4.4 Linear Dielectrics
r
F
stable
θ
stable
magnetic dipole
superconductor
F
image dipole
4.4.1 Susceptibility, Permitivility, Dielectric Constant
r
If E is not too strong, the polarization is proportional to the field.
r
r r r
r
r
P = χ eε 0 E (since D = ε 0 E + P , D is electric displacement from ρ free )
r
r
Æ ∇× D = ∇× P = 0
χ e is called the electric susceptibility.
χe =
M
P
, χm =
ε0E
H
r
r r
r
r
D = ε 0 E + P = ε 0 (1 + χ e )E = εE , ε is called the permittivity.
εr =
ε
= 1 + χ e is called the relative permittivity, or dielectric constant.
ε0
Material
Dielectric Constant Material
Dielectric Constant
Vacuum
1
Benzene
2.28
Helium
1.000065
Diamond
5.7
Air
1.00054
Water
80.1
Nitrogen
1.00055
KTaNbO3
34,000
What is high K material?
+++++++++
+++++++++
+++++++++
εr = 2
E
E
Vacuum
------------------------It may discharge for an electric field higher than E.
Example: A metal sphere of radius a carries a charge Q. It is
surrounded, out to radius b, by linear dielectric material of
permittivity ε . Find the potential at the center.
r
r
r
Q
Q
Q
rˆ and b > r > a , E =
D=
rˆ Æ r > b , E =
rˆ
2
2
4πε 0 r
4πr
4πεr 2
a
b
V=
Q
4π
a
b
 Q 1
1
1
1
1
∫−

+
− 2 dr  =
− + 
dr
∫
 ε r2
εr
b
∞ 0
 4π  ε 0b εb εa 
r
r χεQ
b > r > a , P = χ eε 0 E = e 0 2 rˆ
4πεr
r
ρb = −∇ ⋅ P = 0
r
r
χεQ
χεQ
r = a , σ b = P ⋅ (− rˆ ) = − e 0 2 and r = b , σ b = P ⋅ (rˆ ) = e 0 2
4πεa
4πεb
If the space is entirely filled with a homogeneous linear dielectric, then
r
r
E
∇ ⋅ D = ρ f and ∇ × D = 0
vacuum P
When crossing the boundary:
r
r
∇× D = ∇× P ≠ 0
Example: A parallel-plate capacitor is filled with insulating
material of dielectric constant ε r . What effect does this
----------
have on its capacitance?
++++++++++
2 AD = Aσ + Æ D+ =
2 AD = Aσ − Æ D− =
Ein =
σ+
2
σ−
2
, E+ ,up =
=−
σ+
2
σ+
σ
σ
, E+ ,up , out = + , E+ , down = − +
2ε
2ε 0
2ε 0
, E− ,up = −
σ+
σ
σ
, E− , down = + , E− , doun , out = +
2ε 0
2ε
2ε 0
Q
Q
Aε
σ
σ
Æ ∆V = d =
d, C=
=
= ε r Cvacuum
Aε
∆V
d
ε
ε
4.4.2 Boundary Value Problem with Linear Dielectrics
If we place free charge inside a linear dielectric,
r
r
r
 ε0D 
χ
 = − e ρf
ρ b = −∇ ⋅ P = −∇ ⋅ χ eε 0 E = −∇ ⋅  χ e

ε 
1 + χe

(
)
If we place the free charge on the boundary:
D⊥ , above − D⊥ ,below = σ f Æ ε above E⊥ , above − ε below E⊥ ,below = σ f
 ∂Vabove 
 ∂V

 − ε below  − below  = σ f , continuity: Vabove = Vbelow
∂n 
∂n 


ε above  −
+
Example: A sphere of homogeneous linear dielectric material is
r
placed in an otherwise uniform electric field E0 . Find the
E0
electric field inside the sphere.
Method 1:
 ∂V 
 ∂V 
The free charge is on the boundary Æ ε out  − out  − ε in  − in  = σ f = 0
 ∂r 
 ∂r 
Only bound charge exists, no free charge.
r >> R , Vout → − E0 r cosθ
r = R, ε
∂Vin
∂V
= ε 0 out and Vin = Vout
∂r
∂r
Vout = − E0 r cosθ +
A
cosθ , Vin = Br cosθ
r2
A
3
2A 
ε − ε0

= BR , εB = ε 0  − E0 − 3  Æ B = −
E0 , A = R 3
2
R
R 
2 + ε / ε0
2ε 0 + ε

r
3
3
3
∂V
Vin = −
E0 r cosθ = −
E0 z Æ E = −
zˆ =
E0 zˆ
2 + εr
2 + εr
2 + εr
∂z
− E0 R +
Method 2:
r
Uniformly polarized sphere with polarization P = Pzˆ may produce electric field
r
P
E=−
zˆ inside.
3ε 0
r 
P 
 zˆ inducing the polarization:
Total field inside: E =  E0 −
3ε 0 

r
r

3
3
3χ e
P 
 zˆ Æ P =
Pzˆ = χ eε 0 E = χ eε 0  E0 −
ε 0 E0 Æ E =
E0 zˆ =
E0 zˆ
3
2
3
ε
3
+
+
χ
+
ε
χ
0 
e
e
r

Example: Suppose the entire region below the plane z = 0 is
filled with uniform linear dielectric material of susceptibility χ e .
Calculate the force on a point charge q situated a distance d
above the origin.
Considering q without dielectric material:
q
d
1
E z ,below = −
2
2
2
4πε 0 r + d
r + d2
r
σ
Considering dielectric material without q: σ b = P ⋅ zˆ Æ E z , above = b and
2ε 0
(
)
E z ,below = −
σb
2ε 0

r
σ b 
qd
−
Total effect: P ⋅ zˆ = σ b = χ eε 0 Ez ,below,total = χ eε 0  −
3

2
2
2ε 0 
 4πε 0 r + d

σb = −
1 χe
qd
2
2π χ e + 2 r + d 2
(
)
3/ 2
Total bound charge:
 χe 
q
Qb = ∫ σ b da = −
 χe + 2 
 χe 
q at z = -d to solve the problem.
Use image charge Qb = −
χ
+
2

 e
V=
1 
q
+
4πε 0  x 2 + y 2 + ( z − d )2

Qb
x 2 + y 2 + (z + d )
2




z>0

q + Qb
1 

2
2
2


4πε 0
 x + y + (z − d ) 
r
1 qQb
zˆ
attractive force on q: F =
4πε 0 (2d )2
V=
z<0
4.4.3 Energy in Dielectric Systems
To charge up a capactor, it takes energy of
Q
Q2 1
dQ =
= CV 2
C
2C 2
Change of capacitance in dielectric materials:
Cdielectric = ε r Cvacuum Æ increase the energy because of an increase of charge
dW = VdQ Æ W = ∫
(
)
( )
r
r
r
∆W = ∫ ∆ρ f Vdτ , ρ f = ∇ ⋅ D Æ ∆W = ∫ ∆ ∇ ⋅ D Vdτ = ∫ ∇ ⋅ ∆D Vdτ
(
)
( )
( )
(
)
r
r
r
r
r
r r
∇ ⋅ ∆DV = ∇ ⋅ ∆D V + ∆D ⋅ ∇V Æ ∇ ⋅ ∆D V = ∇ ⋅ ∆DV + ∆D ⋅ E
Choose V=0 at r → ∞ .
r r 1 r r
r r
∆W = ∫ ∆D ⋅ Edτ for a linear dielectric ∆D ⋅ E = ∆ D ⋅ E
2
( )
(
)
W=
1 r r
D ⋅ Edτ
2∫
l
4.4.4 Forces on Dielectrics
x
fringing field
Cvacuum =
Q wlσ ε 0 wl
=
=
V σ d
d
ε0
Cdielectric = ε r Cvacuum Æ C =
ε 0w
d
x + εr
ε 0w
d
(l − x ) = ε 0 w (ε r l − χ e x )
d
Assume that the total charge on the plate is constant ( Q = CV ), W =
F =−
1 Q2
2 C
dW
Q 2 dC 1 2 dC
ε χw
=
= V
= − 0 e V 2 attractive force
2
2d
dx 2C dx 2
dx
If you start from constant voltage, the work supplied by a battery must be included.
F =−
d 1
dQ
2
 CV  + V
dx  2
dx

If the voltage is constant and the charge is varying, you must include the force due to
the battery for maintaining a constant voltage.
++++++++
+ + + + ++++
Exercise: 4.18, 4.23, 4.28
1. Dielectric material Æ increase the capacitance of a capacitor Æ store much more
charges
2.
P
E+
P
3ε 0
E
P
E
E + P / ε0
E
P
−
P
ε0
−
P
3ε 0
3. Ferroelectricity: BaTiO3
ferroelectric
Transition temperature?
Curie-Weiss law?
antiferroelectric