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Chapter 4 Electric Fields in Matter
--
4.1 Polarization
-
4.1.1 Dielectrics
Electric fields in matter: conductors and insulators (dielectrics)
In conductors charges will be pushed to the boundary by
external electric field.
In dielectrics charges are attached to atoms or molecules.
+
E
F
+
+
+
+
+
E
-+
-+
-+
-+
-+
-+
-+
4.1.2 Induced Dipoles
The induced dipole moment or the total polarization of an atom is approximately
proportional to the field:
r
r
p = αE , where α is the atomic polarizability.
r
r
r
Total polarization: p , for an atom: p = αE
r pr
r
r
The polarization per unit volume: P = , for linear dielectrics: P = ε 0 χ e E
V
What is the effect of induced dipoles?
Eexternal
Eexternal
- +
- +
Edipole, outside
Edipole, inside
Edipole, inside
Edipole, outside
α
= a 3 , in units of 10-30 m3 as the unit of atomic polarizabilities.
We use
4πε 0
H
He
Li
Be
C
0.667
0.205
24.3
5.6
1.76
The model to estimate atomic polarizabilities: (Please do not use the picture:
r
r 4π 3
r
r
4π 3
p = VP => αE =
a ε 0 χ e E -> α =
a ε 0 χ e . You got a wrong answer.)
3
3
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.
ρr ρd
q
d
qd
E=
=
=
=
= Eexternal
3
3ε 0 3ε 0 4πa / 3 3ε 0 4πε 0 a 3
a,-q
+q
d
Eexternal
Æ p = qd = αE = 4πε 0 a 3
The first step generalization:
Anisotropy: They polarize more readily in some directions
r
r
r
p = α ⊥ E⊥ + α // E//
p
E
The second step generalization:
pi = α ij E j
Ref: Phys. Rev. 133, A629 (1964).
Shock-Tube Measurement of the Polarizability of Atomic Hydrogen
Wayland C. Marlow*
Lockheed Missiles and Space Company, Palo Alto, California
Daniel Bershader
Stanford University, Stanford, California
Lockheed Missiles and Space Company, Palo Alto, California
Received 3 September 1963; revised 4 November 1963
The electric polarizability of atomic hydrogen has been measured at optical frequencies by shock heating
a gaseous mixture of argon and molecular hydrogen to temperatures causing nearly complete
dissociation of the latter. Application of Mach-Zehnder interferometry to the shock-heated gas gave
measurements of fringe shift across the shock front which were subsequently reduced to obtain the
polarizability of the ground state of atomic hydrogen. The resultant value is alpha H=(4.61±0.07)a03, for
a wavelength of 5870 Å, where a0 is the first Bohr radius and the error assignment is purely statistical.
The shock waves employed were in the Mach number range 10 to 12 in a 2 / 3 Ar-1 / 3 H2 mixture (by
volume) and produced temperatures of the order of 4600°K at densities high enough to ensure thermal
equilibrium. Under these conditions there was negligible ionization or excitation of the argon diluent. In
contrast to earlier experiments reported in the literature, the present result is in good agreement with the
predictions of the quantum-mechanical time-dependent perturbation theory which gives a value of alpha
3
H=(4.66±0.01)a0
for the same wavelength as that used in the experimental work.
4.1.3 Alignment of Polar Molecules
The uniform electric field can move a charge. Can the uniform electric field move a
dipole? What can it do? What kinds of electric fields can move a electric dipole?
+q
If the field is uniform, it can rotate the dipole.
r r r r
r r
r r
r
N = r+ × qE + r− × − qE = qd × E = p × E
(
)
E
-q
If the field is nonuniform, it can move the dipole.
Scalar function Φ ( x, y, z ) ? Vector function?
r
V ( x, y, z ) = Φ1 ( x, y, z )xˆ + Φ 2 ( x, y, z ) yˆ + Φ 3 ( x, y, z )zˆ
r
d
Φ(x+dx,y+dy,z+dz)
Φ(x,y,z)
r
r
r
F = ( p ⋅ ∇ )E
(
)
r
∆Φ = d ⋅ ∇ Φ
r r
r
r
r
r
F = Fat + q + Fat − q = qEat + q − qEat − q = q∆E
r
∂E
∂E
∂E
∆Ex = (∇Ex ) ⋅ d = d x x + d y x + d z x
∂z
∂x
∂y
r
r
r
r
r
F = qd ⋅ ∇ E = ( p ⋅ ∇ )E
(
)
4.1.4 Polarization ÅÆ Magnetization
r pr
The polarization P = , dipole moment per unit volume,
V
r
r
for linear dielectrics: P = ε 0 χ e E
The electric field of a dipole:
r
r
1
ρ (r ')dτ '
V (r ) =
4πε 0 ∫ r 2 + r '2 −2rr ' cosθ
r
r
V (r ) =
r
r
1 1
1 1
ρ (r ')dτ ' +
r ' cosθ ' ρ (r ')dτ ' + ...
2 ∫
∫
4πε 0 r
4πε 0 r
r
1 p cosθ
1 rˆ ⋅ p
=
A dipole: Vdip (r ,θ ) =
4πε 0 r 2
4πε 0 r 2
Er = −
∂Vdip
∂V
1 2 p cosθ
1 p sin θ
=
, Eθ = − dip =
3
4πε 0
r
r∂θ
4πε 0 r 3
∂r
r
E = Er rˆ + Eθθˆ =
r
p ⋅ rˆ = p cosθ
p
4πε 0 r 3
(2 cosθrˆ + sin θθˆ)
r
θˆ = cosθ cosφxˆ + cosθ sin φyˆ − sin θzˆ Æ p ⋅ θˆ = − p sin θ
r r
r
p = p ⋅ rˆrˆ + p ⋅ θˆθˆ = p cosθrˆ − p sin θθˆ
E
-+
-+
-+
-+
+
-+
-+
-+
r
E=
=
p
4πε 0 r
1
4πε 0 r
3
r
E (r ,θ ) =
3
(2 cosθrˆ + sin θθˆ) = 4πε1 r (2( pr ⋅ rˆ)rˆ − (pr ⋅ θˆ)θˆ)
3
0
(3( pr ⋅ rˆ )rˆ − pr )
1
4πε 0 (2 z )
3
(2 p cosθrˆ + p sin θθˆ)
r r r
N = p× E = ?
magnetic dipole
The problem is similar to that of magnetic dipole on a
superconductor.
superconductor
image dipole
Exercise: 4.2, 4.4, 4.6, 4.7