electromagnetic field theory - hansungkwangho/lectures/emt/2019/... · 2019. 4. 23. · prof....
TRANSCRIPT
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
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Dept. Electronics and Information Engineering
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?
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Dept. Electronics and Information Engineering
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?
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Dept. Electronics and Information Engineering
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
Dept. Electronics and Information Engineering
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
Dept. Electronics and Information Engineering
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
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Dept. Electronics and Information Engineering
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
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Dept. Electronics and Information Engineering
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
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Dept. Electronics and Information Engineering
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
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Dept. Electronics and Information Engineering
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
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Dept. Electronics and Information Engineering
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
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Dept. Electronics and Information Engineering
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
Electric Field Intensity
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Dept. Electronics and Information Engineering
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
Electric Field Intensity
14
E
1E
2E
2a
1a1r r 2r r
r1r
2r
Dept. Electronics and Information Engineering
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
Electric Field Intensity
E
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Dept. Electronics and Information Engineering
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
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Dept. Electronics and Information Engineering
Electric Field Line Pattern
17
(For a positive source charge, the lines will radiate outward.)
Dept. Electronics and Information Engineering
Electric Field Line Pattern
18
(The high density of lines between the charges indicates the strong electric field in this region.)
Dept. Electronics and Information Engineering
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
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Dept. Electronics and Information Engineering
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
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Dept. Electronics and Information Engineering
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
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Dept. Electronics and Information Engineering
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
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Dept. Electronics and Information Engineering
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
Electric Fields due to Charge Distributions
vdQ dv
V
r r
r
rOrigin
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Dept. Electronics and Information Engineering
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,
Electric Fields due to Charge Distributions
dE
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Dept. Electronics and Information Engineering
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
Electric Fields due to Charge Distributions
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Dept. Electronics and Information Engineering
Example of Line Charge Distribution
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Dept. Electronics and Information Engineering
Example 5.7: Surface ChargeConsider a infinite sheet of charge in the xy-plane withThen, associated with element 1 is
S
Electric Fields due to Charge Distributions
dE
27
z
x
y
R
Dept. Electronics and Information Engineering
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
Electric Fields due to Charge Distributions
3 3/ 22 20 04 4
zS S a hadS RdE d dhR
a
E
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Dept. Electronics and Information Engineering
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]
Electric Fields due to Charge Distributions
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Dept. Electronics and Information Engineering
Electric Fields due to Charge Distributions
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
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Dept. Electronics and Information Engineering
[TV tube with electron-deflecting charged plates (orange)]
Electric Fields due to Charge Distributions
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Dept. Electronics and Information Engineering
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)
Electric Fields due to Charge Distributions
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Dept. Electronics and Information Engineering
[Configuration of Microstrip Lines]
rElectric fields
Electric Fields due to Charge Distributions
33
Dept. Electronics and Information Engineering
[Microwave Oven]
Electric Fields due to Charge Distributions
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
Dept. Electronics and Information Engineering
Electric Fields due to Charge Distributions
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
Dept. Electronics and Information Engineering
Electric Fields due to Charge Distributions
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 !
Dept. Electronics and Information Engineering
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
Electric Fields due to Charge Distributions
37
Dept. Electronics and Information Engineering
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
Electric Fields due to Charge Distributions
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Dept. Electronics and Information Engineering
(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
Electric Fields due to Charge Distributions
39
Dept. Electronics and Information Engineering
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
Electric Fields due to Charge Distributions
0z
40
Dept. Electronics and Information Engineering
, 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
Electric Fields due to Charge Distributions
41
( )2
2
( )( )
ln ( ) ( )
f x dxf x a
f x f x a
¢
+
= + +
ò
Dept. Electronics and Information Engineering
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
Electric Fields due to Charge Distributions
2x 3 y 210 nC/m215 nC/m
10 nC/mE
E
42
1E
2E
3E
Dept. Electronics and Information Engineering
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
Electric Fields due to Charge Distributions
43
R
a
Dept. Electronics and Information Engineering
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
Electric Fields due to Charge Distributions
44
Dept. Electronics and Information Engineering
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
Dept. Electronics and Information Engineering
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
Dept. Electronics and Information Engineering
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
Dept. Electronics and Information Engineering
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
Dept. Electronics and Information Engineering
Here, Dashed-Lines are Gaussian Surfaces
Applications of Gauss’s Law
49
Dept. Electronics and Information Engineering
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
Applications of Gauss’s Law
50
Dept. Electronics and Information Engineering
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
Applications of Gauss’s Law
51
Dept. Electronics and Information Engineering
Example 5.13: Infinite Sheet of ChargeSuppose the infinite sheet of uniform charge lies on z=0plane. Determine at a point P
SD
Applications of Gauss’s Law
Surface charge
C/m2S
D
P
dS
x
yz
D
D
D
dS
52
Dept. Electronics and Information Engineering
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
Applications of Gauss’s Law
53
Dept. Electronics and Information Engineering
Example 5.14: Suppose two infinite sheets of uniform charge ,as shown in figure. Determine everywhere in spaceSolution:
S
D
Applications of Gauss’s Law
SD ( Between plates )
D
D
D
D
D
S S S S
SS
54
Dept. Electronics and Information Engineering
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
Applications of Gauss’s Law
Gaussian surface
r
D
ra
dS
na
343enc v vQ dv r
24r rS
D dS D dS D rpY = = =ò ò
55
Dept. Electronics and Information Engineering
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
Applications of Gauss’s Law
3vr
3
23vRr
r
D
56
Dept. Electronics and Information Engineering
Applications of Gauss’s Law
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Dept. Electronics and Information Engineering
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
Dept. Electronics and Information Engineering
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 !)
Dept. Electronics and Information Engineering
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
Dept. Electronics and Information Engineering
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]
Dept. Electronics and Information Engineering
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-
-
´´
´= =
Dept. Electronics and Information Engineering
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
Dept. Electronics and Information Engineering
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
Dept. Electronics and Information Engineering
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
Dept. Electronics and Information Engineering
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
Dept. Electronics and Information Engineering
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+
= ´ - ´
´
Dept. Electronics and Information Engineering
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
Dept. Electronics and Information Engineering
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
Dept. Electronics and Information Engineering
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
Dept. Electronics and Information Engineering
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
Dept. Electronics and Information Engineering
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
Dept. Electronics and Information Engineering
Electric Potential
73
Inside cavity is “shielded” from all external electric fields! “Faraday Cage effect”
Dept. Electronics and Information Engineering
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.
Dept. Electronics and Information Engineering
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
+ ++ + +
Dept. Electronics and Information Engineering
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
Dept. Electronics and Information Engineering
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
Dept. Electronics and Information Engineering
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
Dept. Electronics and Information Engineering
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
Dept. Electronics and Information Engineering
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
Electric Dipole and Flux Lines
( ) ( ) ( ) ( )2cos ,r r d r r rq- + + --
E
80
Dept. Electronics and Information Engineering
Electric Dipole and Flux Lines
Electric flux line
Equipotential surface
( ) 0V r
( ) 0V r
( ) 0V r
81
Dept. Electronics and Information Engineering
Electric Dipole Antenna
82
Dept. Electronics and Information Engineering
Electric Dipole Antenna
83
Dept. Electronics and Information Engineering
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 Dipole and Flux Lines
Electric flux line
Equipotential surface
AB
84
Dept. Electronics and Information Engineering
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
Electric Dipole and Flux Lines
85
Dept. Electronics and Information Engineering
Equipotential lines on the surface of the human body reflect the electric dipole nature of the heart
Electric Dipole and Flux Lines
86
Dept. Electronics and Information Engineering
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
Dept. Electronics and Information Engineering 89
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
Homework Assignments
2 cm
312 nC/m ,1 2
0 ,otherwisev
D
2 1 sinV z
Dept. Electronics and Information Engineering 90
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.
Homework Assignments
5 nC/mza 9 nC/mza
0,0, 2 0,0,3
,
0 ,
ov
r r aa
r a