Polarization of Dielectrics
Dielectrics are materials which have no free charges; all electrons are bound and associated with the nearest
atoms. An external electric field causes a small separation of the centres of the electron cloud and the
positive ion core so that each infinitessimal element of volume behaves as an electric dipole. Dielectrics
may be subdivided into two groups :
Non-Polar
Polar
which behave as above
in which the molecules or atoms possess a permanent dipole moment which is
ordinarily randomly oriented, but which become more or less oriented by the
application of an external electric field.
(applied)
E
The induced dipole field opposes the applied field. In the diagram
shown opposite the volume element indicated could represent an atom,
a molecule, or a small region.
The type of polarization on a microscopic scale is determined by the
material. Most materials exhibit polarization only in the presence of an
external field. A few however show permanent polarization:
-
+
-
+
-
+
-
+
-
+
-
+
Vol element
Ferroelectric
crystals exhibit spontaneous permanent polarization.
Electrets
become permanently polarized if allowed to solidify in the presence of a strong
electric field.
The type of polarization may be additionally subdivided into the following categories :
Electronic
Ionic
Orientational
Space-charge
a displacement of the electronic cloud w.r.t the nucleus.
separation of +ve and -ve ions in the crystal.
alignment of permanent dipoles (molecules).
free electrons are present, but are prevented from moving by barriers such as grain
boundaries - the electrons "pile up".
The Electric Polarization P is the dipole moment per unit volume at a given point.
P = Np where p is the average dipole moment per molecule
N is the number of molecules per unit volume.
1.
Any molecule develops a dipole moment which is proportional to the applied field
p = aE where a is the polarizabilty.
2.
Example The electronic polarizability of a simple atom.
With an applied field the electron cloud is displaced until the mutual attractive force between it and the ion
core is is just balanced by that produced by the field E
1
r3
F = eE = e ⋅
⋅
e
4πε0 r2 R3
gives p
=
eE
=
e2r
4πε0R3
er
=
4πε0R3E
where ae = electronic polarizabilty = 4pe R3
0
C.G Pearce EM Fields2
16/10/96
1
=
αeE
3.
The Electric Field at an Exterior point
Consider a block of dielectric, P = dipole moment
per unit volume.
We need to calculate the potential V at an exterior
point.
d t'
dV
=
V
=
P ⋅ ˆr
dτ′
4πε0 r2
1
1
4πε0
∫τ P
⋅ ∇′
r
4.
1
( r ) dτ′
5.
dV
P
integrating over the block
Now
∇ ⋅
A)
A
f
fA
=
f
f
=
( r ) and P = A
V
=
so, putting
(∇
⋅
+
6.
⋅ ∇
1
1
∫τ
4πε0
(∇′
P
dτ′
r)
⋅
−
1
( ∇′ ⋅
4πε0
P)
r
d τ′
7.
By the divergence theorem this becomes
V
1
=
P
4πε0
da′
⋅
∫ s′
1
−
r
P
∇′ ⋅
∫ τ′
4πε0
r
dτ′
8.
for the potential due to the dipoles at the external point.
Both of the terms in equation 8 have the form of potentials produced by charge distributions ;
i.e.
and
a surface charge density sb = P • n where n is the outward unit normal vector
a volume charge density rb = -Û'•P
Thus
V
=
V
=
1
4πε0

∫
 s
σb da
r
+
∫τ
ρb d τ 
due to the dielectric only


r
9.
and
E
=
−∇
1
4πε0

∫
 s′
σb da′ ˆ r
r
2
+
∫ τ′
ρb d τ′ ˆ r
r2



10.
The Macroscopic Field
To the above must be added the effects of the external charge distributions that are responsible for the
polarization. These are simply additive.
V
=
E
=
1
4πε0
1
4πε0
(

∫
 s
(

∫
 s
σf
σf
σ b)
+
r
+
σ b) ˆ r
r2
da
da
+
+
ρ
( f +
∫τ
r
ρ
( f +
∫τ
ρb)
d τ
11.

ρb) ˆ r
r2
d τ

12.
This expression is completely general.
The effect of these expressions is that for the purposes of calculation the dielectric may be replaced by the
equivalent bound charge distributions sb and rb without affecting the field outside the dielectric.
Example cylindrical electret.
C.G Pearce EM Fields2
16/10/96
2
The Bound Charge Densities rb and sb
We now demonstrate how the displacement of charges within
the dielectric results in net volume and surface charge
densities.
Consider a small volume inside the dielectric, where the
electric field E is the resultant of an external field and the
field due to the dipoles. The positive and negative charges
are separated by an average distance s due to the influence of
E. Consider the element of surface da and the charge which
has crossed it. If we fix the origin in the negative charges we
need only consider the movement of the positive charges.
E
+
+
-
da
+
+
-
+
+
+
-
+
-
Then, the amount of charge dQ crossing da is just the amount of positive charge within the volume
dt = s @ da
i.e.
dQ = NQ s @ da Qs = p , NQs = P
i.e.
dQ = P @ da
13.
If da is on the surface of the material, this charge accumulates there in a layer of thickness s @ n ( which is
small, of molecular dimensions) and the charge can be treated as a surface layer with density
s
b
= dQ/da = P @ n
14.
We can similarly show that -Û @ P represents a real volume density of charge as follows.
The net charge flowing out of a volume t across the elementary area da of its surface is P @ da as found
above in (13). Thus the total charge flowing out of the surface bounding t is the integral of this over the
surface, i.e.
Q
=
∫s P
⋅
da
and the net charge remaining within is - Q.
If the density of this remaining charge is rb then
∫ τ ρb d τ
=
Q
−
ρb
hence
=
−
∫s P
=
−
∫ τ (∇
=
−∇ ⋅
⋅
da
⋅
15.
P) d τ
P
16.
i.e. the bound charge density is numerically equal to minus the divergence of the polarization. An
important consequence of this is that if the polarization is uniform within a region and its divergence is zero,
then so is the bound charge density rb .
div E in Dielectrics
Gauss's law states
∫s E
⋅
da
=
∫τ ∇
⋅
E dτ
=
Q
ε0
17.
Q here is the total charge enclosed, which for dielectrics must include free as well as bound charges. i.e.
Q
=
∫ τ ( ρf
+
ρb) d τ
where the integration is intended to cover both volume and surface distributions. Hence substituting (18)
into (17) and equating the integrands of the volume integrals then
C.G Pearce EM Fields2
16/10/96
3
18.
∇ ⋅
E
=
ρ
ρb)
( f +
19.
ε0
or
∇ ⋅
E
=
ρt
ε0
20.
This is Gauss's law in its more general form and is one of the four Maxwell's Equations.. In obtaining it
we have implicitly assumed that space derivatives of E exist. This is not the case at the interface between
media and where this applies we must use the integral form i.e we must integrate E @ da over a closed
surface.
N.B. Since E = -ÛV this leads to Û2V = - rt/e0 Poisson's equation.
The Electric Displacement
Now
∴
or
∇ @
∇ ⋅
ε
( 0E
+
ρb
=
E
=
P)
=
P
−∇ @
1
ε0
ρ
( f
21.
− ∇ @
P)
ρf
23.
i.e. the vector (eoE + P) has a divergence that depends only on rf .
This is called the Displacement, D = (eoE + P)
i.e.
Û
22.
@ D = rf
24.
25.
This is the equivalent Maxwell's equation for a dielectric. Note it does not contain the permittivity e and is
thus independent of the medium. It can be regarded as Gauss's law for D. In its integral form it becomes
∫s D
⋅
da
=
∫ τ ρf
dτ
26.
Note that both divD and ∫s D @ da are unaffected by bound charges.
From the definition of displacement (D =e0E + P) we have
E
D
=
ε0
−
P
27.
ε0
i.e. the electric field inside the dielectric is the resultant of two fields,
D/e0 associated with free charges,
since ∇ ⋅ (D/e0) = rf / e0
28.
and - P/e0 associated with bound charges,
29.
since
∇ ⋅
( - P/e0) = rb / e0
NB Lines of D begin and end only on free charges.
Lines of E begin and end on either free or bound charges.
In writing down expressions for the divergence of E and P we have implicitly assumed their existence. It
should be noted that the space derivatives do not exist at a point charge or at the interface between two
media. In such cases the integral form of Gauss's law must be used.
C.G Pearce EM Fields2
16/10/96
4
The Susceptibility
Provided that P ¶ E , which it is in practice for moderate fields
since D = e0E + P
= e0E(1 + ce )
c
e = P/e0 E
30.
31.
is the electric susceptibility
i.e.
P = e0 ce E indicates the relative ease of polarization.
32.
c
e
The Permittivity
+ ce )E
E
e is the relative permittivity
0 r
r
or dielectric constant
D = eE
in vacuum c = 0 ,
D =
=
where
=
e
e
0(1
33.
34.
e e
e
r
or k = 1
is the permittivity.
e e
0 r
35.
36.
or k are usually in the range 1 - 7, but some non-linear materials have k as high as 105.
Pure water has k ~ 80. For all materials k is a function of frequency.
e
r
In class A dielectrics D
¶
E, and then since er is a constant for class A, then
@ D = rf
@ E = rf /e
V =
If the material is not class A then
-r
/e Poisson's equation for class A dielectrics
37.
38.
39.
V =
-r
/e
40.
Û
and if
e
r
«
f(x,y,z)
Û
\
Û
2
2
Û
f
t
=
0
( rf
+
b)
r
/ e
0
The Relationship between rf and rb
In a class A dielectric
P = D − ε 0E =
∴
or
The total charge density is
∇ ⋅
−
P
=
(
εr − 1
εr
ρb
=
(1
−
ρt
=
ρf
+
)
1
εr )
ρb
εr − 1
∇ ⋅
D
41.
εr
D
=
(1
−
1
εr )
ρf
ρf
=
42.
43.
ρf
− ∇ ⋅
P
=
ρf
εr
44.
This is smaller than rf . rf and rb have opposite signs.
Note also that if rf is zeri (in class A dielectrics) then so is rb. This is nearly always the case. Thus the
bound charges will nearly always be located only on the surface of the dielectric.
The Surface Charge Densities sb and sf
At the interface between dielectric and conductor there is :
a bound charge density sb on the dielectric
a free charge density sf on the conductor.
In the steady state E = 0 inside the conductor, and inside the dielectric, from Gauss's Law
e E = s + s
0
f
b
C.G Pearce EM Fields2
16/10/96
5
45.
and
D = eE =
Hence
s
f
+
s
b =
e e
r 0
s
f
E =
s
f
/er
from Gauss
46.
(cf 44.)
47.
Boundary Conditions
The calculation of electric field variation across a boundary between two media needs a knowledge of the
boundary conditions. These are :
At the boundary between media
•
V1 = V2
V is continuous
(else the field is infinite)
•
The normal component of D is continuous
or discontinuous by the free charge density at the interface
Dn1 = Dn2 ( +sf)
This can be shown by applying Gauss's Law to a small pillbox at the boundary. The net outward
displacement = free charge enclosed. Note that between dielectrics the free charge density sf is usually
zero. At a dielectric / conductor interface, if E is constant, then D = 0 inside the conductor, and Dn = sf in
the dielectric.
•
Et1 = Et2
The tangential component of E is continuous.
This can be shown by evaluating the line integral éE · dl around a thin rectangular loop lying
parallel to the boundary. If the boundary is between a dielectric and a conductor, then E = 0 in the
conductor, and hence Et = 0 in both media. Therefor E must be normal at the surface of a conductor.
Forces on Dielectrics
A dipole in a uniform field experiences a torque, but no net force
τ
=
F
=
(p
¯
or force per unit volume F
=
N (p
p× E
A net force is experienced only in a non-uniform field
¯
F
=
(
ε
⋅ ∇)
−
E
⋅ ∇)
E
ε 0) ( E
=
(P
⋅ ∇)
and P
=
E
=
⋅ ∇)
=
(
E
ε
−
ε 0) E
1
(ε
2
−
εr
1
−
εr
ε0) ∇E2
∇
1
( 2 εE )
2
Ex. Calculate the force in the dielectric of a coaxial cable, radii R1, R2 permitivitty e with charge per unit
length l.
R2
λ
λ R2
By Gauss
,
V = ∫ E dρ =
ln
E =
R1
2περ
2 πε R1
E
∴
C.G Pearce EM Fields2
16/10/96
F
τ
=
λ
2περ
=
1
ε0 ( εr
2
=
−
V
ρ ln ( R2 / R1)
1)
V2
ln2 (R2 / R1)
1
| ρ2 |
∇
6
=
−
ε0 ( εr
1) V 2
ln2 ( R2 / R1) ρ3
−
, radial, inwards.