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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
VE
ED v
Depending on the geometry:
We use appropriate coordinates:
cartesian:
cylindrical:
spherical:
vV 2
v
z
V
y
V
x
V
2
2
2
2
2
2
v
z
VVV
2
2
2
2
2
11
vV
r
V
rr
Vr
rr
2
2
2222
2 sin
1sin
sin
11
Procedure for solving eqs.
1. Choose Laplace (if no charge) or Poisson
2. Solve by Integration if one variable or by
3. Separation of variables if many variables
4. Apply B.C.
5. Find V, then E=-DV, D=eE, J=sE
6. 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
x
V
2
2
BAxa
xV o
6
3
xo
x aAa
xa
x
VE ˆ
2ˆ
2
0A
6
2aB o 33
3
6xa
a
xV o
xo a
a
xE 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(fo=45)=Vo=50
a
Va
VE
o
o ˆˆ1
01
2
2
22
V
V
02
2
V
o
oVV
ED o
ons
VD
)0(
)/ln(1
0
abLV
dzdV
dSQo
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 1.5
Applying B.C. V(0)=0,
V(fo=45)=Vo=50)/ln( abL
V
QC
oo
)/ln( abLV
Qo
o
45o
a b
)/ln( abLV
QC
oo
mmmm
ao
226.52/45sin
2/4 nCVpFCVQ
F
mm
mmm
V
QC
o
o
o
22)50(444
10444
226.5
1000ln)5(
4/
5.1
12
detail
45o
a b
a
brecha
hipotenusa
opuestoo 2/
2
45sin
mmmm
ao
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)
oVaybxV
ybxV
aybxV
ayxV
),0(
0)0,0(
0)0,(
0)0,0(
y
xb
aV=Vo
V=0
V=0V=0
02
2
2
2
y
V
x
V
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
Y
Y
X
X ""
0"
0"
YY
XX
einseparablVaYxXaxV
YYxXxV
bXyYbXybV
XyYXyV
o
)()(),(
0)0(0)0()()0,(
0)(0)()(),(
0)0(0)()0(),0(
Let’s examine 3 Possible Cases
.A l=0
.B l<0
C. l>0
Case A: If l=o
BAxX
0
0
A
B
0)()(
0)(
yYxXV
xX
y
xb
a
V=Vo
V=0
V=0V=0
0"
0"
YY
XX
This is a trivial solution, therefore l cannot be equal to zero.
0)(
0)0(
..
bX
X
CB
Case B: l<o
xBxBX
or
eAeAX xx
sinhcosh
:issolution general
21
21
0
0sinh
0
2
2
1
B
bB
B
2 0" 2 XX
This is another trivial solution, therefore l cannot be equal to zero.
0)(
0)0(
..
bX
X
CB
Case C: l>o
xgxgX
or
eCeCX
o
xjxjo
sincos
:issolution general
1
1
0sin
0
1
bg
go
2 0" 2 XX
0)(
0)0(
..
bX
X
CB
4,3,2,1
0sin
nb
n
b
22
sin)(
:solutions of seriesA
b
n
b
xngxX nn
Case C: l>o
yhyhY o sinhcosh
:issolution generalwith
10oh0)0(
..
Y
CB
0" 2 YY 4,3,2,1
nb
n
xhyYn sinh)( 1
b
yn
b
xnhgyYxXyxV nnnnn
sinhsin)()(),(
b
yn
b
xncyxV n
nn
sinhsin),(
1
By superposition, the combination is also a solution:
Cont.
b
an
b
xncVaxV n
non
sinhsin),(
1
dxb
xn
b
xm
b
ancdx
b
xmV
b
nn
o
b sinsinsinhsin
010
nm
nmdxmxnx
2/
0sinsin
0
B.C. at y=a
If we multiply by sin factor and integrate on x:
dxb
xm
b
ancdx
b
xnV
b
no
b 2
00
sinsinhsin
dxb
xn
b
anc
b
xn
n
bV
b
n
b
o
2cos1
2
1sinhcos
00
Orthogonality property of sine and cosine:
2
sinhcos1b
b
ancn
n
bVn
o
dxb
xn
b
anc
b
xn
n
bV
b
n
b
o
2cos1
2
1sinhcos
00
ban
n
byn
bxn
VyxV
n
o
sinh
sinhsin4),(
5,3,1
evenn
oddn
ban
n
V
c
o
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
ban
n
ban
ban
VaaV
n
o
sinh
2sinhsin4
)2
,(5,3,1
Find E at (a,a/2)
yx
oddn
o ab
yn
b
xna
b
yn
b
xn
banb
VyxE ˆcoshsinˆsinhcos
sinh
14),(
yx ay
Va
x
VVE ˆˆ
V/mˆ25.99
ˆ...)0074.0035.01703.08192.9411.3127.1912.115(
ˆ2/
coshsinsinh
1400ˆ0
y
y
yoddn
x
a
a
ab
an
b
an
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
I
VR
P.E. 6.8 find Resistance of disk of radius b and central hole of radius a.
S
oo
SdE
V
I
VR
011
2
2
2
2
2
z
VVV
01
V
BAV lnoVbV
aVBC
)(
0)(:
aab
VV o
ln/ln
ˆd
dVVE
ˆ
/ln ab
Vo
)/ln(
2
ab
tVSdEI o
S
ˆddzSd
t
ab
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
V
QC
l
S
Capacitance
1. Parallel plate
2. Coaxial
3. Spherical
Parallel plate Capacitor Charge Q and –Q
orx
s
xsn
aE
aD
ˆ
ˆ
Dielectric, e
Plate area, S
S
Qs
d
S
V
QC
S
Qddx
S
QldEV
dd
00
SESdEQ x
Coaxial Capacitor
Charge +Q & -Q
LESdEQ 2
abL
V
QC
ln
2
Dielectric, e
Plate area, S
S
Qd
S
QdSEV
dd
00
++
+
+
+
-
-
-
-
--
-
-
-
c
a
b
L
Qd
L
QldEV
a
b
ln2
ˆˆ2
Spherical Capacitor
Charge +Q & -Q
24 rEdSEQ r
baV
QC
114
ba
Qrdrr
r
QldEV
a
b
11
4ˆˆ
4 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 connection
Series
Parallel
21 CCC
21
111
CCC
Resistance
S
SdE
ldE
I
VR
ldE
SdE
V
QC S
Recall that:
Multiplying, we obtain the Relaxation Time:
Solving for R, we obtain it in terms of C:
RC
CR
So In summary we obtained:Capacitor C R= /e sC
Parallel Plate
Coaxial
Spherical
ba11
4
S
d
ab
L
ln
2d
S
Lab
2
ln
4
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
21
2121 ||
RR
RRRRR
21
1ln
La
b
R
They are connected in parallel
P.E. 6.10a Two concentric spherical capacitors with e1r=2.5 in its outer half and another material with e2r=3.5 in its inner half. The inner radius is a=1mm, b=3mm and c=2mm . Find their C.
e1
pFC 53.0
e2
c
We have two capacitors in series:
21
21
CC
CCC
ba
C
ba
C oror
114
114 2
21
1
P.E. 6.10b Two spherical capacitors with e1r=2.5 in its
upper half and another material with e2r=3.5 in its lower half. Inner radius is a=1mm and b=3mm. Find their C.e1
e2
2121 || CCCCC
ddErdSEQ sin2
We have two capacitors in parallel:
22 ErQ
ba
VQC oro 11
2/ 1
1
pF
ba
C 5.011
2 21
ba
Qrdrr
r
QldEV
a
b
11
2ˆˆ
2 2
Method of Images
Whenever 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 plane
22
11
ˆ2
ˆ2
o
L
o
LEEE
2222 )(
ˆ)(ˆ
)(
ˆ)(ˆ
2 hzx
zhzxx
hzx
zhzxxE
o
L
),0,(),,0(),,(
),0,(),,0(),,(
2
1
hzxhyzyx
hzxhyzyx
VVV
2
1ln2
o
Ldlo
L
o
L
2
21
1 2
ˆˆ
2
22
22
)(
)(ln
2 hzx
hzx
o
L
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