Download Chapter 5 Electric Field in Material Space

Fall 2016
ECE 305 Electromagnetic Theory
Chapter 5 Electric Fields In Material Space
Qiliang Li
Dept. of Electrical and Computer Engineering,
George Mason University, Fairfax, VA
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§ 5.1 Introduction
Free Space / Vacuum: ε0
conductors
Real material space
nonconductors
Nonconductors – insulators or dielectrics
Susceptibility, permittivity, linearity,
isotropy, homogeneity, dielectric
strength, and relaxation time
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§ 5.2 Properties of Materials
• Conductivity, σ
unit: Siemens per meter ( S/m )
metals: σ >> 1
Insulators: σ << 1
in between  Semiconductors
As T  0K, some conductors have σ ∞
Called: superconductors
What is “Semiconductor”?
3
Examples
MOSFET
conductor
Semiconductor
Optical fiber
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§ 5.3 Convection and Conduction
Current
• The current (in amperes) through a given
area is the electric charge passing
through the area per unit time. 1 A = 1
C/s
𝑑𝑄
𝐼=
𝑑𝑡
Current density 𝐽 =
∆𝐼
∆𝑆
or 𝐼 =
𝑱 ∙ 𝑑𝑺
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• Different between convection current
and conduction current
• Convection current does not involve
conductors, therefore, does not satisfy
Ohm’s law. It occurs when current flows
through an insulating medium such as
liquid, rarefied gas or a vacuum
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ΔS
ρV (charge density)
z
y
Δy
x
∆𝑄 𝜌𝑉 ∆𝑉 𝜌𝑉 ∆𝑆∆𝑦
∆𝐼 =
=
=
= 𝜌𝑉 ∆𝑆 𝑢𝑦
∆𝑡
∆𝑡
∆𝑡
∆𝐼
𝐽𝑦 =
= 𝜌𝑉 𝑢𝑦
∆𝑆
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• In general: 𝑱 = 𝜌𝑉 𝒖
So, this equation is the general form of
both convection and conduction current
density
For conduction current, 𝜌𝑉 can be
expressed by electron density – see next
page
8
• Conduction current requires a conductor
𝑭 = −𝑒𝑬
From momentum Ft = mu:
𝑚𝒖
= −𝑒𝑬
𝜏
τ: the average time interval between
collisions – electron travel in a real materials
space
e
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• Therefore, 𝒖 =
𝑒𝜏
− 𝑬
𝑚
If there are n electrons per unit volume
𝜌𝑉 = −𝑛𝑒
So
𝑛𝑒 2 𝜏
𝑱 = 𝜌𝑉 𝒖 =
𝑬
𝑚
or
𝑱 = 𝜎𝑬
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§ 5.4 Conductors
• A perfect conductor (𝜎 = ∞) cannot contain
E-field within it.
Ee
Ei
Ee
ρV=0
E=0
Ee is applied external field, Ei is internal field
--isolated conductor: V= constant
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• For a conductor whose ends are
maintained at potential difference V
𝑬 ≠ 0 inside the conductor
𝑉
𝐸=
𝑙
𝐼
𝑉
𝐽 = = 𝜎𝐸 = 𝜎
𝑆
𝑙
𝐼
𝑆
=
𝑉
𝜎
𝑙
𝑅 =
𝑉
𝐼
=
𝑙
𝜎𝑆
or 𝑅 =
𝜌𝑐 𝑙
𝑆
Where ρc = 1/σ is the resistivity
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• For a conductor with nonuniform cross
section
𝑉
𝑅= =
𝐼
𝑉
𝑆
𝑬 ∙ 𝑑𝒍
𝜎𝑬 ∙ 𝑑𝑺
Power P (in watts): rate of change of energy
W (in Joules)
𝑃=
𝜌𝑉 𝑑𝑉 𝑬 ∙ 𝒖 =
𝑉
𝑬 ∙ 𝜌𝑉 𝒖𝑑𝑉 =
𝑬 ∙ 𝑱𝑑𝑉
𝑉
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Power density wP (in W/m3)
𝑑𝑃
𝑤𝑃 =
= 𝑬 ∙ 𝑱 = 𝜎|𝑬|2
𝑑𝑉
For a conductor with uniform cross-section
𝑃=
𝑬 ∙ 𝑑𝒍
𝐿
𝑱 ∙ 𝑑𝑺 = 𝑉𝐼
𝑆
Or
𝑃 = 𝐼 2 𝑅 --- the common
form of Joule’s law
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Example 5.1: If 𝑱 =
1
𝑟3
2𝑐𝑜𝑠𝜃𝒂𝒓 + 𝑠𝑖𝑛𝜃𝒂𝜽 𝐴/𝑚2
Calculate the current passing through
(a) A hemispherical shell of radius = 20 cm, 0 <
𝜋
𝜃 < ; 0 < 𝜙 < 2𝜋
2
𝐼=
Solve:
𝑆
𝑱 ∙ 𝑑𝑺
where 𝑑𝑆 = 𝑟 2 𝑠𝑖𝑛𝜃𝑑𝜙𝑑𝜃𝒂𝒓
So
𝐼=
4𝜋 𝑠𝑖𝑛2 𝜃
(θ
0.2 2
𝜋
2
0
2𝜋 1
2𝑐𝑜𝑠𝜃
0 𝑟3
𝑟 2 𝑠𝑖𝑛𝜃𝑑𝜙𝑑𝜃|𝑟=0.2 =
from 0 to π/2)=31.4 (A)
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Continue example 5.1
(b) A spherical shell of radius 10 cm
4𝜋 𝑠𝑖𝑛2 𝜃
𝐼=
(θ from 0 to π)=0
0.1 2
Example 5.4: A lead (𝜎 = 5 × 106 𝑆/𝑚) bar of
square cross section has a circular hole. L=4 m
Find the resistance.
Solve: 𝑅 = 𝑙/𝜎𝑆
b/c: 𝑆 = 𝑑 2 − 𝜋𝑟 2
1 cm
So, 𝑅 =
4
5×106 ×
𝜋
9−
4
×10−4
3 cm
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§ 5.5 Polarization in Dielectrics
Dipole moment p= 𝑄𝒅
Polarization in dielectrics 𝑷 = lim
𝑁
𝑘=1 𝑄𝑘 𝒅𝒌
∆𝑉→0
∆𝑉
This is dipole moment per unit volume
So, potential 𝑑𝑉 =
𝑷∙𝒂𝒓 𝑑𝑉 ′
4𝜋𝜖0 𝑅 2
where P dV’ is the dipole moment (dV’ is diff vol.)
b/c: 𝛻
So
′
1
𝑅
=
𝒂𝑹
𝑅2
𝑷∙𝒂𝑹
𝑅2
= 𝜵′ ∙
𝑷
𝑅
−
𝜵′ ∙𝑷
𝑅
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• Therefore, 𝑉 =
𝑷∙𝒂 ′
𝑹
𝑆 ′ 4𝜋𝜖0 𝑅
𝑑𝑆 ′ +
−𝛻 ′ ∙𝑷
′
𝑑𝑉
𝑉 ′ 4𝜋𝜖0 𝑅
Compared to Eq. (4.68) and (4.69)
The total positive bound charges on surface S:
𝑄𝑏 =
𝑷 ∙ 𝑑𝑺
The charge that remains inside S is
−𝑄𝑏 = −
𝛻 ∙ 𝑷𝑑𝑉
𝑉
Total volume charge density ρt:
𝜌𝑡 = 𝜌𝑉 + 𝜌𝑝𝑉 = 𝛻 ∙ 𝜖0 𝐸
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Therefore, 𝜌𝑉 = 𝛻 ∙ 𝜖0 𝑬 − 𝜌𝑝𝑉 = 𝛻 ∙ (𝜖0 𝑬 + 𝑷)
From this equation, we have:𝐃 = (𝜖0 𝑬 + 𝑷)
For some dielectrics, 𝑷 = 𝑋𝑒 𝜖0 𝑬
where Xe is electric susceptibility
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§ 5.6 Dielectric Constant and Strength
𝑫 = 𝜖0 1 + 𝑋𝑒 𝑬 = 𝜖0 𝜖𝑟 𝑬
Or
𝑫 = 𝜖0 𝜖𝑟 𝑬 = 𝜖𝑬
where 𝜖 = 𝜖0 𝜖𝑟 , the permittivity
𝜖𝑟 is the dielectric constant: 𝜖𝑟 = 1 + 𝑋𝑒 =
𝜖
𝜖0
Dielectric strength is the maximum electric field
that a dielectric can tolerate or withstand
without electric breakdown.
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§ 5.7 is not required
§ 5.8 Continuity equation and
relaxation time
𝑑𝑄𝑖𝑛
𝑱 ∙ 𝑑𝑺 = −
=−
𝑑𝑡
𝐼𝑜𝑢𝑡 =

𝜵∙𝑱=
𝜕𝜌𝑉
−
𝜕𝑡
b/c: 𝐽 = 𝜎𝐸 , and 𝛻 ∙ 𝐸 =
So, 𝛻 ∙ 𝜎𝐸 =
𝜎𝜌𝑉
=
𝜖
−𝑡/𝑇𝑟
So, 𝜌𝑉 = 𝜌𝑉0 𝑒
𝑉
𝑑𝜌𝑉
𝑑𝑉
𝑑𝑡
𝜕𝜌𝑉
−
𝜕𝑡
𝜌𝑉
𝜖

𝜕𝜌𝑉
𝜌𝑉
=
𝜎
− 𝜕𝑡
𝜖
, where 𝑇𝑟 = 𝜖/𝜎
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§ 5.9 Boundary Conditions
• Use Maxwell’s Equations
𝑬 ∙ 𝑑𝒍 = 0
𝑫 ∙ 𝑑𝑺 = 𝑄𝑒𝑛𝑐
A. Dielectric-Dielectric
E1n
Boundary Conditions
Consider L: abcd
𝑆
𝑬 ∙ 𝑑𝒍 = 0
E1t
E2
E2n
ε1
E1
E2t
Δh
ε2
d
a
ΔW
b
c
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𝑬 ∙ 𝑑𝒍 = 0
∆ℎ
∆ℎ
∆ℎ
= 𝐸1𝑡 ∆𝑤 − 𝐸1𝑛
− 𝐸2𝑛
− 𝐸2𝑡 ∆𝑤 + 𝐸2𝑛
2
2
2
∆ℎ
+ 𝐸1𝑛
2
𝐷1𝑡
𝐷2𝑡
𝐸1𝑡 = 𝐸2𝑡 
=
𝜖1
𝜖2
Use cylindrical Gaussian surface
∆𝑄 = 𝜌𝑆 ∆𝑆 = 𝐷1𝑛 ∆𝑆 − 𝐷2𝑛 ∆𝑆
 𝐷1𝑛 − 𝐷2𝑛 = 𝜌𝑆
If 𝜌𝑆 = 0 (no free surface charge), 𝐷1𝑛 = 𝐷2𝑛
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therefore, 𝜖1 𝐸1𝑛 = 𝜖2 𝐸2𝑛
From 𝐸1𝑡 = 𝐸2𝑡 and 𝜖1 𝐸1𝑛 = 𝜖2 𝐸2𝑛
𝑡𝑎𝑛𝜃1

𝑡𝑎𝑛𝜃2
=
𝜖𝑟1
𝜖𝑟2
θ1
B. Conductor-Dielectric
θ2
Boundary Conditions
① No electric field may exist within a conductor
𝜌𝑉 = 0 ,
𝐸=0
②b/c E = −𝛻𝑉 = 0  there can be no potential
difference between any two point in a conductor
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③ 𝐷𝑡 = 𝜖0 𝜖𝑟 𝐸𝑡 = 0, 𝐷𝑛 = 𝜖0 𝜖𝑟 𝐸𝑛 = 𝜌𝑆
E-field must be external to the conductor
C. Conductor-Free space boundary conditions
𝐷𝑡 = 𝜖0 𝐸𝑡 = 0, 𝐷𝑛 = 𝜖0 𝐸𝑛 = 𝜌𝑆
Example 5.9 and 5.10 will discussed in the class.
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