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Chapter 3
Conductors in electrostatic field &
capacitors
《物理学》中册，马文蔚改编
P56-99
Contents in this chapter:
* 1)Conductors in electrostatic equilibrium
导体静电平衡
2) Electric properties of conductors and dielectrics
in E filed. 电场中导体和电介质的电学性质
* 3) Gauss’s law with dielectrics
有介质时的高斯定理
* 4) Properties and calculation of capacitors.
电容器的性质和计算
5) Energy of electrostatic field
静电场的能量
§3.1 conductors in electrostatic field
recall
Induction 感应
Induction process of a conductor in E field
导体在静电场中的感应过程
+
E
(a) Conductor without
external E field
(b) Conductor in an external E field
(at the beginning)
Einside  Eexternal  Einduced  0
Eexternal
Induced charge
Einduced
+
+
+
+
+
+
+
+
+
(c) Conductor in an external E field (at equilibrium)
Induced charge
Properties of an isolated conductor in electrostatic
equilibrium
孤立导体静电平衡时的性质
1)The E field is zero everywhere inside the conductor.
导体内部电场处处为零。
Reasoning:
The conductor is in electrostatic equilibrium.
2) If the isolated conductor carries a net charge, the net
charge resides entirely on its surface.
如果导体本身带有净电荷，那么这些净电荷将全部分布在
导体表面。
 
Reasoning:
Gauss’s law
 
 P S





E 0

  
内
S is an arbitrary Gaussian surface
inside the conductor.
Both the net charge and the induced charge reside on the
Surface.净电荷和感应电荷都分布在表面上。
3) The E field on the surface of the charged conductor
is perpendicular to the conductor surface and has a
magnitude  /  0 , where  is the surface charge
density at that point.
导体表面的电场与表面垂直，大小为 /  0 。
Reasoning: Gauss’s law
S is a small Gaussian surface,
whose end faces parallel to the
surface of the conductor.
En





S




E 0

  
内
The E field is perpendicular to the surface because the
conductor is in electrostatic equilibrium.
The magnitude of the E field on the conductor surface is:

E
0
Reasoning: Gauss’s law
 
qin A
 E   E  dS  EA 

0

E
0
0
Example
E
++
+
+
+
++

E
one uniform E field
the E field after a metal
ball was put into the field
(in equilibrium)
4) On an irregularly shaped conductor, the surface
charge density is highest at locations where the radius
of curvature of the surface is smallest.
不规则形状导体曲率半径最小的地方表面电荷密度最大。
越尖的地方感应电场越强。
尖端放电
5) The surface of the conductor in electrostatic equilibrium
is an equi-potential surface and the conductor is an
equi-potential body.
处于静电平衡的导体的表面是一等势面，导体是一等势体。
Reasoning:
Take one arbitrary path from point A to point B inside the
conductor, the E potential difference between the two points
is:
V  VB  VA   E  dr   0dr 0
AB
AB
Applications of properties of conductors in electrostatic
Equilibrium
导体静电平衡性质的应用
1) Discharge 放电
Losing charges of charged particle/object.
带电物体失去电荷。
Link: the atmospheric E field & lightening
了解：大气电场和闪电
There is a huge E field in the atmosphere of the Earth.
The Earth is a good conductor.
The average magnitude of the E field at the surface of the
Earth is:

2
E   10 N / C
0
Very strong E fields in the thundercloud cause lightening
discharge between the cloud and the ground.
Typical E fields during a thunderstorm are as high as
25000N/C.
2) Electrostatic shielding 静电屏蔽
Cavity conductor空腔导体
a)No net charges inside the cavity conductor
空腔导体无净电荷
The external E field has no
effect on the space inside the
cavity. 外电场对空腔内部无影
响。可将仪器等放入空腔内部
起到屏蔽作用。






b) There is net charges inside the cavity conductor
空腔内有净电荷
* The external E field has no effect on E field inside the
cavity. 腔内电场不受外电场影响。
Gauss’s law
Q
Q  q
q







q








** The external E field does not depend on the E field
inside the cavity. 腔外电场也不受腔内电场的影响。It
does not relate to the charge distribution inside the
cavity, but the charge quantity. 外电场与腔内电荷分布无
关，但与腔内带电体电量有关。


Q

Q q


q

由于导体内表面上电量与腔内电荷等量异号，



q




在＋ｑ发出的电场线全部终止在内表面上，则
＋ｑ及－ｑ在腔外产生的合场强为０。



The external E field is the
superposition of the E
fields due to Q+q.导体外
的电场是Ｑ＋ｑ产生的电场的
叠加。
*** If connect the external surface of the cavity to the
ground, the external E field does not relate to E field
inside the cavity. 如果腔外表面接地，则外场不受腔内电
场影响。
q



q



外表面电荷Ｑ＋ｑ全部流入地下，
导体外部由Ｑ＋ｑ产生的电场随
之消失。
Summary: the charge distribution of a conductor in
electrostatic equilibrium
小结： 静电平衡导体的电荷分布
1) No charges inside the conductor 导体内部无电荷









  
Net charges entirely on the external surface.静电荷
全部位于外表面
2) cavity conductors with charges空腔导体带电荷Q
a) There is no charges inside the cavity conductor空腔导体
内部无电荷
q




q






b) There is charges inside the cavity conductor空腔导体内部
有电荷
Q
Q  q
q







q








Example 3-1
Q
As shown in the figure, there are one
ball A and its coaxial spherical shell
B
B. The radius of the ball is R1. The
inner and outer radii of the spherical
shell are R2 and R3 respectively. The
charge on the ball is q and Q on
external surface of the shell.
q
A R1
O
R2
R3
Find: 1) the charge in the space.
2) the E field in the space.
3) the potential in the space.
4)if connect A and B with a wire, calculate 1), 2) and 3).
Solution:
1) since the ball and the shell will
arrive electrostatic equilibrium, we
can get the charge distribution in the
space as following:
q: on the surface of the core
Q
-q: on the inner surface of the shell
Q+q: on the outer surface of the shell
q
B
q
A R1
O
q
R2
R3
Qq
2) according to Gauss’s law, the E
field in the space is:
0
r  R1
q
4 0 r
Qq
4 0 r 2
2
R2  r  R3
R1  r  R2
Q
q
B
q
A R1
O
q
R2
R3
Qq
r  R3
3) the potential of the core:
Q q

Vcore
 
  E  dr
B
0
R1
R2
R3

0
R1
R2
R3
R2

R1
R3
  E1dr1   E2dr2   E3dr3   E4dr4
  E2dr2   E4dr4
1
1
1 qQ

(

)
4 0 R1 R2
4 0 R3
q
q
q
A R1
O
R2
R3
3) the potential of the shell:
Q q

Vshell
Qq
  Edr 
40 R3
R3
B
q
q
A R1
O
R2
R3
4) if connect A and B with a wire,
Qq
the charge on the surface of the core
q q
and the charge on the inner surface
B A R1 R2
of the shell are neutralized. There is
O
R3
no charge inside the shell.
1) The charge distribution is:
0: inside the shell
Q+q: on the outer surface of the shell
2) The E field distribution is:
E  0 r  R3
Qq
E
2 r  R3
4 0 r
Q q
B
A R1
O
R2
R3
3) the potential distribution:
r  R3
Q q
B

Qq
V1   E1dr1   E2dr2 
40 R3
0
R3
R3
r  R3

qQ
V2   Edr 
40 r
r
A R1
O
R2
R3
§3.2 capacitance &capacitor
电容 电容器
Capacitor is one kind of electrical components.
In general, a capacitor consists of two conductors
of any shape, which have charges of equal
magnitude and opposite sign. 两个含有等值异号的
任意形状的导体组成的系统叫做电容器。
The capacitor stores charges.
A capacitor stores energy as well as charge.
Capacitance C of a capacitor is defined as:
Q
C
V
Q: the charge on the capacitor.
V : the potential difference across the capacitor.
The unit of capacitance： coulomb per volt (farad F)
Capacitance C is always a positive quantity.
1) The parallel-plate capacitor平行板电容器
q
A
d
B - - - - -- - - - - q
The E field between the plates:

Q
E 
0 0 A
The potential difference between the plates:
Qd
V  Ed 
0 A
The capacitance of parallel-plate capacitor is:
Q
Q
0 A
C


V Qd
d
0 A
A: the area of the plate.
d: the distance between the plates.
2) The cylindrical capacitor 圆柱形电容器
b
L
A
a
Q
B
Q
L
r
Gaussian surface
l
RA
RB
The potential difference between the two cylinders:
b
Vb  Va    E  ds
a
From the previous result, we know:
b
b
a
a
Vb  Va    Er dr  2ke 
dr
b
 2ke ln  
r
a
The capacitance of cylindrical capacitor is:
Q
C

V
Q
l

Q b
b
2ke ln   2ke ln  
l a
a
3) The spherical capacitor 球形电容器
The potential difference between the two spherical
shells:
VA  VB 
RB
 4 r
RA
q
q
2
dr
q
Gaussian surface
r
0
1
1

(

)
40 RA RB
q
RA
RB
The capacitance of spherical capacitor is:
q
40 RA RB
C

uA  u B
RB  RA
if RB  RA or RB  
C  40 RA
Combinations of capacitors 电容器的并联和串联
1) Parallel combination 并联
The maximum charges on the two
capacitors Q1 and Q2
The total charge Q stored by the
two capacitors is:
Q  Q1  Q2
 C1V  C2 V  Ceq V
The equivalent capacitance of the
two capacitors is:
Ceq  C1  C2
C1
Q1
C2
Q2
V
V  V1  V2
The equivalent capacitance of a parallel combination of
capacitors is the algebraic sum of the individual
capacitances and is larger than any of the individual
capacitances.
并联电容器的等效电容等于这些电容器的电容的代数和，
并且比单个电容器的电容都大。
Ceq  C1  C2  C3  ...
2) series combination 串联
both capacitors must have the
same charge Q. ？？？
V  V1  V2
Q Q


C1 C2
C1
C2
V1
V2
V
Q  Q1  Q2
The equivalent capacitance of the two capacitors in
series combination is:
1
1
1
 
Ceq C1 C2
The inverse of the equivalent capacitance is the
algebraic sum of the inversed of the individual
capacitances and the equivalent capacitance of a
series combination is always less than any individual
capacitance in the combination.
串联电容器的等效电容的倒数等于这些电容器电容的倒
数的代数和。等效电容总是小于任意单个电容。
1
1
1
1
 
  ...
Ceq C1 C2 C3
§3.3 capacitors with dielectrics
有电介质的电容器
One parallel-plate capacitor
with distance d and charge Q.
The potential between the
plates is V0


d
If fill some uniform dielectric into the two plates, keep
the distance and the charge unchanged. The potential
is found to be
V0
V
r 1
r
The capacitance of the capacitor with the dielectric is:
Q Q
C    r   r C0   C0
V V0

dielectric constant
r
C0 is the capacitance of the capacitor without dielectric.
 r will change with different dielectrics.
 r is a physical quantity to describe the electrical
properties of dielectric.
 r is called relative permittivity of dielectric.电介质的
相对电容率
C
r 
C0

   0 r
is permittivity of dielectric.电介质的电容率
In this case, V 
V0
r
E
E0
r
That means, for the identical charge distribution, the E field
in a certain dielectric is only 1  r that in free space. 对于同一
电荷分布，在电介质中产生的场强仅仅是真空中的1  r
Polarized charge density  & free charge density  0
极化电荷面密度 与自由电荷面密度  0 的关系
Parallel-plate capacitor in free space.
Q ',  
The charge on the plate： Q
The E field between plates： E0
Q ',  

E
The free charge density： 0
Q,  0
E0
Q,  0
Filling a dielectric into the plates
The polarized charge is Q '
The E field due to polarized charge is
The polarized charge density： '

E .
The total E field of the
capacitor with dielectric:
 

1 
E  E0  E   E0
r
0
E0 
0
 ＝（1-
r
Q ',  
Q ',  

E
E0
Q,  0

E 
0
1
Q,  0
） 0
0  1 0
 
0 0 r 0
0   0

0
 r 0
The capacitance of parallel-plate capacitor is:
Q0  0 A
C0 

V
d
The capacitance when the capacitor is filled with a
dielectric is:
Q
 0 r A  A
C


V
d
d
Q
 0 r A  A
C


V
d
d
Therefore, we can increase the capacitance by:
1) decreasing the distance d. But d is limited.
2)increasing the area of the capacitor.
3) filling a dielectric
The advantage of dielectrics:
1)It increases the capacitance of a capacitor.
2) It increases the maximum operating voltage of a
capacitor.
3) It may provide mechanical support between the
conducting plates.
§3.4 Gauss’s law at presence of dielectrics
有电介质时的高斯定理
Recall: Gauss’s law in free space 真空中的高斯定理
 
 e   E  dS
S


1
0
1
0
 Q  分立 
i
 dQ  连续 
But actually, in many cases, there are dielectrics in E
field.
What will Gauss’s law be when there is dielectric in E
field?
Get the E field between the
plates when filling with a
dielectric using Gauss’s law:
A cylindrical Gaussian surface
Q ',  
Q ',  
Q,  0

E
E0
Q,  0
From Gauss’s law:



E  dS 
Q  0   S
0
0
0   0

We already have:
0
 r 0
S

S

S
   0S
E  dS 
 0 r
 
 r  0 E  dS   0 S  Q0


We define D   0 r E as electrical displacement.
Then we get Gauss’s law with dielectric as:


 D  dS  Q0
S


where D   0 r E
电介质中的高斯定理
The net electric displacement flux through any closed
surface is equal to the net free charge inside the
surface.
在静电场中通过任意闭合曲面的电位移通量等于该
闭合曲面包围自由电荷的代数和。


 D  dS  Q0
S


D   0 r E
The polarized charge does not appear in this equation,
but it is included in D or  r
此式没有显示极化电荷，但极化电荷的影响已经通过 r
表现出来。


D   0 r E
1) It is true only in the uniform isotropic dielectrics.
仅在均匀各向同性电介质中成立
2) The electric displacement has no obvious physical
meaning.
电位移矢量没有明显的物理意义。
3) The electric displacement D does relate to the free
charge and the polarized charge, but the electric
displacement flux only relates to the free charge.


 D 由极化电荷和自由电荷共同决定，不能认为
D 仅由

自由电荷决定，但 D 的通量仅由自由电荷决定。
E field lines & electric displacement lines
+Q
r
E 线
Start from positive charges, and end
at negative charges.
+Q
r
D 线
Start from free positive charges, and
end at free negative charges.
Example 3-2
As shown in the figure, there are one
metal ball with charge Q ，radius R1
and one uniform isotropic dielectric
spherical shell with outer radius R2
relative permittivity  r in the space.
 
Find：D、E、V at point A, B and C.
 
C
R1
R2
B A Q
r
Solution: the direction of D、E is along the direction of
radium.

the magnitude of D can be obtained from Gauss’s
law with dielectrics:


 D  dS  Q0
S


 D  dS  Q0
S
DA  0
since there is no free charge
inside the conductor
C
R1
R2
B A Q
Q
DB  Dc 
2
4 r
because the free charges inside the Gaussian
surfaces surrounding B or C are both Q.
r


because D   0 r E
E
D
C
 0 r
r
therefore
Q
EB 
2
4  0 r r
EA  0
EC 
R1
R2
B A Q
Q
4 0 r 2
The E potential at point A
 

VA   E  dr
r
R 
R 
 



  E A  dr   E B  dr   E C  dr
1
2
r
R1
R2
Q  1
1
1 





4 0   r R1  r R2 R2 
The E potential at point B
VB  
R2
r

 


EB  dr   EC  dr
R2
Q  1
1
1 





4 0   r r  r R2 R2 
C
R1
R2
B A Q
r
The E potential at point C
VC  

r



EC  dr
C
R1
R2
B A Q
r
Q
4 0 r
由 D1   0 r E1 得
1

E1 
 0 r1
 r1  r2
Exercise 3-1
As shown in the figure, one parallel
plate capacitor formed by
conducting plate A and B is filled
with two dielectrics with distance d1
and d2, relative permittivity  r ,  r2
1
 
Find：D, E in dielectric 1 and 2.
A
d1
d2
B
Exercise 3-2
One metal ball with net charge Q and radius R is placed
into an infinity dielectric with relative permittivity  r

Find: 1) D at any point outside the metal ball.
2) the E potential of the conducting ball.