electrostatic boundary value problems
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Electrostatic Boundary value
problems
Sandra Cruz-Pol, Ph. D.INEL 4151 ch6
Electromagnetics IECE UPRM
Mayagüez, PR
Last Chapters: we knew either V or charge distribution, to find E,D.
NOW: Only know values of V or Q at some places (boundaries).
Some applications Microstrip lines capacitance Microstrip disk for microwave equipment
To find E, we will use:
Poisson’s equation:
Laplace’s equation: (if charge-free)
They can be derived from Gauss’s Law
02 V
vV 2
VEED v
Depending on the geometry:
We use appropriate coordinates:cartesian:
cylindrical:
spherical:
vV 2
v
zV
yV
xV
2
2
2
2
2
2
v
zVVV
2
2
2
2
2
11
vV
rV
rrVr
rr
2
2
2222
2 sin1sin
sin11
Procedure for solving eqs.
1. Choose Laplace (if no charge) or Poisson 2. Solve by Integration if one variable or by3. Separation of variables if many variables4. Apply B.C.5. Find V, then E=-DV, D=E, J=sE6. Also, if necessary:
SnD S
dSJIdSQ S
P.E. 6.1 In a 1-dimensional device, the charge density is given by
If E=0 at x=0 and V=0 at x=a, find V and E.
Evaluating B.C.
vV 2
axov /
v
xV
2
2
BAxaxV o
6
3
xo
x aAaxa
xVE ˆ
2ˆ
2
0A
6
2aB o 333
6xa
axV o
xo a
axE ˆ
2
2
axov /
45o
P.E. 6.3 two conducting plates of size 1x5m are inclined at 45o to each other with a gap of width
4mm separating them as shown below. Find approximate charge per plate
if plates are kept at 50V potential difference and medium between them has permittivity of 1.5
Applying B.C. V(0)=0,V(o=45)=Vo=50
aVaVE
o
o ˆˆ1
012
2
22
VV
02
2
V
o
oVV
ED o
ons
VD )0(
)/ln(1
0
abLVdzdVdSQo
oL
z
b
ao
os
1m
BAV
P.E. 6.3 two conducting plates of size 1x5m are inclined at 45o to each other with a gap of width
4mm separating them as shown below.
permittivity of 1.5Applying B.C. V(0)=0,V(o=45)=Vo=50
)/ln( abLVQ
Coo
)/ln( abLVQo
o
45o
a b
)/ln( abLVQ
Coo
mmmma o 226.52/45sin
2/4 nCVpFCVQ
F
mmmmm
VQ
C
o
o
o
22)50(44410444
226.51000ln)5(
4/5.1
12
detail
45o
a b
abrecha
hipotenusaopuestoo 2/
245sin
mmmma o 226.52/45sin
2/4
P.E. 6.5 Determine the potential function for the region inside the rectangular trough of
infinite length whose cross section is shown. The potential V
depends on x and y. Vo=100V, b=2a=2m,
find V and E at:a) (x, y)=(a, a/2)b) (x, y)=(3a/2, a/4)
oVaybxVybxV
aybxVayxV
),0(0)0,0(0)0,(0)0,0(
y
xb
aV=Vo
V=0
V=0V=0
02
2
2
2
yV
xV
P.E. 6.5 (cont.) Since it’s 2 variables, use Separation of Variables
y
xb
aV=Vo
V=0
V=0V=0
)()(),( yYxXyxV
0"" XYYX
YY
XX ""
0"0"
YYXX
einseparablVaYxXaxVYYxXxV
bXyYbXybVXyYXyV
o
)()(),(0)0(0)0()()0,(0)(0)()(),(0)0(0)()0(),0(
Let’s examine 3 Possible Cases
A. =0B. <0C. >0
Case A: If =o
BAxX
00
AB
0)()(0)(
yYxXVxX
y
xb
a
V=Vo
V=0
V=0V=0
0"0"
YYXX
This is a trivial solution, therefore cannot be equal to zero.
0)(0)0(
..
bXX
CB
Case B: <o
xBxBXor
eAeAX xx
sinhcosh
:issolution general
21
21
00sinh
0
2
2
1
BbB
B
2 0" 2 XX
This is another trivial solution, therefore cannot be equal to zero.
0)(0)0(
..
bXX
CB
Case C: >o
xgxgXor
eCeCX
o
xjxjo
sincos
:issolution general
1
1
0sin0
1
bggo
2 0" 2 XX
0)(0)0(
..
bXX
CB
4,3,2,1
0sin
nb
nb
22
sin)(
:solutions of seriesA
bn
bxngxX nn
Case C: >o
yhyhY o sinhcosh:issolution generalwith
1 0oh0)0(..Y
CB
0" 2 YY 4,3,2,1
nb
n
xhyYn sinh)( 1
byn
bxnhgyYxXyxV nnnnn
sinhsin)()(),(
byn
bxncyxV n
nn
sinhsin),(1
By superposition, the combination is also a solution:
Cont.
ban
bxncVaxV n
non
sinhsin),(1
dxb
xnb
xmb
ancdxb
xmVb
nn
o
b sinsinsinhsin010
nmnm
dxmxnx2/
0sinsin
0
B.C. at y=a
If we multiply by sin factor and integrate on x:
dxb
xmb
ancdxb
xnVb
no
b 2
00
sinsinhsin
dxb
xnb
ancb
xnnbV
b
n
b
o
2cos121sinhcos
00
Orthogonality property of sine and cosine:
2
sinhcos1 bb
ancnn
bVn
o
dxb
xnb
ancb
xnnbV
b
n
b
o
2cos121sinhcos
00
bann
byn
bxn
VyxVn
o
sinh
sinhsin4),(5,3,1
evennoddn
bann
V
co
n
0
sinh
4
Flux linesEquipotential lines
V=Vo
V=0
V=0
Find V(a,a/2) where Vo=100V, b=2a=2m
V
nn
ynxn
Vn
51.442
1sinh
2sinh
2sin400
5,3,1
Flux linesEquipotential lines
V=Vo
V=0
V=0
bann
ban
ban
VaaVn
o
sinh
2sinhsin4)
2,(
5,3,1
Find E at (a,a/2)
yx
oddn
o ab
ynb
xnab
ynb
xn
banb
VyxE ˆcoshsinˆsinhcossinh
14),(
yx ayVa
xVVE ˆˆ
V/mˆ25.99
ˆ...)0074.0035.01703.08192.9411.3127.1912.115(
ˆ2/coshsinsinh
1400ˆ0
y
y
yoddn
x
a
a
aban
ban
banb
aE
Resistance and Capacitance
Resistance If the cross section of a conductor is not
uniform we need to integrate:
Solve Laplace eq. to find VThen find E from its differentialAnd substitute in the above equation
S
l
SdE
ldE
IVR
s
P.E. 6.8 find Resistance of disk of radius b and central hole of radius a.
S
oo
SdEV
IVR
s
0112
2
2
2
2
zVVV
01
V
BAV lnoVbV
aVBC
)(0)(
:
aabVV o ln
/ln
ˆddVVE
ˆ
/ln abVo
)/ln(2
abtVSdEI o
S
ss
ˆddzSd
s tab
o2)/ln(
a
t
b
Capacitance Is defined as the ratio of
the charge on one of the plates to the potential difference between the plates:
Assume Q and find V (Gauss or Coulomb)
Assume V and find Q (Laplace)
And substitute E in the equation.
FaradsldE
SdE
VQC
l
S
Capacitance
1. Parallel plate2. Coaxial 3. Spherical
Parallel plate Capacitor Charge Q and –Q
orx
s
xsn
aE
aD
ˆ
ˆ
Dielectric,
Plate area, S
SQ
s
dS
VQC
SQddx
SQldEV
dd
00
SESdEQ x
Coaxial Capacitor Charge +Q & -Q
LESdEQ 2
abL
VQC
ln
2
Dielectric,
Plate area, S
SQd
SQdSEV
dd
00
++
+
+
+
--
-
--
-
-
-
-
c
ab
LQd
LQldEV
a
b
ln2
ˆˆ2
Spherical Capacitor Charge +Q & -Q
24 rEdSEQ r
baVQC
114
ba
Qrdrrr
QldEVa
b
114
ˆˆ4 2
What is the Earth's charge?
The electrical resistivity of the atmosphere decreases with height to an altitude of about 48 kilometres (km), where the resistivity becomes more-or-less constant. This region is known as the electrosphere. There is about a 300 000 volt (V) potential difference between the Earth's surface and the electrosphere, which gives an average electric field strength of about 6 V/metre (m) throughout the atmosphere. Near the surface, the fine-weather electric field strength is about 100 V/m.
The Earth is electrically charged and acts as a spherical capacitor. The Earth has a net negative charge of about a million coulombs, while an equal and positive charge resides in the atmosphere.
Capacitors connectionSeries
Parallel
21 CCC
21
111CCC
Resistance
S
SdE
ldE
IVR
s
ldE
SdE
VQC S
Recall that:
Multiplying, we obtain the Relaxation Time:
Solving for R, we obtain it in terms of C:
s
RC
CR
s
So In summary we obtained:Capacitor C R=/sC
Parallel Plate
Coaxial
Spherical
ba11
4
Sds
ab
Lln
2dS
Lab
s2
ln
s4
11
ba
P.E. 6.9 A coaxial cable contains an insulating material
of s1 in its upper half and another material with s2 in its lower half. Radius of central wire is a and of the sheath is b. Find the leakage resistance of length L.
s1
s2
Lab
RLab
R2
21
1
lnln
ss
21
2121 ||
RRRRRRR
21
1ln
ss
Lab
R
They are connected in parallel
P.E. 6.10a Two concentric spherical capacitors with 1r=2.5 in its outer half and another material with 2r=3.5 in its inner half. The inner radius is a=1mm, b=3mm and c=2mm . Find their C.
1
pFC 53.0
2
c
We have two capacitors in series:
21
21
CCCCC
ba
C
ba
C oror
114
114 2
21
1
P.E. 6.10b Two spherical capacitors with 1r=2.5 in its
upper half and another material with 2r=3.5 in its lower half. Inner radius is a=1mm and b=3mm. Find their C.1
2
2121 || CCCCC
ddErdSEQ sin2
We have two capacitors in parallel:
22 ErQ
ba
VQC oro 11
2/ 11
pF
ba
C 5.0112 21
ba
Qrdrrr
QldEVa
b
112
ˆˆ2 2
Method of ImagesWhenever the is a charge in the presence
of a conductor. The conductor serves as a mirror.
Substitute the conductor for a plane at V=0 and the image.
The solution will be valid only for the region above the conductor.
Line charge above ground plane2
21
1
ˆ2
ˆ2
o
L
o
LEEE
2222 )(ˆ)(ˆ
)(ˆ)(ˆ
2 hzxzhzxx
hzxzhzxxE
o
L
),0,(),,0(),,(),0,(),,0(),,(
2
1
hzxhyzyxhzxhyzyx
VVV
2
1ln2
o
Ldlo
L
o
L
2
21
1 2ˆˆ
2
22
22
)()(ln
2 hzxhzx
o
L
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