Transcript Document

Electromagnetism
Zhu Jiongming
Department of Physics
Shanghai Teachers University
Reference Books
1. 《电磁学》 赵凯华、陈熙谋
高等教育出版社 1978
2. 《电磁学》（第二版） 贾启民、郑永令 等
复旦大学出版社 2002
3. 《 Fundamentals of Physics 》（sixth edition）
David Halliday etc. 2001
The important thing is
never to stop
questioning.
--------- Albert Einstein
Electromagnetism
Chapter 1
Chapter 2
Chapter 3
Chapter 4
Electric Field
Conductors
Dielectrics
Direct-Current Circuits
Chapter 5
Chapter 6
Chapter 7
Chapter 8
Chapter 9
Magnetic Field
Electromagnetic Induction
Magnetic Materials
Alternating Current
Electromagnetic Waves
Chapter 1 Electric Field
§1.
§2.
§3.
§4.
§5.
§6.
Electric Charge
Coulomb’s Law
Electric Field
Gauss’s Law
Electric Field Lines
Electric Potential
§1. Electric Charge
1. Positive and negative charges：
positive
repel or attract each other
charges
negative
2. Principle of conservation of charge：The algebraic
sum of all the electric charges in any closed system
is constant.
3. Quantization of charge：
basic charge e = 1.6  10 19 C
1
2
（quark charge；  e ， e ）
3
3
4. Conductors, insulators and semiconductors
§2. Coulomb’s Law
1. Coulomb’s Law
2. Unit of Charge
3. Vector Form
4. Principle of Superposition
5. Examples
1. Coulomb’s Law
Point charge：charged bodies，very small
（ ideal model，size << distance between them ）
 Coulomb’s Law  experiments
the electric force between two static point charges in a
vacuum (equal in magnitude，opposite in direction)
● direction ：along the line joining them， repel or
attract each other
● magnitude ：proportional to charges，inversely
proportional to the square of the distance

expression： F  k
q1q2
r2
Electric Force
Repulsion for same sign
Attraction for opposite sine
q2
r
q1
Repulsion for Same Sign
Repulsion for same sign
Attraction for opposite sine
q2
r
q1
Attraction for Opposite Sine
Repulsion for same sign
Attraction for opposite sine
q2
r
q1
2. Unit of Charge
Gauss System（from CGS ）
k=1
when r = 1 cm， let q1= q2 ，F = 1
define q1= q2 = 1 Static Coulomb
 International System（from MKS ）
（SI or MKSA ）
unit of electric current ：A（ampere）
1 Coulomb = 1 A • s
when r = 1 m， q1 = q2 = 1 C， F = k N
k = 9  10 9 N • m2 • C 2
write k = 1 / 40 ， 0 = 8.9  1012
 Unit conversion： 1 C = 3  10 9 Static C

3. Vector Form


a
a=|a|
ea= a／| a | | ( or
q1 exert on q2
q2 exert on q1

vector（bold, italic or a ）
magnitude
â ) unit vector（ | â | = 1）
1
q1q2
F12 
 2 er12 ( er 12 1  2)
40 r
1
q1q2
F21 
 2 er 21 ( er 21 2  1)
40 r
same sign, q1q2 > 0，F12 same direction with er 12
opposite sign, q1q2 < 0，F12 opposite to er 12
Vector Form
same sign, q1q2 > 0, F12 same direction with er 12
opposite sign, q1q2 < 0, F12 opposite to er 12
er 21
F21
q1
er 12
F21
F12
q2
F12
4. Principle of Superposition
the total force exerted on a charge by many point
charges = the vector sum of the forces that the
charges would exert individually
Example： the force that q2 、q3
exert on q1
F1 = F21 + F31
F1
F31
q2
q1
q3
F21
Example 1
Point charge q1= 1  10 5 C is placed at ( 0, 1 ) and
q2=  1  10 5 C at ( 1, 0 ). Find the total ( net ) force
exerted on Q0= 1  10 4 C located at ( 1, 1 ) .
Solution：
q1Q0 ˆ
F1 
 2 i  9iˆ N
40 r1
1
q2Q0 ˆ
F2 
 2 j  9 ˆj N
40 r2
1
F = F1 + F2= 9i  9j N
=9(ij) N
或：F  9 2 N
y
Q0
q1
0
q2
Direction：  = - 45o
x
Example 2
Three identical spheres of charge q are placed at corners
of a triangle, another of q’ is placed at the center, all in
equilibrium. Find the relation between q and q’ . A
Solution： FO= FAO+ FBO+ FCO
1
3qq'

 2 (eAO  eBO  eCO )  0
40 a
FA= FBA+ FCA+ FOA
3eOA
q eBA  eCA

(
q  2 q' )  0
2
40
a
a
∴ q’ =  q /
3
O
B
C
FCA
FOA
FBA
Exercises
p.38 / 1-2- 2,
4
§3. Electric Field
1. Electric Field
2. Field Intensity
3. Electric Field Calculations
● Field of a Point Charge
● Field of a Collection of Charges
● Field of Continually Distributed Charges
4. Examples
1. Electric Field
Field：distribution of a physical quantity in space
● Physical quantity ： scalar、vector
Ex. ：temperature field，
velocity field，gravitational field， etc.
● Electric field E
at any point around a point charge Q ，
a force is exerted on q0 by Q

Q  Electric Field E  q0
2. Field Intensity
 At
any point around a point charge Q，a force is
exerted on q0 by Q
1 Qq0
F
 2 er
40 r
physical quantity describing electric field should not
depend on test charge q0
definition ： E = F / q0
Field intensity
● vector ：
E = Ex iˆ  E y ˆj  Ez kˆ
● function of r ： E = E ( x, y, z ) = E ( r )
● unit ： N / C
 Force exerted on q in a field E： F = q E
 Uniform field ：E is independent of ( x, y, z )
Electric Field of A Point Charge
1
Q
E
 2 er
40 r
● Q：electric charge
● r ：distance from Q to field point P
● e ：unit vector from Q to P
r
E
P
r
Q>0
Q<0
Q
er
Field of A Collection of Charges
F  Fi
1
E

  Ei 
q0
q0
40
i
Qi
i r 2 ei
i
Principle of superposition of electric field：
The total electric field at P caused by a collection of
charges is the vector sum of the fields at P due to
each point charge
Ex.1
qi
1
E
e

2 i
（ see p.9）
4
r
0
i
i
Continuously Distributed Charges
Electric field due to the charge dq（infinitesimal，as
a point charge ）
1 dq
dE 
 2 er
40 r
Electric field due to the continuously distributed
charge Q
dE
1
dq
E   dE 
e
2 r

er
40 r
P
r
dq
Q
Element Charge
volume charge：
dq = dV
 surface charge ：
dq = dS
 linear charge ：
dq = dl
（charge density  、、 ）

r
dS
P
er
r
dl
P
er
Example 1
Electric charge q is distributed uniformly along a line
with length L. A point P is a distance a apart from the
line, 1 and 2 are known. Find the electric field at P。
dE
Solution：
P
dq = dl = ( q/L )dx
a
r
q/L
1

2
dE 
dx
2
40 r
0 x
dx
1 q/L
dE x 
dx cos 
relation of r, x,  ？
2
40 r
r = a / sin
1 q/L
x =  a ctg
dE y 
d
x
sin

40 r 2
dx = ( a / sin2 )d
1
x
r = a / sin
x =  a ctg
dx = ( a / sin2 )d
1 q/L
1
dE x 
dx cos  
2
40 r
40
1
1 q/L
dE y 
dx sin  
2
40
40 r
q/L
cosd
a
q/L
sin d
a
2
1 q
1 q/L
(sin  2  sin 1 )
Ex 
cosd 

40 aL
40 a 1
2
1 q
1 q/L
( cos  2  cos 1 )
Ey 
sin d 
2 
40 aL
40 r 1
Discussion
(1) L >> a （ or L  ，q   ， remains ）
1 = 0，2 = 
Ex = 0，Ey =  / 20a
(2) L << a （point charge）
sin 2  sin 1 = 0
Ex = 0，
cos 1  cos 2 = L / a
Ey = q / 40 a2
Example 2 （p.12/[Ex.2]）
Find the electric field caused by a disk of radius R with
a uniform surface charge density  , at a point along
the axis of the disk.
dq
r
Sol.：dq =  dS =  r dr d
P
z
1 rdrd
o
dE 
2
2
40 r  z
dE
Symmetry E = 0， E = Ez
z
1 zrdrd
dE z  dE

2
2 3/ 2
2
2
4

(
r

z
)
r z
0
R

|z|
1
rdr
(1 
)
E   dEz 
z 2  2 2 3 / 2 
2
2
2

40
(
r

z
)
R

z
0
0
Discussion
(1) R >> z （ close to disk or big disk ）

1

E  lim
(1 
) 
2 0
2 0
1 R2 / z 2
(2) R << z （ far from disk or small disk ）
1
2
1

1
R
 (1  R 2 / z 2 ) 2  1 
1 R2 / z 2
2 z2
1 q
 R2
1 R 2

E

2
2
40 z 2
2 0 2 z
40 z
（point charge）
Exercises
p.38 / 1- 3 - 1,
6, 7, 8, 9
§4. Gauss’s Law
1. Electric Flux
● Flux
● Electric Flux
2. Gauss’s Law
3. Application of Gauss’s Law
4. Examples
1. Electric Flux
Flux：volume flow through dS per unit time
n
—— flux through dS
d = dS vn = dS v  n  v  dS
dS
vn
v
  v  dS


S

to any vector A ： = A  dS
S
 Electric Flux
 E   E  dS
S
dS
E、n： vector、f (r)，but not for E
outward forv
n：perpendicular to dS （2 way）
closed surface
S
v
2. Gauss’s Law

Spherical surface with q at center： E = q /  0
 Any
E = q /  0
closed surface surround q ：

Closed surface enclosing no charge ：E = 0

Charge distribution：
 E  dS  q
in
S
0
E = qin /  0
qin 
q
i
inS
Spherical Surface with q at Center
Electric field at any point on the spherical surface of
radius r
1 q
E
e
dS  dSer
2 r
40 r
1 q
dS
 E   E  dS  
2
40 r
S
S
1
q

40 r 2
q
q
2
S dS  4 r 2 4r   0
0
1
The result is independent of radius r .
Any Closed Surface Enclosing q
q at center，projection
dS1 ：on spherical surface r1
dS2
dS1
dS
dS ：on any closed surface
q r
r2
dS2 ：on spherical surface r2
1
er  en  cos dS cos  dS2
1 q
d E1 
dS1
2
dS1 dS 2
40 r1
 2
2
r1
r2
1 q
1 q
d E 
e  en dS
dS 2
2 r
2
40 r2
40 r2
d E  d E1
q
 E   E  dS   E  dS 
S
S1
0
Closed Surface Enclosing No Charge
E 1+ E2 = q / 0
E 3’+ E2 = q / 0

E 1= E3’
and
E 3’=  E 3

E = E 1+ E3
en1
q
en2
= E 1  E 3’ = 0
（ Notice： E 1= E3’ ≠ 0 ）
n3
S2
S 3’
S3
S1
Charge Distribution
 E   E  dS    Ei  dS    Ei  dS    Ei
S
S
i
qi outside S ： E i = 0
 qi inside S ： E i = qi /  0
i
S

E 

1
0
q内
q  
i
S内
0
Many point charges： q内   qi
S内

Continuous distribution：q内   dq
Conclusion（Gauss’s Law）
S内
i
3. Application of Gauss’s Law
Gauss’s Law ：  E   E  dS 
qin
0
The total electric flux through a closed surface is equal
to the net charge inside the surface , divided by  0 .
 symmetry：sphere、cylinder、plane etc.
 suitable closed surface —— Gaussian Surface
electric field at any point on the Gaussian Surface is a
constant or zero
● zero：E is perpendicular to dS, then
E  dS  0
● constant：symmetry，E and dS in same direction
then
E  dS  EdS  E dS  ES
S



Example 1（p.19/[Ex.1]）
Find the electric field caused by a thin, flat, infinite
sheet with a uniform surface charge  .
S1
Sol.：select Gaussian surface
S2
E = E1+ E2+  side wall
en1
en2
symmetry，E  sheet plane
P

 side wall = 0
E constant on S1 、S2 ，opposite direction
E  en1  E1n  E
E  en 2  E2 n  E
E = E1+ E2 = ES1 + ES2 = 2ES = q内 /  0 =  S /  0
 E =  / 2 0

E
en
2 0
Discussion
(1) compare with p.17/[Ex.2] R >> z ，( same )
(2) infinite，when P is very close to the sheet
(3) en sheet，outward from the charged sheet
for  > 0 ，E is in same direction with en
 < 0 ，E in opposite direction to en
(4) E = Ein + Eout
E  dS  q / 

内
0
S
if qout = 0, Eout = 0, qin unchanged
 E
内
S
 dS  q内 /  0
both right, but the 2nd equ. is useless
Example 2（p.40/1-4-3）
Electric charge is distributed uniformly along an infinite
long thin wire, with linear charge density  . Find E.
Sol.：select Gaussian surface
r
S E  dS  E  2r  l  0  0  l /  0 l
P

E
er
 E =  / 2 0 r
20 r
Discussion ：
(1) infinite long ，when P is very close to the wire
(2) for  > 0 ，E is in same direction with er
 < 0 ，E in opposite direction to er
Example 3（p.21/[Ex.2]）
Electric charge q is distributed uniformly on the
surface of a sphere of radius R. Find E.
E
Sol.：select Gaussian surface
P

q
/


E

4

r
E

d
S
0
S
1 q
 Eout 
e
2 r
40 r
r
2
2
E

d
S

E

4

r
0
inside： 
O R
E
S
 Ein  0
o
R
r
Example 4（p.22/[Ex.3]）
Electric charge q is distributed uniformly throughout
the volume of a sphere with radius R. Find E.
Sol.：select Gaussian surface
3
R
q r
2
S E  dS  E  4r   0 R 3
r P
O
1 qr
 Ein 
e
3 r
40 R
2
E

d
S
 q / 0

E

4

r
outside： 
S
 Eout
q

e
2 r
40 r
E
1
o
R
r
Exercises
p.40 / 1- 4 - 1,
5, 6, 7, 8, 9
§5. Electric Field Lines
1. Electric Field Line—— curve with direction
● Tangent at any point is in the direction
of the electric field at that point
● Lines dense where field strong
2. Properties of Electric Field Lines
● Away from positive charge（or ）
toward negative charge（or ）
Gauss’s : E > 0 ,  q > 0
● Never closed
（high potential toward low）
Electric Field Line Density
Number of field lines —— Electric flux
number of field lines N through S
N = K(S  E) = KS Ecos
S S
= KSE
( S= Scos perpendicular to E )
E = E  S = N （ take K = 1 ）
 Field line density —— Electric field intensity
N/S= KE = E
（ take K = 1 ）

E
§6. Electric Potential
1. Closed Loop Law of Electric Field
● Work done by the force of a point charge
● Work done by any electric field
● Closed loop integral of electric field is zero
2. Electric Potential and Potential Difference
● Potential , Potential Difference
● Potential of a point charge
3. Calculating Electric Potential ( 2 ways ), Examples
4. Equipotential Surfaces
5. Potential Gradient and Electric Field
Work Done by a Point Charge
Work done on q by the force of Q
as q moves d l
1
Qq
 2 dl cos
dA  F  dl 
40 r
P1
E
q
dl
dr
r
r’
r1
Qq

 2 dr
40 r
Q
r2
P2
Qq 1 1
Qq dr

(  )
A   dA 
2

40 r1 r2
40 r1 r
P1
P2
r2
1
Caution ：The work depends only on the end points
r1 and r2 ，not on the details of the path.
Work Done by Any Electric Field
Principle of superposition of electric field
P2
A   F  dl  q  E  dl  q  L Ei  dl  q   Ei  dl
L
L
i
i
P1
Caution ：
The work depends only on the end points P1 and P2 ，
not on the details of the path L .
—— conservative force
Electric field due to a static charge distribution is
conservative.
Closed Loop Integral of Electric Field
Unit charge，q = 1，F = qE = E
 E  dl 
L
B
 E  dl 
A( L1 )

A
 E  dl
B
L2
B ( L2 )
B
B
A( L1 )
A( L2 )
 E  dl   E  dl
B
B
  E  dl   E  dl  0
A
A
A
Closed Path Law :
 E  dl  0
L
L1
Electric Potential
The work depends only on the end points. Take one of
them P0 as a reference point. Work done on a unit
charge by the electric field, as the unit charge moves
from P to P0, depends only on P.
Define electric potential at P as the work done on a
unit charge by electric field as the unit charge moves
from point P to P0 . （ Unit：Volt ）
P0
A
1 P0
UP 
  F  dl   E  dl
P
q
q P
( Reference point ：U P0  0 )
Potential Difference
Potential difference between points A and B
P0
P0
U AB  U A  U B   E  dl   E  dl   E  dl
A
B
B
A
Work done on q as it moves from A to B
B
A   qE  dl  q(U A  U B )  qU AB
A
Unit ：Volt
 Potential ：1 point，relative to some reference point
 Voltage ： 2 points，difference
Potential of a Point Charge

charge distribution is within finite area ：
take infinity to be reference point
UP  

P
1

dr
1
1
Q 2 
Q(  )
E  dl 
P
40
r
40
r r
q
 UP 

40 r
P
1


r
Q
charge distribution extends to infinity ：
can not take infinity to be reference point
3. Calculating Electric Potential

Point charge ( Principle of superposition )
( charge distribution is within finite area )
● Collection of point charges
qi
UP 

40 i ri
1
●
r
dq
Q
Continuously distributed charges
1
dq
UP 
40  r

Definition of electric potential U 
P
( E is given or found )

P
P0
P
E  dl
Example 1 （ p.32/[Ex.1]）（1）
Find the potential at a point along the axis of a disk
of radius R with a uniform surface charge density  .
dq
Sol.：dq =  dS =  r dr d
r
P
1 rdrd
dU 
z
o
2
2
40 r  z
U   dU 
1
40
R
 2 
0
rdr
r2  z2
U


( R 2  z 2  | z |)
2 0
0
z
Example 1 （ p.32/[Ex.1]） （2）
Find the potential at a point along the axis of a disk
of radius R with a uniform surface charge density  .

z
解： p.17/[Ex2] ( z > 0 )
E
(1 
)
2 0
R2  z2
U 
P0
P


z
(1 
)dz
E  dl  
2
2
2

R

z
0
z



(z  R2  z 2 )
z
2 0


( R 2  z 2  z)
2 0
Example 2 （p.32/[Ex.2]）
Electric charge q is distributed uniformly on the
surface of a sphere of radius R. Find the potential.
Sol.：from Gauss’s Law
1 q
R
O
Eout 
e
E

0
2 r
in
40 r

P0
1 q
1 q
dr 
U out   E  dl  
2
P
40 r
40 r
r
U in 

P0
P
E  dl 


R
r

Ein  dl   Eout  dl
R
1 q
q
dr 
0
2
40 R
40 r
R
1
Discussion
(1) Both ways are workable，
but the 2nd is better
(2) Uin = constant，
equipotential everywhere
within the sphere
(3) Charge is on the surface
E not continual
U continual，but not smooth
O R
E
o
R
r
R
r
U
o
Example 3 （SJTU/p.45/[Ex.10-12]）
Find the potential at a distance r from a long line
with linear charge density  .
Sol.：from Gauss’s E =  / 2 0 r
r
r0
P0
r0

 dr

ln
U   E  dl  
l
P
20 r
20 r
r
if take r0 =  ，then U = 
（charge distributed to infinity）

 if take r0 = 1 ，then
U 
ln r
20
 same for infinite charged plane
（can not take  as reference ）

P
4. Equipotential Surfaces
A surface on which the electric potential is the same
at every point —— Equipotential Surface
（see p.34 / Fig. 1-36 ）
 point charge ：concentric spherical surfaces
 charged long line：coaxial cylindrical surfaces
 charged big plate ：parallel plates
Electric field lines  equipotential surfaces
Show：if E||  0，
then U =  E d l  0 on equipotential surfaces
5. Potential Gradient and Electric Field
P2
U1  U 2   E  dl
P1
P2
P2
1
1
U  U 2  U1  P dU  P  E  dl
 dU   E  dl  Edl cos
dU
E cos  
dl
Directional derivative
for  = 0 ， E in the same direction with d l ， with
maximum | dU |
dU
E
en
dn
Potential gradient
Exercises
p.41 / 1- 6 - 2,
3, 4, 5, 8