Transcript Document

Chapter 6
Conductors and Dielectrics
in Static Electric Fields
Conductors and Dielectrics in Static Electric Fields
6-1 Conductors in Static Electric Fields
6-2 Dielectrics in Static Electric Fields
6-3 Electric Displacement, Gauss Theorem at the
Presence of the Dielectric
6-4 Capacity, Condenser
6-5 Energy of Static Electric Fields, Energy Density
* 6-6 The Charge and Discharge of the Condensers
* 6-7 Application of Electrostatics
6-1 Conductors in Static Electric Fields
1. Condition of Electrostatic Equilibrium
2. The Charge Distribution of a Conductor at
Electrostatic Equilibrium
3. Electrostatic Screening
1. Condition of Electrostatic Equilibrium
1. Electrostatic Induction
++
+++ + +
+
+
+
Induced
charges
1. Condition of Electrostatic Equilibrium
2. Electrostatic Equilibrium ：there is no motion of charges in
a definite direction internally of a conductor

E0
＇
E

E 0
+
+
+
+
+
+
+
+

E0
1. Condition of Electrostatic Equilibrium
Conditions：
●
（1）Ein= 0 ，everywhere in the conductor
（2）Electric field at the surface
direction  surface

E
1. Condition of Electrostatic Equilibrium
A conductor is an equipotential body

E  0
U AB  
AB

en 
 
E  dl  0
E
+
The surface is a equipotential surface


 E  dl
U AB  
AB
+
+
 
E  dl  0
A
+

dl
+

eτ
B +
2. Charge Distribution on a Conductor


Charge Distribution Determined by：
●
Charges Outside
●
Shape of the Conductor
Charge Distribution on the Surface of an Isolated
Conductor （p.47/Fig.2-5 ）
●
convex（positive curvature）
sharp： larger
●

flat：
 smaller
concave（ negative curvature ）  smallest
Point discharge （ Ex. lightning rod ）
sharp   , E large  air ionized（breakdown）
2. Charge Distribution on a Conductor
(1). A charged solid conductor：the charges can only be
distributed on the surface of the conductor and there is no net
charge internally of the conductor.

E  0
 
q
SE  dS  0  ε0
q  0
(2). A conductor with a cavity：
internal surface ？External surface？
Gauss
surface
+
+ +
+
+
+
S
+
+
+ +
2. Charge Distribution on a Conductor
(2). A conductor with a cavity：
If there is no charged body in cavity，the charges must
distribute on the external surface, and no charge distributed in the
internal surface inside of the cavity
   qi
SE  dS  ε0  0
If there are charges on the internal
surface, they must be charges with an
equal but opposite charges
U AB  
AB
 
E  dl  0
Gauss
surface
+
+ +
+
+
＋
+
A
－
+
B
S+
+ +
It contradicts with that a conductor is a equipotential body
2. Charge Distribution on a Conductor
If there is a charged body with
+q in cavity, the internal surface
will be induced and charged
with –q, and the charges
distributed on the external
surface should be calculated by
Principle of Conservation of
Charge.
 
 E  dS  0
S
q
i
0
Gauss surface
q
+q
-q
S
2. Charge Distribution on a Conductor
（3）. The relationship between the surface charge density and
neighboring electric field of the surface of a charged conductor
Create a thin cylindrical Gaussian
surface as shown in the right Fig.
 
 E  dS  ES  σS / ε0

S
σ
E
ε0
So electric field at the surface
direction  surface（equipotential）
magnitude   （by Gauss’s）
S
+
+
+

E 0
+
  
E  en
0
+
+ +
2. Charge Distribution on a Conductor
(4). The Charge distribution law on
the surface of a conductor
σ
E
ε0
σ  E ; σ , E 
+
+
+
+
+
+
+
++
Charge Distribution on a Conductor
Phenomenon of Sharp Point Discharge
E near the sharp point is so
strong that it can ionize the
nearby air and make it a
conductor and electric
discharge will appear in the
form of vehement fireball
explosion.
σE
Charge Distribution on a Conductor
< electric wind experiment >
+++
++
+ + +
+
+
Charge Distribution on a Conductor
+ +
+
+
+ Charged cloud
The principle of lightning rod
+
+
－
－－ － － －
Electrostatic Induction
corona discharge
Grounded reliably
3. Electrostatic screening
1. Screening the external electric field

E
Screening the external electric field
by a cavity conductor
Electric field inside the cavity is not affected by charges
outside the conductor（no matter grounded or not）
3. Electrostatic screening
2. Screening the internal electric field
+
q
+
+
q
+
+
+
+
q
+
3. Electrostatic screening
3. the grounded cavity conductor screen the internal field
q
q
Electric field outside the grounded conductor
is not affected by charges inside the cavity
Example 1（p.201/[Ex.]）
There is a spherical metal shell with an external R1=10 cm and internal
radius R2=7cm ，in the shell is a concentric metal ball with radius R3 = 5cm. If
the shell and ball both have a positive charge of q =10-8 C, how are the charges
distributed between the two sphere? What is the potential at the center of the
ball?
Sol.：Take a sphere with radius r as the
Gaussian surface
q
q
E1  0 (r  R3 )
The charges of the inner ball distributes
on the surface R3
 
q
S2 E2  dS  ε0
q
E2 
( R3  r  R2 )
2
4 π ε0 r
R3
R2
R1
Example 1（p.201/[Ex.]）
E3  0 ( R1  r  R2 )
 
 E3  dS  0
q
q
S3
the internal surface of the shell is
charged with –q, and the charges
of external surface is 2q
  2q
S4 E4  dS  ε0
2q
E4 
(r  R1 )
2
4 π ε0 r
R3
R2
R1
Example 1（p.201/[Ex.]）
E1  0
(r  R3 )
q
E2 
( R3  r  R2 )
2
4 π ε0 r
E3  0
( R2  r  R1 )
2q
E4 
(r  R1 )
2
4 π ε0 r
R1=10 cm，R2=7 cm
R3=5 cm，q=10-8 C
 
Vo   E  dl
0
 R2   R1     
R3 
  E1  dl   E2  dl  R E3  dl  R E4  dl

0
R3
2
1
q
1
1
2

[0  (  )  0  (  0)]
4 π ε0
R3 R2
R1
 2.3110 V
3
Example 2（[supplement ]）
A point charge with q is located outside of a grounded conductive
sphere , and the distance from the center of the sphere is l > R，
What is the induced charges q’ on the conductive sphere?
Sol. ： q’ distributes on the surface of the sphere
R
Grounded，equipotential body，Uo = 0
O
l
Uo = Uo1 + Uo2
q ：U  1  q
O1
40 l
1
dq
1
1 q'

dq 

q’ ：U O 2 


40
R 40 R
40 R
q q'
R
 0
 q'   q
l R
l
q
Homework
p.229 / 6-
8, 9, 13
6-2 Dielectrics in Static Electric Fields
1.
The influence of dielectric on electric field,
the relative permittivity
2.
The polarization of dielectric
3.
The intensity of polarization
4.
The relationship between polarized charges
and free charges
Introduction
General Laws （Chapter 5，Vacuum） applied to
 Conductors （Chapter 6-1）
 Dielectrics （Chapter 6-2）
Microscopically ，vacuum inside materials
Coulomb’s Law is correct（10-13 cm）
Macroscopic, statistical average of microscopic
Study Dielectrics with General Laws
1.
The influence of dielectric on electric field, the relative permittivity
Dielectric is filled in between the two Parallel-plates E gets smaller
+++++++
σ
- - - - - - - σ
σ
E0 
ε0
relative permittivity
+++++++
εr
σ
- - - - - - - σ
E0
E
εr
εr  1
permittivity
ε  ε0 ε r
2. The polarization of dielectric
dielectric
 Polar Molecules ：（H O、organic glass etc）
Non-polar Molecules：（H2、CH4、paraffin etc）
2
2. The polarization of dielectric
Two kinds of molecules —— different ways
 Non-polar Molecules —— distance
-+
E=0, p=0
E 0,
+
p 0
Polar Molecules —— direction
（ |  p |  |E |， all p tend to line up with E ）

E=0, p=0
E 0,
p
 0
3. The Intensity of electric Polarization P
There are m dipoles within the volume element 
m
p
i 1
i
0
m
polarized
p
i 1
i
0
Polarization vector P ：
m
P   pi V
i 1
P at a point ：V  0
Uniform ： Same P at every point
Unit：
C/m 2
3. Polarization P
Polarization P and the area density of the polarized charges

P

p
V
S
+++++++++++

p ：The electric dipole of molecule  ' - - - - 

r P
P ：The intensity of electric
l
Polarization
σ ' ： The area density
of
the polarized charges
 p σ ' Sl

P
 σ'
Sl
V
 ' + + + + +
-----------
4. The relationship between polarized charges and free charges
E0
E  E0  E ' 
εr
εr  1
E' 
E0
εr
 '  E' ,
 0  E0
εr  1
σ'
σ0
εr
εr  1
Q' 
Q0
εr
+++++++++++
- - - - -
d

 r E0
+
+

E'
+
+

E
+
-----------
4. The relationship between polarized charges and free charges
The relationship of polarization P and electric field intensity E
εr  1
σ'
σ0
εr
E0  σ 0 / ε0
E  E0 / εr
P  σ'


P （εr  1)ε0 E
  εr  1 polarizability


P  ε0 E
+++++++++++
- - - - -
d

 r E0
+
+

E'
+
+

E
+
-----------
6-3 Electric Displacement Gauss’s Law in Dielectrics
  1
'
SE  dS  ε0 (Q0  Q )
εr  1
Q 
Q0
εr
  Q0
SE  dS  ε0 εr
 
  0 r E  dS  Q0
'
S
+++++++++++
 ' - - - - -
r
 ' + + + + +
-----------
S
permittivity
ε  ε0 ε r
 
 εE  dS  Q0
S
6-3 Electric Displacement Gauss’s Law in Dielectrics
 
 εE  dS  Q0
S



Electric Displacement Vector
D   0 r E  E
 
The flux of electric displacement
 D  dS
S
Gauss’s Law at the presence of the dielectric
n
 
 D  dS   Q0i
S
i 1
Example 1（p.209/[Ex.1]）
Put a piece of dielectric with r =3 in between the two parallel charged
plates with the plate distance d =1 mm. Before putting the dielectric
into the two plates is 1000 V. If the surface charge density remains
uncharged after the dielectric inserted, what are the E, P, the surface
charge densities of the plates and the dielectric, and the D in the
+++++++++++
dielectric?
Solution:
U
E0   103 kV  m 1
d
E  E0  r
εr
U
d
-----------
 3.33 10 kV  m
2
1
P  ( r  1) 0 E  5.89 106 C  m- 2
r =3
d=10-3 m
U=103 V
0 = 8.85  10-12
Example 1（p.209/[Ex.1]）
 0   0 E0
6
 8.85 10 C  m
2
E0 = 103 kV· m-1
P = 5.8910-6 C · m-2
0 = 8.85  10-12
 ' P
 5.89 106 C  m 2
D   0 r E   0 E0   0  8.85 106 C  m- 2
Example 2（p.209/[Ex.2]）
As shown in the right Fig. a cylindrical condenser is made up
of a thin and long cylinder conductor of radius R2 and a long
solid cylinder of radius R1in the center, between them the
dielectric with the relative permittivity r is filled. Assume that
the charge per unit length of the outer and inner cylinders are
+ and - ，What are：(1)E，D, P in the dielectric.；
(2)The surface densities of the polarized charges in
the inner and outer surfaces of the dielectric.
Sol. ：(1)
 
 D  dS  l
S


( R1  r  R2 )
R2
D
D2 π rl  l
2πr
D



1
r
E

P  ( r  1) 0 E 

ε0 ε r 2 π ε0 ε r r
2 π  rr

R1
Example 2（p.209/[Ex.2]）

(2) E 
2 π  0 r r

E1 
2 π  0 r R1

E2 
2 π  0 r R2


(r  R1 )
(r  R2 )
( r  1)
 1 '  ( r  1) 0 E1  
2 π  r R1
( r  1)
 2 '  ( r  1) 0 E2 
2 π  r R2
R2
R1
Example 3（supplement）
A conducting sphere of radius R , with electric charge
q0 , is in a uniform infinite dielectric of  . Find the
electric field and bound charge density on surface.
Sol.： D  dS  D  4r 2  q0

S
q0
 D
rˆ
2
4r
D
1 q0
E 
rˆ
2
 4 r
- +
+
- +
+
- +
1 q0
rˆ  (rˆ)
 '  P  nˆ   0 E  nˆ   0 
2
4 R
   0 q0
 r 1

0
2 
 4R
r
+ +
R
O
+ +
-
+
’
+ +
-
Discussion
   0 q0
 r 1
 ' 

0
2
 4R
r
(1) ∵  >0 ( r =1 +  >1 )
(2)
∴ ’ , q0 opposite sign
  0
q'  4R  '  
q0

2
∴ | q’ | < | q0 |
and q = q0 + q’ = q0 0/  = q0/ r < q0
q0
E0 
40 r 2
E = E0/ r < E0
(3) Vacuum
∴
1
Homework
p.230 / 6-
19
6-4. Capacitor and Capacitance
1. Capacitance of a Single Isolated Conductor
2. Capacitors and Capacitance
● Spherical Capacitor
● Parallel-Plate Capacitor
● Cylindrical Capacitor
3. Connections of Capacitors
● Parallel Connection
● Series Connection
1. Capacitance of an isolated conductor
Isolated conductor，potential proportional to charge
1
V Q
C
1
a constant，depend on size、shape
C
1 Q
Ex.：conducting sphere V 
or ：C  40 R
40 R
Capacitance ：C
Unit ：Farad（F）
1 F = 1 C/1 V
F：1 F = 10 -6 F
pF：1 pF = 10 -12 F
2. Capacitors and Capacitance
Isolated
Conductor：one conductor，potential
Capacitor ：two conductors，potential difference
Properties of Capacitors
● Charge distributed uniformly
● Same magnitude, opposite sign of charges
● Voltage proportional to charge
Q
Q
Capacitors often seen
C

VA  VB U
● Spherical capacitor
● Parallel-plate capacitor
● Cylindrical capacitor
Capacitors
The steps for calculating the Capacitance of Capacitors
（1）Assume the two plates are charged with  Q
（2）Calculate the E between the two plates
（3） Calculate the electric potential difference between the
two plates U
（4）Calculate the C from C = Q /U
Parallel-Plate Capacitor （p.213/[Ex.1]）
Symmetry —— Charge distributed uniformly
 Gauss’s Law —— Same magnitude, opposite sign
between two plates ：
 -
Q

D 

E
 0 r  0 r S
S
B
Qd
U   E  dl  Ed 
d


S
0 r
A

A
Q
S
 C
  0 r
U
d
Q  0 r S
C 
U
d
B
Cylindrical Capacitor （p.213/[Ex.2]）
Symmetry —— Charge distribution
 Gauss’s Law ——  
R1 < r < R2 ：

Q
E

20 r r 20 r rl
R2
Q
R2
Q 2 π ε 0 r l
ln
U   E  dl 
C 
2


l
R
0
r
1
R2
R1
U
ln
20 r l
Q
R1
 C 

U
When
ln( R2 / R1 )
2Rl
S
R2  R1  R1： C   0 r
  0 r
R2  R1
d
Spherical Capacitor （p.214/[Ex.3]）
Symmetry —— Charge distribution
 Gauss’s Law ——  
1 q
R1 < r < R2 ：
E
e
2 r
40 r

R2
q
R2
R1
1
1
q
(  )
( R2  R1 )
U   E  dl 
40 R1 R2
40 R1 R2
R1
q
40 R1 R2
 C

U
R2  R1
4R 2
S
 0
When R2  R1  R1： C   0
R2  R1
d
Example（p.215/[Ex.4]）
Assume there are two infinitely long parallel
straight wires each with radius R , the distance
between their centers is d，and d  R, What is the
capacity per unit length?
Assume the charge line density of two wires  
E  E  E
λ
λ


2 π  0 x 2 π  0 (d  x)
U 
d R
R
2R

E

P
o
x
dx
Edx


2 π0

d R
R
1
1
( 
)dx
x dx

d
x
Example（p.215/[Ex.4]）
 d R 1
1
U
)dx
R ( 
2 π0
x dx

d R
d R

[ln x R  ln( d  x)
]
R
2 π0

d

d R

ln

ln
π0 R
π0
R
 π ε0
C 
d
U
ln
R
2R

E


P
o
x
dx
d
x
3. Connections of Capacitors
●
Parallel Connection（U same） q = q1 + q2
C1
q
q1  q2
C

 C1  C2
U
U
C2
—— Increase capacitance
●
U
Series Connection（q same） U = U1+ U2
1
1
1 U U1  U 2
 


C1 C2
C q
q
—— Increase working voltage
C1
C2
U
6-5 Energy of Static Electric Fields, Energy Density
1. Energy in a Capacitor
2. Energy of Static Electric Fields, Energy Density
1. Energy in a Capacitor
A parallel-plate air capacitor is charged—— Charges are
moved from cathode to anode
dq


U fixed when connected to a battery
Q fixed when the battery is disconnected
+++++++++
+

E
U
---------
q
dW  Udq  dq
C
Q2
1 Q
W   qdq 
C 0
2C
Q
Q2 1
1
C
2
W


QU

CU
e
U
2C 2
2
Example
A uniformly charged sphere of radius R , with electric charge
Q ,. Calculate the energy of Static electric field W.
1
Q
1
Qr
Sol.： E 
e
E内 
e
外
2 r
3 r
40 r
40 R
1 Q
1 Q
2
2
U内 
(
3
R

r
)
U外 
3
40 2 R
40 r
R
1
3
Q
1
Q
1
2
2
2

(
3
R

r
)
4

r
dr
W   UdV
3
3 
2 4R 40 2 R 0
2
2
3Q

6
160 R
R
2
3Q
0 (3R  r )r dr  200 R
2
2
2
2. Energy of Static Electric Fields, Energy Density
Where is the energy stored ? —— charge？field？
（ Theories and experiments support the latter. ）
Ex. ：electromagnetic waves （carry energy，not charge）
 Energy density w ：energy per unit volume
For parallel-plate capacitor： W  1 CU 2
2
（charges on plates，energy in between them）
E
W CU 2 / 2 U 2 1 2

w
Sd


Sd
 E
2
2d
2
1
w  D  E （Valid for any electric fields）
2

The total energy stored in the whole electric field
1 2
space

We   we dV  
V
V
2
εE dV
-
Example 1（p.219/[Ex.1]）
The inner and outer radii of a spherical condenser are R1 and R2，respectively,
and their charges Q， respectively. If a dielectric with permittivity  is filled in
between the two shells, what is the electric field energy stored in this condenser.
Solution ： R1< r < R2
-Q
1 Q
E
Q
2
d
r
4πε r
r
Q2
1 2
R
1

we  εE
2
4
32
π
εr
2
2
Q
2
dr
dWe  we dV we 4r dr 
2
8 π εr
Q 2 R 2 dr
Q2 1
1
We   dWe 

(  )
2

8 π ε R1 r
8 π ε R1 R2
R2
Example 1（p.219/[Ex.1]）
2
Q
1
1
Discussion
We 
(  )
8 π ε R1 R2
Q2
（1） We 
2 C　R2 R1
（ spherical condenser ）
C  4πε
R2  R1
Q 2 （ isolated conductive Sphere ）
（2） R2   We 
8 π εR1
Example 1（p.219/[Ex.2]）
As shown in the right Fig. , there is a cylindrical
condenser, in between the cylinder is air and the
breakdown electric field intensity is Eb=3106 V·m-1，
the outer radius of the condenser is R2= 10-2 m. Under
the circumstances that the air is not breakdown, what
is the maximum value of R1 so that the energy stored in
the condenser reaches maximum?
Solution ：

E
2 π ε0 r
Eb 
l -
( R1  r  R2 )
max
2 π ε0 R1

R2
 R2 dr

ln
U
R1
2 π ε0 R1
2 π ε0
r
+
+
+
+
R1
R2
_
_
_
_
_
+ ++ _
+ +
+++ _
_
Example 1（p.219/[Ex.2]）

R2
U
ln
2 π ε0 R1
the electric field energy stored per unit length
l
2
1

R2
We  U 
ln
2
4πε
R
Eb 
0
max
+
+
+
+
1
R1
R2
_
_
2 πε0 R1
  max  2 π ε0 Eb R1
R2
We  π ε0 E R ln
R1
2
b
2
1
_
_
_
+ ++ _
+ +
+++ _
_
Example 1（p.219/[Ex.2]）
R2
We  π ε0 E R ln
R1
dWe
R2
2
 π ε0 Eb R1 (2 ln
 1)  0
dR1
R1
R2
R1 
 6.07 103 m
e
R2 Eb R2
U max  Eb R1 ln

 9.10 103 V
R1 2 e
2
b
2
1
Eb=3106 V·m-1 ，R2= 10-2 m
Homework
p.230 / 6-
18, 21, 25, 31, 32
The end