Electromagnetism
Physics 15b
Lecture #22
Maxwell’s Equations in Dielectrics
Purcell 10.11–10.15
What We Did Last Time
2p cos θ
A dipole generates electric field Er =
r3
A dipole in an electric field receives:
  Torque N = p × E
  If E is non-uniform, net force F = (p ⋅ ∇)E

, Eθ =
p sin θ
r3
Dipoles are attracted to stronger E field
Density of polarization P = Np
Small volume dv of dielectric looks like a dipole Pdv
A cylinder parallel to P looks like charge density ±P
on the ends
  Average electric field inside the cylinder is 〈E〉 = −4πP


A uniformly polarized sphere


External field looks like produced by a dipole VP
Internal field E = −(4π/3)P
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Today’s Goals
Connect polarization P and the dielectric constant ε
Continue discussion of polarized sphere
Boundary condition at the surface
  How to make a sphere polarized uniformly

Examine the effects of non-uniform polarization


Non-uniform P creates “bound” charge distribution
Charge screening, electric displacement
Discuss time-dependent E field in dielectrics
Modified Maxwell’s equations
  Electromagnetic waves in dielectrics

Dielectric Constant
How does P relate to the dielectric constant ε?
Consider the filled capacitor again

−Q
+Q
Electric field is reduced by factor 1/ε
E=
4πσ
= 4πσ − 4π P
ε
Field from charge
on the plates
Field from polarization
P ε −1
=
≡ χ e electric
susceptibility
E
4π
d
Polarization density P is related to the average electric field
E that causes it by:
ε −1
E = ε E − 4π P

This is an empirical law, and
is “correct” within limits
P = χ eE =
4π
E
2
Uniformly Polarized Sphere
A dielectric sphere is uniformly polarized along +z

It contains dipoles p = qs with density N
P = Np = Nqs

This can be seen as two overlapping spheres
  Charge densities are +Nq and −Nq
  Centers are separated by distance s

From outside, each sphere looks like a point
charge (recall Gauss) +NqV and −NqV
R
Field outside is identical to that generated by a single
dipole moment NqVs = V P

We know this field
V P ⋅ r̂ VP cos θ
=
from the last lecture: ϕ (r > R) =
2
2
r
r
Uniformly Polarized Sphere
Inside the sphere, there is no net charge
 The field must obey Laplace’s eqn.

We also need the boundary condition,
i.e., values of φ at the surface,
which we know from the outside field
ϕ (r = R) =

VP cos θ V
4π
= 3 Pr cos θ =
Pz
2
3
R
R
A uniform electric field along +z works
ϕ (r < R) =
4π
Pz
3
E(r < R) = −
4π
P
3
The field due to a uniformly polarized sphere is
Inside: E = −(4π/3)P
  Outside: identical to the field generated by a dipole (4π/3)R3P

3
Boundary Conditions
Electric potential φ is a continuous function of space

Otherwise there would be infinite electric field
As a result, electric field has to satisfy the following
conditions on the surface of the dielectric
E|| parallel to the surface is continuous
  E⊥ perpendicular to the surface
may be discontinuous

Check this with the uniformly
polarized sphere:
⎧
⎧
8π
4π
P cos θ
P cos θ
⎪⎪Er =
⎪⎪Er = −
3
3
Outside ⎨
Inside ⎨
⎪E = 4π P sin θ
⎪E = 4π P sin θ
θ
⎪⎩
⎪⎩ θ
3
3
Dielectric Sphere in E Field
How does a dielectric sphere get uniformly polarized?
Try the simplest way — put it in a uniform external field E0
  Suppose this leads to a uniform polarization P
4π
  P generates a uniform field inside: E′ = −
P
3
4π
  Total field inside is Einside = E0 + E′ = E0 −
P
3
  Resulting polarization is
ε − 1⎛
4π ⎞
P = χ eEinside =
E0 −
P
⎜
4π ⎝
3 ⎟⎠

We can solve this to find
⎛ 3 ⎞
E′ = ⎜
E
⎝ ε + 2 ⎟⎠ 0

and
P=
3 ⎛ ε − 1⎞
E
4π ⎜⎝ ε + 2 ⎟⎠ 0
Uniform external field polarizes the sphere uniformly
4
Bound Charge
Consider a small area da inside a dielectric
da
Polarization P = Nqs
  How many dipoles “straddle” this area?

s
P

Charge qNs⋅da = P⋅da is split from the corresponding negative
charge by the area da
Integrate this over a closed surface S

da
How much (negative) charge remains inside?
P
Q = − ∫ P ⋅ da
S

Use the Divergence Theorem
charge distribution
= − div P “Bound”
due to non-uniform polarization
ρ
Free and Bound Charges
Inside dielectric, two “types” of charges may exist:

“Bound” charge ρbound belongs to the dielectric material
  Appears only when E field polarizes the dielectric
ρbound = − div P

P=
ε −1
E
4π
“Free” charge ρfree is brought in from outside
Both “bound” and “free” charges create E field
div E = 4π ( ρfree + ρbound )

For a given ρfree distribution and a constant ε
4πρfree
ε
E follows Gauss’s Law with ρfree except for a factor 1/ε
div E = 4πρfree − (ε − 1) div E
div E =
5
Screening
A point charge Q is inside a dielectric
  Q
E
is the only “free” charge here
E field at distance r from the charge is
r
P
Q
r̂ Coulomb reduced by 1/ε
εr 2
  Polarization density P due to this field is
ε −1
(ε − 1)Q
P=
E=
r̂
4π
4πε r 2
⎛ r̂ ⎞
(ε − 1)Q
div ⎜ 2 ⎟
  This creates bound charge density ρbound = div P =
4πε
⎝r ⎠
  This div is zero everywhere except at the origin
  Integrate inside a very small sphere around Q
ε −1
∫V ρbounddV = − ∫S P ⋅ da = − ε Q Negative charge surrounds Q
E=
Polarization creates a “screen” around free charges
Electric Displacement
Electric displacement D is defined by D ≡ E + 4π P

For electric field inside an isotropic dielectric
ε −1
P=
E
D = εE
4π

D satisfies Gauss’s Law with free charge:
div D = 4πρfree
Importance of D is more historic than practical
It’s easy to calculate only in linear, isotropic dielectric
  In that case, writing D instead of εE saves only a little ink
  In a more complex medium, it’s safer to use E and 4πP, and keep
track of how the material reacts to E

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Bound-Charge Current
In a non-static system, P may vary with time
  Imagine
  Charges
s changing in response to changing E
+q and −q move  “bound” current
s
⎛ dr
dr ⎞
ds ∂P
Jbound = N ⎜ q + − q − ⎟ = Nq
=
dt ⎠
dt
∂t
⎝ dt
Jbound adds to the “free” current J in generating B
∇×B =
  For
4π
c
⎛
∂P ⎞ 1 ∂E 4π
1 ∂
⎜⎝ J + ∂t ⎟⎠ + c ∂t = c J + c ∂t E + 4π P
(
= D if you like
a linear isotropic dielectric,
∇×B =
)
4π
ε ∂E
J+
c
c ∂t
Electromagnetic Waves
Inside a dielectric with no free charge and no free current
1 ∂B
∇ ⋅E = 0 ∇ × E = −
c ∂t
ε ∂E
∇ ⋅B = 0 ∇ × B =
c ∂t
  Same
technique used in Lecture #18 turn them into
∇ 2E =
  Solutions
ε ∂ 2E
ε ∂ 2B
2
and
∇
B
=
c 2 ∂t 2
c 2 ∂t 2
are waves propagating with speed
c
ε <c
EM waves travel slower in a dielectric by factor n =

ε
n is the index of refraction of the material
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Electromagnetic Waves
Plane wave solutions of the Maxwell’s eqns. are
E = E0 sin(k ⋅ r − ω t) B = B0 sin(k ⋅ r − ω t)

Wave equation:
∇ 2E =

ε ∂ 2E
c 2 ∂t 2
Divergence:
∇ ⋅E = 0

Curl:
∇ ⋅B = 0
k2 =
ε 2
ω
c2
k ⊥ E0
ω
c
=
k
ε
k ⊥ B0
ω
B
c 0
Similar to the vacuum solutions except:
∇×E = −
1 ∂B
c ∂t
k × E0 =
k̂ × E0 =
1
ε
B0
Propagation velocity is reduced by 1 ε
  |E| is smaller than |B| by the same factor

Frequency Dependence
Discussion so far applies to any dielectric = any insulator
All insulators are transparent, with n = ε
  Index of refraction of water is 80 = 8.9

Wrong!
When E changes, it takes time for dielectrics to polarize

Dielectric “constant” is constant only for static/slowly-changing field
Especially true for liquid of polar molecules, e.g. water
Molecules must rotate  Takes ~10−11 seconds
  ε large up to 1010 Hz, then drops to an “ordinary” value of 1.78
  Index of refraction of water for visible light is 1.33

8
Ordinary Dielectrics
Inside insulators are electrons bound to atoms
They behave as mass-spring oscillators
EM waves drive them with F = −eE
  If the frequency ω is close to the resonance
frequency ω0, the electrons oscillate strongly
  They absorb the incoming waves


Typical ω0 for bound electrons are
−e
atom
light
E
1015–1016
ω0 =
k
me
Hz
In the visible to ultraviolet (UV) region
  Loosely bound electrons (small k  small ω0) make material
opaque in visible light

Positively charged hydrogen atoms in molecules oscillate at
lower ω0 because of larger m

Microwave is absorbed by water and organic compounds
Summary
Electric susceptibility of a dielectric P =

At boundaries of dielectrics,
E|| is continuous, but E⊥ may not be
χ eE =
ε −1
E
4π
If P is not uniform, bound charge ρbound = − div P appears
4πρfree
div E = 4π ( ρfree + ρbound ) =
for linear, isotropic dielectric
ε
4π
ε ∂E
In such a medium, Maxwell’s
∇×B =
J+
equation is modified as:
c
c ∂t


Wave solutions propagate with a reduced speed c
ε
Frequency dependence of ε makes the solutions more complex
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