Electromagnetic Field Theory - HANSUNGkwangho/lectures/EMT/2019/... · 2019. 4. 23. · Prof....

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

Electromagnetic Field Theory(Chapter 5: Electrostatic Fields)

Dept. Electronics and Information Engineering

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

Outline

2


Electrostatic Phenomena:Leave glass rod which has been rubbed with silkBring another glass rod close to that one. Two rods separate furtherBring a plastic rod close to that one. Two rods approach each other

(Electric Forces)

(Movement of charges)

What is electrostatics?

3


How do we interpret the results?Explanations: There exist two kinds of charges 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 devices, ...

What is electrostatics?

4


Two fundamental laws governing electrostatic fields:Coulomb’s law Experimental law formulated in 1785

by Charles Coulomb, who is French army engineer

Gauss’s lawLaw of electrical force:

Force law that governed the interactions between two charge objectsUsing two small electrically charged spheres, he deduces the force is proportional to the product of the two charges and followed the inverse square law of distance between them

Run Animation !

Suspension head

Fiber

1Q

2QR

Coulomb’s Law

5

1 22

Q QFR


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

where the measured value k is

The permittivity (or dielectric constant) of medium is

In air, it becomes

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

1 22

Q QF kR

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

k

1 2and Q Q

1 2 and r r

2Q 1Q

medium r o

9

0 0101.0006 F/m36medium

Coulomb’s Law

6

2Q

1Q

2Q

1Q

R F

F


Here, and

Combining two equations, we haveSome useful notes:

The force on due to is

and must be point charges and at rest!

12

1 212 2

04

R

Q QF aR

2 1 12 R r r R

12

12

R

RaR

1 2 2 112 3

0 2 14

Q Q r rF

r r

21 12 F F

Like charges repel;Unlike charges attract

2Q1Q

Coulomb’s Law

2Q1Q

1Q 2Q

1r

2r

12 2 1 R r r

21

F 12

F

7


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 exerted by the charges

Example 5.1:Consider point charges and located at (3,2,1) and (1,1,4), respectively Calculate the electric force on a charge located at (0,3,1)

1 2, , , NQ Q Q1 2, , Nr r rr

1 2, , , NQ Q Q

3

104

Nk k

k k

Q r rQFr r

2 (1 mC)Q 3(2 mC)Q

Q

1(10 nC)Q

1r

r

1r r

2r r

Nr r

NF

2F

1F

Coulomb’s Law

8


Solution: Using the following equation

we have

where

1 31 21 12 132 2

0 12 13

14

Q QQ QF a aR R

1 3.797 7.149 0.637 mN

x y zF a a a

12 1 2 13

131212 13

12 13

0,3,1 3,2, 1 , 0,3,1 1, 1, 4 ,

3,1,2 1,4, 3,

9 1 4 1 16 9

R r r R

RRa aR R

Coulomb’s Law

Origin

1r

2r

3r

12F

13F

1F

9


Example 5.2:Calculate the electric force on a charge +2QSolution:

2

1 12 20 0

2

2 22 20 0

2

3 32 20 0

2 1,11 1 ,4 4 22

2 21 1 4 1,0 ,4 4

21 1 2 0, 14 4

Q Q QF aaa

Q Q QF aa a

Q Q QF aa a

1 2 3

0.169 0.0465 [ ]

x y

F F F Fa a N

Coulomb’s Law

12

3

71 105

Q Ca cm

Q Q

2Q2Qx

y

10


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!

Visual EMT using MatLab

11


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

Place a positive test charge at arbitrary point P, and measure the force acting on it

tQ

Electric Field Intensity

12


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

Thus,

For N point charges located at , the electric field intensity at point is

32

0 04 4

tR

t

Q r rQE aR r r

Ra

rtr

E2 2

0 0

14 4

tR R

t t

QQF QE a aQ Q R R

3

10

14

Nk k

k k

Q r rE

r r

1 2, , , NQ Q Q1 2, , Nr r rr


13


Example 5.3:Find the total electric field intensity at P due to Q1and Q2

Solution:

1 2

1 1 2 23 3

0 1 0 24 4

E E EQ r r Q r r

r r r r


14

E

1E

2E

2a

1a1r r 2r r

r1r

2r


Example 5.4: 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) (b) Find the electric field at (1,-3,7)

Solution: (a)

(b)

9 99

3 39

3/ 2 3/ 2

5 10 (1, 3,7) (2,0,4) 2 10 ( 3,0,5) (1, 3,7)1010 (1, 3,7) (2,0,4) ( 3,0,5) (1, 3,7)436

45( 1, 3,3) 18( 4,3, 2) 1.004 1.284 1.4 [ ]19 29 x y z

F

a a a nN

1.004 1.284 1.4

x y z

F VE a a aQ m


E

15


How can we visualize the electric field intensity?Done by drawing continuous lines from the charge which are everywhere tangent toThese lines are called Electric Flux (or field) Lines

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

Electric Field Line Pattern

E

16



17

(For a positive source charge, the lines will radiate outward.)



18

(The high density of lines between the charges indicates the strong electric field in this region.)


What is the electric flux line?When 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 charges)

Electric Flux Lines

19


Example 5.5: The figure shows the electric field(flux) lines for a system of two point 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? orweak?

Electric Flux Lines

20


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 top

The one on the bottom is positive; field lines leave it. The one on 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, but this is for clarity. The field is weakest to the upward of the top charge

Electric Flux Lines

21


Various charge distributions and charge elements:

where represent the line charge, surface charge, and volume charge density, respectively.

2 3C/m , C/m , and C/mL S v

What is the Electric Fields due to these charge distributions ?

Electric Fields due to Charge Distributions

22


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

Integrating over volume V, we have

vdQ dv

3 3

0 0

( )4 4

vr r r rdvdQdE rr r r r

( )3

0

1( )4

v

V

r rE r dv

r r

r

pe

¢-=

¢-ò

( )3

0

1( )4

S

S

r rE r dS

r r

r

pe

¢-=

¢-ò

( )3

0

1( )4

L

L

r rE r dL

r r

r

pe

¢-=

¢-ò

Surface Line


vdQ dv

V

r r

r

rOrigin

23


x y

zdE

R

Example 5.6: Line ChargeConsider a line charge with uniform charge density extending from A to B along z-axisThen, associated with element dl at is

where

L

304

Ldl RdER

0,0, z

22

2 2

,, , 0,0,

1 tan sec

x y z

z

dl dzR x y z z

xa ya z z aa z z a

R z z

Cartesian:

Cylindrical:

Thus,


dE

24


Considering , and integrating from A to B, it becomes

For infinite line charge, that is, , it is

2tan , secz z dz d

( )

( )2

1

2

1

3/2220

2 3

3 30

0

4

sec cos sin4 sec

cos sin4

zL

zL

Lz

a z z aE dz

z z

a ad

a a d

r

ar

aa

r

a

rrpe r

r a a ara

pe r a

ra a a

pe r

¢+ -¢=

é ù¢+ -ê úë ûé ù+- ê úë û=

- é ù= +ê úë û

ò

ò

ò

2 1 2 10

sin sin cos cos4

L

zE a a

1 2/ 2, / 2

02

LE a

Usingsintan ,cos

1seccos


25


Example of Line Charge Distribution

26


Example 5.7: Surface ChargeConsider a infinite sheet of charge in the xy-plane withThen, associated with element 1 is

S


dE

27

z

x

y

R


Due to the symmetry of charge distribution, the contribution along cancelsThus, the total electric field intensity is

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

2

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

1/22 20 0

0

24 4

12 2

S Sz z z

S Sz z

hh d d dE dE a ah h

h a ah

p

f r r

r rr f r r rp

pe per r

r re er

¥ ¥

= = =

¥

= = =é ù é ù+ +ê ú ê úë û ë û

ì üï ïï ï-ï ï= =í ýï ïé ù+ï ïê úë ûï ïî þ

ò ò ò ò


3 3/ 22 20 04 4

zS S a hadS RdE d dhR

a

E

28


E

E

E

E

E

E

0 E E E 0

E E E

0 0

22

S Sz z

E E E

a a

z

Practical Application Examples:[Configuration of A Real Capacitor]


29



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

xV

yV

[Configuration of Cathode Ray Oscilloscope]

xV

AV

y

z

x

cathodeanode

deflectionplates

screenyV

30


[TV tube with electron-deflecting charged plates (orange)]


31


0V

[Configuration of Color Ink-Jet Printer]

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 ink

droplets are collected by gutter (at old version)


32


[Configuration of Microstrip Lines]

rElectric fields


33


[Microwave Oven]


Electricity flows from the wall through fuses The controller sends power to the high voltage transformer (About 3000-

4000 V) The magnetron tube transforms the high voltage into electromagnetic energy A waveguide guides the 2.45GHz microwaves into the cooking chamber A stirring blade spreads the microwaves evenly

34



Water molecules are unusually polarAn electric field orients water moleculesA fluctuating electric field causes water molecules to fluctuate in

orientationWater molecules orient back and forthLiquid water heats due to molecular “friction”Food’s liquid water content heats the food

35



Oscillating electric field

time

36

Microwave frequency of 2.45 GHz is ideally suited for the time it takes to flip a water molecule around !


Example 5.8: A circular disk of radius a is uniformly charged . If the disk lies on the z=0 plane,

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

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

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

2[ / ]S C m

2 20

12

Sz

hE ah a

a hE

E

S

r

dSa


37


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

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

dS d d dS

3 3/22 2

0 04 4

S zSdS a hadS rdE

r h

( ) ( )

2

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

2 2 2 20 0

0

4 2

1 12 2

a aS S

z

a

S Sz z

hh d d dEh h

h h E E ah a h

p

r f r

r rr f r r rpe er r

r re er

= = =

= =+ +

æ ö é ù÷-ç ÷ ê úç= = - =÷ç ÷ ê úç ÷ç + +ê úè ø ë û

ò ò ò


38


(b) As

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

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

0

,2

S

za E a

1/22

2 2 2

2 2

11 1 1 1

1

1 11 12 2

h aha h a

h

a ah h

-é ùæ öê ú÷ç- = - = - + ÷çê ú÷çè ø+ æ ö ê úë û÷ç+ ÷ç ÷çè ø

é ùæ ö æ öê ú÷ ÷ç ç- - =÷ ÷ç çê ú÷ ÷ç çè ø è øê úë û

2 2

2 2 20 0 04 4 4

S S

z z za a QE a a a

h h h


39


Example 5.9: A square plate described by carries a surface charge density

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

(b) Using

2 2, 2 2, x y212 [ / ]y mC m

S

r

dS

r r

r

2 2

2 22

0

12

12(4) 2 192

S SQ dS y dxdy

ydy mC

r- -

= =

= =

ò ò ò

ò

320 0

( )4 4

S Sr

dS dS r rE ar r r

r rpe pe

¢-= =

¢-ò ò


0z

40


, in whichwe have

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

( )

( )

2 2 3

3/22 20 2 2

2 2 27

3/22 22 0

27

2 22

22

7

22

(12 10 ) ( , ,10)14 100

(1/ 2) ( )(108 10 ) 2100

1 1( 216 10 )104 100

104( 216 10 ) ln100

z

z

z

y x y dxdyE

x y

d ya dxx y

a dxx x

x xax x

pe

-

- -

-

-

-

´ - -=

+ +

é ùê ú

= ´ ê úê ú+ +ê úë ûé ùê ú= - ´ -ê ú+ +ê úë û

é ùê ú+ +

= - ´ ê úê ú+ +êë û

ò ò

ò ò

ò

16.46 [ / ]za MV m=ú

By the symmetry, and 21/ 2ydy d y


41

( )2

2

( )( )

ln ( ) ( )

f x dxf x a

f x f x a

¢

+

= + +

ò


Example 5.10:Planes and , respectively, carry charges and If the line carries charge , calculate at P(1,1,-1) due to the three charge distributionsSolution:The contributions to at

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

0, 2x z

1

2

10

20

180 ,2

2702

Sx x

Sy y

E a a

E a a


2x 3 y 210 nC/m215 nC/m

10 nC/mE

E

42

1E

2E

3E


Since , the unit vector gets as

Then, the electric field is

The total field can be obtained by

3

z xR a a

110, 310

x zRR a a aR

30

9

9

210 10 1 3 18 3

10 10236

L

x z x z

E a

a a a a

1 2 3 180 270 18 3162 270 54 [ / ]

x y x z

x y z

E E E E a a a aa a a V m


43

R

a


Volume Charge:Consider the volume charge distribution withThen, it is not so easy to calculate the total electric field intensity, and the result can be obtained through tedious mathematical procedureIf so, what is an easier way to get the result?The answer is Gauss’s law, which will be discussed later in details

v


44


Definition: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 is called the electric flux density, which is defined as

20 [C/m ]rD E E

Electric Flux Density

+

D

45


Thus, when the orientations of the surface defined and the electric flux lines are different, the total electric flux through the surface S can be evaluate by

S

D dSY = ò

Electric Flux Density

D

dS

D

dS

dSdS

46


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

Applying divergence theorem to the middle term in equation above

vS V

Q D dS dvr= =ò ò

S V

D dS Ddv= ò ò

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

( 0)

( 0)

( 0)

( 0) No enclosed charge!

Gauss’s Law

47


we have

Gauss' law is a powerful tool for the calculation of electric fields when they originate from charge distributions of sufficient symmetry to apply itPart 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 situation

vD r =

Applications of Gauss’s Law

48


Here, Dashed-Lines are Gaussian Surfaces


49


Example 5.11: Point ChargeSuppose a point charge is located at the origin Determine at a point P

Solution:Since is everywhere

normal to the Gaussian surface applying Gauss’s law gives

Thus, the electric flux density is

D

r rD D a

24

rQD ar

24r rS S

Q D dS D dS D rp= = =ò ò


50


Example 5.12: Infinite Line ChargeSuppose the infinite line of uniform charge lies along z-axis Determine at a point P

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

2

LD a

r

LD

2

LS

S

l Q D dS

D dS D rlr r

r

p

= =

= =

ò

ò

Arranging

Line charge C/mL

PD

l


51


Example 5.13: Infinite Sheet of ChargeSuppose the infinite sheet of uniform charge lies on z=0plane. Determine at a point P

SD


Surface charge

C/m2S

D

P

dS

x

yz

D

D

D

dS

52


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

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

Note that evaluated on the sides of the box is zeroThus, the electric flux density is

z zD D a

( )S z zS top botton

A Q D dS D dS dS D A Aré ùê ú= = = + = +ê úê úë û

ò ò ò

, z>02

, z<02

Sz

Sz

aD

a

D dS


53


Example 5.14: Suppose two infinite sheets of uniform charge ,as shown in figure. Determine everywhere in spaceSolution:

S

D


SD ( Between plates )

D

D

D

D

D

S S S S

SS

54


Example 5.15: Uniformly Charged Sphere Consider a sphere of radius R with a uniform charge Determine everywhere

Solution:Construct Gaussian surfaces

for and , respectivelyFor , the total charge

enclosed by the surface of radius r is

And, the electric flux is

3[C/m ]vD

r Rr R

r R


Gaussian surface

r

D

ra

dS

na

343enc v vQ dv r

24r rS

D dS D dS D rpY = = =ò ò

55


Thus, using Gauss’s law, we have

For , since total charge enclosed by the surface is

from Gauss’s law the electric flux density becomes

3 24, 43enc v rQ r D r

for 03

v r

rD a r R

r R

34 ,3enc v vQ dv Rr r p= =ò

3 24, 43enc v rQ R D r

3

2 for 3 v rRD a r Rr


3vr

3

23vRr

r

D

56



57


Electric Potential

Similarity between gravitational and electrical potential energy :

Work done by gravitational force is

Then, the gravitational potential energy decreases, and is equal to the negative of work done as follows:

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

GW mg L mgL= =

GU mg L mgLD =- =-

L L

Q

Q

La La

58


Electric Potential

field, it experiences an electric force and the work done by the electric field becomes

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

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

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

EW F L QE L QEL= = =

EU QE L QELD =- =-

EUV E LQ

DD = =-

(Potential energy is negative !)

59

(Potential energy done by +1C test charge !)


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

The potential difference between points A and B, , is defined as the change in potential energy (final minus initial value) of a charge Q, moved from A to B, divided by the charge

Electric Potential

E

A B

EAB B A

UV V VQ

E L

D= - =

=-

E

B AV V

60


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

If , 10 J of work is required to move a 1 C of charge between two points that are at potential difference of 10 V (See slide 69 !)

Example 5.16:A proton moves from rest in an electric field of along the +x axis for 50 cm. Find (a) the change in the electric potential, (b) the change in the electrical potential energy, and (c) the speed after it has moved 50 cm.

10 [V] 0ABV

Electric Potential

,A BV V

, ,

,E A B

A B

UV

Q

61

48 10 [V/m]


Solution: (a) (b) (c)

Now, let’s consider a chargeQ moving from pt A to pt Balong arbitrary pathin electric fieldThen, the potential energy in displacing the charge by is

Electric Potential

dl

F

A

Q

B

dl

E

EdU F dl QE dl=- =-

62

4 4(8 10 [V/m])(0.5 m) 4 10 [V]ABV Ed=- =- ´ =- ´19 4 15(1.6 10 [C])( 4 10 [V]) 6.4 10 [J]E ABU QV - -D = = ´ - ´ =- ´

2, / 2i i f f Ef p i fKE PE KE PE KE m v PE PE U+ = + = =-D= -(Since 0)iKE =( )15

627

10 [J]10 [m/s]

102 6.4

2.81.67 [kg]

v-

-

´´

´= =


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

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

Is the potential difference independent of path taken?Consider the case of constant vector fieldThen, the potential difference at the direction along path A-B is

B

EA

U Q E dl=- ò

BE

AB B AA

UV V V E dlQ

= - = =-ò

A

B C

rhE

dl

Electric Potential

E

B

AB B AA

V V V E dl Eh= - =- =ò

63


E

dr dl

QA Ar

Br

r

B

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

Integral does not depend upon the path chosen to move from Ato B so that the integral is the same for BOTH paths (conservative field)

Example 5.17:Assume that the electric field due to a point charge Q located at the origin, we have

( )( )sin 0 sinC B C

AB B AA C A

V V V E dl E dl E dl Er Ehq q= - =- - =- - - = =ò ò ò

Electric Potential

204

rQE a

r

64


Then, determine the potential difference between A and BSolution:

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

20 0

1 14 4

B

A

r

AB B A r rB Ar

Q QV V V a drar r rpe pe

é ùê ú= - =- = -ê úë û

ò

0AV Ar Br r

04

rQV E dlrpe

¥

= =-ò

Choose infinity as referencebecause the potential at infinityis supposed to be zero (Ground) !

Electric Potential

65

sinrdl dra rd a r d a Using


For N point charges located at , the potential at is

For continuous charge distributions, the potential at becomes

10

1( ) (point charges)4

Nk

k k

QV rr r

1 2, , , NQ Q Q 1 2, , Nr r rr

( )

( )

( )

0

0

0

1( ) (line charge)4

1( ) (surface charge)4

1( ) (volume charge)4

L

L

S

S

v

V

rV r dl

r r

rV r dS

r r

rV r dv

r r

r

pe

r

pe

r

pe

¢¢=

¢-

¢¢=

¢-

¢¢=

¢-

ò

ò

ò

Electric Potential

r

66


Example 5.18:Find the electric potential at the point P. Solution:

Thus, applying the superposition principle gives

Electric Potential

67

1 5Q C 2 2Q C

69 2 2 4

1

69 2 2 3

2 2 2

5.0 10(8.99 10 / ) 1.12 10 ,4.0( 2.0 10 )(8.99 10 / ) 3.6 10

(3.0 ) (4.0 )

CV Nm C Vm

CV Nm C Vm m

-

-

´= ´ = ´

- ´= ´ =- ´

+

1 24 3

3

1.12 10 [V] 3.60 10 [V]=7.6 10 [V]

PV = V V+

= ´ - ´

´


Example 5.19:Charge is uniformly distributed within a rod.Find the electric potential on the perpendicular bisector of the charged rodSolution:Since , the potential

due to at a point P becomes

Thus, the potential due to the rod is

Electric Potential

L

L

LdQ dydQ

2 20 0

1 14 4

LdydQdVr x y

2 2

2 2 2 20 0

ln4 4

aL L

a

dy a x aVx y a x a

r rpe pe

-

æ ö+ + ÷ç ÷ç= = ÷ç ÷ç ÷ç+ - + +è øò

68


Example 5.20:Charge is uniformly distributed within a spherical shell of radius a. Then, the electric field is

Find the potential everywhere For ,

For

v

3

20 0

for 0 , for 3 3

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

r

r a

( )3 3

20 03 3

rv v

out r ra aV r a dra

r rr r

e e¥

=- =ò

0 ,r a

( ) ( )0

22

0

3

2 3

rv

in out r ra

v

rV r V r a a dra

ra

re

re

= = -

æ ö÷ç ÷= -ç ÷ç ÷çè ø

ò

( )outV r

( )inV r

a

E

v

Electric Potential

69


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 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 makes it give off light

Electric Potential

70

D AA

Bulb


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

to flow from high voltage to low voltage

Electric Potential

71


But 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 legs

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

Electric Potential

72


Electric Potential

73

Inside cavity is “shielded” from all external electric fields! “Faraday Cage effect”


Electric Potential

74

Have you experienced this before? Inside elevator, your mobile phone does not work. A cage made of conducing material can effectively shield all

electromagnetic waves from entering into the cage, thus preventing any signal from the service provider to reach your mobile phone.

Observe what the demonstrator is doing with the radio.


Electric Potential

75

A hollow metal box is placed between two parallel charged plates.

The conducting box is an effective device for shielding!

The internal electric field is exactly equal and opposite external field.

Net result is zero electric field inside conducting sphere

+ ++ + +


Since the potential difference is independent of the path taken, we haveThat is,

Applying Stokes’s theorem to the equation above, it becomes

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

BA ABV V

0BA ABV V E dl+ = =ò

( ) 0E dl E dS= ´ =ò ò

0 E

Relationship between E and V

76

E


B AV V dV

Q

E

L

AV

dl

BV

Now, let’s define the relationship between and V, satisfying the conservative property

Consider a point charge in an electrostatic fieldThen, the potential difference between two points is

Rearranging for , we have

If is and -axis, it becomes

( )

/

cosE

L

dV dU Q

E dl E dl E dlq

=

=- =- =-

LE

(V/m)LdVEdl

Q

dl ,x y z

, ,x y zdV dV dVE E Edx dy dz

Relationship between E and VE

77


Then, the vector at any point is given by

Finally, we have

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

, the electrostatic field must be a conservative field

(V/m)

x y zE a a a V V

x y z

0V

(V/m)

x x y y z z

x y z

E E a E a E a

V V Va a ax y z

Relationship between E and VE

78


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

Now, use the potential to calculate the 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

Electric Dipole and Flux Lines

E

79


Rewriting this for special case it becomes

becauseCalculating in spherical coordinates, we have

( )r d

20

cos( )4Q dV r

r

30

1

2cos sin4

r

r

V VE V a ar r

Qd a ar

0 0

1 1( )4 4

r rQ QV rr r r r


( ) ( ) ( ) ( )2cos ,r r d r r rq- + + --

E

80



Electric flux line

Equipotential surface

( ) 0V r

( ) 0V r

( ) 0V r

81


Electric Dipole Antenna

82


Electric Dipole Antenna

83


Equipotentials of a charged sphere:The electric field of the charged sphere has spherical symmetryThe potential depends only on the distance from the center of the sphereAn equipotential surface is a surface on which all points are at the same potentialThe electric field at every point

on an equipotential surface is perpendicular to the surface


Electric flux line

Equipotential surface

AB

84


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

equipotential !)

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

(Orthogonal relation between two factors)

0,

0

B

B AA

V V E dl

E dl

- =- =

=

ò


85


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


86


Draw the potential due to two opposite point charges in three-dimensional space, and confirm the equipotential lines from the resulting shape

Run potential.m!

Visual EMT using MatLab

87

Dept. Electronics and Information Engineering 88

Problem 5.1:Let us illustrate the use of the vector form of Coulomb’s law by locating a charge of at

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

Problem 5.2:Find at caused by four identical charges located at and , as shown in Fig.

Homework Assignments

41 3 10 [C]Q

42 10 [C]Q

2Q 1Q

E

3[nC]

1 2 3, , ,P P P

4P

1,1,1P

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


Problem 5.3:Find the total charge contained in a length of the electron beam shown in Fig.

Problem 5.4:Given that

Determine everywhereProblem 5.5:

Determine the electric field due to the potential


2 cm

312 nC/m ,1 2

0 ,otherwisev

D

2 1 sinV z


Problem 5.6:Two dipoles with dipole moment and are located at points and , respectively. Find the potential at the origin.

Problem 5.7:A spherical charge distribution is given by

Find V everywhere.


5 nC/mza 9 nC/mza

0,0, 2 0,0,3

,

0 ,

ov

r r aa

r a

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