Post on 28-Dec-2015
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
4
(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
6
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
7
+- 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
9
+ -
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
10
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
11
Example (cont.)
0
0
1ˆ V/mV
E
12
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
13
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!)
14
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
15
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,
16
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:
17
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
18
+ - r1 h1
h2 xr2V0
0V
0
h
++++++++++++++++
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