Today’s lecture
IV. Electrostatics in matter
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Electric displacement
Gauss’s law in Dielectrics
Susceptibility, Permittivity, Dielectric constant
Gauss’s law in dielectrics
r
The net effect
r of a dielectric on the total
r field E is to
r r
increase D inside by the amount of P : r
D := ε 0 E + P
r
The total field E decreases by the polarization P due to
the influence of the dielectric material.
The total field is always weakened by the dielectric material.
The field due to the free charges D is called the electric
displacement (or electric flux density).
Gauss’s law becomes in dielectrics:
r r
∇ ⋅ D = ρ free
r r
or ∫∫ D ⋅ da = Q fenc
Gauss’s law only connects electric displacement and free
charges!
Example: (Griffith 4.31)
r r
r
A dielectric cube of side a carries a polarization P (r ) = k ⋅ r
where k is constant. Find all bound charges and check
whether they add up to zero.
The bar electret
r
r r
D := ε 0 E + P
electric total Polarization
displacement field
Example: (4.11) Bar electret (electric analog to a magnet)
A short cylinder, radius a, length L carries a “frozen-in”
uniform polarization P parallel to its axis.r
r
ρ b = −∇P = 0
Find the bound charge
r r
σ b = P ⋅ un = ± P
Sketch the electric field for
L>>a
L<<a
r
P
P, E and D:
r
E
L≈a
r
P
r
P
r
D
r
P
No free charges anywhere! E terminates on charge, D on free charge!
Important facts about the electric displacement
• The example of the bar electret shows that although there
r r
are no free charges anywhere but still:
D≠0
D is not exclusively determined by the free charge!
• This means in particular there is no Coulomb law for D :
r r
D(r ) ≠
r
ρ free (r ') r
1
u dτ '
2 r
∫∫∫
4πε 0
(r − r ')
r r r r r r r
r r
• In general: ∇ × D := ε 0∇ × E + ∇ × P = ∇ × P ≠ 0
• This means in particular there is no scalar potential for D
r r r
since:
∇×D ≠ 0
r
Boundary conditions for D
General conditions at boundary of surface charge σ:
1. Component perpendicular to the surface:
r r
⊥
⊥
D
⋅
d
a
=
Q
⇒
D
−
D
∫∫
fenc
above
below = σ f
S
r r Qenclosed
σ
⊥
⊥
E
⋅
d
a
=
⇒
E
−
E
=
∫∫
above
below
S
ε0
ε0
Normal component of E is discontinuous at any boundary!
2. Component parallel to the surface
r r r r
∇×D = ∇×P
⇒
r ||
r ||
r ||
r ||
Dabove − Dbelow = Pabove − Pbelow
r
r ||
r ||
Eabove − Ebelow = 0
Linear dielectrics
r r
rP ∝ E
r
P =: ε 0 ⋅ χ e ⋅ E
For linear dielectrics:
or
χe : electric susceptibility (dimensionless)
Usually one cannot use this equation to calculate P
because it affects E! It would be appropriate to start with
D, if it can be deduced from the distribution of free charges.
The key question for any material is how
r to deduce
r r r the
equation that describes the material: D = D E , P
For linear materials one can define:
[
]
r
r
r
D := ε 0 E + ε 0 χ e E
r
r
= ε 0 (1 + χ e ) ⋅ E =: ε ⋅ E
ε := ε 0 (1 + χ e )
ε is the permittivity of the material. The special case of
Vacuum (χe=0 ⇒ ε= ε0) gives the name to the permittivity
of free space.
One uses widely the relative permittivity or dielectric
constant εr:
ε
ε r :=
ε0
= 1+ χe
More definitions
r r
r r
r r
In linear dielectrics: P ∝ E and D ∝ E and P ∝ D
In homogenous materials ε does not vary in a region
ε(x,y,z)=const.
In isotropic materials
r rD and E have the same properties
in all directions D || E
In anisotropic materials D, E, and P are not parallel and ε
or χe becomes a tensor (single crystals, magnetized plasma).
One can express these definitions in terms of ε. A material
is called
• linear, if ε does not change with E:
• homogenous, if ε does not change from point to point
• isotropic, if ε does not change with direction.
A free charge q embedded in a large dielectric will be
surrounded by bound charges and therefore shielded by the
polarization of the medium and the field will be
r r
1 q r
u
E (r ) =
2 r
4πε r