prof. d. wilton ece dept. notes 22 ece 2317 applied electricity and magnetism notes prepared by the...
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Prof. D. WiltonECE Dept.
Notes 22
ECE 2317 ECE 2317 Applied Electricity and MagnetismApplied Electricity and Magnetism
Notes prepared by the EM group,
University of Houston.(used by Dr. Jackson, spring 2006)
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Uniqueness TheoremUniqueness Theorem
, ,x y z
Given:
is unique
2 v
B
inside (known charge density)
on boundary (known B.C.)
Note: We can guess the solution, as long as we verify that Poisson’s equation and the BC’s are correctly satisfied !
, , (known)v x y z
B on boundary
S
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ExampleExample
0v
Guess:
B = V0 = constant
Check:
Hollow PEC shell0
Prove that E = 0 inside a hollow PEC shell (Faraday cage effect)
0, ,x y z V r V
2 20
0
0 in
on S
V V
V
Therefore: the correct solution is
V
0, ,x y z V
S
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Example (cont.)Example (cont.)
0v
Hollow PEC shell
0 V
0, ,x y z V
S
Hence E = 0 everywhere inside the hollow cavity.
0, , 0E x y z V
B = V0 = constant
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Image TheoryImage Theory
x
z
h
(x,y,z)
q
PEC
Note: the electric field is zero below the ground plane (z < 0).
= 0
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Image Theory (cont.)Image Theory (cont.)
Image picture:
x
z
h
q
h
-q
Note: there is no ground plane in the image picture!
The original charge and the image charge together give the
correct electric field in the region z > 0. They do NOT give the
correct solution in the region z < 0.
original charge
image charge
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z > 0:
x
z
h
q
h
-q
#1
0 1 0 2
1 2
4 4
q q
R R
#2
rR1
R2
1 2
Image Theory (cont.)Image Theory (cont.)
21,2
21,2 1,2
0
ˆ0, ,
ˆ ˆ,
r hz
qdV ndS r hz
divthm.
Note :
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x
z
h(x,y,z)
q
PEC
To see why image theory works, we construct a closed surface S with a large hemispherical cap (the radius goes to infinity).
S = Sh+S0
Sh
S0
Image Theory (cont.)Image Theory (cont.)
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x
z
h(x,y,z)
q
PEC
On Sh: = 0 (the surface is at infinity).
On S0: = 0 (the surface is on a perfect electric conductor at zero volts).
Sh
S0
Image Theory (cont.)Image Theory (cont.)
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:r V
z > 0:
x
q
-q
#1
#2
Sh
V
S0
S = S0+ Sh
correct charge in V
Image Theory (cont.)Image Theory (cont.)
2
0
2
0
ˆ0,
ˆ ˆ ˆ,
V r hz
qdV n dV D n dV q r hz
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0 :r S 0 since R1 = R2
z > 0:
x
q
-q
#1
#2
Sh
V
S0
S = S0+ Sh
:hr S 0 since r =(correct B. C. on S)
Image Theory (cont.)Image Theory (cont.)
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x
q
-q
#1
#2 SV
z < 0:
Wrong charge inside !
Image Theory (cont.)Image Theory (cont.)
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Final SolutionFinal Solutionz > 0:
z < 0:
2 22 2 2 20 04 4
q q
x y z h x y z h
0
x
z
h
q
h
-q
#1
#2
rR1
R2
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High-Voltage Power LineHigh-Voltage Power Line
High-voltage power line (radius a)
earth
h
V0
The earth is modeled as a perfect conductor
0r
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Line-charge approximation:
earth
h
0
High-Voltage Power Line (cont.)High-Voltage Power Line (cont.)
The power line is modeled as a line charge at the center of the line. This is valid (from Gauss’s law) as long as the charge density on the surface of the wire is uniform.
0r
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Image theory:
0
h0
h
Note: the line-charge density can be found by forcing the
voltage to be V0 at the radius of the wire (with respect to the earth).
High-Voltage Power Line (cont.)High-Voltage Power Line (cont.)
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Image theory:
0
h
0
h
Set VAB = V0
A
B
The point A is selected as the point on the bottom surface of the power line.
0
ln2
b
Line-charge formula:
Note: b is the distance to the arbitrary reference “point” for the single line-charge solution.
High-Voltage Power Line (cont.)High-Voltage Power Line (cont.)
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Image theory:
0 0
0 0
ln ln2 2 2
b bA
a h a
0 0
0 0
ln ln 02 2
b bB
h h
00
0
2ln
2AB
h aV V
a
0
h
0
h
Set VAB = V0
A
B
High-Voltage Power Line (cont.)High-Voltage Power Line (cont.)
b
bNote: b is chosen so that it serves as reference point for both line potentials
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Image theory:
0 0 0 00
2 22 2
ln ln
V Vh a ha a
Hence:
0
h
h
0
High-Voltage Power Line (cont.)High-Voltage Power Line (cont.)
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0 0 0 00
2 22 2
ln ln
V Vh a ha a
Hence:
High-Voltage Power Line (cont.)High-Voltage Power Line (cont.)Image theory:
0
h
h
0
20 0 0
0 1 0 2 0 1
2220 0
220 01
ln ln ln2 2 2
ln ln2 2
b b
x z h
x z h
xz
(x,0,z)1
2
b
b
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Point charge over a semi-infinite dielectric region:
Image Theory for Dielectric RegionImage Theory for Dielectric Region
h
r
q
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Solution for air region:
Image Theory for Dielectric Region (cont.)Image Theory for Dielectric Region (cont.)
1'
1r
r
q q
h
q
hq´
0
0
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Solution for dielectric region:
2' '
1r
r
q q
h
q´´
0 r
0 r
Image Theory for Dielectric Region (cont.)Image Theory for Dielectric Region (cont.)
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Flux Plot
Note: the flux lines are straight lines in the dielectric region. q
0 r
0
Image Theory for Dielectric Region (cont.)Image Theory for Dielectric Region (cont.)