Electrostatic
Boundary value
problems
Sandra Cruz-Pol, Ph. D.
INEL 4151 ch6
Electromagnetics I
ECE 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)
v
 V 

2
 V 0
They can be derived from
Gauss’s Law
2
  D    E   v
E  V
Depending on the geometry:
v
 V 

2
We use appropriate coordinates:
cartesian:  2V  2V  2V
v
x
cylindrical:
spherical:
2

y
2

z
1   V
 
   
2


 1  2V  2V
v
  2
 2 
2
z

  
v
1   2 V 
1
 
V 
1
 2V

r
 2
 sin 
 2 2
2
2
r r  r  r sin   
  r sin  

Procedure for solving eqs.
1.
2.
3.
4.
5.
6.
Choose Laplace (if no charge) or Poisson
Solve by Integration if one variable or by
Separation of variables if many variables
Apply B.C.
Find V, then E=-DV, D=E, J=sE
Also, if necessary:
Q    S dS
Dn   S
I   J  dS
S
P.E. 6.1 In a 1-dimensional device, the
charge density is given by v  o x / a
If E=0 at x=0 and V=0 at x=a, find V and E.
v
 2V

2
x

o x3
V 
 Ax  B
6a
v  o x / a
Evaluating B.C.
v

 o x 2

V
E
aˆ x  
 A aˆ x
x
 2a

A0
o a 2
B
6
 2V  
o x3 3 3

V
a x 
6a
  o x 2 
aˆ x
E  
 2a 
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,
Vo
V(o=45)=Vo=50
V 
2
1

V
2
V 2
0
2
 
45o
1m
 2V
0
2

V  A  B
o
 Vo 
1 V
aˆ
ˆ
E
a  
 
 o 
Vo
D  E  s  D n (  0)   
Q    s dS  
o
Vo
o
L
b
 
z 0  a
1

dzd  
Vo
o
L ln( b / a)
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, Q   Vo L ln( b / a )
o
V(o=45)=Vo=50
C
C
Vo

Vo


L ln( b / a)
o

L ln( b / a)
o
1.5 o
 1000mm 
C

(5m) ln 

Vo  / 4
 5.226mm 
Q
45o
b
a
4mm / 2
a
 5.226mm
o
sin 45 / 2

Q
Q

 444 10 12 F
Q  CVo  444 pF (50V )  22nC
detail
45o
a
4mm / 2
a
 5.226mm
o
sin 45 / 2
 45o 
opuesto
brecha / 2
 
sin 

a
 2  hipotenusa
b
P.E. 6.5 Determine the potential function for
the region inside the rectangular trough of
infinite length whose cross section is shown.


a)
b)
The potential V
depends on x and y.
Vo=100V, b=2a=2m,
find V and E at:
(x, y)=(a, a/2)
(x, y)=(3a/2, a/4)
V V
 2 0
2
x
y
2
2
y
V=Vo
a
V=0
V=0
V=0
b
V ( x  0,0  y  a )  0
V ( x  b,0  y  a )  0
V ( 0  x  b, y  0)  0
V (0  x  b, y  a )  Vo
x
P.E. 6.5 (cont.) Since it’s 2 variables, use
Separation of Variables
V ( x, y )  X ( x)Y ( y )
y
V=Vo
a
X "Y  XY "  0
V=0
V=0
X " Y"



b
V=0
X
Y
V (0, y )  X (0)Y ( y )  0  X (0)  0
X "X  0 V (b, y)  X (b)Y ( y)  0  X (b)  0
Y "Y  0 V ( x,0)  X ( x)Y (0)  0  Y (0)  0
x
V ( x, a )  X ( x)Y (a )  Vo  inseparabl e
Let’s examine 3 Possible Cases
A.
B.
C.
=0
<0
>0
Case A: If =o
X "X  0
Y "Y  0
X  Ax  B
B.C.
X ( 0)  0
X (b)  0
B0
A0
X ( x)  0
V  X ( x)Y ( y )  0
y
V=Vo
a
V=0
V=0
V=0
b
This is a trivial solution, therefore
 cannot be equal to zero.
x
Case B: <o
2
X " X  0
   2
general solution is :
X  A1ex  A2 e x
or
B.C.
X (0)  0
X (b)  0
X  B1 cosh x  B2 sinh x
This is another trivial solution,
therefore  cannot be equal to
zero.
B1  0
B2 sinh b  0
 B2  0

Case C: >o
2
X "  X  0
2
general solution is :
B.C.
X  Co e jx  C1e  jx
X ( 0)  0
X (b)  0
or
X  g o cos x  g1 sin x
A series of solutions :
nx
X n ( x)  g n sin
b
 n 
  

 b 
2
2
go  0
g1 sin b  0
sin b  0
n

b
n  1,2,3,4
Case C: >o
2
Y " Y  0
with general solution is :
Y  ho cosh y  h1 sinh  y
Yn ( y)  h1 sinh x
B.C.
Y ( 0)  0
n

b
n  1,2,3,4
ho  0
nx
ny
Vn ( x, y )  X n ( x)Yn ( y )  g n hn sin
sinh
b
b
By superposition, the combination is also a solution:

Vn ( x, y )  
n 1
nx
ny
cn sin
sinh
b
b
Cont.
B.C. at y=a
nx
na
cn sin
sinh
b
b

Vn ( x, a )  Vo  
n 1
If we multiply by sin factor and integrate on x:
b

0

mx
na
mx
nx
Vo sin
dx   cn sinh
sin
sin
dx

b
b 0
b
b
n 1
b
Orthogonality property
of sine and cosine:
b

0


0
mn
 0
sin nx sin mx dx  
 / 2 m  n
nx
na
2 mx
Vo sin
dx  cn sinh
sin
dx

b
b 0
b
b
b
nx
na 1
 Vo
cos
 cn sinh
n
b 0
b 2 0
b
b
2nx 

1

cos

dx
b 

b
nx
na 1 
2nx 
 Vo
cos
 cn sinh
1  cos
dx

n
b 0
b 20 
b 
b
b
Vob
1  cos n   cn sinh na b
n
b 2
4Vo


na
cn   n sinh
b

0

V ( x, y ) 
4Vo



n 1, 3, 5
n  odd
n  even
nx
ny
sin
sinh
b
b
na
n sinh
b
V=Vo
V=0
Equipotential
lines
Flux lines
V=0
Find V(a,a/2) where
Vo=100V, b=2a=2m

4V
a
V ( a, )  o
2

V

n 1, 3, 5

400


n 1, 3, 5
 44.51V
sin
na
na
sinh
b
2b
na
n sinh
b
V=Vo
nx
ny
sin
sinh
2
2
n 1
n sinh
2
V=0
Equipotential
lines
Flux lines
V=0
Find E at (a,a/2)

V
V
E  V  
aˆ x 
aˆ y
x
y

4Vo
E ( x, y ) 
b
1
nx
ny
nx
ny 

cos
sinh
aˆ x  sin
cosh
aˆ y 


na 
b
b
b
b

n odd
sinh
b


400 
1
na / 2 
 na
E  0aˆ x 
sin
cosh
aˆ y



b n odd sinh na 
b
b 
b
 (115.12  19.127  3.9411  .8192  0.1703  0.035  0.0074  ...)aˆ y
 99.25aˆ y V/m
Resistance and Capacitance
Resistance

If the cross section of a conductor is not
uniform we need to integrate:
 
E  dl

V
R  l  
I  sE  dS
S
Solve Laplace eq. to find V
 Then find E from its differential
 And substitute in the above equation

P.E. 6.8 find Resistance of disk of
radius b and central hole of radius a.
1   V
 
   
 1  2V  2V
  2
 2 0
2
z
  
V  A ln   B
dV
E  V  
ˆ 
d
1   V
 
   
V (   a)  0
BC :
V (   b)  Vo V 
Vo
ˆ
 ln b / a 
Vo
Vo

 
I
 sE  dS
S
ln( b / a )

2 ots
Vo

ln
ln b / a  a

dS    dz d ˆ
  2tVos
I   sE  dS 
ln( b / a)
S
R

  0

b
a
t
Capacitance
 
  E  dS
Q
C  S 
V
 E  dl
l
Farads

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.


Capacitance
1.
2.
3.
Parallel plate
Coaxial
Spherical
Parallel plate Capacitor

Charge Q and –Q
Q
s 
S

Plate area, S

Dn   s aˆ x
 s
E
aˆ x
or
Dielectric, 

 
Q    E  dS  E x S
0
 
Q
Qd
V   E  dl   dx 
S
S
d
d
0
Q S
C 
V
d
Plate area, S
-
Coaxial Capacitor
V
Charge +Q & -Q
 
Q    E  dS  E 2L
a
 
V   E  dl  
b
Q
-
-

-
Dielectric,
 d
+d +
  E+ dS +
0 +
0
c


-
Q
b
ˆ  dˆ 
ln
2L
2L a
Q 2L
C 
b
V
ln
a
Q
Qd

S
S
Spherical Capacitor

Charge +Q & -Q
Q 

E  dS  Er 4r 2
a
 
Q
Q 1 1
V    E  dl  
rˆ  dr rˆ 
 
2

4r
4  a b 
b
Q
4
C 
V 1 1
 a  b 
What is the Earth's
charge?
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.
 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.
Capacitors connection

Series

Parallel
1
1
1


C C1 C2
C  C1  C2
Resistance

Recall that:
 
E  dl
V

R 
 
I  sE  dS
S
 
E  dS

Q S
C   
V
 E  dl

Multiplying, we obtain the Relaxation Time:

Solving for R, we obtain it in terms of C:

RC 
s

R
sC
So In summary we obtained:
Capacitor
Parallel Plate
Coaxial
C
R=/sC
S
d
sd
S
2L
ln
Spherical
b
a
4
1 1
 a  b 
b
a
2sL
ln
1 1
 a  b 
4s
P.E. 6.9
s1
s2

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.
b
ln
a
R1 
s1 L
b
ln
a
R2 
s2 L
They are
connected in
parallel
R  R1 || R2 
b
ln
R a
L
1
s 1  s 2 
R1 R2
R1  R2

P.E. 6.10a
1
2
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.
We have two capacitors in series:
c
C1 
4r1 o
1 1

a b
C2 
C  0.53 pF
4r 2 o
1 1

a b
C
C1C2
C1  C2

P.E. 6.10b
1
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.
We have two capacitors in parallel:
C  C1 || C2  C1  C2
2
Q    E  dS     Er 2 sin dd
Q  2Er 2
a
 
Q
Q 1 1
ˆ
ˆ
V    E  dl  
r

dr
r

 
2

2r
2  a b 
b
2r1 o
C1  Q / Vo 
1 1
 a  b 
2  1   2 
C
 0.5 pF
1 1

a b
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
E  E  E 
L
 L
ˆ1 
ˆ 2
2o 1
2o  2
1  ( x, y, z )  (0, y, h)  ( x,0, z  h)
 2  ( x, y, z )  (0, y,h)  ( x,0, z  h)

E L
2o
 xxˆ  ( z  h) zˆ xxˆ  ( z  h) zˆ 
 x 2  ( z  h) 2  x 2  ( z  h) 2 


  L  1 
 L
  L ˆ 2 
V  V  V     2  ˆ1  2    dl  2 ln   
o
 2
o 1
o 2 

  L  x 2  ( z  h) 2 
ln  2

2o  x  ( z  h) 2 