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ECE 307: Electricity and Magnetism
Fall 2012
Instructor: J.D. Williams, Assistant Professor
Electrical and Computer Engineering
University of Alabama in Huntsville
406 Optics Building, Huntsville, Al 35899
Phone: (256) 824-2898, email: [email protected]
Course material posted on UAH Angel course management website
Textbook:
M.N.O. Sadiku, Elements of Electromagnetics 5th ed. Oxford University Press, 2009.
Optional Reading:
H.M. Shey, Div Grad Curl and all that: an informal text on vector calculus, 4th ed. Norton Press, 2005.
All figures taken from primary textbook unless otherwise cited.
8/17/2012 2
Chapter 6: Electrostatic Boundary Value Problems
• Topics Covered
– Poisson’s and Laplace’s Equations
– Uniqueness Theorem
– General procedures for solving Poisson’s or Laplace’s equations
– Resistance and Capacitance
– Method of Images (quick look re-look into Coulomb’s law)
• Homework: 3, 10, 14, 16, 19, 23, 24, 30, 32, 43, 45, 46, 47,
All figures taken from primary textbook unless otherwise cited.
8/17/2012 3
Resistance
• Recall from Chapter 5, that we defined Resistance as R = L/S
• We can also define it using Ohm’s law as
• The actual resistance in a conductor of non-uniform cross section can be
solved as a boundary value problem using the following steps
– Choose a coordinate system
– Assume that Vo is the potential difference between two conductor terminals
– Solve Laplace’s Eqn. to obtain V. Then Determine E = -V and solve I from
– Finally, R = Vo/I
SdE
ldE
I
VR
SdEI
8/17/2012 4
Capacitance • Capacitance is the ratio of the magnitude of charge on two separated plates
to the potential difference between them
• Note that The negative sign is dropped in the definition above
because we are interested in the absolute value of the voltage drop
• Capacitance is obtained by one of two methods
– Assuming Q, determine V in terms of Q
– Assuming V, determine Q in terms of V
• If we use method 1, take the following steps
– Choose a suitable coordinate system
– Let the two conducting plates carry charges +Q and –Q
– Determine E using Coulomb’s or Gauss’s Law and find the magnitude of
the voltage, V, via integration
– Obtain C=Q/V
ldEV
ldE
SdE
V
QC
8/17/2012 5
Parallel Plate Capacitor • Assume two parallel plates separated by a distance, d, with
charges, +Q and –Q, on them
• The charge density on each plate is
00
0
ˆˆ
ˆ
C
C
d
S
V
QC
S
Qddx
S
QldEV
aS
QaE
aD
r
d
x
xxS
xS
S
QS
QVC
Q
S
dQ
S
SdQdv
S
QW
C
QQVCVW
v
E
E
2
1
22
22
1
22
1
2
1
22
22
2
22
2
22
8/17/2012 6
Coaxial Capacitor • Assume to cylindrical plates of inner radius a and outer
radius b with +Q and –Q on them
• The charge density on each plate is
a
b
L
V
QC
a
b
L
Qd
L
Q
adaL
QldEV
aL
QE
LESdEQ
b
a
b
a
ln
2
ln22
ˆˆ2
ˆ2
2
Coaxial Line Resistance
8/17/2012 7
From the resistance if a coaxial line from the electric potential and
current density
abRG
ab
LI
VR
L
ab
I
VR
ab
VLI
dzdab
VSdJI
L
z
/ln
21
2
/ln1
2
/ln
/ln
2
/ln
0
0
0
0
2
0
0
adzdSd
apply
aab
VEJ
aab
Va
d
dVVE
ˆ
ˆ/ln
ˆ/ln
ˆ
0
0
Conductance
Resistance per
unit length
8/17/2012 8
Spherical Capacitor • Assume two spherical plates of inner radius a and outer radius b with +Q
and –Q on them
• The charge density on each plate is
ba
V
QC
ba
Qdr
r
Q
adrar
QldEV
ar
QE
rESdEQ
b
ar
b
ar
r
r
11
4
11
44
ˆˆ4
ˆ4
4
2
2
2
2
8/17/2012 9
Series and Parallel Capacitance
k
n nCC 1
11
k
n
nCC1
Series Capacitance Parallel Capacitance
8/17/2012 10
d
S
d
SC rr 1010
1
2
2/
d
SC r20
2
2
210
21 rrd
SCCC
21
210
21
21
21
21
0
2
0
21
21
2
2
2
rr
rr
rr
rr
d
S
CC
CCC
d
S
d
S
CC
CCC
d
S
d
SC rr
2
2/ 10101
d
SC r
2
202
Series and Parallel Capacitance
(Example)
8/17/2012 11
Capacitance vs. Resistance
For Common Geometries
RC
ldE
SdE
V
QC
SdE
ldE
I
VR
Parallel Plates
Isolated Sphere a
RaC
baR
ba
C
L
a
b
R
a
b
LC
S
dR
d
SC
4
1,4
4
11
,11
4
2
ln
,
ln
2
,
Between 2 Spheres
Coaxial Cylinders
8/17/2012 12
Poisson’s and Laplace’s
Equations for Electrostatics • Solving for the potential, V, using charge density
• Uniqueness theorem: Although there are many ways to solve Laplace’s equation, there is only one solution for any given set of boundary conditions
• General Procedures for solving:
– Laplace’s Eqn. Poison’s Eqn.
• Direct integration of V is a function of only 1 variable
• If V is a function of 2 or more variables, then apply separation of variables method to reduce dimensionality and solve
– Apply BCs to determine unique solution for V
– Having V, find E,D,J
– Find Q from D: C, from Q and V: and R from V and J
VE
ED v
02 V 02 V
v
v
V
V
2
8/17/2012 13
Uniform Charge Density in
Cylindrical Wire Biased at z=0
BAzz
V
Az
dz
dV
z
V
V
v
v
v
v
2
2
2
2
2
0,
,0
Vdz
VVZ o
General 1-D solution in z
Apply Boundary Conditions
d
VdA
VAdd
dzV
VB
VAzV
ov
ov
o
ov
o
2
20
02
0)0(
2
2
Find V,E,F
areationalcrossS
d
Vdz
zSdvEF
ad
Vdza
z
VVE
Vzd
VdzV
d
ovvvv
zovv
z
oovv
_sec_
22
2
22
0
2
2
8/17/2012
14
1-D Capacitor with Two Dielectrics
axBxAV
axBxAV
ax
BAxV
V
,
,
0
222
111
2
axDD
axVaxV
axBxAV
axBxAV
ax
Vx
Vdx
snn
,
)()(
,
,
0,0
0,
21
21
222
111
2
1
General 1-D solution in x
Apply Boundary Conditions
2211
211
2
22
11
11
)(
0
)0(0
0
AA
aABaAaxV
B
BAxV
dAB
BdAdxV
s
Find E
11
0,
1
ˆ1
ˆ
,
1
ˆˆ
1
21
1
2
1
21
22
1
2
1
21
11
a
d
ax
a
d
aa
d
aAE
dxa
a
d
aaAE
xS
x
xSx
Can solve for V, and F using algebra
Is the weighting fraction of the
dielectric constant for the distance
d+a=1
8/17/2012 15
1-D Capacitor with Two Dielectrics Lets solve the algebra showing that we can get E from V(x=a) and s
2211
21 )()(
AA
aAdaAaxV
s
ax
a
d
aa
d
aAExS
x
0,
1
ˆ1
ˆ
1
2
1
21
22
2211
211
222
1111
)(
0)0(0
0
AA
aABaAaxV
BBAxV
dABBdAdxV
s
1
2
1
211
1
2
1
211
1
211
1211
21
1
1
1
)()(
a
dA
a
a
a
dA
a
daA
Aa
daA
aAdaAaxV
s
dxa
a
d
aaAE xS
x
,
1
ˆˆ
1
2
1
21
11
1
2
1
21
1
2
1
21
2
21
1
2
1
21
1
1
1
1
)()(
1
a
d
a
d
a
d
a
da
A
aAdaAaxV
a
dA
SS
S
8/17/2012 16
1-D Semi-infinite
Intersecting Planes
BAV
VVV
00
12
2
2
2
2
2
mVoltsaE
VoltsV
mVV
aV
aV
VE
/*ˆ600
*600
1,6
,100
ˆ1
ˆ1
00
0
0
General 1-D solution in
Apply Boundary Conditions
Find V and E
0
0
0
0
000 )(
0
)0(0)0(
VBAV
VA
AVV
B
BAV
8/17/2012 17
2-D conducting cones
(1-D Problem)
BA
dA
dAV
yieldsegrationFurther
AV
yieldsegration
VV
rV
2/tanln2/tan
2/tan
sin
_int_
sin
_int
0sin0sinsin
12
2
mV
r
aVE
VV
VV
ar
V
ar
Aa
V
rVE
VV
VVV
/sin
ˆ1.95
1584.0
2/tanln1.95
50,6/,10/
ˆsin
2/tan
2/tanln
ˆsin
ˆ1
2/tan
2/tanln
2/tan
2/tanln
2/tanln2/tan
2/tanln2/tanln
2/tan
2/tanln
021
2
10
12
10
1
2
10
2
10
Apply Boundary Conditions
Find V and E
2/tan
2/tanln
2/tan
2/tanln
2/tanln2/tanln
2/tanln
2/tanln0
2
10
1
20
1202
1
11
VAAV
AAVV
AB
BAV
General 1-D solution in
http://en.wikipedia.org/wiki/List_of_integrals_of_trigonometric_functions
From our Trig Identities we know that
Thus
And again from trig identities
Thus
Integral of 1/sin(a)
8/17/2012 18
csc
1sin
d
d csc
sin
Cd
2tanln
sin
Integral of 1/sin(a) (Method 2)
8/17/2012 19
CCC
Cu
uCuu
duuuu
du
uletdd
2tanln
2
1
cos1
cos1ln
cos1
cos1ln
2
1
1
1ln
2
11ln1ln
2
1
1
1
1
1
2
1
1
cos_,sin
sin
sin
2
2
Using the known trig. identity
From our Trig Identities we know that
Thus
And again from trig identities
Thus
Integral of 1/sin(a)
8/17/2012 20
csc
1sin
d
d csc
sin
Cd
2tanln
sin
Spherical Capacitor Using
Laplace’s Equation
8/17/2012 21
General 1-D solution in r
Br
AV
Adr
dVr
dr
dVr
dr
d
rV
2
2
2
2 01
Apply Boundary Conditions
rbAV
yields
bABBb
A
Vbr
11
/0
0,
abAV
VVar
11
,
0
0
Find E and C
abr
VE
adr
dVVE
ab
brV
V
r
11
ˆ
11
11
2
0
0
ab
VA o
11
ba
V
QC
ba
VQ
ddr
abr
VQ
SdEQ
r
r
r
11
4
11
4
sin11
0
0
00
0
2
0
2
2
00
8/17/2012 22
1-D Laplace’s Eqn. in
Solve for E, J, I, R
Boundary Conditions
BAV
Ad
dV
ln
ab
aVV
aAaABAV
ab
VA
aAbABbAVbV
aABaV
/ln
/ln
/ln)ln(lnln
/ln
lnlnln)(
ln0)(
0
0
0
t
ab
I
VR
ab
VtI
dzdab
VSdJI
adzdSd
apply
aab
VEJ
aab
Va
d
dVVE
t
z
/ln2
/ln2
/ln
ˆ
ˆ/ln
ˆ/ln
ˆ
0
0
0
2/
0
0
0
0
012
d
dV
d
dV
Resistance in a Curved
Metal Bar (1)
8/17/2012 23
1-D Laplace’s Eqn. in z
Solve for E, J, I, R
Boundary Conditions
BAzV
dz
VdV
02
22
zt
VV 0
22
0
22
0
2/
0
0
0
0
4
4
ˆ
ˆ
ˆˆ
ab
t
I
VR
t
abVI
ddt
VSdJI
addSd
apply
at
VEJ
at
Va
dz
dVVE
b
a
z
z
zz
t
VA
AtVtzV
B
BAzV
0
0)(
0
00)0(
Resistance in a Curved
Metal Bar (2)
8/17/2012 24
2-D Rectangular Waveguide
0
2
2
2
22
,0
00,0
00,
00,0
0
VaybxV
ybxV
aybxV
ayxV
y
V
x
VV
Laplace equation for
Cartesian coordinates in 2-D
Separation of Variables
0)()(,
0)0(0)0()(0,
0)(0)()(,
0)0(0)()0(,0
0''
0''
''''
0''''
)()(,
VaYxXaxV
YYxXxV
bXyYbXybV
XyYXyV
YY
XX
Y
Y
X
X
XYYX
yYxXyxV
Boundary Conditions
Where is separation constant
Separated Boundary Conditions
Vo term is inseparable
8/17/2012 25
2-D Rectangular Waveguide
0)(
000
000
0''
xX
AbAbxX
BxX
BAxX
X
X
Case number 1
Case number 2
x
x
eAX
eAX
AxX
dxX
dX
XDX
dx
dD
XD
XX
2
1
1
22
2
2
lnln
0
0''
0
Gives 2 solutions
The second solutions is
The separated set of differential
equations has three possible
solutions:
case 1: =0
case 2: <0
case 3: >0
Applying Boundary Conditions
Integrating twice yields
Yields the trivial Solution
8/17/2012 26
2-D Rectangular Waveguide
0)(
0
sinh0
sinhcosh0
0
0100
sinhcosh)(
)(
2
2
21
1
21
212
211
21
21
xX
B
bB
bBxBbxX
B
BBxX
AAB
AAB
xBxBxX
eAeAxX xx
Case number 2
Again, the trivial solution
where
http://en.wikipedia.org/wiki/Hyperbolic_function
8/17/2012 27
2-D Rectangular Waveguide
Modal periodic
solutions allowed
b
xngxX
nb
n
nbnb
buttrivialgfor
g
bg
bgxgbxX
g
ggxX
nn
sin)(
,...3,2,1,
sin0sin
:_,0_
0
sin0
sincos0
0
0100
1
1
1
10
0
10
Case number 3
101
100
10
10
22
2
2
sincos)(
)(
0
0''
0
jCCg
CCg
xgxgxX
eCeCxX
XjDX
dx
dD
XD
XX
xjxj
http://en.wikipedia.org/wiki/Sine_wave
8/17/2012 28
2-D Rectangular Waveguide
ab
nhV
ahVayY
g
ggyY
yhyhyY
YY
b
n
b
xngxX nn
sinh
sinh
0
0100
sinhcosh)(
0''
0
sin)(
10
10
0
10
10
2
2
222
Case number 3 (cont.)
Solution for Y Case number 3
1
0
1
sinhsin,
sinhsin,
sinhsin,
sinh)(
n
n
n
n
nn
nn
b
an
b
xncVayxV
b
yn
b
xncyxV
b
yn
b
xnhgyxV
b
ynhyY
8/17/2012 29
2-D Rectangular Waveguide
Case number 3 (cont.) Solution for Y
oddn
n
n
n
b
n
b
b
n
b
b
ann
b
yn
b
xn
VyxV
evenn
oddn
b
ann
V
c
n
nn
V
nn
V
b
anc
b
b
ancn
n
bV
dxb
xn
b
anc
b
xnV
dxb
xn
b
ancdx
b
xnV
sinh
sinhsin4
,
,0
,
sinh
4
,...6,4,2,0
,..5,3,1,4
cos12
sinh
2sinhcos1
2cos1
2
1sinhcos
sinsinhsin
0
0
00
0
00
0
0
2
0
0
nm
nmdxnxmx
dxb
xn
b
xm
b
ancdx
b
xmV
b
an
b
xncVayxV
b
b
n
n
b
n
n
,2
,0sinsin
sinsinsinhsin
sinhsin,
0
010
0
1
0
8/17/2012 30
2-D Rectangular Waveguide
Solutions using the solved PDE
b
a
b
y
b
xyxV
n
n
b
ac
b
yn
b
xnc
b
xVayV
n
n
n
3sinh
3sinh
3sin10,
3,0
3,3
sinh
10
sinhsin3
sin10)(1
0
oddn
n
n
n
b
ann
b
yn
b
xn
VyxV
evenn
oddn
b
ann
V
c
b
yn
b
xncyxV
sinh
sinhsin4
,
,0
,
sinh
4
sinhsin,
0
0
1
Given an applied bias condition,
one can find the potential
throughout the waveguide using
the relations presented here
b
a
b
y
b
x
b
a
b
y
b
xyxV
n
b
a
n
n
b
a
c
b
yn
b
xnc
b
x
b
xVayV
n
n
n
5sinh10
5sinh
5sin
sinh
sinh
sin2,
5,5
sinh10
1
5,1,0
1,
sinh
2
sinhsin5
sin10
1sin2)(
1
0
8/17/2012 31
3-D Cylindrical Separation
of Variables
0''
111
0
)()()(,,
011
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
ZZ
z
Z
Z
R
R
z
ZR
RZRZV
zZRzV
z
VVVVV
functionBesselR
cc
zczczZ
guVovides
RRR
Yielding
RV
R
RV
R
_
sincos)(
sinhcosh)(
sin__Pr
0'''
0''
1
01
33
21
2222
2
2
2
222
2
222
Method of Images
• Image theory: A given charge configuration above an infinite grounded perfect
conducting plane may be replaced by a mirror image of the charge configuration and
an equipotential line in place of the conducting plane
• This theory is of significant importance b/c it allows one to significantly simplify
complex problems using symmetry
8/17/2012 32
Method of Images
• Symmetry allows us to examine the total
force at point A (+Q) by placing conducting
ground mirrors half way between A and its
nearest neighbors
• The result is a method of images approach
that allows one to quickly determine the exact
force acting on charge +Q located at point A
given either
– The relative orientation of a grounded
conducting surface near A
– The number and location of charges
near A allowing one to quickly establish
a conducting ground plane evaluation
8/17/2012 33
PA
1
360
o
NNumber of charges
applying force on +Q at
A
2/32222/3222/32222
2
12
1
2cos822
ˆsin2ˆ90cos
90sin90cos
ˆ90sinˆ90cos
22
ˆ2ˆ2
2
ˆ
2
ˆˆ
xyyx
yyxx
yx
yyxx
yx
yyxx
y
y
x
xkq
r
rqkqF
FF
N
i i
ii
N
i
i
x
y
Method of Images
8/17/2012 34
3190
3601
360
o
oo
N
2/32222
2
12
1 22
ˆ2ˆ2
2
ˆ
2
ˆˆ
yx
yyxx
y
y
x
xkq
r
rqkqFF
N
i i
iiN
i
i
