Fall 2014 Notes 23 ECE 2317 Applied Electricity and Magnetism Prof. David R. Jackson ECE Dept. 1.
Transcript of Fall 2014 Notes 23 ECE 2317 Applied Electricity and Magnetism Prof. David R. Jackson ECE Dept. 1.
Fall 2014
Notes 23
ECE 2317 Applied Electricity and Magnetism
Prof. David R. JacksonECE Dept.
1
Boundary Value Problem
2 , , 0x y z
, ,B x y z
, ,x y z is unique.
Uniqueness theorem:
On boundary:
Goal: Solve for the potential
function inside of a region, given the value of the potential function
on the boundary.
(Please see the textbooks for a proof of the uniqueness theorem.)
(no charges)
2
As long as our solution satisfies the Laplace equation and the
B.C.s, it must be correct!
Example: Faraday Cage Effect
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
3
Note: We can make any guess
that we wish, as long as our final solution satisfies
Laplace’s equation and the boundary conditions.
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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(Faraday cage effect)
Example
Solve for (x, y, z)
Assume: , ,x y z x
+ - r h
xV0
0V
0
R
5
Ideal parallel-plate capacitor
Note: We can make any assumptions that we wish, as long as our final solution satisfies the boundary conditions.
Example (cont.)
2 2 2
2 2 20
x y z
Hence
2 0
Solution:
1
1 2
x C
x C x C
2
20
x
x
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Example (cont.)
0 1 2 0 1 0 /h V C h C V C V h
1 2 20 0 0 0 0C C C
Hence we have
01
2 0
VC
hC
0 V, ,V
x y z xh
The solution is then
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+- r h
xV0
0V
0
R
1 2x C x C
0 :x
:x h
Example (cont.)Calculate the electric field:
0ˆ ˆx V
E x xx h
From previous notes: 0
0
B
A
x
h
x
V E dr
E dx
E h
0x
VE
hso
8
0 V/mˆV
E xh
0, ,V
x y z xh
0ˆ V/mV
E xh
Hence
Example
2 2
2 2 2
1 10
z
, , z Assume
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+ -
V0
0V
0
R
2 0 0
rWedge
Insulating gap
Example (cont.)
2
20
0 : 1 2
2
0 0
0
C C
C
0 : 1 0 2 0
01
0
C C V
VC
1 2C C Hence
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Example (cont.)
0
0
, , VV
z
1ˆˆ ˆ
1
E
zz
0
0
1ˆ V/mV
E
Hence
Find the electric field:
We then have
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Example (cont.)
0
0
1ˆ V/mV
E
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Flux plot
+ -
V0
0V
0
r
Insulating gap
Example
1 1 2
2 1 2
c cx
d x d
0 :x
:x h 0 1 2 0h V d h d V
(Four unknowns)
+ - r1 h1
h2 xr2V0
0V
0
h
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Two-layer capacitor
20 0 0c
Example (cont.)
1 1
2 1 0 1 1 0
c x
d x V d h d x h V
Hence:
1 :x h
1 1 1 1 0c h d h h V
so
We need two more equations: use interface boundary conditions.
BC #1
The potential is continuous across the boundary.
2
1
1 2 0r
r
E dr
21
Now there are two unknowns (c1 and d1).
(The path length is zero!)
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1 1 1 2 0c h d h V
Example (cont.)BC #2:
1 2x xD D
To calculate Ex , use E
1 1 2 1r rc d Hence we have
1 :x h
xEx
1 2
1 2r rx x
Therefore
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1 1 2 2r x r xE E
1 1
2 1 0
c x
d x h V
The normal component of flux density is continuous.
Example (cont.)
Therefore
11 1 2 0
2
r
r
c h h V
0 01 1
1 21 2 1 2
2 1
,r r
r r
V Vc d
h h h h
or
Hence we have
Also,
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11 1 1 2 0
2
r
r
c h c h V
11 1
2
r
r
d c
1 1 1 2 0c h d h V
1 1 2 1r rc d
1 1
2 1 0
c x
d x h V
Example (cont.)
01 1
11 2
2
00 12
21 2
1
, 0
,
r
r
r
r
Vx x h
h h
Vh V h x hx
h h
0 0 21 1 1
1 2 2 111 2
2
0 0 12 12
1 2 2 121 2
1
0 0 1 21 2
1 2 2 1
ˆ ˆ, 0
ˆ ˆ ,
ˆ, 0
r
r rr
r
r
r rr
r
r r
r r
V VE x x x h
h hh h
V VE x x h x h
h hh h
VD D x x h
h h
We now find the electric fields and flux density:
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Example (cont.)
0 0 1 21 0
1 2 2 1
0 0 1 22
1 2 2 1
ˆ
ˆ
bot r rs x
r r
top r rs x h
r r
Vx D
h h
Vx D
h h
We now find the surface charge densities on the plates.
ˆs D n Use
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+ - r1 h1
h2 xr2V0
0V
0
h
++++++++++++++++
- - - - - - - - - - - - - - - -