Lecture Note: Chapter 6 (ppt file)

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Transcript Lecture Note: Chapter 6 (ppt file)

Chapter 6 Electrostatic
Boundary-Value Problems
Lecture by Qiliang Li
1
§6.1 Introduction
The goal is to determine electric field E
• We need to know the charge distribution
– Use Coulomb’s law
𝑬=
𝑑𝑄 𝒂𝑹
4𝜋𝑅 2
or 𝑬 =
– Use Gauss’s law
𝜓=
𝑆
𝑑𝑄 𝑹
4𝜋𝑅 3
𝑫 ∙ 𝑑𝑺 = 𝑄𝑒𝑛𝑐
• Or We need to know the potential V
𝑬 = −𝜵𝑉
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§6.1 Introduction
• However, we usually don’t know the charge
distribution or potential profile inside the
medium.
• In most cases, we can observe or measure the
electrostatic charge or potential at some
boundaries
 We can determine the electric field E by
using the electrostatic boundary conditions
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§6.2 Poisson’s and Laplace’s Equations
• Poisson’s and Laplace’s equations can be
derived from Gauss’s law
𝛻 ∙ 𝐷 = 𝛻 ∙ 𝜖𝐸 = 𝜌𝑉
Or
𝜌𝑉
2
𝛻 𝑉=−
𝜖
This is Poisson’s Eq.
If 𝜌𝑉 = 0, it becomes Laplace’s Eq.
𝛻2𝑉 = 0
4
Continue 6.2
The Laplace’s Eq. in different coordinates:
𝜕2𝑉 𝜕2𝑉 𝜕2𝑉
+ 2+ 2 =0
2
𝜕𝑥
𝜕𝑦
𝜕𝑧
1 𝜕
𝜕𝑉
1 𝜕2𝑉 𝜕2𝑉
𝜌
+ 2
+ 2 =0
2
𝜌 𝜕𝜌
𝜕𝜌
𝜌 𝜕𝜙
𝜕𝑧
1 𝜕
𝜕𝑉
1
𝜕
𝜕𝑉
2
𝑟
+ 2
𝑠𝑖𝑛𝜃
2
𝑟 𝜕𝑟
𝜕𝑟
𝑟 𝑠𝑖𝑛𝜃 𝜕𝜃
𝜕𝜃
1
𝜕2𝑉
+ 2 2
=0
2
𝑟 𝑠𝑖𝑛 𝜃 𝜕𝜙
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§6.3 Uniqueness Theorem
Uniqueness Theorem: If a solution to
Laplace’s equation can be found that
satisfies the boundary conditions, then the
solution is unique.
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§6.4 General Procedures for Solving
Poisson’s or Laplace’s Equations
1. Solve L’s Eq. or P’s Eq. using (a) direct
integration when V is a function of one
variable or (b) separation of variables if
otherwise.  solution with constants to be
determined
2. Apply BCs to determine V
3. V  E  D  𝑱 = 𝜎𝑬
4. Find Q induced on conductor 𝑄 = 𝜌𝑆 𝑑𝑆,
where 𝜌𝑆 = 𝐷𝑛 C=Q/V  𝐼 = 𝐽𝑑𝑆R
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Example 6.1 (page 219)
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9
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Example 6.2
Details in P222
Example 6.3
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§6.5 Resistance and Capacitance
R’s definition (for all cross sections)
𝑬 ∙ 𝑑𝒍
𝑉
𝑅= =
𝐼
𝜎𝑬 ∙ 𝑑𝑺
Procedure to calculate R:
1. Choose a suitable coordinate system
2. Assume V0 as the potential difference b/w two ends
3. Solve 𝛻 2 𝑉 = 0 to obtain V, find E from 𝑬 = −𝛻𝑉,
then find I from I = 𝜎𝑬 ∙ 𝑑𝑺
4. Finally, obtain 𝑅 = 𝑉0 /𝐼
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(continue)
Capacitance is the ratio of magnitude of the
charge on one of the plates to the potential
difference between them.
𝑄 𝜖 𝑬 ∙ 𝑑𝑺
𝐶= =
𝑉
𝑬 ∙ 𝑑𝒍
Two methods to find C:
1. Assuming Q and determining V in terms of Q
(involving Gauss’s law)
2. Assuming V and determining Q interms of V
(involving solving Laplace’s Eq.)
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(continue)
First methods QV, procedure:
1. Choose a suitable coordinate system
2. Let the two conductor plates carry Q and –Q
3. Determine E by using Gauss’s law and find V
from 𝑉 = − 𝑬 ∙ 𝑑𝒍. (Negative sign can be
ignored. We are interested at absolute value
of V)
4. Finally, obtain C from Q/V
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(continue)
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(continue)
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(continue)
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(continue)
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(continue)
Example 6.8: a metal bar of conductivity σ is
bent to form a flat 90o sector of inner radius a,
outer radius b, and thickness t as shown in
Figure 6.17. Show that (a) the resistance of the
bar between the vertical curved surfaces at ρ=a
2𝑙𝑛𝑏/𝑎
and ρ=b is 𝑅 =
𝜎𝜋𝑡
(b) the resistance between
the two horizontal surface
at z=0 and z=t is
4𝑡
′
𝑅 =
𝜎𝜋(𝑏 2 − 𝑎2 )
x
z
b
a
y
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(a) Use Laplace’s Eq. in cylindrical coordinate system:
1 𝑑
𝑑𝑉
2
𝛻 𝑉=
𝜌
=0
𝜌 𝑑𝜌
𝑑𝜌
𝑉 = 𝐴𝑙𝑛𝜌 + 𝐵
𝑉 𝜌 = 𝑎 = 0 → 0 = 𝐴𝑙𝑛𝑎 + 𝐵 𝑜𝑟 𝐵 = −𝐴𝑙𝑛𝑎
𝑉0
𝑉 𝜌 = 𝑏 = 𝑉0 = 𝐴𝑙𝑛𝑏 + 𝐵 𝑜𝑟 𝐴 =
𝑏
𝑙𝑛
𝑎
𝑉0
𝜌
So, 𝑉 = 𝐴𝑙𝑛𝜌 − 𝐴𝑙𝑛𝑎 = 𝑏 𝑙𝑛 ,
𝑙𝑛
𝑬 = −𝛻𝑉 = −
𝐼=
𝑱 ∙ 𝑑𝑺 =
𝑉0
𝑏
𝜌𝑙𝑛
𝑎
𝑎
𝑎
𝒂𝝆 , 𝐽 = 𝜎𝑬, 𝑑𝑺 = −𝜌𝑑𝜙𝑑𝑧𝒂𝝆 ,
𝜋 𝑡𝑉0 𝜎
,
2 𝑙𝑛 𝑏
𝑎
𝑅=
𝑉0
𝐼
=
2𝑙𝑛
𝑏
𝑎
𝜎𝜋𝑡
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(b) Use Laplace’s Eq. in cylindrical coordinate
system:
2
𝑑
𝑉
2
𝛻 𝑉= 2 =0
𝑑𝑧
𝑉 = 𝐴𝑧 + 𝐵
𝑉 𝑧 = 0 = 0 → 0 = 0 + 𝐵 𝑜𝑟 𝐵 = 0
𝑉0
𝑉 𝜌 = 𝑡 = 𝑉0 = 𝐴𝑡 + 𝐵 𝑜𝑟 𝐴 =
𝑡
𝑉0
So, 𝑉 = 𝑧,
𝑡
𝑬=
𝐼=
𝑉0
−𝛻𝑉 = − 𝒂𝒛 , 𝐽 = 𝜎𝑬, 𝑑𝑺 = −𝜌𝑑𝜙𝑑𝜌𝒂𝒛 ,
𝑡
𝑉0 𝜎𝜋(𝑏2 −𝑎2 )
𝑱 ∙ 𝑑𝑺 =
, R=V0/I=?
4𝑡
(please also use conventional method for (b))
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Example 6.9: a coaxial cable contains an insulating
material of conductivity σ. If the radius of the
central wire is a and that of the sheath is b, show
that the conductance of the cable per unit length is
𝑏
𝐺 = 2𝜋𝜎/𝑙𝑛
𝑎
Solve:
Let V(ρ=a)=0 and V(ρ=a)=V0
𝐽 = 𝜎𝑬 =
𝐼=
2𝜋𝐿𝜎𝑉0
𝑙𝑛𝑏/𝑎
−𝜎𝑉0
,
𝜌𝑙𝑛𝑏/𝑎
𝑑𝑺 = −𝜌𝑑𝜙𝑑𝑧𝒂𝝆
R per unit length =V/I/L
G=1/R= …
a
b
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Example 6.10: find the charge
in shells and the capacitance.
Solve:
Use Laplace’s Eq. (spherical)
And BCs, 𝑉 = 𝑉0
1 1
( − )
𝑟 𝑏
1 1
( − )
𝑎 𝑏
V0=100V
b
a
GND
Єr
So E=-dV/dr ar, Q=?, C=Q/V0
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Example 6.11: assuming V and finding Q to derive
𝑄
𝜖𝑆
Eq. (6.22): 𝐶 = =
𝑉
𝑑
Solve:
From Laplace’s Eq.:𝛻 2 𝑉 = 0
𝑑2
𝑉
2
𝑑𝑥
= 0𝑉 = 𝐴𝑥 + 𝐵
x
d
V0
0
0
From BCs: V(0)=0 and V(x=d)=V0
𝑉0
𝑉= 𝑥
𝑑
𝑉0
So, 𝐸 = −𝛻𝑉 = − 𝒂𝒙 , the surface charge: 𝜌𝑠 =
𝐷 ∙ 𝒂𝒏 =
𝜖𝑉0
− 𝑄
𝑑
𝑑
= 𝜌𝑠 𝑆
𝜖𝑉0
= S,
𝑑
so: 𝐶 =
𝜖𝑆
𝑑
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Example 6.12: determine the capacitance of
each of th Єr1=4 e capacitors in Figure 6.20. Take
Єr2=6, d=5mm, S=30 cm2.
Solve:
d/2
Є
(do it by yourself)
Є
d/2
r1
r2
Єr1
Єr2
w/2
w/2
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Example 6.13: A cylindrical capacitor has radii
a=1cm and b=2.5cm. If the space between the
plates is filled with and inhomogeneous
dielectric with Єr=(10+ρ)/ρ, where ρ is in
centimeters, find the capacitance per meter of the
capacitor.
Solve:
Use Eq. 6.27a, set the inner shell with +Q and
𝑎
𝑄
outer shell with –Q. 𝑉 = − 𝑏
𝑑𝜌 =
𝑄
10+𝑏
𝑙𝑛
2𝜋𝜖0
10+𝑎
2𝜋𝜖0 𝜖𝑟 𝜌𝐿
C=Q/V=…=434.6 pF/m
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§6.6 Method of Images
• The method of images is introduced by Lord
Kelvin to determine V, E and D, avoiding
Poison’s Eq.
• Image Theory states that a given charge
configuration above an infinite grounded
perfect conducting plane may be replaced by
the charge configuration itself, its image, and
an equipotential surface in place of the
conducting plane.
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Conducting plane
grounded
Image charge
So that the
potential at the
plane position =
0V
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Equipotential
V=0
Perfect conducting surface grounded
In applying the image method, two conditions must
always be satisfied:
1. The image charge(s) must be located in the
conducting region (satify Poisson’s Eq.)
2. The image charge(s) must be located such that on the
conducting surface(s) the potential is zero or constant
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A. A point charge above a grounded conducting
plance
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