ... · Coulomb’s law and electric field intensity Electric fields due to continuous charge distributions Study electric flux density

EElectromagnetic lectromagnetic FField ield TTheoryheorygg yy[Chapter[Chapter 4: Electrostatic Fields]4: Electrostatic Fields]

Prof. Kwang-Chun [email protected]: 02-760-4253 Fax:02-760-4435Tel: 02-760-4253 Fax:02-760-4435

OutlineOutline

What is electrostatics?Coulomb’s law and electric field intensityElectric fields due to continuous charge distributionsStudy electric flux density due to electric fieldGauss’s law and the first one of Maxwell’s equationsProcedure for applying Gauss’s law to calculate the electric fieldBasic concept of electric potential known as voltageRelationship between electric field and potentialWhat is an electric dipole and flux lines?Energy density in electrostatic fields

Dept. Information and Communication Eng. 2

What is electrostatics?What is electrostatics?

Electrostatic Phenomena:L l d hi h h b bb d ith ilkLeave glass rod which has been rubbed with silkBring another glass rod close to that one. Two rods separate furtherfurtherBring a plastic rod close to that one. Two rods approach each other

(Movement of charges)

(Electric Forces)

(Movement of charges)

(Electric Forces)


What is electrostatics?What is electrostatics?

How do we interpret the results?Explanations: There exist two kinds of charges Unlike charges attract; Like charges repel Unlike charges attract; Like charges repel May exist electric fields due to their divergence around charges

Definition of electrostatics:Study the effect of static (or time-invariant) electric fields, due to charges at rest

Practical Applications:Oscilloscope, Ink-jet printers, almost all computer peripheral d idevices, ...


CoulombCoulomb’’s Laws Law

Two fundamental laws governing electrostatic fields:C l b’ lCoulomb’s law Experimental law formulated in 1785

by Charles Coulomb, who is French i

Run Animation !Run Animation !

Suspension army engineerGauss’s law

Law of electrical force:

Suspension head

Force law that governed the interactions between two charge objectsUsing two small electrically charged

Fiber

Using two small electrically charged spheres, he deduces the force is proportional to the product of the

1Qp p ptwo charges and followed the inverse square law of distance between them

1Q

2QR



1 22

Q QF

If we insert a proportionality constant, Coulombs law may be written as

2R

Q Qwritten as

where the measured value k is

1 22

Q QF kR

The permittivity (or dielectric constant) of medium is

9 2 2 19 10 N m /C m/F4 o

k

The permittivity (or dielectric constant) of medium is In air, it becomes

medium r o

9

0 0101.0006 F/m36medium

If point charges are located at points having position vector , then the force on due to is

1 2and Q Q

1 2and r r 2Q 1Q

36



1 2 Q QF a Q Q

Here and

1212 204

RF aR

R r r R

1Q 2Q

1r

12 2 1 R r r

21

F 12

F

Here, and2 1 12 R r r R

12

12

R

RaR

1r2r

Combining two equations, we have

Some useful notes:

12 R 1 2 2 1

12 30 2 14

Q Q r rF

r rSome useful notes:

The force on due to is 21 12 F F2Q1Q

and must be point charges and at rest!

Like charges repel;Unlike charges attract

2Q1Q and must be point charges and at rest!2Q1Q



Principle of superposition:If there are N charges located, respectively, at points with position vectors , the force on a charge located at point is the vector sum of the forces

1 2, , , NQ Q Q1 2, , Nr r rrQlocated at point is the vector sum of the forces

exerted by the chargesr

1 2, , , NQ Q Q

N Q r rQ

Q

1r

1r r

3104

Nk k

k k

Q r rQFr r

1

2r r

NF

Example 4.1:Consider point charges and 2 (1 mC)Q 3(2 mC)Q

r Nr r 2F

1F

p glocated at (3,2,1) and (1,1,4), respectively Calculate the electric force on a charge located at (0,3,1)

2 ( )Q 3( )Q

1(10 nC)Q



Solution:U i th f ll i ti

13F

F

Using the following equation

1 31 21

Q QQ QF a a

1F

we have

1 21 12 132 2

0 12 134

F a a

R R1r

2r12F

we have

1 3.797 7.149 0.637 mN

x y zF a a a Origin

3r

where

12 1 2 130,3,1 3,2, 1 , 0,3,1 1, 1, 4 , R r r R

131212 13

12 13

3,1,2 1,4, 3,

9 1 4 1 16 9

RRa a

R R



Example 4.2:C l l t th l t i f h 2QCalculate the electric force on a charge +2QSolution: y

Q Q

71 105

Q Ca cm

12 2Q2 2Q2Q

x

3Dept. Information and Communication Eng. 10

Visual EMT using MatLabVisual EMT using MatLab

Will run the MATLAB program, which computes the force between point charges

Plots the position of each charge Displays the net force on each charge

Run Coulomb.m!Run Coulomb.m!


Electric Field IntensityElectric Field Intensity

Physical Meaning:Th i t f t d b th h h i

yy

There exists forces exerted by the charge everywhere in space surrounding a chargeHow can we detect the forces?How can we detect the forces? Answer:

Place a positive test charge at arbitrary point P, and measure th f ti it

tQthe force acting on it



Mathematical Meaning:

yy

The force per unit charge (i.e. the force acting on test charge of +1C) is called Electric Field Intensity, which yields

1 QQF Q

Thus RaE

2 20 0

14 4

tR R

t t

QQF QE a aQ Q R R

Thus,

320 0

4 4t

Rt

Q r rQE aR r r

Ra

tr

For N point charges located at , the electric field intensity

r1 2, , , NQ Q Q

1 2, , Nr r rat point is

310

14

Nk k

k k

Q r rE

r r

r



Example 4.3:

yy

Find the total electric field intensity at P due to Q1and Qand Q2

Solution:

E E E 2r r 2r

1 2

1 1 2 23 34 4

E E EQ r r Q r r

r r r r 1E

1a1r r

r1r

0 1 0 24 4 r r r r2a

1

E

2E



Example 4.4: P i t h 5 C d 2 C l t d t (2 0 4) d ( 3 0 5)

yy

Point charges 5 nC and -2 nC are located at (2,0,4) and (-3,0,5), respectively (a) Determine the force on a 1 nC point charge located at (1,-3,7)(a) e e e e o ce o a C po c a ge oca ed a ( , 3,7) (b) Find the electric field at (1,-3,7)

Solution:

E

(a)

(b)



How can we visualize the electric field intensity?

yy

Done by drawing continuous lines from the charge which are everywhere tangent toThese lines are called Electric Flux (or field) Lines

E

These lines are called Electric Flux (or field) Lines

The spacing of lines is inversely proportional to the The spacing of lines is inversely proportional to the The spacing of lines is inversely proportional to the The spacing of lines is inversely proportional to the strength of the field !strength of the field !


Electric Field IntensityElectric Field Intensityyy


Electric Field IntensityElectric Field Intensityyy


Electric Fields due to Charge DistributionsElectric Fields due to Charge Distributions

Various charge distributions and charge elements:

where represent the line charge surface charge and volume charge density

2 3C/m , C/m , and C/mL S v charge, surface charge, and volume charge density, respectively.

What is the Electric Fields due to these charge distributions ?What is the Electric Fields due to these charge distributions ?gg



Consider a differential volume charge :Electric field intensity at a point P due to that charge then is

vdQ dvElectric field intensity at a point P due to that charge then is

3 3( )

4 4

vr r r rdvdQdE rr r r r

Integrating over volume V,

0 04 4 r r r r

vdQ dvg g ,

we have 1( )

v r rE r dv

r r

r3

0

( )4

V

E r dvr r

f i

V rOrigin

3

0

1( )4

S

S

r rE r dS

r r

30

1( )4

L r r

E r dLr r

Surface Line

04 S r r 04 L r r



Example 4.5: Line ChargeC id li h ith if h d itConsider a line charge with uniform charge density extending from A to B along z-axisThen, associated with element dl at is

L

0,0, z

dE,

where

304

Ldl RdER

z

dE

where

,

, , 0,0,

dl dzR x y z z R

, , 0,0,

x y z

z

R x y z zxa ya z z a

a z z aCartesian:

Cylindrical:

22

2 21 tan sec

R z zThus,


x y


Considering , and integrating from A to B it becomes

2tan , secz z dz d from A to B, it becomes

3/ 22204

zLa z z a

E dzz z

Usingsint

2 2 3

3 30

sec cos sin4 sec

zL

a ad

tan ,cos

1seccos

1

2

1

0

0

cos sin4

Lza a d

cos

10

2 1 2 10

sin sin cos cos4

L

zE a a

For infinite line charge, that is, , it is1 2/ 2, / 2

2

LE a

02



Example 4.6: Surface ChargeC id i fi it h t f h i th l ithConsider a infinite sheet of charge in the xy-plane withThen, associated with element 1 is

SdE

z

RR

y


x


zS S a hadS RdE d d

Due to the symmetry of charge distribution, the contribution

3 3/ 22 20 04 4

S SdE d dhR

y y g ,along cancelsThus, the total electric field intensity is

2

a

2

3/ 2 3/ 22 2 2 20 00 0 0

24 4

S S

z z zhh d d dE dE a a

h h

1/ 22 20 0

12 2

S Sz z

h a ah

Finally, we can note that if the charge is in the xy-plane,has only a component normal to the sheet

0 00

h

E y p



Practical Application Examples:[Configuration of A Real Capacitor][Configuration of A Real Capacitor]

z

E

E

E

E

E

E

0 E E E 0

E E E

2 S S

E E E

a a

0 0

22

S Sz za a



[Configuration of Cathode Ray Oscilloscope][Configuration of Cathode Ray Oscilloscope]

xVx

cathodeanode

y

z

d fl tiscreen

V

AV

ydeflectionplates

yV

controls the deflection of electron along x-axis controls the deflection of electron along y-axis

xV

yV


Electric Fields due to Charge DistributionsElectric Fields due to Charge Distributions[TV tube with electron[TV tube with electron--deflecting deflecting

charged plates (orange)]charged plates (orange)]



[Configuration of Color Ink[Configuration of Color Ink--Jet Printer]Jet Printer]

0V

Nozzle vibrating at ultrasonic frequency sprays ink in the form of dropletsThese droplets acquire charge proportional to the character to be printed

while passing through a set of charged platesVertical displacement of an ink droplet is proportional to its charge Blank space between characters is achieved by having no charge, then the

i k d l t ll t d b tt ( t ld i )ink droplets are collected by gutter ( at old version)



[Configuration of Microstrip Lines][Configuration of Microstrip Lines]

rElectric fields



[Microwave Oven][Microwave Oven]


Example 4.7: A circular disk of radius a is uniformly h d If th di k li th l2charged . If the disk lies on the z=0 plane,

(a) Show that at point (0,0,h)

2[ / ]S C m

h

(b) From this derive the electric field

2 20

12

Sz

hE ah a

E(b) From this, derive the electric field due to an infinite sheet of charge on the z=0 plane S

ron the z 0 plane.

(c) If , show that is similar to the field due to a point charge

a hE

S

dSa

the field due to a point charge



Solution(a) Consider an element of area of the disk The contribution due to is dS d d

dS

dS h 3 3/ 22 2

0 04 4

S zSdS a hadS rdE

r h

Since the sum of the contribution along gives zero, we have2

3/ 2 3/ 2

a aS Shh d d dE

3/ 2 3/ 22 2 2 20 00 0 04 2

1

z

a

Eh h

h h2 2 2 2

0 00

1 12 2

S Sz z

h h E E ah a h



(b) As

Sa E a(b) As

(c) For very large h, using a binomial series we have0

,2

za E a

1/ 21/ 22

2 2 2

11 1 1 1

1

h aha h a

2 2

1

1 11 1

h

a a

Then, the electric field becomes similar to a field due to a point

1 12 2

h h

charge 2 2

2 2 20 0 04 4 4

S S

z z za a QE a a a

h h h



Example 4.8: A square plate described by 2 2, 2 2, x y

carries a surface charge density (a) Find the total charge on the plate

212 [ / ]y mC m0z

(b) Find the electric field intensity at P(0,0,10) Solution: (a)

2 2

12Q dS d d (a)

r r r

2 22

12

12(4) 2 192

S SQ dS y dxdy

ydy mC

(b) Using S

r

0

12(4) 2 192ydy mC

( ) d ddS

r32

0 0

( )4 4

S Sr

dS dS r rE ar r r



, in which (0,0,10) ( , ,0) ( , ,10) r r x y x y

we have

32 2

3/22 2

(12 10 ) ( , ,10)14 100

y x y dxdyE

2 20 2 2

2 2 27

4 100

(1/ 2) ( )(108 10 ) 2

x y

d ya dx

By the symmetry, and 21/ 2ydy d y

3/22 22 0

27

(108 10 ) 2100

1 1

za dxx y

2

( )( )

f x dxf x a

7

2 22

22

1 1( 216 10 )104 100

104

za dxx x

2ln ( ) ( )f x f x a

27

22

104( 216 10 ) ln 16.46 [100

z zx xa a Mx x

/ ]V m



Example 4.9:2Planes and , respectively, carry charges

and If the line carries charge calculate0 2x z

2x 3 y 210 nC/m215 nC/m

10 nC/mEIf the line carries charge , calculate

at P(1,1,-1) due to the three charge distributions

0, 2x z 10 nC/m E

Solution:The contributions to at

point P(1 1 1) due to the

E

point P(1,1,-1) due to the infinite sheets are

1E

2E


3E


Since , the unit vector gets as

Then, the electric field is

R

The total field can be obtained by

a



Volume Charge:C id th l h di t ib ti ithConsider the volume charge distribution withThen, it is not so easy to calculate the total electric field intensity, and the result

v

e s y, a d e esucan be obtained through tedious mathematical proced reprocedureIf so, what is an easier way to get the result?y gThe answer is Gauss’s law, which will be discussed laterdiscussed later in details



What is the electric flux line?When a test charge is at one point in electric field it movesWhen a test charge is at one point in electric field, it moves along a certain path by force acting on the test chargeThis path is called a line of force or electric flux line

Electric flux lines between chargesElectric flux lines between charges



Example 4.10: Th fi h th l t i fi ld li f t f tThe figure shows the electric field lines for a system of two point charges

What are the relative magnitudes of the charges?What are the relative magnitudes of the charges?What are the signs of the charges?In what regions of space is the electric field strong? or

k?weak?



Solution: There are 32 lines coming from the charge on the bottom, while

there are 8 converging on that on the top. Thus, the one on the bottom is 4 times larger than the one on the topg p

The one on the bottom is positive; field lines leave it. The one on the top is negative; field lines end on iton the top is negative; field lines end on it

The field is strong near both charges. It is strongest on a line connecting the charges Few field lines are drawn there butconnecting the charges. Few field lines are drawn there, but this is for clarity. The field is weakest to the upward of the top charge


Electric Flux DensityElectric Flux Density

Definition:Let’s assume that a point charge is enclosed in an imaginary

yy

Let s assume that a point charge is enclosed in an imaginary sphere, which is called Gaussian surfaceElectric flux lines pass perpendicularly and uniformly through the surface of sphereThen the flux lines per unit area isThen, the flux lines per unit area is called the electric flux density, which is defined as

20 [C/m ]

D E +

D


Electric Flux DensityElectric Flux Density

Thus, when the orientations of the surface defined and the electric flux lines are different the total electric flux through

yy

electric flux lines are different, the total electric flux through the surface S can be evaluate by

S

D dSS

dS

D

dS dSdS

D

dS dSdS


GaussGauss’’s Laws Law

Definition:The total electric flux through any closed surfaceThe total electric flux through any closed surface (Gaussian surface) is equal to the total charge enclosed by that

surface. That is,

vQ D dS dv( 0)

( 0)

Applying divergence theorem to

S V ( 0)

the middle term in equation above

D dS Ddv ( 0) ( 0) No enclosedNo enclosed

S V

Gauss' law is a form of one of Maxwell's equations, Gauss' law is a form of one of Maxwell's equations, the four fundamental equations for electricity and the four fundamental equations for electricity and magnetism magnetism

( 0) No enclosedNo enclosedchargecharge!!

magnetism. magnetism.


Applications of GaussApplications of Gauss’’s Laws Law

we have

G ' l i f l t l f th l l ti f l t i

vD

Gauss' law is a powerful tool for the calculation of electric fields when they originate from charge distributions of sufficient symmetry to apply ity y pp yPart of the power of Gauss' law in evaluating electric fields is that it applies to any surface It is often convenient to construct an imaginary surface called a Gaussian surface to take advantage of the symmetry of the physical situationphysical situation



Here, Dashed-Lines are Gaussian Surfaces



Example 4.11: Point ChargeSuppose a point charge is located at the origin Determine at a point P

S l ti

D

Solution:Since is everywhere

normal to the Gaussian

r rD D a

surface applying Gauss’s law gives

2

D

Thus, the electric flux density is

24 r rS S

Q D dS D dS D r PQ

, y

24

rQD ar



Example 4.12: Infinite Line ChargeSuppose the infinite line of uniform charge lies along z-axis Determine at a point P

S l ti

LD Line charge C/mL

Solution: Choose a cylindrical surface containing

P to satisfy symmetry condition is constant on and normal to

the surface Thus, the electric flux density is

D D a

l P, y

2

L

S

l Q D dS

D dS D l

D

LD a

2 S

D dS D rlArranging

2 r


Applications of GaussApplications of Gauss’’s Laws LawExample 4.13: Infinite Sheet of Charge

Suppose the infinite sheet of uniform charge lies on z=0SSuppose the infinite sheet of uniform charge lies on z 0plane. Determine at a point P

SD

Surface charge

C/m2S

D

dS

DD

dS

DP

x

D

D D

yz



Solution:Choose a cylindrical box, cutting symmetrically by the sheet of

charge. Then, we have two surfaces parallel to the sheetThen, we have two surfaces parallel to the sheet is normal to the surface

z zD D a

A Q D dS D dS dS D A A

Note that evaluated on the sides of the box is zero

S z zS top botton

A Q D dS D dS dS D A A

D dS

Thus, the electric flux density is

z>0

S a , z>02

, z<02

z

Sz

aD

a2



Example 4.14: Suppose two infinite sheets of uniform charge ,as shown in figure. Determine everywhere in space

S

DDetermine everywhere in space

Solution:D

SS

S S S S

SS

D DD

D

D

SD ( Between plates )



Example 4.15: Uniformly Charged Sphere C id h f di R ith if h 3[C/ ]Consider a sphere of radius R with a uniform charge Determine everywhere

Solution:

3[C/m ]vD

D

dSSolution:Construct Gaussian surfaces

for and , respectivelyr R r R

Gaussian surfaceD

ra

na

For , the total charge enclosed by the surface of radius r is

r R

r

343enc v vQ dv r

And, the electric flux is24

r rD dS D dS D r

S



Thus, using Gauss’s law, we have3 24 3 24, 4

3enc v rQ r D r

for 03

v r

rD a r R 3vrD

For , since total charge enclosed by the surface is

3r R

3

3

2

vR

from Gauss’s law the electric

34 ,3

enc v vQ dv R23r

from Gauss s law the electric flux density becomes

3 24, 4Q R D r

r

, 43enc v rQ R D r

3

2 for 3

v rRD a r ar




Electric PotentialElectric Potential

Similarity between gravitational and electrical potential energy :

Work done by gravitational force is La Lagravitational force is

Then, the gravitational GW mg L mgL

L L

Q, gpotential energy decreases, and is equal to the negative of work

Q

Qto the negative of work done as follows:

GU mg L mgL

Similarly, as a positive charge moves in the direction of an electricdirection of an electric



field, it experiences an electric force and the work done by th l t i fi ld bthe electric field becomes

Then the positive charge loses the electric potential energy andEW F L QE L QEL

Then, the positive charge loses the electric potential energy and is equal to the negative of the work

EU QE L QEL (Potential energy is negative !)(Potential energy is negative !)

because it moves from a point of higher potential to a point of lower potential

The electric potential difference between two points isThe electric potential difference between two points is defined as the change in potential energy of a charge Qdivided by the chargey g

EUV E LQ



Suppose a charge Q is moved from pt A to pt B through a region of space described byregion of space described by electric field

The potential difference between

E

E

The potential difference between points A and B, , is defined as the change in potential energy (f ) f

B AV V

(final minus initial value) of a charge, Q, moved from A to B, divided by the charge

A Bdivided by the charge

EAB B A

UV V VQ

QE L



where the potentials at pt A and pt B are defined as the potential energy per unit charge:

,A BV VUpotential energy per unit charge:

If , 10 J of work is required to move a 1 C of 10 [V] 0ABV

, ,

,E A B

A B

UV

Q

, qcharge between two points that are at potential difference of 10 V

Now let’s consider a charge Q

[ ]AB

Now, let s consider a charge Qmoving from pt A to pt Balong arbitrary path

Aalong arbitrary pathin electric fieldThen, the potential energy in

dlQ

E

, p gydisplacing the charge by is

dl

B

dl

EdU F dl QE dl

F

F Dept. Information and Communication Eng. 58


And, the total potential energy in moving from pt A to pt B isB

Thus the potential difference between A and B is denoted by

B

EA

U Q E dl

Thus, the potential difference between A and B is denoted byB

EAB B A

UV V V E dlQ

Is the potential difference independent of path taken?C id th f t t t fi ld

AQ

EConsider the case of constant vector field

Then, the potential difference at the direction along path A-B is

B C

hE

E

at the direction along path A-B is

A

rh E

dl

B

AB B AA

V V V E dl EhA



For long way round along path A-C-B, it becomesC B C

I t l d t d d

sin 0 sin

C B C

AB B AA C A

V V V E dl E dl E dl Er Eh

Integral does not depend upon the path chosen to move from Ato B so that the integral is

E

Bgthe same for BOTH paths

Example 4.16:

dr dl

Br

Assume that the electric field due to a point charge Q located at the origin we have

QA Ar

r

at the origin, we have

204

rQE a

r



Then determine the potential difference between A and BThen, determine the potential difference between A and BSolution:

2

0 0

1 14 4

B

A

r

AB B A r rB Ar

Q QV V V a drar r r

Here, if as , the potential at any point due to a point charge Q located at the origin is

0AV Ar Br r

4

rQV E dlChoose infinity as referenceChoose infinity as referencebecause the potential at infinitybecause the potential at infinity

04 ris supposed to be zero (Ground) !is supposed to be zero (Ground) !



For N point charges located at , the t ti l t i

1 2, , , NQ Q Q 1 2, , Nr r rpotential at is

1( ) (point charges)4

NkQV r

r

For continuous charge distributions, the potential at b

104 k kr r

rbecomes

0

1( ) (line charge)4

L r

V r dlr r

0

0

4

1( ) (surface charge)4

L

S

S

r r

rV r dS

r r

0

0

1( ) (volume charge)4

S

v

V

rV r dv

r r0 V


Electric PotentialElectric PotentialExample 4.17:

Charge is uniformly distributedCharge is uniformly distributed within a rod.Find the electric potential on

L

pthe perpendicular bisector of the charged rodSolution:Since , the potential

due to at a point P becomesL LdQ dy

dQdue to at a point P becomesdQ

2 20 0

1 14 4

LdydQdVr x y

Thus, the potential due to the rod is2 2

2 2 2 2ln

4 4

a

L Ldy a x aV2 2 2 2

0 04 4

a x y a x a



Example 4.18:Ch i if l di t ib t d ithi h i l h ll fCharge is uniformly distributed within a spherical shell of radius a. Then, the electric field is

v

3

f f r a

Find the potential everywhere

20 0

for 0 , for 3 3

v r v rr aE a r a E a r a

r

Find the potential everywhere For , r a

3 3

2

r

v vout r r

a aV r a dra v

For

20 03 3

out r rr r0 ,r a

r

vrV V d

( )outV r( )inV r

E

02

2

3v

in out r ra

v

V r V r a a dra

ra

a

02 3a



Practical Application: battery"D-cell" or "AA-cell" has a rating of 1.5 volts, which means that every charge moving from the negative side of the cell to the positive side will do 1 5 Joules worth of workthe positive side will do 1.5 Joules worth of workThe difference between the D-cell and the AA-cell is that the D-cell has more charges, so it will last longer As shown in figure, every negative charge that passes through the light bulb does 1.5 joules worth of work which

k i i ff li hmakes it give off light Bulb

D AA


Electric PotentialElectric PotentialCuriosity

Question: How can a bird (like this bluebird) stand onQuestion: How can a bird (like this bluebird) stand on a high voltage line without getting zapped?Answer: Because there is no differenceAnswer: Because there is no difference in Voltage across his feet!When does the current flow?When does the current flow? If there are voltage or potential difference, then the current starts

to flow from high voltage to low voltage

Dept. Information and Communication Eng.

66




B t h ll bi d it th li b th f t thBut when a small bird sits on the power line, both feet are on the same voltage line! (no potential difference and no current flow !)

If one leg B of a chicken is on the ground and the other one A is on the power line, then there are potential difference between these two legsthese two legs

Therefore, there is a flow of charge and eventually the chicken , g ywill be barbecued, as shown in the figures


Relationship between E and VRelationship between E and V

Since the potential difference is independent of the path t k htaken, we haveThat is,

BA ABV V E

Applying Stokes’s theorem

0

BA ABV V E dl

pp y gto the equation above, it becomes

0

E dl E dS 0 E

It is called the second Maxwell’s equationThe vector field is said to be conservative Thus, the electrostatic field is a conservative field



Now, let’s define the relationship between and V, ti f i th ti t

E

satisfying the conservative propertyConsider a point charge in an electrostatic fieldTh th t ti l diff

QThen, the potential difference between two points is

/EdV dU QE

Rearranging for , we have cos

E

L

Q

E dl E dl E dl

LEQ

L

dlg g ,L

(V/m)LdVEdl

Q dl

BV

If is and -axis, it becomesdl ,x y z

, ,x y zdV dV dVE E Ed d d

V V dV

AV

x y zdx dy dz B AV V dV



Then, the vector at any point is given byE

(V/m)

x x y y z zE E a E a E a

V V Va a a

Finally, we have

(V/m) x y za a a

x y z

(V/m)

x y zE a a a V V

x y z

Note that: Since the curl of gradient of scalar function is always zero Since the curl of gradient of scalar function is always zero

, the electrostatic field must be a conservative field 0V


Electric Dipole and Flux LinesElectric Dipole and Flux Lines

What is an electric dipole?Two point charges of equal magnitude but opposite sign are separated by a small distanceby a small distance

Now, use the potential to calculate the field of a dipole

Ethe field of a dipole

Remember how messy the direct calculation was?The potential is much easier to calculate than the fieldsince it is an algebraic sum of 2 scalar terms given as



1 1( )

r rQ QV r

Rewriting this for special case r d

0 0

( )4 4

Q QV rr r r r

it becomes

2

cos( )4Q dV r

r

becauseCalculating in spherical

04 r

2cos , r r d r r r

ECalculating in spherical

coordinates, we have1

V VE V a a

E

3 2cos sin4

r

r

E V a ar r

Qd a ar04 r



Electric flux line

( ) 0V r ( ) 0V r

( ) 0V r

Equipotential surface ( ) 0V r


Electric Dipole AntennaElectric Dipole Antenna


Electric Dipole AntennaElectric Dipole Antenna



Equipotentials of a charged sphere:The electric field of the charged sphere has spherical symmetryThe potential depends only on Electric flux lineThe potential depends only on the distance from the center of the sphere

Electric flux line

AB

An equipotential surfaceequipotential surface is a surface on which all points are t th t ti lat the same potentialThe electric field at every point

on an equipotential surface q pis perpendicular tois perpendicular to the surface

Equipotential surface



Why??Why?? Along the surface, there is NO change in V (it’s an

equipotential !)

0,B

B AA

V V E dl

((Orthogonal relation between two factors)Orthogonal relation between two factors)

0E dl

Thus, no work is required to move a charge at a constant speed on an equipotential surface

((Orthogonal relation between two factors)Orthogonal relation between two factors)

speed on an equipotential surface



Equipotential lines on the surface of the human body reflect the electric dipole nature of the heart



D th t ti l d t t it i tDraw the potential due to two opposite point charges in three-dimensional space, and confirm the equipotential lines from theconfirm the equipotential lines from the resulting shape

Run potential.m!Run potential.m!


Energy Density in Electrostatic FieldsEnergy Density in Electrostatic Fields

In electrostatic fields, we have, in a mathematical sense, t t f fi ld

y yy y

two types of fieldVector fields (electric fields) and scalar fields (potential)

H h i i d l i t th ki fHowever, when we inquire more deeply into the workings of the universe, we find that the fundamental nature of reality is the presence of energyis the presence of energy

The electric field is primarily an energy density fieldThe electric field surrounding a charge stores energy in spaceThe electric field surrounding a charge stores energy in spaceElectric potential does the pulling apart of the positive and negative layers of space and the pulled-apart-ness stores and transfers energy



Consider a system with just two charges

y yy y

Assume is the electric potential due to charge at a point PWhen a second charge moves from infinity to P without

21V 1Q

QWhen a second charge moves from infinity to P without acceleration, the potential energy isSince the potential at point P due to is P

2Q21 2 21U Q V

1Q

2Q

P

12r

1Q

1 1 221 21

0 12 0 12

1 1, 04 4

Q Q QV Ur r

1Q

12

Because the second charge moves from lower potential to higher potential !

Now, add a third charge to the system(That is, a third charge moves from infinity to a point without acceleration)

3Q



Then, since the potential at the point due to and is

y yy y

1Q 2Q

th t ti l i d fi d2Q

P1 231 32

0 13 23

1 ,4

Q QV Vr r

the potential energy is defined as

12r 23r1 3 2 3

31 321

4Q Q Q QU Ur r

The total potential energy required to assemble the charges is

1Q3Q13r

0 13 234 r r

to assemble the charges is1 3 2 31 2

21 31 320 12 13 23

14E

Q Q Q QQ QU U U Ur r r

Rewriting the potential energy as , we have

12E E EU U U


Energy Density in Electrostatic FieldsEnergy Density in Electrostatic Fieldsy yy y

1 3 2 3 1 3 2 31 2 1 21 1 1Q Q Q Q Q Q Q QQ Q Q QU

0 12 13 23 0 12 13 23

3 32 1 1 2

2 4 4

1

EUr r r r r r

Q QQ Q Q QQ Q Q

1 2 30 12 0 13 0 12 0 23 0 13 0 232 4 4 4 4 4 4

1

Q Q Qr r r r r r

QV Q V Q V

where V1 , V2 and V3 are the total potentials at Q1 , Q2 and Q3

1 1 2 2 3 32QV Q V Q V

In general, if there are n point charges, it becomes1 n

1

12E k k

k

U Q V



If, instead of point charge, it is a continuous charge di t ib ti th ti b i t ti

y yy y

distribution, the summation becomes integration 1 1 1 1

2 2 2 2E vvol vol vol vol

U Vdv D Vdv D Vdv D Edv

vol vol vol vol

D V VD D V

2 3large

1 1 1, , 0r

V D VDr r r

Developing further, we can define electrostatic energy density as

22

00

1 12 2 2

EE

dU Du D E Edv



Draw the energy density due to one point charge with negative value in three dimensional spacewith negative value in three-dimensional space

Run energyDensity m!Run energyDensity m!Run energyDensity.m!Run energyDensity.m!


Homework AssignmentsHomework Assignments

Problem 4.1:Let us illustrate the use of the vector form of Coulomb’s law by locating a charge of at

d h f t i

41 3 10 [C]Q

4 and a charge of at in a vacuum. Find the force exerted on by

42 10 [C]Q

2Q 1Q 1,2,3M 2,0,5N

Problem 4.2:

Find at caused by four identical charges located at

E

3[nC]P P P

1,1,1P

charges located at and , as shown in Fig.

1 2 3, , ,P P P

4P

87


Problem 4.3:Find the total charge contained in a length of the electron beam shown in Fig.

2 cm

Problem 4.4:Given that

312 nC/m ,1 2

0 ,otherwisev

Determine everywhereProblem 4.5:

D

Determine the electric field due to the potential 2 1 sinV z

88


Problem 4.6:Two dipoles with dipole moment and are located at points and , respectively. Find the potential at the origin

5 nC/mza 9 nC/mza

0,0, 2 0,0,3Find the potential at the origin.

P bl 4 7Problem 4.7:A spherical charge distribution is given by

,

0

ov

r r aa

r a

Find V everywhere.

0 , r a

89

Documents

Electromagnetic Field Theory [Chapter[Chapter 4 ... · Coulomb’s law and electric field intensity Electric fields due to continuous charge distributions Study electric flux density