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Electromagnetism
Physics 15b
Lecture #20
Dielectrics
Electric Dipoles
Purcell 10.1–10.6
What We Did Last Time
Plane wave solutions of Maxwell’s equations
⎧⎪E = E0 sin(k ⋅ r − ω t)
ω = kc, E0 = B0 , k̂ = Ê × B̂
⎨
⎩⎪B = B0 sin(k ⋅ r − ω t)
Propagates along k
with speed c
  Two possible polarizations
for the same k
 
Energy flow given by the Poynting vector S =
 
For plane waves, average energy flow is S =
c
E×B
4π
c 2
c 2
E k̂ =
E k̂
4π
8π 0
1
Goals Today
Introduce dielectric material
Synonym for “insulator”
Effects on electric field expressed by dielectric constant
  Universally found in capacitors
 
 
Look into the microscopic origin of dielectric
Dipole moment of small charge distribution
  Electric field generated by a dipole moment
  Force on a dipole moment due to electric field
 
Next lecture: construct dielectric from dipoles
Material in a Uniform E Field
A slab of insulator is in a uniform E field
E
+ and − charges feel the force F = qE
F
  They can’t flow, but they do move slightly from
the natural (equilibrium) positions
  Excess + and − charges appear on the surfaces
  Additional electric field E′ appears inside
E ′ = 4πσ and opposite to E
 
It’s reasonable to assume σ ∝ E
 
Total electric field inside the insulator is
1
Ein = Eout + E′ = (1− k)Eout = Eout
ε
ε = the dielectric constant of the material
 
+σ
E′
−σ
ε > 1, i.e., the electric field is weaker inside
2
Dielectric Constant
Dielectric constant ε depends
on the density
Gases very close to 1
  Liquids and solids 2 to 10
 
Molecules can rotate to align with
electric field
  Prime example: water
 
+
 
−
O −
H
−
O
−
H +
Condition
ε
Air
Gas, 0°C
1.0006
Methane
Gas, 0°C
1.0009
HCl
Gas, 0°C
1.0046
Gas, 110°C
1.0126
Water
Liquids and gases of polar
molecules have large ε
+ H
Substance
E
H +
See textbook §10.6 and 10.12
Liquid, 20°C
80.4
Benzene
Liquid, 20°C
2.28
Methanol
Liquid, 20°C
33.6
Ammonia
Liquid, −34°C 22.6
Mineral oil
Liquid, 20°C
2.24
NaCl
Solid, 20°C
6.12
Sulfur
Solid, 20°C
4.0
Silicon
Solid, 20°C
11.7
Polyethylene Solid, 20°C
2.25–2.3
Porcelain
6.0–8.0
Solid, 20°C
Dielectric in Capacitor
Capacitor is filled with a dielectric
−Q
+Q
4π Q
area A
Charge ±Q on the plates create E0 =
A
  Dielectric polarizes and creates
surface charges
E
4π Q
  Actual electric field inside is E = 0 =
ε
εA
  Potential difference is
 
V=
∫
right
left
E ⋅ds =
4π Qd
εA
C=
Q
εA
=
= ε ⋅C0
V 4π d
Dielectric increases the capacitance
by a factor ε
d
capacitance
w/o dielectric
Real-world capacitors use various dielectric materials
  paper, ceramics, mica, oil, liquid electrolyte, etc.
  Issues: field (voltage) tolerance, frequency response, temperature
dependence, polarity, long-term stability
 
3
Half-Filled Capacitor
A rectangular capacitor is partially filled with a dielectric
 
If it were empty, the capacitance
would be
ab
C0 =
4π d
Consider it as two capacitors:
a(b − x)
4π d
ε ax
=
4π d
 
Empty part: Cempty =
 
Filled part: Cfilled
a
ε
x
d
b
Ctotal =
a{b + (ε − 1)x}
4π d
connected
in parallel
Suppose there is charge Q in this capacitor
 
dU
2π dQ 2 (ε − 1)
Q2
2π dQ 2
=−
<0
=
dx
2C a{b + (ε − 1)x}
a{b + (ε − 1)x} 2
  Increasing x decreases the potential energy
Energy is U =
Electrostatic force pulls the dielectric into the capacitor
Small Charge Distribution
So far we had uniform E field causing uniform polarization
 
More general case (non-uniform E) requires a better framework
Consider an arbitrary charge distribution of a small size,
and the electric field far away from it
 
 
A
Electric potential at point A is
ρ(r ′)dv ′
Integral over volume where ρ ≠ 0
ϕ=∫
V
r − r′
Since r′ << r, we use Taylor expansion
r − r′
−1
= (r − 2rr ′ cos θ + r ′ )
2
2 −1 2
E
r
ρ≠0
r′
⎞
1⎛
r′
r ′ 2 (3 cos2 θ − 1)
= ⎜ 1+ cos θ + 2
+ ⎟
r⎝
r
2
r
⎠
4
Moments
We can now express the potential at A as
ϕ=
1
1
ρ(r ′) dv ′ + 2
∫
r V
r
K0
∫
V
ρ(r ′)r ′ cos θ dv ′ +
1
r3
∫
V
ρ(r ′)r ′ 2
(3 cos2 θ − 1)
dv ′ + 
2
K1
K2
/r2
At large distance, K0/r >> K1 >> K2 >> …
  We must consider higher-order terms only if the preceding terms
happen to be zero
 
/r3
K0 is the net charge of the source
 
 
K0/r is the familiar Coulomb potential
For a small charged object, the Coulomb force due to the net charge
outweighs everything else
If the object (e.g. a molecule) is net neutral, the K1 term
becomes important
Dipole Moment
We can rewrite the K1 term as
K1
r
2
=
1
r2
∫
V
r̂
⋅ ρ(r ′)r ′ dv ′
r 2 ∫V
ρ(r ′)r ′ cos θ dv ′ =
a vector determined by
the charge distribution
Define the dipole moment of a charge distribution by
p≡
∫
V
ρ(r ′)r ′ dv ′
then the electric potential due to it is
ϕ=
r̂ ⋅ p
r2
Q: Doesn’t this definition depend on where the origin of the
coordinate system is?
  A: It does. But that’s OK
 
5
Coordinate Origin
Let’s move the origin by Δr
 
It’s still inside the charge distribution
r → r − Δr
r ′ → r ′ − Δr
The dipole moment becomes
p → ∫ ρ(r ′)(r ′ − Δr)dv ′ = p − QΔr
Δr
V
 
 
A
E
r
r′
If Q = 0, no change to p
If Q ≠ 0, we now have a different p
  Re-calculate the first two terms of the potential
ϕ=
 
Q r̂ ⋅ p
Q
(r − Δr) ⋅ (p − QΔr)
+ 2 →
+
3
r
r
r − Δr
r − Δr
Taylor expand by Δr/r and take the leading terms
⎛ 1 r̂ ⋅ Δr
⎞ r̂ ⋅ (p − QΔr)
Q ⎜ + 2 + ⎟ +
+
⎝r
⎠
r
r2
Extra terms cancel between the
Coulomb and the dipole parts
Simple Electric Dipole
Two charges, +q and –q, separated by a distance s
p≡
∫
V
This is a useful model for any neutral object
(e.g. a molecule) with a dipole moment p
 
 
s
ρ(r ′)r ′ dv ′ = +qs − q0 = qs
−q
+q
Electric field at large distances will be identical
We know how to draw field lines around this
6
Dipole Electric Field
⎛ r̂ ⋅ p ⎞
The electric field due to a dipole p is E = −∇ϕ = −∇ ⎜ 2 ⎟
⎝ r ⎠
  Use spherical coordinates, the polar axis parallel to p
ϕ=
p cos θ
r2
⎧
∂ϕ 2p cos θ
=
⎪⎪Er = −
∂r
r3
⎨
⎪E = − 1 ∂ϕ = p sin θ
⎪⎩ θ
r ∂θ
r3
p
E field decreases with 1/r3
 
Faster than Coulomb, as
expected
Force on Dipole
An electric dipole p is in an external field E
Force qE acts on each charge q
  Net force is zero if total charge Q is zero
+q +qE
 
Forces on +q and –q may be equal and
opposite, but not on the same line
 
s
E
−qE
−q
Combined, they produce a torque
N = ∑ r × F = r+ × qE + r− × (−qE) = q(r+ − r− ) × E = qs × E = p × E
 
The torque N rotates the dipole so that it will line up with E
Also: the net force F might not be zero if E is not uniform
F = qE(r+ ) − qE(r− ) = q(s ⋅ ∇)E = (p ⋅ ∇)E
 
This one is tricky, so let’s look at a simple example
7
Dipole and a Charge
A dipole p is near a charge Q
 
 
Assume s << r
p is lined up with E
Q
−qE
−q
s
+q +qE
r
Net force is
⎛ Q
Q⎞
2qQ
F = qE(r + s) − qE(r) = q ⎜
−
≈− 3
2
2⎟
⎝ (r + s)
r ⎠
r
 
i.e., the dipole is attracted to the charge
What if Q is negative?
 
 
Torque will rotate p so that it points toward the charge
Net force F will point toward the charge again
Electric dipole is generally pulled toward stronger E field
 
This is why neutral objects (e.g. dust particles) are attracted by static
electricity
Summary
1
ε
Electric field inside dielectric is reduced Ein = Eout
ε = dielectric constant
  Capacitance is increased by factor ε
 
Far field due to small charge distribution is determined by:
 
 
First, the net charge Q  Coulomb field
If Q = 0, then the dipole moment p ≡ ρ(r ′)r ′ dv ′
V
s
  Modeled by
∫
p = qs
−q
ϕ=
r̂ ⋅ p
r2
+q
A dipole in an electric field receives:
  Torque N = p × E
  If E is non-uniform, net force F = (p ⋅ ∇)E
 
Dipoles are attracted to stronger E field
8