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Chap 22-24 Electric field
First circular accelerator in the world
Easy chair
Charged hair
Electricity and Magnetism
 Electric forces hold atoms and molecules together.
 Electricity controls our thinking, feeling, muscles and
metabolic processes.
 Electricity and magnetism underpin much of our
current technology (e.g. computers).
 Electricity and magnetism are linked on a
fundamental level.
Key terms:
electric charge
electrostatics
electron
proton
neutron
nucleus
positive ion
negative ion
ionization
conductor
insulator
point charge
electric dipole
closed surface
electric flux
surface integral
Gauss’s Law
Gaussian surface
Faraday’s icepail
experiment
electric potential energy
electric potential
voltage
electron volt
equipotential surface
gradient
cathode-ray tube
§1 charges and couloumb’s law
1. 1 charges
1) Negative and positive
2) quantified (millikan)
Q  ne
19
e  1.6  10 C
Fraction charge
1
 e
3
3) conservation (Franklin)
4) Invariance of charges
2
 e
3
1.2 Couloumb’s law
1) Model:point charge
2) Couloumb’s law

Q1Q2 
Q1Q2 
1
F  k 2 r0 
r0
2
4

r
r
0
k=9×109Nm2/c2，
0
=8.85×10-12c2/Nm2
q1

r21
r
q2

F21
Caution:
1)valid for point charge in air
2) obey newton’s law
3) in general
Henry Cavendish Experiment
Fe  Fg
Quiz
1. Even though electric forces are very much stronger than
gravitational forces, gravitational forces determine the
motions in the solar system. Why?
2. When we approach another person, we are not aware
of the gravitational and electric forces between us. What are
the reason in each case?
§2 Electric field E
2.1 Electric field
Electric source

 F
E
q0
Electric field
Electric field for point charge

F
1 qq0  0
r
2
4 0 r

 F
1 q 0
E

r
2
q0 4 0 r
Collection of point charge

E

 Fk
k
q0

1 qk  0
  Ek  
r
2 k
k
k 4 0 rk
superposition of electric field
Electric charge

1 dq  0
Charge distribution dE 
r
2
4 0 r


r dE
dl
dq 
dS
P
dq
dV

E
dq  0
r
2
4 0 r
Step to solve E:

dq  dE 
dE x
dE y

E x   dE x
E y   dE y
Positive electric charge Q is distributed uniformly
along a line with length 2a, lying along the y-axis
between y=-a and y=+a. Find the electric field at point
P on the x-axis at a distance x from the origin.
y
a
r  x2  y2
O
Q
-a
x
P

x

dE
Solution:
1 dQ
Q
dy
Q
dE 

dQ  dy 
dy
2
2
2


4

r
4

2
a
x

y
0
0
2a
 Q

dy
x
Q
xdy
dE x  dE cos   

2
2 
2
2 32
2
2


4

2
a
x

y
4

0
0 2a x  y 

 x y
 Q

dy
y
Q
ydy
dE y  dE sin    

2
2 
2
2
40 2a x 2  y 2
 40 2a x  y  x  y
1 Qx a
dy
Q
1
Ex 

32

2
2

a
40 2a
40 x x 2  a 2
x y


1

Q a
ydy
Ey  
0
3
2

2
2
40 2a a x  y 

Q
1
E 
iˆ
40 x x 2  a 2



32
Discussion:
When x is much larger than a (that is, x>>a),
a)


Q
1
1 Qˆ
ˆ
E
iE
i
2
2
2
40 x x  a
40 x
This is same as the field of a point charge.
b)
c)
When a is much larger than x (that is, a>>x),

1

 ˆ
   Q 2a
ˆ
E 
i
i
2
20 x x a   1
20 x
Infinite line of charge

E
20 r
Field due to a Power Line
A ring conductor with radius a carries a total charge Q
uniformly distributed around it. Find the electric field at a
point P that lies on the axis of the ring at a distance x from
its center.
y
dQ
r  x2  a2
a
O
Q
x
P

x

dE
Solution:
1
dQ
dE 
40 x 2  a 2
1
dQ
dEx  dE cos  
40 x 2  a 2
Ex  
1
xdQ
40 x  a
2

2 32

1
x
x a
2
Qx
2

40 x 2  a 2 3 2
Because the symmetry , Ey=0.

1
Qx
ˆ
ˆ
 E  Exi 
i
40 x 2  a 2 3 2
1
1
40 x 2  a 2 3 2
x  0 ，E  0
Discussion:
1
q
2). x  R ，E 
 2
4 0 x
3).
2
x
R ，E  Emax
2
z
dl
E
2
 R
2
O 2
2
R
x
R
O
y
r dE x
x
P q dE
Find the electric field caused by a disk of radius R with a
uniform position surface charge density , at a point along
the axis of the disk a distance x from its center.
R
O
dQ
x
r
dr
Q
x
Solution:
dQ  dA   2rdr   2rdr
dE x 
Ex  
2rdr x
1
2rdr x 
40 x 2  r 2 3 2
R
0
1

40 x 2  r 2

32
x
2 0
rdr
R
 x
0
2
r

2 32
x

2 0

1 



2
2
R x  1 

1

 dE y  dE z  0
 E y  Ez  0
Discussion:
If R is much larger than x while
  cons



1
1 
 E
(an infinite sheet of charge)
2
2
2 0

R x  1 
The electric field produced by an infinite plane sheet of
charge is independent of the distance from the sheet.
Thus the field is uniform. Its direction is everywhere
perpendicular to the sheet, away from it.
x
Ex 
2 0


Sheet 2
Example

x
d
Sheet 1

Two infinite plane sheets are placed parallel to each other,
separated by distance d. The lower sheet has a uniform
positive surface charge density  , and the upper sheet has
a uniform negative surface charge density   with the same
magnitude. Find the electric field between the two sheets,
above the upper sheet, and below the lower sheet.
Solution:

E1  E2 
2 0

At all points, the direction of E1 is away from the positive

charge of Sheet 1, and the direction of E2 is towards the
negative charge of sheet 2.
0
   
E  E1  E2    0 ˆj
0

above the upper sheet
between the sheets
below the lower sheet
Example: Find the intensity of electric field of dipole in P

Solution: E 

q


i
E E
l
4 0 ( x  l 2) 2

 q
q
O q
P
x
E  
i
2
4 0 ( x  l 2)
 

 Electric dipole moment
q  2 xl
i
E  E  E 

2
2
2

p  ql

 4 0 ( x  l 4)
2 xp
E


2
2
2
4 0 ( x  l 4)
P
E

E
Along perpendicular bisector
q
E  E 
r
2
2

4 0 (r  l 4) 

P

q

q
l
E
E  2 E cos
4 0 r 3
Example
Find the torque exert on dipole
Solution:to pivot 0
1
1
M  F  l sin θ  F  l sin θ
2
2
 qlE sin θ
   

M  ql  E  p  E
q
O

F
l θ
q
Discussion

(1) θ 
2
(2)   0
(3)
 
Maximum torque
Torque=0 stable
Torque=0,not stable equilibrum

P

E

F
§ 3 Gauss law
Mathematics & Physics
& astronomy
  Qenclosed
 E  dA 
0
3.1 Electric field line
1) rule:
The tangent to the field line express the direction
of field
the density of the line measure the intensity of
the field
2) Essence of Electric field
d
E 
ds
1) originating from the positive to negative；
(2) never intercross；
(3) no close line.
+q
-q
Field Lines
3.2 flux
1) In uniform field

n
d e  EndS  E cosdS
 EdS


dS  dSn
 
de  E  dS

En

E
dS
dS

n
2) Non uniform field
 
de  E  dS
dS
 
 e   d e   E  dS
S

E
E
To closed surface
 
 e   d e   E  dS
S
Discussion

(1) Direction of S
Open surface
Closed surface
n

E
Concave:positive,
Outward: positive
3.3 Guass’s law
1) suppose a point charge is in the center of a sphere surface
 
Q
Q
Q
   E  ds   k 2 ds  k 2 4a 2 
S
S
0
a
a
2) suppose a point charge is in any interior place of
a sphere surface

S
  Q
E  ds 
0
+q
 q outside the surface
 e   e1   e 2  0
+q
 many charges
  

E  E1  E2  ...  E5

 
 

 e   E  dS   ( E1  E2  ...  E5 )  dS


 


  E1  dS   E2  dS  ...   E5  dS
q1 q2 q3
  
0 0 0
Caution:

E
produced by all charges
e only related to internal charge
S1
S2
S
q3
q1
q2
q4
q5
Guass law
e  
S
  1
E  dS 
0
 qi
i
  1
 e   E  dS 
S
Meaning:
（separated charges)
dV

0 V
Source field
（distribution charge)
Caution:
(1) E in Guass’s law is the E in Guassian surface,it is
produced by all the charges in the space
(2) the flux only depends on the charges inside
(3) Zero flux does’t mean the zero field.
4)to moving charge,coulumb’s law don’t work,but
guass’s law works
3.4 Application of guass law: find E
A.symmetry distribution of charges
B.intensity in guass surface is uniform or piecewise
uniform
Example:What is the electric flux through a
cylindrical surface? The electric field, E, is uniform
and perpendicular to the surface. The cylinder has
radius r and length L
A) E 4/3 p r3 L
B) E r L
C) E p r2 L
D) E 2 p r L
E) 0
Example:In a model of the atom the nucleus is a
uniform ball of +e charge of radius R. At what
distance is the E field strongest?
A) r = 0
B) r = R/2
C) r = R
D) r = 2 R
E) r = 1.5 R
Conclusion:
Step to find E with guass law:
(1) analysis symmetry
(2) from above
 guass surface must be closed one
 the field point must in guass surface
 E is easy to take out from integral function
Find the E of the following charged ball with
uniform density
Solution:
+
( r  R)
r
+
R
r' +
3

1 q 0
 R 0
E
r

r
2
2
4 0 r
3 0 r
+
rR
 
14 3
1
2 
SE  dS  E  4r  0 3 r    0 q' E

E
r
3 0
R
O
r
Find E of infinite long line

dS
Solution:
r
From symmetry:
 
 e   E  dS
S
 
   
  s E  dS   t E  dS  b E  dS
  s EdS  E  s dS  E  2r  l

E
P

dS

E
l
E
E  2r  l 
1
0

E
2 0 r
l
O
r
Infinite plane or Sheet
Charge per unit area  Cm-2. Mirror symmetry about and
perpendicular to plane,find E
Solution:
From symmetry
 
 e   E  dS
S
 
 
 
 s E  dS  ls E  dS  rs E  dS
 0  ES  ES  2 ES
2 ES 
1
0
S

E
2 0
Ex
O
x
k

r
Example: find E
rR
Solution:
R
 
1 k 4 3
2
 E  ds  E 4r   0 r 3 r
k
E
3 r 0
?
  Q
 E  ds 
k
Q    dV  
4r  2 dr   2kr 2
V
V r
0
2
2

kr
k
2
No relation with r
E 4r 
E
0
2 0
r  R　q   dq   4kr dr   2kR 2
R
0
4r 2 E 
2kR 2
0
kR 2
E
2 0 r 2
Inversely proportion to r2
dr
r
r
Example:find E in p
b
Solution:

(1) fill in the hollow ball E1 
a
Q1 
r
2 0
4 0a
(2) E produced by filling parts:

 

r0 Q1 Q2
E  E1  E 2 
(  )
4 0 a 2 b 2

E2 
Q2 
r
2 0
4 0b
Q1  Q2 
4
 ( R3  r 3 )
3
P
Problem to deal with guass’s law
•
sphere symmetry
•
mirror symmetry
•
cylindrical symmetry
R
§ 4 potential
4.1 work done by electric force
Aa b  
b
a

 
b 
b
F  dl  q 0  E  d l  q 
qQ

4 0
a

b
a
a
Q
cosdl
2
4 0 r
dr qQ  1 1 
    Wa  Wb

2
r
4 0  ra rb 
kQq
wa 
r
a

q
Q
Potential energy
l
b
F
l
b
Discussion:
a.conservative force
 
 F  dl  0
C
b.potential of point charge
 kQ
 kQ
 
A
U a    2 dr   E dl 
a r
a
q
ra
c.potential difference
U ab  U a  U b  
b
a
  kQ kQ
E  dl 

ra
rb
a

q
Q
F
Caution:
(1) belong to q0 and field source
(2) with related value
(3) rule for choose zero potential energy
• when sources is limited,choose infinite place
• in contrary, a specific place is choose
• in practice,earth and the out shell of equipment
3、the calculation of potential
4.2 calculation of potential
1)potential of charged particle
  q
q
 
U p   p E  dl  r p
dr

4 0 r 2
4 0 rp
2)collection of charge particles: U ( p)  
i
4 0 ri
3)continuous charge distribution
dU 
dq
40 r
U (r )  
V
dq
40 r
qi
 dV

dq   ds
 dl

Find the potential of dipole
Solution:
r r
q
Ua  U  U  k 1  k
 kq  
r
r
r r
r>>l，
 
2
 0
r r  r
2
Ua  0
U a  U max
r  r  l cos
q
l

-q
pe cos
Ua 
r2
Example:find potential of rod with l=15.0cm, =2.0 10-7c/m
Solution:
(1)
  dx

al
3
U1  

ln

2
.
5

10
V
0 4   ( l  a  x )
4



a
0
0
l
y
(2) in perpendicular bisector
l/2
  dx
l / 2
4   0 ( x 2  b 2 )1 / 2
U 2   dU  

b2  l 2 / 4  l / 2

ln 2 2
 4.3  10 3V
4   0
b  l /4l/2
dx
P
x
l
a
l
x
Example:Find potential in axis
z
dl
solution：1) dq  dl
dq
1  dl


4 0 r 4 0 r
2R dl

1
q
U



0
4 0
r 4 0 x 2  R 2
R

2 0 x 2  R 2
q
Uo 
In center x  0
4 0 R
dU 
1

R
O
y
r dE x
x
P q dE
q
4 0 R
Ｕ
O
x
§5
relationship between field and potential
5.1 basic concept
Equivalent potential surface：
U ( x, y, z )  C
Character:
(1)
 
 E  dl
(2)the density of equal potential surface reflect
the intensity of electric field.
(3)the direction of the line of force is the descended
direction of potential
5.2 the relation between electric field and potential

U ˆ U ˆ U ˆ
 E  U  { i 
j
k)
x
y
z

u  u  u
E  ( er 
e  )
r
r
z
Negative of the gradient of U
Example:find the E and U of the following disk
x
Ex P
Solution: take a circular strip
dq   2rdr
dU 
dq
4 0 r  x
2
U   dU  
R2
R1
R2
R1

2 1/ 2

rdr
2 0 r  x
2
rdr
2 0 r 2  x

2 1/ 2
R1 r

2 1/ 2


2 0

R2
R 2  x 2  R12  x 2
2

U r 
x
x

E  Ex 


1
/
2
1
/
2
x 2 0  R 2  x 2 
R 2  x 2  
1
2

dr