Download Document 8930369

Electrostatics of dielectrics
Polarization and bound charge density
A dielectric material contains permanent charge distributions. We approximate the
charges which are bound to the ions and are not free to move around, by tiny dipole
moments. The potential created by those dipole moments is found as follows.



The potential at a point at r , due to an electric dipole p at a point r ' is
  

p  (r  r ' )
(r )    3
| r  r '|
The potential due to the superposition of all dipoles in the matter is
 
 

 P( r ' )  ( r  r ' )
(r )   dr '
 
| r  r ' |3
where
 
P (r ) is the mean dipole moment density, and is the polarization field.
We now transform the integrand
 

 
 
 P(r ' ) 'P(r ' )
P( r ' )  ( r  r ' )   
1
 P(r ' )  '    '     
 
| r  r ' |3
| r  r '|
| r  r '| | r  r '|
to obtain (using the Gauss theorem)
  

 


  b (r ' )
b
1
'P(r ' )
(r )   da' nˆ  P(r ' )     dr '     da'     dr '  
| r  r '|
| r  r '| S
| r  r '|
| r  r '|
S

The bound surface charge density is
 b  nˆ  P
and the bound charge density is
b (r )    P(r )

  
Exercise: What is the averaged electric field at the center of an empty sphere of radius a
having bound surface charge density?
1

Solution: Let us assume that the polarization P is along the z-direction, and use spherical
coordinates. From the above expression for the potential we find
 
 

 r  r '
b
r  r'
E (r )   da'     da'  b   3   da' nˆ  P   3
| r  r '|
| r  r '|
| r  r '|
S



We now put here r=0. Since nˆ  P  P cos ' , we find that the integration over the surface
leaves only the z-component of the field at the center to be non-zero. That component is

Ez (0)  2  d ' sin  ' (cos  ' ) 2 P 
0
4
P
3
The displacement field
Inside the matter, the charge density consists of the free (to move) charges and the
bound charges. The total charge density is



  
 (r )   f (r )  b (r )    E (r ) / 4
It follows that (when the bound charges are approximated by the dipoles)

  
  
 f (r )    P(r )    E (r ) / 4 


 
1   
1   
  E (r )  4P(r ) 
  D( r )
4
4
and hence by Gauss's theorem
 
  D  4 f



ˆ
D

n
da

4

d
r

(
r
)
f


S
The boundary conditions on a surface between two different media are: The tangential
components of


E are continuous, the normal component of D jumps by 4 f .


E
P
In linear media there is a linear relation between
and
,
Pi    ij E j ,
i, j  x, y, z
j
2
where
 ij
is the dielectric susceptibility tensor. In highly symmetric materials
 ij   ij  .
As a result, there is also a linear connection between the electric and the displacement
fields, given by the dielectric tensor
 ij ,
Di    ij E j    ij  4ij E j
j
j
Again, in highly symmetric materials
 ij   ij
.
Exercise: A dielectric sphere of radius a, made of a linear material of dielectric constant
 , is placed in a uniform electric field. Find the potential everywhere in space.
Solution: Let us take the z-direction as the direction of the applied electric field,


E0  zˆE0 . Such a field, as we know, is created by a potential  0 (r )   zE0  r cos E0 .
 
Since there are no free charges in the system we have that   D  0. Because the
   
 
 


D




E




E

0

 E  0.
medium is linear, this equation becomes
, or simply,

The electric field E is created by the potential that we need to find,  , namely,


E   . In this problem, this potential satisfies the Laplace equation,  2   0 .
Obviously, we try to solve the equation using spherical coordinates. Since there is no
dependence on the angle  , it follows that the general solution is

 in (r ,  )   An r n Pn (cos  )

n 0
1 

 out (r ,  )    Bn r n  Cn n1  Pn (cos  )
r 
n 0 
As r
that
  , the effect of the dielectric sphere tends to zero, and therefore we expect
(r  ,  )  r cos E0
3
B1   E, Bn1  0
Hence
Let us now apply the boundary conditions on the surface of the sphere,
tangential electric field there is continuous, namely

r  a . The
C
1  in (a,  )
1  out (a,  )

 An  Bn  2 nn1
a

a

a
The perpendicular displacement field is continuous, namely

 in
r

r a
 out
C
 nAn  nBn  (n  1) 2 nn1
r
a
n  1 , for which Bn  0 , this two equations have no non-trivial
solution. Hence we find that n=1, and solve for the coefficients A1 and
C1 in terms of E0 . The solution is
It is easy to see that for
3
E r cos 
 2 0
  1  a3 
 out (r ,  )   E0 r cos  
E   cos 
  2 0  r 2 
 in (r ,  )  
Inside the sphere, the potential is
 in (r ,  )  
3
E z
 2 0
4
and consequently the electric field is along the z-direction, and its magnitude is smaller
than that of the applied field (since   1 ). Outside the sphere the potential can be
  1  a3 
 out (r ,  )   E0 z 
E  z
  2 0  r 3 
written as
We see that the field outside the sphere consists of the applied field (the first term) and
a second term,


  1 3( z / r )rˆ  zˆ 
  1 3 cos rˆ  zˆ
Eout  E0  E0 a 3
 E0  E0 a 3
3
 2
r
 2
r3
The second term is the field due to a dipole of the following value
  1 3
p
a E0 zˆ
 2
This means that the applied field polarizes the sphere and makes it into a dipole. We can
calculate the polarization field, which is the volume density of the dipole. In our case, the
result is simple,
  1 3
P
E zˆ
  2 4 0
In this example, the polarization field is uniform. Therefore, there is no bound charge
 

 P  0 ), but there is bound surface charge density, given by
density (since

 b  nˆ  P , which in our case becomes
b 
 1 3
E cos 
  2 4 0
One can check that the total surface bound charge vanishes.
Exercise: An electric field line is passing through a planar interface between two
insulating media with dielectric constants
1
and
 2 . Find the angle it makes with the
normal to the interface.
Solution: By the boundary conditions, we have
5
Et1  Et 2
1 En1   2 En 2
and hence
Et 1
Et 2
tan 1 tan  2



 1 En1  2 En 2
1
2
The Clausius-Mossotti formula—the dielectric susceptibility
In this model the atoms or the molecules forming the medium are treated as little
conducting spheres, of radius a. The radius of each sphere, a, is chosen such that the
4a 3 / 3 is equal to the mean atomic volume, that is, to the inverse
3
density, 1 / n . In other words, n 4a / 3  N, the number of atoms in the material (or the
corresponding volume,
number of molecules).
We know that when a sphere is placed in an electric field it acquires a dipole moment, and
we expect that this dipole moment is proportional to the electric field. Let us denote the
proportionality constant by  (  is called the atomic polarizability). Namely,


p  E
What is the electric field that acts on each sphere? That field consists of the applied

field, E , and the field exerted by all other spheres in the system (which are polarized,
and therefore produce a field). We now attempt to find the latter field.
We focus on a single sphere, and draw a fictitious big sphere around it. That fictitious
sphere is filled with other spheres, each one exerting the field of a dipole. When all these
6
dipole fields are added up, the result is zero. For example, if all dipoles are along the zdirection, their total field is
Edip
3zi2  ri 2
2 zi2  xi2  yi2
 p
 p
0
5
5
r
r
i
i
i
i
since on the average we cannot distinguish between x, y, and z. Hence the field due to the
dipoles in the fictitious sphere averages to zero. The contribution of all other (far away)
dipoles is approximated by some average polarization
4 
P . Hence we find that
gives rise to the field
3

P , which (as we have found above)
 4 

p   (E 
P)
3
The average polarization field

P on the other hand, is just


P  np
Therefore,

 4 
P  n ( E 
P)
3
Solving this equation,
 
PE
  1 
n
 E 
E
1  n (4 / 3)
4
It follows that the atomic polarizability is given by

3  1 1
4   2 n
which is the Clausius-Mossotti formula.
7
Origin of atomic (or molecular) polarizability
Polar molecules


p
E
The energy of a polar molecule of dipole moment
in an external electric field
is
 
U   E  p . The average total dipole moment of a system of polar molecules is given by
(take the electric field along the z-direction)


p  pz  p
e
0
U / kT

e
sin  cos d
U / kT
sin d
1

 p coth a   , a  pE / kT
a

0
(Langevin's equation).
[ Note that

 d sin e
a cos
1
  de a 
0
1


1 a
2
e  e a  sinh a ]
a
a
At low enough temperatures this gives just p; at high temperatures
pz
p2
 pa / 3 
E,
3kT
at low temperatures
pz  p .
Induced dipole moments
Let us represent each molecule by harmonically-bound charges. For each charge e


m02 r  eE
so that

e2 
p
E
m02
8
The molecular polarizability is then (the sum is over all the charges composing the
molecule)
 
j
e 2j
m j 2j
The polarization field is (n is the number density of the molecules)

e 2j 
P  n
E
2
j m j j
The displacement field is
 
e 2j 
D  E  4n
E
2
j m j j
and the dielectric constant is given in terms of the plasma frequencies
  1  4n
j
e 2j
m j 2j
 1 
j
 2j
 2j
Induced dipole moments by time-dependent fields
Let the applied field be


E(t )  E0eit
Neuton second law for a charge e is then




d 2r
dr
m( 2  
  02 r )  eE
dt
dt
where we have added dissipation, characterized by phenomenological damping.
The solution is
9

 e
1
r
E
m 02   2  i
and the dipole moment is

 e2
1
p
E
m 02   2  i
When the charge belongs to free electrons, we can relate

to the conductivity,
,
ne 2
 
m
and we may assume that there is no restoring force (namely,
0  0 ). When the charge
belongs to an ion, we can neglect the damping. It follows that for a material made of equal
density, n, of ions and free electrons, the polarization field is



e2
1
e2
1
Pn
E

n
E
M 02   2
m   2  i
The (frequency-dependent) dielectric function is
2
ion
4ne 2 / M
4ne 2 / m
4
 ( )  1  2


1


i
0   2
i (1  i /  )
02   2
1  i / 
It is a complex function!
The electrostatic energy in dielectric media
We have found that in free space, the energy needed to establish a charge density
W 
 is

1  
d
r

(
r
)

(
r
)
2
 
Since in the matter   D  4 , we may transform this expression
10
W 
1
8
  

1
d
r


D
(
r
)

(
r
)


8


  

1
d
r


D
(
r
)

(
r
)


8

 
 
d
r
D
(
r
)



(r )


Applying Gauss’s theorem, the first term vanishes, while the second yields
W
   
1
d
r
D( r )  E ( r )
8 
Let us now find the energy it costs to insert a dielectric body into a uniform field
(assuming the dielectric body is linear). The fields before the body is inserted are
 
 
 
E0 (r ), D0 (r )  E0 (r ) , and therefore the electrostatic energy then is
   
1
W0 
dr D0 (r )  E0 (r )
8 
After the body has been inserted, the energy is as given above, and hence the energy cost
is
1
W  W0 
8
   
1
d
r
D
(
r
)

E
(
r
)


8
   
 dr D0 (r )  E0 (r )
We may now write the integrand in the form
   
 
 
 
 
D  E  D0  E0  ( E  E0 )  ( D  D0 )  E  D0  E0  D
The volume integral of the first term here vanishes, since one may write
 

E  E0  F , because the rotor of the fields vanishes, and therefore
1
8
 
  
1
d
r
(



F
(
r
))(
D

D
)

0

8
    
d
r
 F (r )  ( D  D0 )  0


D
Because both the divergence of 0 and of D are given by the same charge density that
creates the uniform field. The volume integral of the second term,
W  W0 
1
8
   
    
1
d
r
(
E
(
r
)

D
(
r
)

E
(
r
)

D
(
r
))

0
0

8
 
 
 
d
r
E
(
r
)

(
E
(
r
)

D
(r ))
 0
11
gives
W  W0  
1    
dr P(r )  E0 (r )
2
and, in fact, the integration is over the dielectric body alone, since only there exists the
polarization field.
Example: Let us take our results above for the dielectric sphere, and calculate the energy.
We have found

Eout

   1
Ein  E0 1 
   2 

  1 a3
 E0  E0
2 cos rˆ  sin ˆ
  2 r3



Din  Ein



Dout  Eout
  1 3 
P
E
  2 4 0
and
Let us now find the energy cost.
1    
1
3 2  1
1
W  W0    dr P(r )  E0 (r )  
E

 4 0   2 8
2
2 sphere
2
 E0
sphere
 1
(3)
 2
Exercise: Find the potential energy of a point charge in vacuum placed at a distance d from
a semi-infinite dielectric medium of dielectric constant  .
Solution: We use cylindrical coordinates, with the z-axis passing through the point charge.
Let us denote by
 b ( )
the bound surface charge density (we assume that the medium
does not carry free charges).
12
Consider the normal components of the electric field and the displacement right at z=0.
En1  

At z  0 we have
while at
z  0
En 2  


qd
2
 d2
qd
2
 d2


3/ 2
3/ 2
 2 d (  )
 2d (  )
Therefore, by continuity of the corresponding displacement vectors, we find
 d ( ) 
1 
qd
2 (1   )  2  d 2


3/ 2
The electric field the bound charge distribution is applying at the point charge is normal
(by symmetry) and its magnitude is given by
E   dd d (  )

d
2
d

2 3/ 2

1 

qd 2  d
1 
2  d2


3

1  q
1   4d 2
The force on the point charge is just Eq. The work required to bring the point charge from
infinity to a distance d from the medium is hence
1   q dz 1   q
W    dzF ( z )  

2

1


4
z
1   4d


d
d
Exercise: Find the potential created by a dipole moment located at the center of a sphere
of radius a and dielectric constant
2 .
1
which is placed in a medium of dielectric constant
Solution: Let us take the dipole moment along the z-direction and use spherical
coordinates. The general solution of the potential is
13
 1
  r 

a


 in (r ,  )    A    B   P (cos  )
 a
 r  
 0 

 1
  r 

a


 out (r ,  )    C    D   P (cos  )
 a
 r  
 0 

At the center of the sphere there is only the potential of the dipole. Therefore, as r
p cos 
1 r 2
p
B1  2 . In addition, C  0
a 1
0,
 in (r ,  ) 
and hence all
B  0 , except
to ensure a well-behaved
potential at infinity. The boundary condition on the tangential component of the electric
field at r=a then gives that
A  D  0 except for D1
and A1 . The boundary condition on
the normal component of the displacement field gives
1 ( A1  2B1 )   2 2D1
Applying again the boundary condition on the tangential component of the electric field
gives
A1  B1  D1
Solving these equations,
 in (r ,  ) 
p cos   1
2(   ) 
 2  3 1 2 r 
1  r
a (1  2 2 ) 
 out (r ,  ) 
3 p cos 
1
2
r
 1  2 2
14