Download Review of Electrostatics Electric Field Electrostatic Potential

Review of Electrostatics
The Coulombic force on charge j due to charge i is:
Fj =
1 q iq j r
4πε 0 r ij2 ij
Electric Field
The electric field is is the force per unit charge. The most
precise statement is that it is the force per unit charge in
the limit that the charge is infinitesimally small:
The Coulombic force is additive. The combined force is
a superposition. The force on charge k due to a number
of charges with the index j is:
Fk =
q jqk
1
r jk
2
4πε 0 jΣ
≠k r
jk
Ej =
When applied to the Coulomb force the electric field
becomes:
Ek =
The constant ε0 is the permittivity of vacuum. In MKS units
the value is ε0 = 8.854 x 10-12 C2 N-1 m-2. In the cgs-esu
unit system the permittivity of free space is 1/4π and the
constant 1/4πε0 does not appear in the Coulomb force.
Electrostatic Potential
The potential at a distance r from a charge is:
φ= –
q
4πε 0r
The electric field represents the force per unit charge.
The potential is the work per unit charge.
W12 = φ(q2 – q1)
Potential and Field due to a Dipole
φ(r) =
µ⋅r
4πε 0r 3
The assumption in this equation is that the distance between
the charge and dipole, r, is large relative to the separation of
charges in the dipole, d, r >> d.
The electric field due to a dipole is:
E=
1 3(µ ⋅ r)r – µ
4πε 0
r3
r5
Using the expression W = -µ.E we can calculate the interaction
energy of two dipoles.
In MKS units the potential has units of V where 1 V = 1 J/C.
W=
Example: Effect of Dipole Orientation
Consider two dipoles, which have the orientations below that
we can call aligned
r jk
1
qj 2
4πε 0 jΣ
≠k
r jk
The potential due to a dipole is:
The electric field is the negative gradient of the scalar
potential:
E = – ∇φ
∂F j
∂q
1 µ 1 ⋅µ 2 – 3(µ 1 ⋅ r)(µ 2 ⋅ r)
4πε 0 r 3
r5
Electric moments
The potential due to discrete charge distribution is:
φ(R) =
and head-to-tail
qk
1
k |R – r |
4πε 0 Σ
k
If the distance R of the test charge is large relative to the
distances between the charges then the expansion:
1
= 1 + r k ⋅∇ 1 + 1 r k⋅r k :∇∇ 1 + ...
2
R
R
|R – r k | R
can be made
Aligned:
µ1. µ2 = µ2 , µ1. r = µ r , µ2. r = µ r , W = -2µ2/4πε0r3
Head-to-tail: µ1. µ2 = -µ2 , µ1. r = 0 , , µ2. r = 0 , W = -µ2/4πε0r3
rk
R
1
Electric moments
The potential is then given by:
4πε 0φ(R) =
q
+ µ⋅∇ 1 + 1 Θ:∇∇ 1 + ...
R
R
R
2
This is the multipole expansion. The terms are the charge (also
called the monopole), q, the dipole, µ, the quadrupole, Θ, and
higher order terms.
q = Σ qi =
i
ρ(r)dr
µ = Σ q ir i =
i
ρ(r)rdr
Θ = Σ Σ q ir ir j =
j
i
The interaction of a collection of charges subjected to an
electric field is given by:
W = qφ – µ⋅E + 1 Θ:∇E + ...
2
The picture is that of a charge interacting with the potential,
the dipole interacting with a field, the quadrupole interacting
with the field gradient etc.
An electric field can exert a force:
F =Σ qiE(r i)
ρ(r)rrdr
T =Σ r i × qiE(r i)
i
on a collection of charges.
Polarizability
In the presence of an externally applied electric field the
eipole moment of the molecule can also be expressed
as an expansion in terms of moments:
µ = µ 0 – α⋅E + 1 β :EE + ...
2
The leading term in this expansion is the permanent dipole
moment, µ0. The polarizability is a tensor whose components
can be described as a follows:
∂µ x
∂Ey
i
or a torque:
q is a scalar,
µ is a vector (a first rank tensor)
Θ is a matrix (a second rank tensor)
α xy =
Interaction of electric moments
with the electric field
0
Where the 0 subscript refers to the fact that the derivative is
Evaluated at zero field. The β tensor is called the hyperpolarizability and is third ranked tensor.
Properties of the polarizability tensor
Like the quadrupole moment, the polarizability can be
made diagonal in the principle axes of the molecule.
In the laboratory frame of reference the polarizability
depends on the orientation of the molecule.
The average polarizability is independent of orientation.
It is given by the Trace, which is written Tr α.
Tr α = 1 α xx + α yy + α zz
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The polarizability increases with the number of electrons in
The molecule or with the volume of the charge distribution.
Classically, for a molecule of radius a, α = 4πε0a3.
Polarizability as second rank tensor
The dipole moment components each can depend on
as many as three different polarizability components as
described by the matrix:
α xx α xy α xz
µx
µ y = α yx α yy α yz
µz
α zx α zy α zz
Ex
Ey
Ez
If a molecule has a center of symmetry (e.g. CCl4) then
The polarizability is a scalar (i.e. the induced dipole moment
Is always in the direction of the applied field). However, for
non-centrosymmetric molecules components can be induced
in other directions. The directions are often determined by
the directions of chemical bonds, which may not be aligned
with the field. This is the significance of the tensor.
Lorentz (classical) polarizability
The Lorentz model is given here. An electron in
one dimension is subject to an electric field that results in
a displacement from the center of charge. For a
displacement in the x direction (this is the polarization of the
electric field, i.e. the electric vector of radiation) the induced
moment is given by:
µ ind = – ex = αE
The restoring force is:
F = – kx = – mω0 2x
The force is balanced by the force due to the applied electric
field F = eE. The balance of forces is:
– mω0 2x = eE
2
Lorentz polarizability
Thus x is:
x = – eE 2
mω0
and the dipole moment is:
e2 E
mω0 2
µ ind =
Frequency dependence of polarizability
The quantity fj is the oscillator strength. The above treatment
is for a static field. In the case where E(t) is a sinusoidal field
the equation resembles that of a driven harmonic oscillator.
F = m d x2 = – eE(t) – mω0 2x(t) – Γ dx
dt
dt
2
The classical Lorentz polarizability is:
e2
mω0 2
α=
The expression can be generalized by letting there be N
electrons on a molecule. Each electron has an intrinsic
harmonic oscillator frequency. The fraction of the electrons
in each of the j modes is fj. The polarizability for all of the
electrons is:
2
α=
In addition to the electric force and the harmonic restoring
force we have added a frictional force. The time dependent
electric field must have the form E(t) = E0xexp[i(ky - ωt)].
Therefore, we try a solution of the form x(t) = x0exp[i(ky - ωt)].
dx = – iωx exp i ky – ωt
0
dt
d x = – ω 2x exp i ky – ωt
0
dt 2
f
e
Σ j
m j ωj2
2
Frequency dependence of polarizability
Frequency dependence of polarizability
The equation becomes:
e E(t) – ω 2x(t) + iωΓ x(t)
– ω 2x(t) = – m
0
m
Collecting terms in x(t) solving for x(t) we have:
e E(t)
–m
x(t) =
ω0 2 – ω 2 – iωΓ
m
2
fj
e
α ω = mΣ
ω j 2 – ω 2 – iωΓ
m
We can also define the polarizability α = -ex(t)/E(t) so the
polarizability has real and imaginary parts. These are
obtained by multiplying both numerator and denominator by
ωj2 - ω2 + iωΓ/m.
The complex polarizability is:
2
αω =e
mΣ
j
ω j 2 – ω 2 + iωΓ
m
iωΓ
2
2
ω j 2 – ω 2 – iωΓ
m ωj – ω + m
2
=e
m Σ fj
j
fj
ωj 2 – ω2
ωj 2 – ω2
2
+ ωΓ
m
2
+i
ωΓ
m
2
ω j 2 – ω 2 + ωΓ
m
2
j
Kramer’s Kronig Relations
The real and imaginary parts are related to one another
through the Kramers-Kronig relations
∞
α′ ω = 2
π
0
sα′′ s ds
s2 – ω2
∞
α′′ ω = –2ω
π
0
α′ s ds
s2 – ω2
Clearly the polarizability consists of real and imaginary parts.
We can write it as
α ω = α′ ω + iα′′ ω
Oscillator Strength
The oscillator strength is a classical formalism. The oscillator
strength fij is proportional to the intensity of a transition i Æ j.
It is a number less than one and in fact the sum of all of the
oscillator strengths in the molecule equals one. Thus,
Σj fij = 1
Comparison of the quantum treatment with the Lorentz
formulation gives
2
f0 j =
2m µ 0 j ω j0
3e h
2
The factor of 1/3 arises because only one polarization of
electromagnetic radiation is considered here.
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