Transcript (Susceptibility),電容率(Permittivity),介電常數(Dielectric Constant)

Ch4. 物質中的電場(Electric Fields in Matter)
4.1 極化(Polarization)
4.2 極化物體的場(The Field of a Polarized Object)
4.3 電位移(The Electric Displacement)
4.4 線性電介質(Linear Dielectrics)
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.1.1.1 極化(Polarization) —電介質(Dielectrics)
Electric fields in matter: conductors and insulators (dielectrics)
In conductors charges will be pushed to the boundary by external field.
In dielectrics charges are attached to atoms or molecules.
The electric field can distort the charge distribution of a dielectric atom or
molecule by two principal mechanisms: stretching and rotating.
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.1.1.2 極化(Polarization) —電介質(Dielectrics)
Many non-conducting solids have permanent dipole moments,
or become polarized when immersed in an external electric field.
Materials such as these are known as dielectrics.
Normally, the dipole moment is zero on large scales since atomic
dipoles are oriented in random directions.
Immersion of a dielectric in an electric field polarizes atoms and
tends to align the atomic dipoles.
The induced dipole moment or the total polarization of an atom is
approximately proportional to the field:
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.1.2.1 感應出的電偶極(Induced Dipoles)
e : electric susceptibility of the medium
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.1.2.2 感應出的電偶極(Induced Dipoles)
Example : A primitive model for an atom consists of a
point nucleus (+q) surrounded by a uniformly charged
spherical cloud (-q) of radius a. Calculate the atomic
polarizability of such an atom.
In the presence of an external field E, the nucleus will be shifted slightly to the
right and the electron cloud to the left (extremely small).
The electric field at a distance d from the center of a uniformly charged sphere
is
E
The atomic polarizability
  40 a 3
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.1.2.3 感應出的電偶極(Induced Dipoles)
Carbon dioxide
Polarizability :
4.5  10 40 C 2  m / N
2  10 40 C 2  m / N
When the field is at some angle to the axis, you must resolve it into parallel and
perpendicular components, and multiply each by the pertinent polarizability :



p    E    // E //

The induced dipole moment may not be in the same directions as E .
For
asymmetrical molecule, the most general linear relation between

 a completely
E and p is :
p x   xx E x   xy E y   xz E z
p y   yx E x   yyE y   yzE z
p z   zx E x   zy E y   zz E z
The set of nine constants  ij constitute the polarizability tensor for the molecule.
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.1.2.4 感應出的電偶極(Induced Dipoles)
Problem 4.4
A point charge q is situated a large distance r from a neutral atom
of polarizability . Find the force of attraction between them.
r̂
The electric field produced by the charge q :

E
1 q
r̂
2
40 r
The induced dipole moment of the atom :
r


 q
p  E 
r̂
2
40 r
The electric field produced by the dipole, at location of q :

E
1 1 2 q
(
)r̂
3
2
40 r 40 r
The force on q due to this field :


q 2 1
F  qE  2 (
)
r̂
40 r 5
attractive
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.1.3.1 極化分子的排列(Alignment of Polar Molecules)

p (Permanent dipole moment)
H+
H+
1050
Polar molecule
O-

F
+q


Force : F   F



F
d
-q

E
The electric field is uniform


 
  


 
Torque : N  ( r  F )  ( r  F )  [(d / 2)  (qE)]  [(  d / 2)  ( qE)]  qd  E
  
N  p E
 
  
 

If the field is non-uniform, F  F  0
F  F  F  q(E   E  )  q(E)
Assuming the dipole is very short

E x  (E x )  d


E z  (E z )  d
E y  (E y )  d
 

E  ( d   ) E
  
 



F  F  F  q(E   E  )  q(E)  (p   )E
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.1.3.2 極化分子的排列(Alignment of Polar Molecules)
Problem 4.6
A (perfect) dipole p is situated a distance z above an infinite grounded
conducting plane. The dipole makes an angle  with the perpendicular
to the plane. Find the torque on p. If the dipole is free to rotate, in
what orientation will it come to rest?

Using image dipole :

p
p
Redraw
z
z
pi
2z
pi 

p
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.1.3.3 極化分子的排列(Alignment of Polar Molecules)

The electric field produced by pi, at location of p: E i 
1
p
(2 cos  r̂  sin  ˆ )
3
40 (2z)

The dipole moment p can be expressed by p  p cos  r̂  p sin  ˆ
The torque on p is
  
N  p  Ei 
p
[( p cos  r̂  p sin  ˆ )  (2 cos  r̂  sin  ˆ )]
3
40 (2z)
2
p2
p
sin  cos  ˆ
ˆ
ˆ

(cos  sin    2 sin  cos  ) 
( )
3
3
40 (2z)
40 (2z)
p 2 sin 2 ˆ

( )
3
640 z
Out of the page
For 0 <  < /2, the torque tends to rotate p counterclockwise
For /2 <  < , the torque tends to rotate p clockwise
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.1.3.4 極化分子的排列(Alignment of Polar Molecules)
Problem 4.9
A dipole p is a distance r from a point charge q, and oriented so that p
makes an angle  with the vector r form q to p. What is the force on p ?
p
q
r
 

F  (p   )E

E

1 q
q
xx̂  yŷ  zẑ
r̂

40 r 2
40 ( x 2  y 2  z 2 )3 / 2






q
x
 py
 p z )E x  (p x
 py
 pz )
x
y
z
x
y
z 40 ( x 2  y 2  z 2 )3 / 2

  
q p x 3x
q p 3r (p  r )

[ 3  5 (p x x  p y y  p z z)] 
[ 3
]
5
40 r
r
40 r
r
x
Fx  (p x

F


q
[
p

3
(
p
 r̂ )r̂ ]
3
40 r
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.2.1.1極化物體的場(The Field of a Polarized Object)– 束縛電
荷(Bound Charges)
dipole
What can it be if the dipole is not placed at the origin?
Assume that the dipole is placed at


and
the
observed
is
sitting
at
r'
r.
The dipole moment in each volume element
 
p  P d '
d'
So the potential is
 

1
ˆ  P( r ' )
V( r ) 
d'
2

40 V 
  
where   r  r '
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.2.1.2極化物體的場(The Field of a Polarized Object)– 束
縛電荷(Bound Charges)
1
ˆ
' ( )  2


Note that the differentiation is with respect to the source
coordinates r '
 
 

1
ˆ  P( r ' )
1
1
V( r ) 
d' 
P( r ' )   ' ( )d'
2


40 V 
40 V

Integrating by parts



1
P
1
V( r ) 
[   '( )d'   ( 'P)d']
40 V


V
Using the divergence theorem

V( r ) 

1
1 
1
[  P  da '   ( 'P)d']
40 S 

V
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.2.1.3極化物體的場(The Field of a Polarized Object)– 束
縛電荷(Bound Charges)

V( r ) 

1
1 
1
[  P  da '   ( 'P)d']
40 S 

V
The first term looks like the potential of a surface charge
surface charge density :

b  P  n̂
Where
n̂
is the normal unit vector
While the second term looks like the potential of a volume charge
volume charge density :

b    P

V( r ) 
b
b
1
[
da '   d']
40 S 

V
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.2.1.4極化物體的場(The Field of a Polarized Object)– 束
縛電荷(Bound Charges)
Example 4.2
Find the electric field produced by a uniformly polarized sphere of
radius R
z
n̂

P

R


Since P is uniform  b    P  0

b  P  n̂  P cos 
azimuthally symmetry and spherically symmetrical problem
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.2.1.5極化物體的場(The Field of a Polarized Object)– 束
縛電荷(Bound Charges)
Boundary conditions
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.2.2.1束縛電荷的物理意義(Physical Interpretation of
Bound Charges)
The dipole moment
A
d
+
++
+++
++
b 
n̂

P

+
 

 bd    P  da    (  P)d
V
+
+
+
q PA

P
A
A
If the polarization is non-uniform we get
accumulation of bound charge within
the material as well as on the surface.
+
- - - - -
+
q = PA

q
q
PA cos 
b 


 P cos   P  n̂
A end A / cos 
A
+
+
qd = PAd
S

b    P
V
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.2.2.2束縛電荷的物理意義(Physical Interpretation of
Bound Charges)
There is another way of analyzing the uniformly polarized sphere: (radius : R)
There are two spheres of charge : a positive sphere and a negative sphere
Without polarization the two are superimposed and cancel completely.
But when the material is uniformly polarized,
+++
+
++
+
+
+
+
d
-- - --
Bound surface charge b
The electric field in the region of overlap between two uniformly charged spheres


1 qd
E
40 R 3
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.2.2.3束縛電荷的物理意義(Physical Interpretation of
Bound Charges)
Proof
The electric field inside the positive sphere

 
E 
r
3 0
The electric field inside the negative sphere

 
E  
r
3 0
Problem 2.18

d
+

r
-

r
  
d  r  r
So the total field
 

 
 
  
E  E  E 
r  (
r ) 
( r  r )
3 0
3 0
3 0


1
q
1 qd

( d)  
3 0 4 R 3
40 R 3
3
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.2.2.4束縛電荷的物理意義(Physical Interpretation of
Bound Charges)
 4


3
p  qd  R P
3


1 qd
E
40 R 3

P

qd
4 3
R
3

1 
E
P
3 0

1 p  r̂
Meanwhile, for points outside, V 
40 r 2
When we calculate the electric fields and potential via bound charges,
we obtain the “macroscopic” field.
the “macroscopic” field : the average field over regions large enough to
contain many thousands of atoms.
Ordinarily, the macroscopic field is what people mean when they speak
of “the” field inside matter.
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.3.1.1 電介質存在下的高斯定律(Gauss’s Law in the Presence
of Dielectrics)
When we consider the “free” charges alone, the Gauss’s law is expressed as
  free
E 
0
  1
S E  da  0 V free d
When we include the “bound” charges for the presence of dielectrics,


1
1
  E tot  (free   b )  (free    P)
0
0


  (0 E tot  P)  free

 
Here we define the electric displacement : D  0 E  P
Hence the Gauss’s law is now expressed as :

  D  free
 
 D  da   free d
S
V
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.3.1.2 電介質存在下的高斯定律(Gauss’s Law in the Presence
of Dielectrics)
Example 4.4
A long straight wire, carrying uniform line charge , is surrounded by
rubber insulation out to a radius a. Find the electric displacement.
L
Drawing a cylindrical Gaussian surface,
of radius s and length L.
s

 
 D  da   free d
S
V
D(2sL)  L
a
therefore


D
ŝ
2s
The formula above for D holds both within the insulation and outside it.




Outside region :
E outside 
ŝ
D  0 E
20s

Inside region : since we do not know P , the electric field cannot be determined!

P0
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.3.1.3 電介質存在下的高斯定律(Gauss’s Law in the Presence
of Dielectrics)
Problem 4.15
A thick spherical shell (inner radius a, outer radius b) is made of dielectric
material with a “frozen-in” polarization
 
k
P( r ) 
r
r̂
Where k is a constant and r is the distance from the center. Find the
electric field in all three regions by two different methods :
(a). Locate all the bound charge, and use Gauss’s law to calculate the
field it produces.

 
 


(b). Use  D  da   free d to find D , and then get E from D  0 E  P
S
V

 P b
P

P
a

P
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.3.1.4 電介質存在下的高斯定律(Gauss’s Law in the Presence
of Dielectrics)

P b
Solution :

1  2k
k
 b    P   2 (r )   2
(a).
r r
r
r
k
 
at r  b

 P  r̂  b
 b  P  n̂   
k
P  ( r̂ )  
at r  a
a


For r < a :
Qenc = 0, so E  0
For r > b :

P

P
a

P

Qenc = 0, so E  0
For a < r < b :
k
k
2
(
)( 4a )   ( 2 )4r 2 dr  4ka  (4k )( r  a )  4kr
a
r
a

1 Q enc
1  4kr
k
E
r̂

r̂


r̂
2
2
40 r
40 r
0r
r
Q enc
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.3.1.5 電介質存在下的高斯定律(Gauss’s Law in the Presence
of Dielectrics)

P b
Solution :
(b).

P
 
 D  da   free d  Qfenc  0
S

P
V

D0
everywhere

 
D  0 E  P  0
For r < a and r > b :
For a < r < b :

P0
 
k
P( r )  r̂
r
a


P
E
0

E0


P
k
E 
r̂
0
0r
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung

P
4.3.1.6 電介質存在下的高斯定律(Gauss’s Law in the Presence
of Dielectrics)
Problem 4.16

Suppose the fieldinside alarge piece of dielectric is E 0 , so that the electric
displacement is D   E  P
0
0 0
(a). Now a small spherical cavity (Fig. a) is hollowed out of the material. Find


the field at the center of the cavity in terms of E 0 and P .


Also find the displacement at the center of the cavity in terms of D 0 and P .

(b). Do the same for a long needle-shaped cavity running parallel to P (Fig. b)

(c). Do the same for a thin wafer-shaped cavity perpendicular to P (Fig. c)
a
b
c

P
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.3.1.7 電介質存在下的高斯定律(Gauss’s Law in the Presence
of Dielectrics)
Solution :

P

(a). The electric field at the center of a sphere with uniformly polarization P :

1 
(see Example 4.2)
E'  
P
3 0
So the electric field at the center of a spherical cavity
 
 
1 
E  E 0  E'  E 0 
P
3 0



 1 
1 
2
D  0 E  0 E 0  P  D0  P  P  D0  P
3
3
3
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.3.1.8 電介質存在下的高斯定律(Gauss’s Law in the Presence
of Dielectrics)
Solution :

P
(b). The electric field of opposite charges at the two far away ends of the needle
is very small :

E'  0
So the electric field at the center of a long needle-shaped cavity
 
 
E  E 0  E'  E 0





D  0 E  0 E0  D0  P
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.3.1.9 電介質存在下的高斯定律(Gauss’s Law in the Presence
of Dielectrics)
Solution :

P
(c). The electric field of a parallel-plate capacitor with upper plate at  = P :

1 
E'   ( )P
0
So the electric field at the center of a thin wafer-shaped cavity
 
 
1 
E  E 0  E'  E 0  P
0



 
  
D  0 E  0 E0  P  D0  P  P  D0
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.3.2 迷惑的相似處(A Deceptive Parallel)
 
E( r ) 
In vacuum
In dielectrics
  free
E 
0

  D  free
1
40

ˆ
V 2 free ( r ' )d'
 

1 ˆ
D( r ) 

(
free r ' )d'
2

4 V 
Why?
Note that while the curl of E is zero, the curl of D is in general non-zero:

 



  D    (0 E  P)  0  E    P    P  0
In general
Therefore D cannot in general be written as the gradient of some scalar
potential (unlike E)
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.3.3 邊界條件(Boundary Conditions)
In vacuum

E above
 E below 
1

0
//
//
E above
 E below
In dielectrics

D above
 D below   free
//
//
//
//
D above
 D below
 Pabove
 Pbelow
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.4.1.1 線性電介質—極化率(Susceptibility),電容率(Permittivity),
介電常數(Dielectric Constant)
K=
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.4.1.2 線性電介質—極化率(Susceptibility),電容率(Permittivity),
介電常數(Dielectric Constant)
Linear dielectrics :
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.4.1.3 線性電介質—極化率(Susceptibility),電容率(Permittivity),
介電常數(Dielectric Constant)
Example 4.5
A metal sphere of radius a carries a charge Q. It is surrounded, out to
radius b, by linear dielectric material of permittivity . Find the potential
at the center. (relative to infinity)
b
For all points r > a :
Q
For a < r < b :
Inside the metal sphere :
For r > b :
  
EPD0
a
The potential at the center :
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.4.1.4 線性電介質—極化率(Susceptibility),電容率(Permittivity),
介電常數(Dielectric Constant)

P0
vacuum
dielectric
 
 P  dl  0

P  0

P0
 
 
 P  d l   (  P)  da  0
S

D  0
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.4.1.5 線性電介質—極化率(Susceptibility),電容率(Permittivity),
介電常數(Dielectric Constant)
Example 4.6
A parallel-plate capacitor is filled with insulating material of dielectric
constant r. What effect does this have on its capacitance?
 
 D  da   free d  Qfenc
S
V
+++++++++
ẑ
2DA  A

---------------
1
1
D up,down     ẑ
D up,up    ẑ
2
2


1
1
E up,up 
  ẑ
E up,down  
  ẑ
2 0
2 0  r
For bottom plate
2DA  A


1
1
1
1
D bottom,down     ẑ    ẑ
D bottom,up    ẑ     ẑ
2
2
2
2


1
1
E bottom,down 
  ẑ
E bottom,up  
  ẑ
2 0
2 0  r
For upper plate
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.4.1.6 線性電介質—極化率(Susceptibility),電容率(Permittivity),
介電常數(Dielectric Constant)
For upper plate

1
D up,up    ẑ
2

1
E up,up 
  ẑ
2 0

1
D up,down     ẑ
2

1
E up,down  
  ẑ
2 0  r
+++++++++
ẑ
----------------
For bottom plate

1
1
D bottom,up    ẑ     ẑ
2
2

1
E bottom,up  
  ẑ
2 0  r
In dielectric
 

E  E up,down  E bottom,up  
1
Qd
V  Ed 
 d 
0 r
A 0  r

1
1
D bottom,down     ẑ    ẑ
2
2

1
E bottom,down 
  ẑ
2 0
1
1
1
  ẑ 
  ẑ  
  ẑ
2 0  r
2 0  r
0 r
A 0  r
Q
C

  r C vacuum
V
d
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.4.1.7 線性電介質—極化率(Susceptibility),電容率(Permittivity),
介電常數(Dielectric Constant)
+++++++++
s
s
ẑ
----------------
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.4.1.8 線性電介質—極化率(Susceptibility),電容率(Permittivity),
介電常數(Dielectric Constant)
(a).
+++++++++
+++++++++
ẑ
s
 
 D  da   free d  Qfenc
S
V
(b).
In slab 1
In slab 2
(c).
In slab 1
In slab 2

D  ẑ

Inside the upper plate : D  0
This is true for both slabs



E1 
ẑ 
ẑ
1
2 0




 2
E2 
ẑ 
ẑ
D   ẑ   2 E 2
2
3 0





P1   0  e E1   0 ( r  1)E1   0 (
ẑ)   ẑ
2 0
2



0
2

P2   0  e E 2   0 ( r  1)E 2  (
ẑ)   ẑ
2
3 0
3


D   ẑ  1E1
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.4.1.9 線性電介質—極化率(Susceptibility),電容率(Permittivity),
介電常數(Dielectric Constant)
(d).

2
7
s
s
s
2 0
3 0
6 0





P1   0  e E1   0 ( r  1)E1   0 (
ẑ)   ẑ
2 0
2
constant



0
2

P2   0  e E 2   0 ( r  1)E 2  (
ẑ)   ẑ
2
3 0
3
V  E1s  E 2s 
(e). In slab 1
In slab 2

b    P  0
For both slabs
In slab 1
In slab 2
top


 b  P1  n̂  
2
top


 b  P2  n̂  
3
bottom


 b  P1  n̂  
2
bottom


 b  P2  n̂  
3
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.4.1.10 線性電介質—極化率(Susceptibility),電容率(Permittivity),
介電常數(Dielectric Constant)

 / 2
 / 2
 /3
 /3

(f).
In slab 1, the total surface charge above :
/ 2
the total surface charge below :
  

   
2 3 3
2
In slab 2, the total surface charge above :
   2
   
2 2 3
3

2
 
the total surface charge below :
3
3


E1 
ẑ
2 0

 2
E2 
ẑ
3 0
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.4.2.1 線性電介質的邊界值問題(Boundary Value Problems with
Linear Dielectrics)
In a homogeneous linear dielectric the bound charge density (b) is
proportional to the free charge density (f).
In particular, unless free charge is embedded in the material,  =0 ,
and any net charge must reside at the surface.
Remember the boundary conditions for dielectric :

D above
 D below   free
//
//
//
//
D above
 D below
 Pabove
 Pbelow
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.4.2.2 線性電介質的邊界值問題(Boundary Value Problems with
Linear Dielectrics)
Example 4.7
A sphere of homogeneous
linear dielectric material is placed in an otherwise

uniform electric field E 0 . Find the electric field inside the sphere.
There is no free charge and only bound charges exist on
the surface.

E0
So the boundary conditions :
Inside the sphere :

Vin   A  r P (cos )

0
Outside the sphere :

B
P (cos )
 1 
0 r
Vout   E 0 r cos   
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.4.2.3 線性電介質的邊界值問題(Boundary Value Problems with
Linear Dielectrics)


B
P (cos )
 1 
0 R
 A  R P (cos )   E 0 R cos   

 0
B
AR 
R  1

A1R   E 0 R 

B1
R2
for
 1
for
 1
(  1)B
P (cos )
 2
R
0

 r  A  R P (cos )   E 0 cos   
0
 1
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.4.2.4 線性電介質的邊界值問題(Boundary Value Problems with
Linear Dielectrics)
(  1) B
 r A  R  
R  2
2B1
 r A1   E 0  3
R
 1
for
 1
for
 1
A   B  0
A1  
3
 1 3
E 0 B1  r
R E0
r  2
r  2
for
 1
for
 1
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.4.2.5 線性電介質的邊界值問題(Boundary Value Problems with
Linear Dielectrics)
Example 4.8
Suppose the entire region below the plane z = 0 is filled with uniform linear
dielectric material of susceptibility  e . Calculate the force on a point charge q
situated a distance above the origin.
z
q
d 
y
r
x
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.4.2.6 線性電介質的邊界值問題(Boundary Value Problems with
Linear Dielectrics)
The surface bound charge on the xy plane is of opposite sign to q, so the
force will be attractive.
Let us first calculate b :

b  P  n̂  Pz  0e E z
Where Ez is the z-component of the total field just outside the dielectric, at z = 0.
The contribution of charge q to the z-component of the total field :
E q ,z
1
q
1
qd

cos   
2
2
40 (r  d )
40 (r 2  d 2 )3 / 2
The contribution of the bound charge to the z-component of the total field :
E  b ,z
b

2 0
Therefore
 b   0  e E z   0  e [
b
1
qd

]
2
2 3/ 2
40 (r  d )
2 0
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.4.2.7 線性電介質的邊界值問題(Boundary Value Problems with
Linear Dielectrics)

1
qd
( e ) 2
2  e  2 ( r  d 2 ) 3 / 2
e
) , this is exactly the same as the induced charge
Apart from the factor (
e  2
Therefore
b  
on an infinite conducting plane under similar circumstances.
q b  (
For z > 0
e
)q
e  2
Using the method of images : qb at the image position z = -d
1
q
V
[

2
2
2
40 x  y  (z  d )
For z < 0
qb
x  y  (z  d)
2
2
2
]
Using the method of images : q+qb at the image position z = d
q  qb
1
V
[
]
2
2
2
40 x  y  (z  d )
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.4.2.8 線性電介質的邊界值問題(Boundary Value Problems with
Linear Dielectrics)
We check the boundary condition for the image solution :
 0 (

V
V
1
qd

) ( e ) 2
 b
2 3/ 2
z z  0 z z  0
2  e  2 ( r  d )
Right!
So the force on q is :

F
e
1 qq b
1
q2
ẑ  
(
) 2 ẑ
2
40 (2d)
40  e  2 4d
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.4.3.1 電介質系統中的能量(Energy in Dielectric Systems)
The energy stored in any electrostatic system :
W
0
2
E
d

2
The energy stored in dielectric systems :
0
1  
2
W    r E d   (D  E)d
2
2
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.4.3.2 電介質系統中的能量(Energy in Dielectric Systems)
Suppose the dielectric material is fixed in position, and we bring in the free charge.
The work done on the incremental free charge :
W   (f )Vd



f    (D) W   (f )Vd   [  (D)]Vd
Since   D   f



  [( D)V]  [  (D)]V  D  (V)


 
W   [  (D)]Vd     [(D)V]d   (D)  Ed

 
   [(D)V]d 0 W   (D)  Ed
If the medium is a linear dielectric, then D  E
 
 
1  
1
2
(D  E)  (E )  (E)  E  (D)  E
2
2
 
1  
1  
W   D  Ed
W   (D)  Ed  (  D  Ed)
2
2
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.4.4.1 作用在電介質的力(Forces on Dielectrics)

d
Fringing field
dielectric

The bound surface charge on the dielectric experiences the forces F
as shown. The dielectric is pulled into the region of the plates.
Let W be the energy of the system. If I pull the dielectric out an infinitesimal
distance dx, the energy is changed by an amount equal to the work done :
dW  Fme dx
Fme   F
So the electric force on the slab is
F
dW
dx
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.4.4.2 作用在電介質的力(Forces on Dielectrics)
Let’s assume that the total charge Q on the plates is held constant, as the
dielectric moves.
2
Now, the energy stored in the capacitor is
W
1
Q
CV 2 
2
2C
So
If the potential on the plates is fixed by connecting it up to a battery,
the battery also does work as the dielectric moves.
dW  Fme dx  VdQ
So
F
dW
dQ
1
dC
dC 1 2 dC
V
  V2
 V2
 V
dx
dx
2
dx
dx 2
dx
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung
4.4.4.3 作用在電介質的力(Forces on Dielectrics)
If the potential on the plates is fixed by connecting it up to a battery,
the battery also does work as the dielectric moves.
dW  Fme dx  VdQ
So
F
dW
dQ
1
dC
dC 1 2 dC
V
  V2
 V2
 V
dx
dx
2
dx
dx 2
dx
Y.M. Hu, Assistant Professor, Department of Applied Physics, National University of Kaohsiung