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Chapter 6 Electrostatic Boundary Value
Problems
Dr. Talal Skaik
2012 1
Introduction • In previous chapters, E was determined by coulombs law or Gauss
law when charge distribution is known, or when
potential V is known throughout the region.
• In most practical applications, however, neither the charge
distribution nor the potential distribution is known.
• In this chapter, practical electrostatic problems are considered
where only electrostatic conditions (charge & potential) at some
boundaries are known, and E and V are to be found.
• Such problems are tackled using Poisson's or Laplace’s equation
or the method of images.
2
E V
6.2 Poisson’s and Laplace’s Equations
3
• They are easily derived from Gauss’s law.
2
2
poisson's e
D
quation
Lapl
E
E
for an inhomogeneous medium
for homogeneous medium ( )
when 0, (for charge free region):
0 ( )ace's equation
V
V
V
V
V
V
V
V
2 2 2
2 2 2
2 2
2 2 2
22
2 2 2 2 2
0
1 10
1 1 1sin 0
sin sin
V V V
x y z
V V V
z
Cartesian
Cylindrical
Spherical
V V Vr
r r r r r
Laplace’s Equation
4
2 0V
2 2 2
2 2 2
2 2
2 2 2
22
2 2 2 2 2
1 1
1 1 1sin
sin sin
V
V
V
V V V
x y z
V V V
z
V V Vr
r r r r r
Cartesian
Cylindrical
Spherical
5
Poisson’s Equation
2 VV
Uniqueness Theorem • There are several methods of solving a given problem (analytical,
graphical, numerical, experimental, etc).
• The question is: If a solution of Laplace’s equation satisfies a given
set of boundary conditions, is this the only possible solution?
• The answer is Yes, there is only one solution, and it is unique.
• This is uniqueness theorem: If a solution of Laplace’s equation
can be found that satisfies the boundary conditions, then the
solution is unique.
6
General Procedure for Solving Poisson's or Laplace's Equation
7
1. Solve Laplace's (if ) or Poisson's ( if ) equation using:
(a) direct integration when V is a function of one variable, or
(b) separation of variables if V is a function of more than one
variable.
The solution at this point is not unique but expressed in terms of
unknown integration constants to be determined.
2. Apply the boundary conditions to determine a unique solution for V.
Imposing the given boundary conditions makes the solution unique.
3. Having obtained V, find E using and D using . E V
0V 0V
D E
Current-carrying components in high-voltage power equipment must
be cooled to carry away the heat caused by ohmic losses. A means of
pumping is based on the force transmitted to the cooling fluid by
charges in an electric field. The electrohydrodynamic (EHD) pumping
is modeled in the Figure. The region between the electrodes contains
a uniform charge ρ0 which is generated at the left electrode and
collected at the right electrode. Calculate the pressure of the pump if
ρ0 =25 mC/m3 C and V0 =22 kV.
8
Example 6.1
•The technology promises to make it easier and more efficient to
remove heat from small spaces, vastly expanding the capabilities of
advanced instruments and microprocessors.
•Unlike traditional thermal-control technologies that rely on
mechanical pumps and other moving parts, EHD cooling uses
electric fields to pump coolant through tube.
•Electrodes apply the voltage that pushes the coolant through tube.
•EHD-based systems are lightweight and consume little power. 9
EHD pumps
V
2
0
2 2 2
2 2 2 2
Since 0, we apply poisson's equation
the boundary conditions V(z=0)=V and V(z=d)=0 show that V depends only on z,
1 1Hence becomes:
Integratin
V
V o
V
V V V V
z z
2
0 0 0
g once gives
Integrating again gives =2
where A and B are integration constants to be determined by applying
the boundary conditions.
when 0, =0 0
when ,
o
o
dV zA
dz
zV Az B
z V V V B B V
z d
2
000 0= or A=
2 2
o od d VV Ad V
d
10
Example 6.1 - Solution
0
0
0
200
0
The electric field is given by
1E= V=
E2
The net force is F= E,
F= E E
2
z
o oz z z
T T v
v
d
v
z
d
oz
V V Va a a
z
z VV da A a z a
z d
Q Q dv
dv dS dz
V zS z dz a
d
0 0
3 3 2
0 0
F= a
Force per unit area (Pressure) is 25 10 22 10 550 N/m
zSV
FSV
S
11
Example 6.1 - Continued
2
0
0
=2
=2
o
o
zV Az B
B V
d VA
d
12
Example 6.2 The xerographic copying machine is an important application of
electrostatics. The surface of the photoconductor is initially charged
uniformly as in Figure 6.2(a). When light from the document to be
copied is focused on the photoconductor, the charges on the lower
surface combine with those on the upper surface to neutralize each
other. The image is developed by pouring a charged black powder
over the surface of the photoconductor. The electric field attracts
the charged powder, which is later transferred to paper and melted
to form a permanent image. We want to determine the electric
field below and above the surface of the photoconductor.
13
Example 6.2
14
Example 6.2
V
22
2
1 2
1 1 1
Since 0, we apply Laplace's equation.
Also potential depends only on x.
0
Integrate twice gives
V=Ax+B
Let the potentials above x=a be V , and below x=a be V .
V =A x+B x>a
V
VV
x
2 2 2
1 1 1 1 1
2 2 2
=A x+B x<a
The boundary conditions at the grounded electrodes are,
V (x=d)=0 0=A d+B B = A d (1)
V (x=0)=0 0=0+B B =0 (2)
15
Example 6.2 - Solution
1 2 1 1 2
1n 2n 1 1n 2 2n
1 21 2 1 1 2 2
At the surface of the photoconductor,
V (x=a)=V (x=a) A a+B A a (3)
=
Since E= V=
Therefore (4)
S Sx a
x y z n
S S
D D E E
dV dV dV dVa a a E
dx dy dz dx
dV dVA A
dx dx
11 1
2 21
1 1
22 2
2 21
1 1
Solve equations (1) to (4) to find the constants.
aE = a a ,
1
1 a
E = a a , 0
1
S xx x
S x
x x
dVA a x d
dx d
a
d
dV aA x a
dx d
a
16
Example 6.2 - Continued
1 1 1
2 2 2
1 1 2
V =A x+B x>a
V =A x+B x<a
B = A d, B =0
17
Example 6.3 Semi-infinite conducting planes at Ф=0 and Ф=π/6 are separated by an infinitesimal insulating gap as in Figure6.3. If V(Ф=0)=0 and (Ф=π/6 )=100 V, Calculate V and E in the region between the planes.
18
Example 6.3 Solution
2 2 22 2
2 2 2 2 2
Since depends only on , Laplace's equation in cylindrical coordinates becomes
1 1 10 0 0
since =0 is excluded due to the insulating gap, we can multip
V
V V V d VV V
z d
22
2
0 00 0 0 0
0 0
0
0
0
ly by to get 0
Integrating twice gives V=A +B.
Apply boundary conditions:
when 0, 0 0=0+B B=0
when , V =A , Hence =
1
Substituting V =100 an
d V
d
V
V VV V A V
VdVE V a a
d
0
2
600 600d / 6 gives: = and E=
check V( 0)=0, V( / 6)=100, 0
V a
V
19
Example 6.4 Two conducting cones (Ѳ=π/10 and Ѳ=π/6) of infinite extent are separated by an infinitesimal gap at r = 0. If V(Ѳ=π/10 )=0 and V(Ѳ=π/6)=50 V, Find V and E between the cones.
2
2
2
Since depends only on , Laplace's equation in spherical
1coordinates becomes: sin 0
sin
since =0 and =0, are excluded, we can multiply by sin to get
s
V
d dVV
r d d
r r
d
d
2
in 0
Integrating once gives: sin sin
Integrating this results in =sin 2cos( / 2)sin( / 2)
cos( / 2) 1 sec ( / 2)
2cos( / 2)sin( / 2) cos( / 2) 2 tan( / 2)
Let = tan
dV
d
dV dV AA
d d
d dV A A
d dV A A
u
2/ 2, =1/ 2sec ( / 2)
= ln( ) ln(tan / 2)
du d
duV A A u B V A B
u
20
Example 6.4 Solution
1 1 1
1
1
2 0 0
Apply boundary conditions to determine the integration constatnts and .
( ) 0 0 ln(tan / 2) ln(tan / 2)
tan / 2Hence ln(tan / 2) ln(tan / 2) ln
tan / 2
Also ( ) l
A B
V A B B A
V A A V A
V V V A
02
1 2
1
0
1
2
1
tan / 2n
tan / 2 tan / 2ln
tan / 2
tan / 2ln
tan / 2Thus =
tan / 2ln
tan / 2
VA
V
V
21
Example 6.4 Continued ln(tan / 2)V A B
22
Example 6.4 Continued
0
2
1
1 2 0
1
sin tan / 2sin ln
tan / 2
Taking , , and 50 gives10 6
tan / 250ln
tan / 2tan / 20= 95.1ln
tan /12 0.1584ln
tan / 20
and
95.1E=
sin
VdV AE V a a a
r d rr
V
V V
r
2
0
V/m
Check: V( /10)=0, V( / 6)=V , 0
a
V
1
0
2
1
0
1
2
1
tan( / 2)ln
tan( / 2)
tan( / 2)ln
tan( / 2)
tan( / 2)ln
tan / 2=
tan( / 2)ln
tan( / 2)
V A
VA
V
V
sin
dV A
d
6.5 Resistance and Capacitance
23
• For conductor with uniform cross section, the resistance is
• If the cross section of the conductor is not uniform, the resistance is obtained from:
lR
S
E. l
E. S
dVR
I d
24
The resistance R ( or conductance G=1/R) of a conducting material
can be found by the following steps:
1. Choose a suitable coordinate system.
2. Assume Vo as the potential difference between conductor
terminals.
3. Solve Laplace's equation to obtain V then determine E
from and find I from .
4. Finally, obtain R as Vo/I.
2 0V
E V E. SI d
Resistance
Capacitance • To have capacitance, we must have two (or more) conductors
carrying equal but opposite charges.
• The conductors may be separated by free space or dielectric.
• The conductors are maintained at a potential difference:
• (note E is always normal to conducting surfaces)
25
1
2
E. lV d
The capacitance C of the capacitor is defined as the ratio of the
magnitude of the charge on one of the plates to the potential
difference between them:
Obtain C for any two conductor capacitors by either:
1. Assume Q and determine V in terms of Q (involves Gauss Law).
2. Assume V and determine Q in terms of V (involves Laplace’s equation)
Method (1) involves the following:
1. Choose a suitable coordinate system.
2. Let the two conducting plates carry charges +Q and -Q.
3. Determine E using Coulomb or Gauss law and find V from
4. Finally, obtain C from C=Q/V. 26
E. S
E. l
dQC
V d
E. lV d
Capacitance
Storing Charges- Capacitors: A capacitor consists of 2 conductors of any shape placed near one another without touching. It is common; to fill up the region between these 2 conductors with an insulating material called a dielectric. We charge these plates with opposing charges to set up an electric field. 27
Capacitors
Capacitance or Capacitance Resistance
28
1
2
E l
E Gauss or
Coulombs law
(in terms of Q)
QC
V
V d
0
2
2
D S
D= E
E=
=0
or = /V
QC
V
Q d
V
V
V
0
2
2
= J S
J= E
E=
=0
or = /V
VR
I
I d
V
V
V
Suppose each plate has an area S and the plates are separated by distance d. Assume plates 1 and 2 carry uniformly distributed charges +Q and –Q 29
Parallel Plate Capacitor
1
2 0
22
E
, or ( )
Hence E l
and thus for a parallel-plate capacitor
1 1The energy stored in a capacitor is W
2 2 2
SS S x x x
d
x x
Q QD a E a a
S S
Q QdV d a dx a
S S
Q SC
V d
QCV QV
C
Assume Conductors 1 and 2 carry uniformly distributed charges +Q and –Q. Apply gauss law to arbitrary Gaussian cylindrical surface of radius ρ such that a< ρ< b, 30
Coaxial Capacitor
1
2
E S= 2 E=2
E l ln2 2
Thus the capacitance of a coaxial cylinder is given by
2C=
ln
a
b
QQ d E L a
L
Q Q bV d a d a
L L a
Q L
bV
a
Assume charges +Q and –Q are on inner and outer spheres. Apply gauss law to arbitrary Gaussian spherical surface of radius r such that a< r< b, 31
Spherical Capacitor
2
2
1
2
2
E. S= 4 E=4
1 1E l a a
4 4
Thus the capacitance of a spherical capacitor is given by
4 C=
1 1
r r
a
r r
b
QQ d E r a
r
Q QV d dr
r a b
Q
V
a b
32
Series and Parallel Capacitors •If two capacitors with capacitance C1 and C2
are in series ( have same charge on them), the
total capacitance is:
•If the capacitors are in Parallel (have same voltage
across them), the total capacitance is:
1 2
1 2 1 2
1 1 1 or C=
C C
C C C C C
1 2C C C
It has been shown that:
Which is the relaxation time Tr of the medium separating the conductors.
33
E l
E S
E S
E l
The product of these expressions yields
dVR
I d
dQC
V d
RC
Note: R in these equations is NOT the resistance of the capacitor plate, but the leakage resistance between the plates. Therefore, σ is the conductivity of the dielectric medium separating the plates. 34
Assuming homogeneous media, the resistance of various
capacitors mentioned earlier can be obtained using RC = ε/σ :
For a parallel-plate capacitor ,
ln2
For a cylindrical capacitor , 2ln
1 14
For a spherical capacitor , 1 1 4
S dC R
d S
bL aC R
b L
a
a bC R
a b
35
Example 6.8
A metal bar of conductivity σ is bent to form a flat 90o sector of inner
radius a, outer radius b, and thickness t as shown in the Figure. Show
that :-
(a) The resistance of the bar between the vertical curved surfaces at r = a
and r = b is:
(b) The resistance between
the two horizontal surfaces
at z=0 and z=t is
2lnb
aRt
2 2
4'
( )
tR
b a
0
(a) Between the vertical curved ends located at =a and =b, the bar has
a nonuniform cross section. Let a potential difference V be maintained
between the curved surfaces at = and =b so that V(a
0
2
= )=0 and
V( = )=V . Solve V using Laplace's equation in cylindrical coordinates,
1since V=V( ), 0
As 0 is excluded, multiply by and
integrate once, , i
a
b
d dVV
d d
dV dV AA
d d
0 0
ntegrating again yields,
V=A ln +B , where A and B are constants of integration.
Use the boundary conditions:
( = )=0 0=A ln +B or ln
( = )= =A ln +B=A ln ln ln or
V a a B A a
bV b V V b b A a A A
a
0
ln
V
b
a36
Example 6.8 Solution
0
0
/2
0 0
0 0
0
Hence,
ln ln a= ln ln
ln
ln
,
2
ln ln
2ln
Thus R=
t
z
VV A A A
ba a
a
VdV AE V a a a
bd
a
J E dS d dza
V tVI J dS dz d
b b
a a
bV
I
as requireda
t37
Example 6.8 Continued
0
V=A ln +B
ln
ln
VA
b
a
B A a
0
0
2
2
(b)
Let be the ptential difference between the two horizontal surfaces
so that V( 0)=0 and V( )=V .
since ( ),Laplace's equation becomes
0, integrating twice gives
, apply bou
V
z z t
V V z
d V
dz
V Az B
00 0
0
0
ndary conditions to get and ,
V( 0)=0 0 0 0
V( )= A
Hence,
=z z
A B
z B or B
Vz t V V At or
t
VV z
t
VdVE V a a
dz t
38
Example 6.8 Continued
39
Example 6.8 Continued
0
/2 2 22
0 0 0
0
0
2 2
J ,
( )J .
2 2 4
4Thus R'=
Alternatively, for this case, the cross section of the bar is uniform
between the horizo
z z
bb
a a
VE a dS d d a
t
V V V b aI dS d d
t t t
V t
I b a
2 22 2
ntal surfaces at z=0 and z=t, then
4R'=
( )4
l t t
S b ab a
A coaxial cable contains an insulating material of conductivity σ . If the radius of the central wire is a and that of the sheath is b, show that the conductance of the cable per unit length is:
40
Example 6.9
2G=
lnb
a
Transmission Line Equivalent circuit: (See Chapter 11)
41
Example 6.9
2G=
lnb
a
G=Conductance per unit length for dielectric.
G=1/(Leakage Resistance of Dielectric)
0
0
2
0
Let V be the potential difference between the inner and outer conductors, so that
( ) 0 and ( )
1Solve 0. From example 6.8, we got:
ln
ln
V a V b V
d dVV
d d
VV
b a
a
0
0
2
0 0
0 0
0
and =
ln
,
ln
2
ln ln
ln1
The resistance per unit length is R= .2
1 2and the conductance per unit length is G=
ln
l
z
VE a
b
a
VJ E a dS d dza
b
a
V L VI J dS dzd
b b
a a
bV a
I L
R
b
a42
Example 6.9 Solution
Conducting spherical shells with radii a = 10 cm and b = 30 cm are maintained at a potential difference of 100 V such that V(r=b) = 0 and V(r=a) = 100 V. Determine V and E in the region between the shells. if εr =2.5 in the region, determine the total charge induced on the shells and the capacitance of the capacitor. 43
Example 6.10
2 2
2
2
2 2
2
10
since 0 in the region of interest, we multiply through by to obtain
0, Integrating once gives: or
Integrating again: (use bo
d dVV r
r dr dr
r r
d dV dV dV Ar r A
dr dr dr dr r
AV B
r
00 0
0
undary conditions to find A and B)
1 1when , 0 0= or B= , Hence
1 1when , , or A=
1 1
1 1
Thus =1 1
A Ar b V B V A
b b b r
Vr a V V V A
b a
b a
r bV V
a b
44
Example 6.10 Solution
0
22
2
20 0 0 0
20 0
2
E1 1
To determine capacitance, first find Q on either shell
4E sin
1 11 1
, Q= For inner shell, Q= [4 ], bu
r r r
r r
S S
VdV AV a a a
dr rr
a b
V VQ dS r d d
ra ba b
Or dS a
20 0 0 0 0 0
2 2
0
t
4 Q= .4
1 1 1 1 1 1
The capacitance is easily determined as
4C=
1 1
S n r a
r r rn nr a r a
D
V V VD E a
a aa b a b a b
Q
V
a b
45
Example 6.10 Continued
0
0
1 1
A=1 1
1 1
=1 1
V Ab r
V
b a
r bV V
a b
0
2 2
9
Substituting =0.1 m, =0.3 m, =100 V yields,
1 10
1 103=100 15
10 10 / 3 3
100 15 V/m
10 10 / 3
104 (2.5)(100)
36Q= 4.167 nC
10 10 / 3
positive charge is induced on the inner s
r r
a b V
rV V
r
E a ar r
9
0
hell, negative charge
is induced on the outer shell.
4.167 10C= 41.67 pF
100
Q
V
46
Example 6.10 Continued 0
1 1
=1 1
r bV V
a b
In Section 6.5, it was mentioned that the capacitance C=Q/V of a capacitor can be found by either assuming Q and finding V or by assuming V and finding Q. The former approach was used in Section 6.5 while we have used the latter method in the last example. Using the latter method, derive capacitance for parallel plate capacitor:
0
0
Assuming the potentail difference between the paralle plates is
so that (x=0)=0 and (x=d)= .
V
V V V 47
Example 6.11
Q SC
V d
22
2
00 0 0
Solve Laplace's equation: 0
Integreating twice gives
At 0, 0 0 0 or 0
At , 0 or / . Hence =
To find the capacitance, first find the charge o
d VV
dx
V Ax b
x V B B
Vx d V V V Ad A V d V x
d
0n
0 0
0 0
n either plate
But D = E a , where E
On the lower plate , So and
On the upper plate , So and
(Q is equal
S
S n x x x
n x S
n x S
Q dS
dV Va V a Aa a
dx d
V V Sa a Q
d d
V V Sa a Q
d d
0
but opposite on each plate). C=Q S
V d
48
Example 6.11 Solution
Determine the capacitance of each of the capacitors in the Figure. Take εr1 =4, εr2=6, d=5 mm, S=30 Cm2.
1 2
0 1 0 1 0 21 2
(a) Since D and E are normal to the dielectric interface, the capacitor in
Fig. (a) can be treated as two capacitors C and C in series.
2 2= , =
/ 2
r r rS S SC C
d d d
49
Example 6.12
Example 6.12 Solution
50
9 4
1 2 0 1 2
3
1 2 1 2
The total capacitor is given by
2 ( ) 10 30 10 4 6= 2. . . =25.46 pF
36 5 10 10
(b)
D and E are parallel to the dielectric interface, the capacitor in
Fig. (b) can be treated a
r r
r r
C C SC C
C C d
1 2
1 2
0 1 0 1 0 21 2
9 4
01 2 1 2 3
s two capacitors and in parallel
(the same voltage across and )
/ 2= , C =
2 2
The total capacitance is
10 30 10= ( ) . .10 26.53 pF
2 36 (5 10 )
r r r
r r
C C
C C
S S SC
d d d
SC C C
d
Example 6.12 Solution Continued
0 0
0 0
E S= 2 . Hence E=2
2 2 2 1010
10ln(10 ) ln
2 2 10
Thus the capacitance per meter is (L=1m)
a a a
o rb b b
a
b
QQ d E L a
L
Q Q d Q dV d
L L L
Q Q bV
L L a
9
02 10 1C= 2 . . 434.6 pF/m
10 12.536ln ln
10 11.0
Q
bV
a
51
Example 6.13 A cylindrical capacitor has radii a = 1 cm and b = 2.5 cm. If the space between the plates is filled with an inhomogeneous dielectric with εr=(10+ρ)/ρ, where ρ is in centimeters, find the capacitance per meter of the capacitor.