Post on 17-Dec-2015
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)
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
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
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
Image TheoryImage Theory
x
z
h
(x,y,z)
q
PEC
Note: the electric field is zero below the ground plane (z < 0).
= 0
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
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 :
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.)
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.)
: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
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.)
x
q
-q
#1
#2 SV
z < 0:
Wrong charge inside !
Image Theory (cont.)Image Theory (cont.)
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
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
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
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.)
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.)
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
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.)
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
Point charge over a semi-infinite dielectric region:
Image Theory for Dielectric RegionImage Theory for Dielectric Region
h
r
q
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
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.)
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.)