Ch2 potential theory examples
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Chapter 2 of an electrostatics text, titled Electrostatics II: Potential Boundary Value Problems, with worked examples. It covers Dirichlet, Neumann and mixed problems, Green's theorem, and the Green's function method. It derives the plane Green's function by images and applies it to a grounded plate with a circular region at potential V. The author is not identified in the excerpt, and the file appears to include oblate spheroidal coordinates later on.
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Chapter 2
Electrostatics II. Potential Boundary
Value Problems
2.1 Introduction
In Chapter 1, a general formulation was developed to
nd the scalar potential (r)and consequent
electric
eld E= rfor a given static charge distribution (r):In a system involving conductor
electrodes, often the potential is speci
ed on electrode surfaces and one is asked to
nd the po-
tential in the space o¤ the electrodes. Such problems are called potential boundary value problems.
In this case, the surface charge distribution on the electrodes is unknown and can only be found
after the potential and electric
eld have been found in the vicinity of the electrode surfaces from
s="0En;(C/m2)
where
En= @
@n;
is the electric
eld component normal to the conducting electrode surface with nthe normal coor-
dinate.
If the potential is speci
ed on a closed surface, the potential o¤ the surface is uniquely deter-
mined in terms of the surface potential. This is known as Dirichlet s boundary value problem and
most problems we will consider belong to this category. Solving Dirichlet s problems is greatly
facilitated by
nding a suitable Green s function for a given boundary shape. However, except
for simple geometries ( e.g., plane, sphere, cylinder, etc.),
nding Green s functions analytically is
not an easy task. For complicated electrode shapes, potential problems often have to be solved
numerically.
1
Specifying the normal derivative @=@non a closed surface also uniquely determines the poten-
tial elsewhere. This category of boundary value problems is called Neumann problem. Physically,
specifying the normal derivative of the potential on a closed surface corresponds to specifying the
surface charge distribution on the surface through
= "0@
@n:
Then, the problem is reduced to
nding the potential due to a prescribed charge distribution as
worked out in Chapter 1.
Specifying both the potential itself and its normal derivative everywhere on a closed surface is
in general overdetermining. However, in some problems, the potential is known in one part of a
closed surface and its normal derivative in the remaining part. This constitutes the so-called mixed
boundary value problem.
By introducing suitable coordinates transformation, some potential problems can be reduced
to one dimensional, that is, the potential becomes a total function of a single coordinate vari-
able. This happens if the Laplace equation and potential are completely separable, (u1;u2;u3) =
F1(u1)F2(u2)F3(u3):There are some 30 known rectilinear coordinate systems developed in the past
for speci
c purposes. As one example, we will study the oblate spheroidal coordinates because of
its wide variety of applications in electrostatics and magnetostatics.
2.2 Dirichlet Problems and Green s Functions
If a charge is given to a conductor, the potential of the conductor becomes constant everywhere after
a short transient time as shown in Chapter 1. Electrostatic state is thus quickly established. Since
the volume charge density should vanish ina conductor, all of the charge given to a conductor
must reside entirely on the conductor surface in the form of singular surface charge density
(C/m2):The corresponding volume charge density involves a delta function
=(n ns);
wherenis the coordinate normal to the surface and nsindicates the location of the surface.
After static condition is established, the volume charge density and the electric
eld in a con-
ductor both vanish. The potential of a conductor thus becomes constant = c=const. If a
chargeqis given to an isolated conductor, the potential of the conductor relative to zero potential
at in
nity is uniquely determined and the proportional constant Cde
nes the self-capacitance of
the conductor,
C=q
c;(F): (2.1)
2
Let us consider a trivial case, a conducting sphere of radius acarrying a charge q:The potential
outside the sphere is given by
(r) =q
4"01
r; ra: (2.2)
The sphere potential is
s=q
4"0a; (2.3)
which determines the self-capacitance of the sphere,
C= 4"0a;(F). (2.4)
The outer potential (r)can be written in the form
(r) =a
rs; r>a; (2.5)
which indicates that the potential is uniquely determined if the sphere potential sis known. In
general, if the potential is speci
ed everywhere on a closed surface, the potential elsewhere o¤
the surface is uniquely determined in terms of the surface potential s(rs)where rsdenotes the
coordinates on the closed surface. This is known as Dirichlet s theorem and
nding a potential for
given boundary potential distribution on a closed surface is called Dirichlet s problem.
The same problem can also be solved in terms of the electric
eld on the sphere surface,
Er=1
4"0q
a2; (2.6)
which can be replaced with a surface charge,
="0Er=q
4a2;(C/m2): (2.7)
The potential due to the uniform surface charge is
(r) =1
4"0I
jr r0jdS
=
4"02a2Z
01p
r2+a2 2arcossind
=q
4"0r;(r>a ) (2.8)
whereis measured from the direction of r:(This is allowed because of symmetry. For r<a; the
integral yields
(r) =q
4"0a=const.;(r<a; interior)
3
which is also an expected result.) Since
Er= @
@r=@
@n; (2.9)
wherenis the normal coordinate on the surface directed away from the volume of interest , the
potential can be rewritten as
(r) =1
4I
S1
jr r0j@
@ndS0: (2.10)
As this simple example indicates, potential boundary value problems can be solved in terms of
either the surface potential sor its normal derivative, @=@n: The latter method may be regarded
as a boundary value problem for the electric
eld.
Let us revisit the potential due to a prescribed charge distribution,
(r) =1
4"0Z(r0)
jr r0jdV0: (2.11)
The potential can be understood as a convolution between the charge density distribution (r)and
the function
G(r;r0) =1
41
jr r0j; (2.12)
which is the particular solution to the singular Poisson s equation
r2G= (r r0); (2.13)
subject to the boundary condition that Gvanish at in
nity. The function Gis called Green s
function. Physically, the Green s function de
ned as a solution to the singular Poisson s equation
is nothing but the potential due to a point charge placed at r=r0:In potential boundary value
problems, the charge density (r)is unknown and one has to devise an alternative formulation
in terms of boundary potential s(r):It is noted that the Green s function in Eq. (2.12) is the
particular solution to the singular Poisson s equation and we still have freedom to add general
solutions satisfying Laplace equation,
G=Gp+Gg; (2.14)
whereGpis the particular solution and Ggis a collection of general solutions satisfying
r2Gg= 0: (2.15)
This freedom will play an important role in constructing a Green s function suitable for a given
boundary shape as we will see shortly. In doing so, we exploit the following theorem:
4
Theorem 1 Green s Theorem: For arbitrary scalar functions and ;the following identity holds,
Z
V
r2 r2
dV=I
S( r r )dS: (2.16)
Proof of this theorem goes as follows. Gausstheorem applied to the function r gives
Z
Vr ( r )dV=I
S( r )dS: (2.17)
The LHS may be expanded as
Z
Vr ( r )dV=Z
V
r r + r2
dV: (2.18)
Therefore, Z
V
r r + r2
dV=I
S( r )dS: (2.19)
Exchanging and ;Z
V
r r + r2
dV=I
S( r )dS: (2.20)
Subtracting Eq. (2.20) from Eq. (2.19) yields
Z
V
r2 r2
dV=I
S( r r )dS; (2.21)
which is the desired identity.
Figure 2-1: s(r0)is the potential speci
ed on a closed surafce S;nis the coordinate normal to the
surface directed away from the volume wherein the potential (r)is to be evaluated.
5
We now apply the formula to electrostatic potential problems. Let be the Green s function
=G;satisfying
r2G= (r r0); (2.22)
and = be the scalar potential satisfying Poisson s equation
r2 =
"0: (2.23)
Then, the terms in the LHS of Eq. (2.21) become
Z
Vr2
r0GdV0= Z
V
r0
(r r0)dV0
= (r);
provided the coordinates rresides in the volume Vwhere we wish to
nd the potential, and
Z
Gr2dV0= 1
"0Z
G(r0)dV0: (2.24)
The RHS of Eq. (2.21) reduces to
I
S
@G
@n G@
@n
dS; (2.25)
where is the potential on the closed surface and nis the coordinate normal to the surface directed
away from the volume of interest as indicated in Fig.2-1. Therefore, the solution for the potential
(r)is given by
(r) =1
"0Z
VG(r0)dV0 I
S
s@G
@n G@s
@n
dS: (2.26)
At this stage, the Green s function is still arbitrary except it should satisfy the singular Poisson s
equation in Eq. (2.22). The
rst term in the RHS allows evaluation of the potential for a given
charge distribution as we saw earlier. The surface integral involves the potential on the closed
surface sand its normal derivative, namely, the normal component of the electric
eld at the
surface.
In usual boundary value problems, the potential on a closed surface is speci
ed as a function
of the surface coordinates. In this case, it is convenient to choose the Green s function so that it
vanishes on the surface,
G= 0 onS:
6
Then the last term in Eq. (2.26) vanishes, and the solution for the potential becomes
(r) =1
4"0Z
V(r0)
jr r0jdV0 I
Ss@G
@ndS0; G = 0onS: (2.27)
In particular, if there are no charges in the region of concern = 0;the potential is uniquely
determined in terms of the surface potential alone,
(r) = I
Ss@G
@ndS0; = 0inV; G = 0onS: (2.28)
We have a freedom to make such a choice for the Green s function that it vanish on the closed
surfaceSthrough adding general solutions to the particular solution of the singular Poisson s
equation. Therefore, solving a potential boundary value problems for a given closed surface Sboils
down to
nding a Green s function satisfying
r2G= (r r0); G = 0onS: (2.29)
Once such an appropriate Green s function is found for a given surface shape S;the potential at
arbitrary point can be found from Eq. (2.28) for a speci
ed potential distribution s(r)on the
surface.
In the following, Green s functions for some simple surface shapes will be found. It is noted
that three dimensional Green s functions have dimensions of 1/length, two dimensional Green s
functions are dimensionless, and one dimensional Greens functions have dimensions of length.
2.3 Examples of Green s Functions
2.3.1 Plane
Suppose that the potential is speci
ed everywhere on an in
nite (x;y)plane, s(x;y):The plane
is closed at in
nity and the method of Green s function is applicable. The Green s function is to
be found as a solution to the equation
r2G= (x x0)(y y0)(z z0); (2.30)
with the boundary condition G= 0; z= 0:Mathematically, the Green s function is equivalent to
the potential due to a point charge placed near a grounded conducting plate that can be readily
7
worked out using the method of image as shown in Fig.2-2,
G(r;r0) =1
4
1p
(x x0)2+ (y y0)2+ (z z0)2 1p
(x x0)2+ (y y0)2+ (z+z0)2!
;(2.31)
where the second term in the RHS is the contribution from the image charge at the mirror point.
Note that the Green s function is reciprocal and remains unchanged against the coordinates inter-
change,
G(r;r0) =G(r0;r):
This is expected from the fact that the delta function in the original singular Poisson s equation is
even,
(r r0) =(r0 r):
In the upper region z>0;
@G
@n= @G
@z0
z0=0
= 1
2z
[(x x0)2+ (y y0)2+z2]3=2: (2.32)
Therefore, for a surface potential s(x0;y0)speci
ed as a function of (x0;y0);the potential in the
regionz>0is given by
(r) =z
2Z1
1dx0Z1
1dy0 s(x0;y0)
[(x x0)2+ (y y0)2+z2]3=2: (2.33)
Figure 2-2: Image charge qfor a large, grounded conducting plate. The potential due to qand
qvanishes at the plate.
8
Figure 2-3: A large conducting plate is grounded except for a circular region which is at a potential
V:
Let us apply this formula to the boundary condition on the (x;y)plane,
s() =(
V; <a
0; >a(2.34)
where=p
x2+y2is the radial distance on the plane as shown in Fig. 2-3. Physically, the
boundary condition describes a large conducting plate which is grounded except for a circular
region of radius awhose potential is maintained at V:The potential on the z-axis can be found
easily,
(z) =zV
2Za
020d0
(02+z2)3=2
=V
1 zp
z2+a2
; z> 0: (2.35)
The axial potential in the lower region z < 0can be found by observing the up-down symmetry
and for both regions,
(z) =V
1 jzjp
z2+a2
: (2.36)
Then, the potential at arbitrary point ( r;)is
(r;) =8
>>>><
>>>>:V
1 r
ajP1(cos)j+1
2r
a3
jP3(cos)j 3
8r
a5
jP5(cos)j+
; r<a
V
1
2a
r2
jP1(cos)j 3
8a
r4
jP3(cos)j+
; r>a(2.37)
9
Note that at r a;the potential is of dipole type,
(r a)_1
r2jcosj: (2.38)
This problem should not be confused with the potential due to an isolated charged conducting disk
which will be discussed later. The potential and electric
eld in the upper half region are identical
to those realized by an ideally thin circular capacitor whose top plate is at a potential Vand the
lower plate at V:The appearance of the dipole potential is thus an expected result. (For a thin
capacitor with plate separation distance , the electric
eld between the plates diverges but the
productE= 2Vremains constant. Such a structure is called a double layer.)
2.3.2 Sphere
Figure 2-4: The image of charge qwith respect to a grounded conducting sphere is q0= qa=r0
located ata2r0=r02where r0is the location of the charge q:
In
nding a Green s function for a given surface shape, the method of images is most conveniently
exploited. In the case of a sphere having a radius a;the Green s function can be found as a solution
for the potential due to a charge qplaced at a distance r0from the center of a grounded conducting
sphere. In the case of a sphere having radius a, an image charge
q0= a
r0q; (2.39)
10
placed at
r00=a2
r02r0; (2.40)
together with the charge q, makes the surface potential vanish. This is illustrated in Fig.2-4. The
potential due to charge qplaced near a grounded conducting sphere is thus equivalent to that due
to two charges, qand its image charge q0;and is given by
(r) =q
4"00
@1
jr r0j a
r0
jr r00j1
A
=q
4"00
@1p
r2+r02 2rr0cos
1q
(rr0=a)2+a2 2rr0cos
1
A; (2.41)
which readily yields the Green s function for a sphere,
G(r;r0) =1
40
@1p
r2+r02 2rr0cos
1q
(rr0=a)2+a2 2rr0cos
1
A: (2.42)
Here
is the angle between the two position vectors r= (r;; )andr0= (r0;0; 0). Its cosine
value is
cos
= coscos0+ sinsin0cos( 0): (2.43)
The Green s function indeed vanishes on the sphere surface r=aorr0=a:Again, the Green s
function is invariant against coordinates exchange, r$r0;that is, Green s functions are reciprocal.
For exterior (r>a )potential problems, the normal gradient @G=@n is
@G
@n= @G
@r0
r0=a
=1
4a r2
a
(r2+a2 2arcos
)3=2; r>a (2.44)
and for interior ( r<a )problems,
@G
@n= +@G
@r0
r0=a
=1
4r2
a a
(r2+a2 2arcos
)3=2; r<a: (2.45)
Note that the normal coordinate nis directed away from the volume of interest. If the surface
11
potential is speci
ed as a function of 0and 0;s(0; 0);and there are no charges, the exterior
potential at an arbitrary point r= (r;; )can be found from
(r) = I
s@G
@ndS
=1
4Ia(r2 a2)
(r2+a2 2arcos
)3=2s(0; 0)d
0
=a(r2 a2)
4Z
0sin0d0Z2
0d 0 s(0; 0)
(r2+a2 2arcos
)3=2: (2.46)
Recalling the expansion of the function 1=jr r0jin terms of the spherical harmonic functions
1
jr r0j=X
l;m4
2l+ 1r0l
rl+1Ylm(; )Y
lm(0; 0); r>r0(2.47)
the exterior potential can be decomposed into multipole potentials,
(r;; ) =X
l;ma
rl+1
Ylm(; )I
s(0; 0)Y
lm(0; 0)d
0; r>a: (2.48)
The interior potential can be found using the expansion
1
jr r0j=X
l;m4
2l+ 1rl
r0l+1Ylm(; )Y
lm(0; 0); r<r0; (2.49)
(r;; ) =X
l;mr
al
Ylm(; )I
s(0; 0)Y
lm(0; 0)d
0; r<a: (2.50)
Example 2 Charge near a Floating Conducting Sphere
To become familiar with the Green s function method, let us consider a somewhat trivial problem
of
nding the potential when a charge qis placed at a distance dfrom the center of a oating
conducting sphere of radius a:The charge qand its image q0= a
dqat(a=d)2dmake the sphere
potential 0 as we have just seen. However, since the oating sphere should carry no net charge, a
charge q0=a
dqmust be placed at the center of the sphere which raises the sphere potential to
s= q0
4"0a=q
4"0d; d>a:
Therefore, the exterior potential can be found by summing contributions from q;its imageq0and
the charge q0at the center,
(r) =1
4"0q
jr dj qa=d
jr (a=d)2dj+qa=d
r
:
12
In this expression, the function
1
41
jr dj a
d1
jr (a=d)2dj
;
is the Green s function which vanishes on the sphere surface, r=a:The last term is in the form
sa
r;
where
s=q
4"0d;(independent of a)
is the surface potential. Indeed,
I
s@G
@ndS =a(r2 a2)
4Z
0sin0d0Z2
0d 0 s(0; 0)
(r2+a2 2arcos
)3=2
= sa(r2 a2)
2Z
01
r2+a2 2arcos03=2sin0d0
= sa
r;
where0is measured from the direction of the vector d. (This is allowed because of the symmetry.)
Example 3 Speci
ed Potential on a Sphere Surface
Figure 2-5: s= +Vfor0<<= 2; Vfor=2<<:
Let us
nd the potential outside a spherical shell of radius awhose top half is maintained at
potential +Vand lower half at V;
s(0) =8
<
:+V; 00
2;
V;
20;(2.51)
13
as shown in Fig.2-5. Because of axial symmetry, only m= 0terms survive the integration over
the azimuthal angle 0:Also, because of up-down antisymmetry, only odd lterms survive the
integration over the polar angle 0:Noting
I
s(0; 0)Y
l0(0; 0)d
0
= 22Vr
2l+ 1
4Z1
0Pl()d; l = 1;3;5; ;
we readily
nd the exterior potential,
(r;) =V3
2a
r2
P1(cos) 7
8a
r4
P3(cos) +
; ra: (2.52)
The interior potential is
(r;) =V3
2r
aP1(cos) 7
8r
a3
P3(cos) +
; ra: (2.53)
The surface charge density on the sphere can be found from the normal component of the
electric
eld,
= "0@
@r
r=a+0
="0V
a
3P1(cos) 7
2P3(cos) +
:
The total surface charge on the upper hemisphere
q= 2a2Z=2
0() sind;
simply diverges (albeit only logarithmically) and it is not possible to de
ne the capacitance of the
hemispheres. This is because of the assumption of ideally small gap separating the two hemispheres.
If a small gap ais assumed, a
nite capacitance containing a factor ln(a=)emerges.
2.3.3 Interior of Cylinder of Finite Length
The Green s function for the interior of a cylinder of radius aand length lshown in Fig.2-6 can be
found as a solution for the following singular Poisson s equation
r2G=@2
@2+1
@
@+1
2@2
@ 2+@2
@z2
G= ( 0)
( 0)(z z0); (2.54)
14
Figure 2-6: Cylinder of a
nite length.
with the boundary condition
G= 0; =a; z = 0andl: (2.55)
Since the Green s function should be periodic with respect to and should also be invariant with
respect to exchange of and 0;the angular dependence can be assumed to be cos[m( 0)]where
mis an integer. Assuming the following separation of variables,
G(r;r0) =X
mRm()Zm(z) cosm( 0); (2.56)
we see that the radial function Rm()and the axial function Zm(z)satisfy, respectively,
d2
d2+1
d
d m2
2+k2
Rm() = 0; (2.57)
d2
dz2 k2
Zm(z) = 0; (2.58)
wherek2is a separation constant which can be either positive or negative.
Let us
rst consider the case k2>0:Solutions for Rm()which satis
es the boundary condition
15
Rm(=a) = 0 is them-th order Bessel function,
Rmn(;0) =Jmxmn
a
Jmxmn0
a
; (2.59)
wherexmnis then-th root ofJm(x) = 0:(The Bessel function of the second kind Nm(x)is discarded
because it diverges on the axis, = 0:)
Solutions for the axial function Zm(z)areekzorsinh(kz)and cosh(kz):The boundary con-
dition forZm(z)is it vanish at z= 0 andl:Therefore, we can construct the axial function as
follows,
Zm(z;z0) =8
>><
>>:sinh(kmnz) sinh[kmn(l z0)]; 0<z<z0<l;
sinh[kmn(l z)] sinh(kmnz0); 0<z0<z<l;(2.60)
wherekmn=xmn=a:A more fancy way to write Zm(z;z0)is
Zm(z;z0) = sinh[kmnmin(z;z0)] sinhfkmn[l max(z;z0)]g: (2.61)
The Green s function may thus be assumed in the form
G(r;r0) =X
m;nAmnRmn(;0)Zmn(z;z0) cos[m( 0)]: (2.62)
The expansion coe¢ cient Amncan be determined from the discontinuity in the derivative of the
axial function Zmn(z;z0)atz0;
d
dzZmn
z=z0+0= kmncosh[kmn(l z0)] sinh(kmnz0);
d
dzZmn
z=z0 0= +kmncosh(kmnz0) sinh[(kmn(l z0)]:
Then, a singularity appears in the second order derivative,
d2
dz2Zmn= kmnsinh(kmnl)(z z0); (2.63)
which is compatible with the delta function in the RHS of the original singular Poisson s equation
in Eq. (2.54). Eq. (2.54) now reduces to
X
mnAmnkmnsinh(kmnl)Jm(kmn)Jm(kmn0) cosm( 0) =( 0)
( 0): (2.64)
16
Multiplying both sides by 0Jm(kmn0) cosm 0and integrating over 0and 0;we
nd
A0n=1
a2kmn1
J2
m+1(kmna) sinh(kmnl); m = 0; (2.65)
Amn =2
a2kmn1
J2
m+1(kmna) sinh(kmnl); m1; (2.66)
where use has been made of the following integral,
Za
0J2
m(kmn)d=a2
2J2
m+1(kmna): (2.67)
The
nal form of the desired Green s function is
G(r;r0) =1
a1X
m=01X
n=1Jmxmn
a
Jmxmn0
a
Zmn(z;z0)
xmnJ2
m+1(xmn) sinh(kmnl)cos[m( 0)]"m; (2.68)
where
"m=(
1; m = 0
2; m1
If one does not like the appearance of "m;the summation over mcan be changed to from 1 to
1;
G(r;r0) =1
a1X
m= 11X
n=1Jmxmn
a
Jmxmn0
a
Zmn(z;z0)
xmnJ2
m+1(xmn) sinh(kmnl)cos[m( 0)]: (2.69)
If it is assumed that k2= 2<0;appropriate general solutions to
d2
d2+1
d
d m2
2 2
Rm() = 0; (2.70)
d2
dz2+2
Zm(z) = 0; (2.71)
are
Rm
;0
=
Km(ma)Im
m0
Im(ma)Km
m0
Im(km); <0<a; (2.72)
Rm
;0
= [Km(ma)Im(m) Im(ma)Km(m)]Im
km0
; 0<<a; (2.73)
withm=m=l and
Zm(z) = sin (mz) sin
mz0
; (2.74)
from which the Green s function can be constructed. Remaining calculation is left for exercise. The
17
reader should appreciate how a delta function ( 0)appears from the term
d2Rm
d2: (2.75)
One may wonder about Green s function for the exterior region of a cylinder of
nite length.
This problem appears to be a di¢ cult one and analytical expressions are not available to the
author s knowledge. It may be the case the problem can only be solved numerically.
2.3.4 Long Cylinder (3-Dimensional)
Three dimensional Green s function for a long cylinder satis
es
r2G=@2
@2+1
@
@+1
2@2
@ 2+@2
@z2
G= ( 0)
( 0)(z z0); (2.76)
which is to be solved for the boundary conditions
G(=a) = 0; G (z=1) = 0: (2.77)
Following the same procedure as in the preceding example, the interior solution for interior ;0<a
may be assumed as
G(r;r0) =X
m;nAmnJm(kmn)Jm(kmn0) cos[m( 0)] exp[ kmnjz z0j]; (2.78)
where
kmn=xmn
a; (2.79)
andxmnis then-th root of Jm(x) = 0:Since
d2
dz2e kmnjz z0j=k2
mne kmnjz z0j 2kmn(z z0); (2.80)
1X
m= 1cos[m( 0)] = 2( 0); (2.81)
we readily
nd the interior Green s function (; 0<a)
G(r;r0) =1
1X
m= 11X
n=1Jm(kmn)Jm(kmn0)
kmnJ2
m+1(kmna)cos[m( 0)]e kmnjz z0j: (2.82)
18
For exterior of a long cylinder, solutions to the equation
@2
@2+1
@
@+1
2@2
@ 2+@2
@z2
G= ( 0)
( 0)(z z0);
can be found in terms of Fourier transform with respect to the z-coordinate. Let G(r;r0)be
G(r;r0) =X
meim( 0)Z
Rm(;0;k)eik(z z0)dk: (2.83)
The radial function Rm(;0;k)satis
es
d2
d2+1
d
d m2
2 k2
Rm(;0;k) = ( 0)
2: (2.84)
Elementary solutions are the modi
ed Bessel functions Im(k)andKm(k)and we can construct
following solutions which remain bounded in the region a<< 1;
Rm(;0;k) =8
>><
>>:A(k)Im(k) +B(k)Km(k); a<<0<1;
C(k)Km(k); a<0<< 1;(2.85)
The boundary conditions are Rm(=a) = 0 andRm()be continuous at =0;
A(k)Im(ka) +B(k)Km(ka) = 0; (2.86)
A(k)Im(k0) +B(k)Km(k0) =C(k)Km(k0): (2.87)
Then,
Rm(;0;k) =8
>>>><
>>>>:A(k)
Im(k) Im(ka)
Km(ka)Km(k)
; a<<0<1;
A(k)1
Km(k0)
Im(k0) Im(ka)
Km(ka)Km(k0)
Km(k); a<0<< 1:
(2.88)
The unknown function A(k)can be found from the discontinuity in the derivative at =0;
d2
d2Rm(;0;k)
=0=kA(k)K0
m(k0)Im(k0) Km(k0)I0
m(k0)
Km(k0)( 0) (2.89)
= A(k)
Km(k0)( 0)
0; (2.90)
19
where again use has been made of the Wronskian of the modi
ed Bessel functions,
I0
m(x)Km(x) Im(x)K0
m(x) =1
x: (2.91)
We thus
nd
A(k) =Km(k0)
2; (2.92)
andRm(;0;k)reduces to
Rm(;0;k) =8
>>>><
>>>>:1
2Km(k0)
Im(k) Im(ka)
Km(ka)Km(k)
; a<<0<1;
1
2
Im(k0) Im(ka)
Km(ka)Km(k0)
Km(k); a<0<< 1:(2.93)
The exterior Green s function of a long cylinder is given by
G(r;r0) =1
2X
meim( 0)Z
Rm(;0;k)eik(z z0)dk: (2.94)
2.3.5 Long Cylinder (2-Dimensional)
Cross-section of a long cylinder. are the line charge and its image, respectively, that together
make the cylinder surface an equipotential surface,
(=a) ==(2"0) ln(a=0):
For boundary value problems in which z-dependence is suppressed, it is convenient to formulate
a two dimensional Green s function. Two dimensional Green s function for a long cylinder is to be
found from
r2G(r;r0) = 2(r r0); (2.95)
20
where2(r r0)is the two-dimensional delta function. In the cylindrical geometry, it is given by
2(r r0) =( 0)
( 0); (2.96)
and the Green s function satis
es
@2
@2+1
@
@+1
2@2
@ 2
G= ( 0)
( 0): (2.97)
In this case, the method of image can be exploited very conveniently. Let us consider a long line
charge(C/m) placed at (0; 0)parallel to a long, grounded conducting cylinder of radius a:A
negative line charge placed at (00; 0)where
00=a2
0; (2.98)
makes the cylinder surface an equipotential surface at a potential
s=
2"0lna
0
: (2.99)
Since we are seeking a potential that vanishes on the cylinder surface =a;the constant potential
scan be subtracted from the potential due to two line charges and ;
(r;r0) =
2"0
ln r r0 ln r r00 + lna
0
; (2.100)
where r r0 =q
2+02 20cos( 0); (2.101)
r r00 =q
2+002 200cos( 0)
=s
2+a2
02
2a2
0cos( 0): (2.102)
The desired Green s function is
G(r;r0) = 1
2ln p
2+02 20cos( 0)p
(0=a)2+a2 20cos( 0)!
: (2.103)
21
For exterior Dirichlet problems, the normal derivative at the cylinder surface is
@G
@n= @G
@0
0=a+0=1
2a 2
a
2+a2 2acos( 0); >a; (2.104)
and for interior,
@G
@n=@G
@0
0=a 0=1
22
a a
2+a2 2acos( 0); <a: (2.105)
If the potential on a long cylindrical surface is speci
ed as a function of the angle ;s( );the
potential o¤ the surface can be calculated from
(; ) = aI
s( 0)@G
@nd 0: (2.106)
For the interior (<a );the potential is given by
(; ) =1
2Z2
0s
0a2 2
a2+2 2acos( 0)d 0
=1
2Z2
0s
0
1 + 21X
m=1
am
cos[m
0
]!
d 0; (2.107)
where use is made of the following expansion,
a2 2
a2+2 2acos( 0)= 1 + 21X
m=1
am
cos[m
0
]:
Example 4
As an example, let us consider a long conducting cylinder consisting of two equal troughs. The
upper half in the region 0< < is at a potential Vand the lower half < < 0is at a
potential Vas shown Fig.2-7. The exterior potential >a is given by
(; ) =V2 a2
2Z
01
2+a2 2acos( 0)d 0 Z0
1
2+a2 2acos( 0)d 0
:
(2.108)
22
Figure 2-7: =Vfor0< < ; = Vfor < < 0on the surface of a long cylinder.
(Example of 2-D Green s function.)
The
rst integral can be e¤ected by changing the variable from 0tothrough 0 =2 =;
Z=2
=21
2+a2+ 2asin( )d
=2
2 a22
664tan 10
BB@(2+a2) tan
2
+ 2a
2 a21
CCA3
775=2
=2
=2
2 a2
tan 1(2+a2)(cot tan ) + 2a
2 a2
+ tan 1(2+a2)(cot + tan ) + 2a
2 a2
=2
2 a2
tan 12asin
2 a2
2
; (2.109)
where use has been made of the identities,
tan
4x
2
= cotxtanx;
tan 1x+ tan 1y= tan 1x+y
1 xy
;
tan 1x=
2 tan 11
x
:
Similarly, the second integral yields
2
2 a2
tan 12asin
2 a2
+
2
; (2.110)
23
and the potential becomes
(; ) =2V
tan 12asin
2 a2
; >a: (2.111)
The interior potential is
(; ) =2V
tan 12asin
a2 2
; <a: (2.112)
2.3.6 Wedge
A wedge is formed by two large plates intersecting at an angle
as illustrated in Fig.2-8.
Figure 2-8: A wedge formed by two large conducting plates intersecting at an angle
:
The potential due to a point charge qat(0; 0;z0)with the boundary conditions = 0 at the
plates = 0and =
, and=1;jzj=1essentially gives the Green s function. We thus seek
a solution to the Poisson s equation
@2
@2+1
@
@+1
2@2
@ 2+@2
@z2
G= ( 0)
( 0)(z z0); (2.113)
subject to the those boundary conditions. As in the case of 3-dimensional Green s function for a
long cylinder, we Fourier transform the Green s function,
G(r; ;z ) =1
2Z1
1g(; ;k )eik(z z0)dk: (2.114)
24
The angular dependence of the Green s function can be assumed to be
sinm
sinm
0
; (2.115)
which indeed vanishes at = 0and =
:We thus assume
g(; ;k ) =X
mAmRm() sinm
sinm
0
; (2.116)
to obtain
X
mAm"
d2
d2+1
d
d 1
2m
2
k2#
Rm() sinm
sinm
0
= ( 0)
( 0):
(2.117)
The radial function can be composed of the modi
ed Bessel functions,
Rm() =8
>><
>>:Im=
(k)Km=
(k0); <0;
Im=
(k0)Km=
(k); 0<:(2.118)
The derivative of the radial function Rm()has discontinuity at =0;and the second order
derivative yields
d2Rm()
d2= 1
0( 0); (2.119)
where the Wronskian of the modi
ed Bessel functions,
I(x)K0
(x) I0
(x)K(x) = 1
x;
has been substituted. The expansion coe¢ cient Amis thus determined as
Am=1
; (2.120)
and the desired Green s function is
G(r;r0) =2
X
mZ1
0Rm(;0) cos[k(z z0)]dksin( ) sin( 0); (2.121)
where
=m
: (2.122)
We will encounter an application of wedge potential in the section of inversion method later in this
25
Chapter.
2.4 Other Useful Rectilinear Coordinates
The familiar three coordinate systems, cartesian, spherical, and cylindrical, are frequently used in
analyzing potential problems. However, there are some 30 known coordinate systems developed for
speci
c problems. For simple electrode shapes, potential problems can be rendered one dimensional
by a suitable choice of coordinates. However, in some coordinates, solutions to Laplace equations
are not always completely separable. We have encountered one such example in Chapter 1, the
toroidal coordinates, in analyzing the potential due to a ring charge. In this section, some coordinate
systems useful for potential problems will be introduced.
2.4.1 Oblate Spheroidal Coordinates ( ; ; )
Figure 2-9: Oblate spheroidal coordinates (;; ): !0degenerates to a thin disk of radius a:
=cons. describes the surface of a hyperboloid.
The oblate spherical coordinates ( ;; )are related to the cartesian coordinates through the
26
following transformation,8
>><
>>:x=acoshsincos
y=acoshsinsin
z=asinhcos(2.123)
A surface of constant is the surface of an oblate spheroid described by
x2+y2
(acosh)2+z2
(asinh)2= 1; (2.124)
as shown in Fig.2-9. In the limit of !0;the surface degenerates to a thin disk of radius a
with negligible thickness, and in the opposite limit 1;the surface approaches a sphere with
a radiusr=acosh'asinh:This coordinate system is convenient if electrode shapes are an
oblate sphere or disk. A surface of constant is a hyperboloid described by
x2+y2
(asin)2 z2
(asin)2= 1: (2.125)
The metric coe¢ cients are
h=s@x
@2
+@y
@2
+@z
@2
=aq
cosh2 sin2; (2.126)
h=s@x
@2
+@y
@2
+@z
@2
=h; (2.127)
h =s@x
@ 2
+@y
@ 2
+@z
@ 2
=acoshsin: (2.128)
The Laplace equation in the oblate spherical coordinates can thus be written down as
1
hhh @
@hh
h@
@
+@
@hh
h@
@
+@
@ hh
h @
@
= 0; (2.129)
which reduces to
1
cosh2 sin2@2
@2+ tanh@
@+@2
@2+ cot@
@
+1
cosh2sin2@2
@ 2= 0: (2.130)
Assuming a separated solution (;; ) =F1()F2()eim ;(m=integer), we obtain
d2
d2+ tanhd
d l(l+ 1) +m2
cosh2
F1() = 0; (2.131)
d2
d2+ cotd
d+l(l+ 1) m2
sin2
F2() = 0; (2.132)
27
wherel(l+ 1) is a separation constant. Eq. (2.132) is the standard form of the Legendre equation
and solutions for F2()are
F2() =Pm
l(cos); Qm
l(cos): (2.133)
Eq. (2.131) can be rewritten as
d2
d2+sinh
coshd
d l(l+ 1) +m2
1 + sinh2
F1() = 0; (2.134)
which is also the Legendre equation with a variable isinh:Therefore, solutions for F1()are
F1() =Pm
l(isinh); Qm
l(isinh); (2.135)
and general solution to Laplace equation can be constructed from these elementary solutions.
If a point charge qis placed at
0;0; 0
;the potential in terms of the oblate spheroidal
coordinates can be found as
(r) =1
4"0q
jr r0j
=q
"0a1X
l=0lX
m= l(l jmj)!
(l+jmj)!(
Pm
l(isinh)Qm
l(isinh0)
Pm
l(isinh0)Qm
l(isinh))
Ylm(; )Y
lm
0; 0
;(
<0
>0)
: (2.136)
Derivation of this expression is left for exercise. The Wronskian of the Legendre functions,
Pm
l(x)d
dxQm
l(x) Qm
l(x)d
dxPm
l(x) =1
x2 1(l+jmj)!
(l jmj)!; (2.137)
should be useful. Furthermore, the Green s function for an oblate spheroidal surface described by
=0can readily be worked out to be:
for0<<0;
G
r;r0
=1
a1X
l=0lX
m= l(l jmj)!
(l+jmj)!Pm
l(isinh)Qm
l
isinh0
Ylm(; )Y
lm
0; 0
1
a1X
l=0lX
m= l(l jmj)!
(l+jmj)!Pm
l(isinh0)
Qm
l(isinh0)Qm
l(isinh)Qm
l
isinh0
Ylm(; )Y
lm
0; 0
; (2.138)
28
and for0<0<;
G
r;r0
=1
a1X
l=0lX
m= l(l jmj)!
(l+jmj)!Pm
l
isinh0
Qm
l(isinh)Ylm(; )Y
lm
0; 0
1
a1X
l=0lX
m= l(l jmj)!
(l+jmj)!Pm
l(isinh0)
Qm
l(isinh0)Qm
l(isinh)Qm
l
isinh0
Ylm(; )Y
lm
0; 0
: (2.139)
Example 5 Charged Conducting Disk
Figure 2-10: A charged conducting disk of radius a. A disk is described by = 0;0:
A thin disk of radius ais described by = 0in the oblate spherical coordinates. If a constant
surface is an equipotential surface, the potential o¤ the surface is a function of only, that is,
the potential problem becomes one dimensional. This is the most advantageous merit of using a
coordinate system most suitable for particular potential problems. The relevant solution which
vanishes at =1is the lowest order Legendre function of the second kind,
() =AQ 0(isinh) +B; (2.140)
whereAandBare constants. Since
Q0(isinh) =ih
tan 1(sinh)
2i
= icot 1(sinh); (2.141)
and the boundary condition is
(= 0) =V(disk potential),
29
we readily
nd the potential at an arbitrary ;
() =2V
cot 1(sinh): (2.142)
Note that cot 1(0) ==2:The far
eld potential at 1orr acan be found from the
asymptotic form of the function cot 1x;
cot 1x'1
x 1
3x3+ ; x 1: (2.143)
The leading far
eld potential is monopole as expected,
( 1)'2V
1
sinh'2V
a
r: (2.144)
Comparing with the standard monopole potential
(r) =1
4"0q
r; (2.145)
we readily
nd the total charge carried by the disk,
q= 8"0aV;
and the self-capacitance of the disk,
C= 8"0a;(F). (2.146)
This expression was
rst found by Cavendish.
The surface charge distribution on the disk is quite nonuniform because like charges repel each
other. Charge is distributed in such a manner that the tangential electric
eld on the disk surface
vanishes. The surface charge density can be found from the normal component of the electric
eld,
="0En="0E; (2.147)
where
E= 1
h@
@
=0
=2V
a1
jcosj: (2.148)
Note that
d
dxcot 1x= 1
1 +x2: (2.149)
30
The surface charge density diverges at the edge of the disk where ==2:The charge residing on
the disk surface can be found from the following surface integral,
q="0Z
0dZ2
0d h h
=0
=2"0aV
Z
0sindZ2
0d
= 8"0aV:
This is consistent with the charge found earlier using the monopole potential.
If one uses a coordinate system other than the oblate spheroidal system, solutions will be much
more involved. Let us employ the cylindrical coordinates ( ; ;z ):Because of axial symmetry,
dependence can be suppressed and we seek a solution in the form of Laplace transform,
(;z) =Z1
0(;k)e kjzjdk: (2.150)
The Laplace equation without dependence
@2
@2+1
@
@+@2
@z2
(;z) = 0; (2.151)
becomes d2
d2+1
d
d+k2
(;k) = 0; (2.152)
which suggests that
(;k) =A(k)J0(k): (2.153)
The boundary conditions are:
(a; z =0) =V(constant):
The following integral has a peculiar property,
Z1
0sinax
xJ0(bx)dx=8
>><
>>:
2; ifa>b;
sin 1(a=b);ifa<b:(2.154)
Exploiting this property, we can construct the following solution for the potential,
(;z) =2V
Z1
0sinka
kaJ0(k)e kjzjdk: (2.155)
31
The potential in the disk plane ( z= 0) is
(; z = 0) =8
>>><
>>>:V; if<a;
2V
sin 1(a=);if>a:(2.156)
Example 6 Dipole Moment of a Conducting Disk in an External Electric Field
Figure 2-11: Conducting disk in an external electric
eld parallel to the disk surface.
If a thin conductor disk is placed perpendicular to an external
eld, the dipole moment is zero
because of negligible thickness of the disk even though charge separation does take place in such
a manner that disk surfaces are oppositely charged. The external electric
eld is little disturbed
by the disk in this case. The maximum disturbance occurs when the disk surface is parallel to the
eld.
We assume a uniform external electric
eld in the x direction and a thin conducting disk placed
in thex yplane with its axis in the z direction as shown in Fig.2-11. The potential associated
with the external uniform electric
eld is
0= E0x
= E0acoshsincos : (2.157)
The radial function coshis actually P1
1(isinh)and the presence of the disk should yield
a perturbation proportional to the Legendre function of the second kind Q1
1(isinh)since the
perturbed potential should have the same angular dependence as 0(;; )to satisfy the boundary
32
condition at the disk :Thus we assume
(;; ) = E0acoshsincos +AQ1
1(isinh) sincos ; (2.158)
whereQ1
1(isinh)is actually a real function,
Q1
1(isinh) = cosh
cot 1(sinh) sinh
cosh2
: (2.159)
The constant Acan be determined from the boundary condition that the disk potential be zero,
that is, (= 0) = 0:We thus
nd
A=2
aE0;
and the potential becomes
(;; ) = aE0
cosh 2
Q1
1(isinh)
sincos : (2.160)
Far away from the disk at r aor 1;the potential approaches
lim
1(;; )! aE0coshsincos +4E0a3
3sincos
r2; (2.161)
where the asymptotic form of Q1
1(isinh);
Q1
1(isinh)'2
31
sinh2=2
3a
r2
; (2.162)
has been substituted. Comparing the dipole term in Eq. (2.161) with the standard dipole potential
dipole =1
4"0pr
r3; (2.163)
we can readily identify the dipole moment induced by the disk,
p= 4"04a3
3E0k; (2.164)
where E0kis the component of the external electric
eld tangential to the disk surface. Note that the
dipole moment is proportional to a3. The moment is equally applicable for low frequency oscillating
electric
eld as long as the wavelength associated with the oscillating
led is much longer than the
disk radius, ka=2
a 1:A resultant scattering cross-section of a conducting disk (sphere too)
placed in a low frequency electromagnetic wave is proportional to a6.
Example 7 Leakage of Electric Field through a Small Hole in a Conducting Plate
33
Figure 2-12: The lower plate of a parallel plate capacitor has a small hole of radius a:The electric
eld leaks throught the hole.
Consider a parallel plate capacitor whose grounded, lower plate has a small circular hole of
radiusaas shown in Fig.2-12. We wish to
nd how the hole perturbs the potential. This problem
has important applications in analyzing leakage of microwaves through a small hole in waveguide
walls.
The unperturbed electric
eld E0between the plates is assumed downward with a corresponding
potential
0(z) =8
>><
>>:E0z; z> 0
0; z< 0(2.165)
wherez=asinhcos:We note
sinh= iP1(isinh): (2.166)
Therefore, the perturbed potential can be sought in term of the Legendre function of the second
kindQ1(isinh)which is equivalent to
Q1(isinh) = sinhcot 1(sinh) 1: (2.167)
We thus assume the following form for the potential in both regions,
(;) =8
>><
>>:aE0sinhcos+A
sinhcot 1(sinh) 1
cos;0<<
2
A
sinhcot 1(sinh) 1
cos;
2<<;
which ensures continuity of the potential at the hole ( = 0):Continuity of the normal component
34
of the electric
eld at the hole requires
@
@ =0
z=+0=@
@ =0
z= 0;
from which we readily
nd the constant A;
A= aE0
:
In the region below the lower plate ( z<0);the potential is
(;) =aE0
sinhcot 1(sinh) 1
cos;
2<<: (2.168)
Its asymptotic form is of dipole nature,
(r a)! E0a3
31
r2cos>0; (2.169)
(note that cos<0in the region below the plate )and the e¤ective dipole moment of the hole is
p=4"0a3
3E0; (2.170)
which is downward . The far-
eld potential in the upper region (z>0)is
'0+E0a3
31
r2cos; (2.171)
in which the dipole term is due to an e¤ective dipole moment upward . The potential at the center
of the hole is
(= 0;= 0or) =aE0
: (2.172)
The results of this example, together with those of Example 14 in Chapter 3 (leakage of magnetic
eld through a hole in a superconducting plate), will have important implications on di¤raction of
electromagnetic waves by an aperture in a conducting plate. Since the e¤ective dipoles are opposite
to each other in the two regions z>0andz<0;it follows in general that
Ez( z) = Ez(z);
that is, the electric
eld normal to the plate is an odd function of z:This means that the surface
charges="0nE(C/m2)induced on both sides of the plate at z= +0 andz= 0are identical,
where nis the unit normal vector at the plate surface. (Note that nchanges its sign from one side
35
to other.) The component tangential to the plate,
Et=nE;
is an even function of z;
Et( z) =Et(z):
Of course, on the surface of the conducting plate, Etvanishes but it does not in the hole. For
magnetic
elds resulting from a hole in an ideally conducting plate, we will see that the normal
component should vanish at the plate surface
Hz= 0;atz=0;
and o¤ the plate, it is even with respect to z;
Hz( z) =Hz(z);
while the tangential component Ht=nHis an odd function of z;
Ht( z) = Ht(z):
It follows that the surface currents
Js=nH;(A/m)
on both surfaces of the plate are identical.
2.5 Method of Inversion
The method of inversion is useful when an electrode has a spherical shape, either complete spheres
(e.g., two spheres touching) or incomplete sphere (e.g., spherical bowl, solid hemisphere, etc.). For
a given sphere of radius awhich we call inverting sphere ;the inverted position of a point at ris
de
ned by
ri=a2
r2r: (2.173)
(See Fig.2-13.) A sphere is inverted into another sphere. If the center of the inverting sphere is
chosen on the surface of a sphere to be inverted, the inverted surface becomes a plane as shown
in Fig.2-14. This is where the merit of method of inversion is found because potential problems of
planar electrodes are often simpler than those involving spheres.
36
Figure 2-13: Point Pat(r;; )is inverted with respect to the sphere of radius atoQat(a2=r;; );
i.e.,at the image position.
Figure 2-14: If an inverting sphere is centered on a surface of a sphere to be inverted, the sphere is
inverted to an in
nite plane.
Consider a charge qplaced at r0= (r0;0; 0):The potential at position r= (r;; )is
=q
4"01p
r2+r02 2rr0cos
; (2.174)
where
cos
= coscos0+ sinsin0cos( 0):
In the inverted space with respect to a sphere of radius a;a chargeq0will appear at
a
r02
r0; (2.175)
37
and the position ris inverted toa
r2
r: (2.176)
The potential at the inverted position is
i=q0
4"01r
a4
r2+a4
r02 2a4
rr0cos
=q0
4"0rr0
a21p
r2+r02 2rr0cos
: (2.177)
In general, if a function (r;; )satis
es the Laplace equation, the potential function
a
ra2
r;;
; (2.178)
also satis
es the Laplace equation.
It should be noted that an equipotential spherical surface is in general not inverted to an
equipotential sphere. However, a spherical surface at zero potential is inverted to a zero potential
spherical surface. Since the reference potential can be chosen arbitrarily without a¤ecting the
electric
eld, one can always choose the potential of an equipotential spherical surface at zero
potential. For example, the potential of a charged conducting sphere of radius ais
s=1
4"0q
a; (2.179)
relative to zero potential at in
nity. However, we can subtract sfrom the potential everywhere
and choose the sphere potential at zero and the potential at in
nity as
1= 1
4"0q
a:
The electric
eld remains unchanged through uniform shift of the potential. If an inverting sphere
is chosen in such a way that it has a radius 2acentered at the surface of the conducting sphere of
radiusa;the conducting sphere is inverted to an in
nite plane touching the both spheres as shown
in Fig. 2-15. Since the sphere potential is chosen at zero, the potential of the plane is also zero.
The potential at in
nity is inverted to
1
4"0q
a2a
r= 1
4"02q
r; (2.180)
whereris the radial distance from the center of the inverting sphere with radius 2a:This is a
38
potential due to a point charge 2q:Therefore, a charge
2q= 8"0sa; (2.181)
appears at the center of the inverting sphere.
Example 8 Capacitance of Touching Spheres
Figure 2-15: Touching spheres are inverted to parallel plates by a sphere of radius 2acentered at
the touching point. Images appear in the inverted space.
Using the method of inversion, we can
nd the capacitance of two conducting sphere touching
each other as shown in Fig.2-15. The potential of the touching spheres is denoted by s:If the
inverting sphere has radius 2aand its center at the touching point, the two spheres become two
parallel planes separated by a distance 4a:A charge
q= 8"0as; (2.182)
appears at the midpoint between the plates after inversion which can be analyzed easily using the
method of multiple images. The following image charges appear: qatjzj= 4a; qatjzj= 8a; q
atjzj= 12a;:The amount of total charge on the surface of the original spheres can be found by
re-inverting the image charges,
Q= 2q2a
4a 2a
8a+2a
12a
=qln 2
= 8"0asln 2: (2.183)
39
Therefore, the self-capacitance of the touching spheres is
C=Q
s= 8"0aln 2: (2.184)
Figure 2-16: Geometry in the inverted space.
The potential i(;z)in the inverted space shown in Fig.2-16can be found in the form of Fourier
transform,
i(;z) =Z1
0A(k) sinh[k(2a jzj)]J0(k)dk; (2.185)
whereA(k)is a weighting function to be determined. It is noted that the elementary solution to
the Laplace equation is
J0(k)ekz; (2.186)
and the assumed form of the potential certainly satis
es the Laplace equation as well. The weighting
functionA(k)can be determined by noting
d2
dz2sinh[k(2a jzj)] = 2kcosh(2ak)(z); (2.187)
and Z1
0kJ0(k)dk=1
(): (2.188)
The charge density of the point charge qat the origin is
c=q
2()(z): (2.189)
40
Then,A(k)can be determined from the Poisson s equation
r2i= c
"0; (2.190)
as
A(k) = q
4"01
cosh(2ak); (2.191)
and the potential in the inverted space is
i(;z) = q
4"01Z
0sinh[k(2a jzj)]
cosh(2ak)J0(k)dk: (2.192)
The potential in the original con
guration can be found by reinverting ithrough the transforma-
tion
z!2a
r2
z; !2a
r2
;
where
r2=2+z2;
is the distance from the center. The result is
(;z) = q
4"02a
r1Z
0sinh"
k
2a 2a
r2
jzj!#
cosh(2ak)J0"
k2a
r2
#
dk; (2.193)
withr2=2+z2:Recalling that we have subtracted s=q=4"0a(the sphere potential) from the
potential everywhere to make the sphere potential vanish, we
nally obtain
(;z) = s2
666641 (2a)2
rZ1
0sinh"
k
2a 2a
r2
jzj!#
cosh(2ak)J0"
k2a
r2
#
dk3
77775: (2.194)
Example 9 Capacitance of Spherical Bowl
As a second example, we consider a hollow spherical bowl of radius awith an angle 2subtended
at the center shown in Fig.2-18. As inverting sphere, one can choose a sphere having a radius 2asin
centered at the edge of the bowl. After inversion, the bowl becomes a semi-in
nite plane as shown
and a charge q= 8"0asinswill appear at the center of the inverting sphere. Potential
41
Figure 2-17: Geometry in the original space.
problems involving a semi-in
nite conducting plate can be analyzed as a limiting case of a wedge.
For a charge qplaced at (0; 0;z0)near a wedge intersecting at an angle , the potential is given
by
(; ;z ) =2q
"0 8
>>><
>>>:P
mR1
0I(k)K(k0) cos[k(z z0)]dksin( ) sin( 0); <0
P
mR1
0I(k0)K(k) cos[k(z z0)]dksin( ) sin( 0); >0(2.195)
where=m= : Noting
Z1
0I(k)K(k0) cos[k(z z0)]dk=1
2p20Z1
e
pcosh coshd; (2.196)
where
cosh=2+02+ (z z0)2
20; (2.197)
and the sum formula1X
m=1pmcos(mx) =1
21 p2
1 2pcosx+p2 1
; (2.198)
42
Figure 2-18: A bowl (radius a;center angle 2)is inverted to a semi-in
nite plane by a sphere of
radius 2asincentered at the edge of the bowl.
we see that the potential reduces to
(r) =q
4"01
p20Z1
sinh
1
cosh(= ) cos[( 0)= ] 1
cosh(= ) cos[( + 0)= ]
1pcosh coshd: (2.199)
For a plate = 2;and this becomes
(r) =q
42"01
Rcos 1
cos[( 0)=2]
cosh(=2)
1
R0cos 1
cos[( + 0)=2]
cosh(=2)
; (2.200)
where
R=q
2+02 20cos( 0);
R0=q
2+02 20cos( + 0):
The potential in the vicinity of the charge qcan be found by letting 0= r; = 0= ;=
r=2asin 1;
(r) =q
4"01
r q
4"01
4asin
1 +
sin
; (2.201)
43
whereris the distance from the charge. The correction due to the presence of the conducting plate
is therefore
= q
4"01
4asin
1 +
sin
; (2.202)
and in the physical space, the far
eld potential due to a charged conducting bowl is in the form
(r a) = q
4"01
2r
1 +
sin
=1
a
r(sin+) s: (2.203)
Comparing with the standard monopole potential
=Q
4"0r;
we
nally
nd the capacitance of the bowl,
C= 4"0a(+ sin): (2.204)
For a sphere, =;we recover C= 4"0a:For a disk of radius R; 'sin'R=a 1;and we
also recover
C= 8"0R:
The capacitance of a solid (or closed) hemisphere can be found in a similar manner and given
by
C= 8"0a
1 1p
3
: (2.205)
This is left for an exercise.
2.6 Numerical Methods
Analytic solutions in potential problems can only be found for a limited number of applications, and
in practice, it is often necessary to resort to numerical analysis. In this section, we will estimate the
capacitance of a square conductor plate of side a:Mathematically speaking, this problem constitutes
an integral equation for the potential ;which is constant at the conductor,
=1
4"0I(r0)
jr r0jdS0= 1
4I1
jr r0j@
@n0dS0=V=constant, (2.206)
where
="0En= "0@
@n;
44
is the unknown surface charge density. The capacitance can be found from
C=1
Z
dS: (2.207)
As a very rough estimate, we recall that the capacitance of a circular disk of radius ais given
by
C= 8"0a; (2.208)
and approximate the capacitance of the square plate by
C= 8"0re¤= 8"00:564a; (2.209)
wherere¤is the radius of a circular disk having the same area as the plate,
r2
e¤=a2; re¤= 0:564a:
The
nite element numerical method given below yields C'8"00:547a.
Figure 2-19: A square conducting plate of side ais divided into 25 sub-areas. Because of symmetry,
the number of unknown potentials is reduced to 6.
The capacitance is the ratio between the total charge Qand the plate potential V,C=Q=V:
We divide the plate into nnsub-areas of equal size each with side a=n. Each sub-plate is at an
equal potential Vbut charges on the sub-plates di¤er. To illustrate the procedure, we choose n= 5
(25 sub-areas) as shown in Fig. 2-19. Because of symmetry, there are 6 unknown charges to be
found. The potential on each sub-plate can be calculated by summing contributions from charges
on all sub-plates including the charge on itself. The self-potential of one unit can be estimated as
45
follows. Consider a square plate of side carrying a uniform surface charge density (C/m2):The
potential at the center of the plate can be found from
=
4"0Z=2
=2dxZ=2
=2dy1p
x2+y2
=
4"04Z=2
0h
lnp
x2+2+
lnxi
dx
=
4"04ln
1 +p
2
=q
4"04 ln
1 +p
2
=q
4"03:5255; (2.210)
whereq=2is the charge carried by the sub-plate. With this preparation, we can write down
the potential of sub-plate Aas follows:
4"0A=
3:5255 +2
4+1
4p
2
qA+
2 +2
3+2p
17+2
5
qB
+
1 +2p
20
qC+1p
2+2p
10+1
3p
2
qD+2p
5+2p
13
qE+1
2p
2qF
= 4:2023qA+ 3:5517qB+ 1:4472qC+ 1:5753qD+ 1:4491qE+ 0:3536qF: (2.211)
Other potentials BF, which are all equal, can be written down in a similar way and we obtain
6 simultaneous equations for qAqFwhich can be solved easily. A resultant total charge is
Q= 1:7434"0;
and the capacitance is
C'0:34864"0a
= 0:5478"0a: (2.212)
Accuracy will improve if a larger number of sub-areas are used.
The method can be applied to estimate the capacitance of a conducting cube as well. With 150
sub-areas (25 sub-areas on each side), the following capacitance emerges,
C'0:654"0a; (2.213)
whereais the side of the cube. An estimate based on a sphere having the same surface area gives
C= 4"0re¤= 0:694"0a; (2.214)
46
where
re¤=r
6
4a= 0:69a:
A well known
nite element method of solving the Laplace equation is based on the fact that the
potential at the center of a cube may be approximated by the average of 6 surrounding potentials
on each face of the cube,
0=1
66X
i=1i:
This follows from the Taylor expansion of the potential,
(x;y;z ) = (x;y;z )@
@x+1
2@2
@x2 ;
(x;y;z) = (x;y;z )@
@y+1
2@2
@y2 ;
(x;y;z) = (x;y;z )@
@z+1
2@2
@z2 ;
Adding these 6 equations, we
nd
(x;y;z ) + (x;y;z) + (x;y;z) = 6(x;y;z ) +r2 +O(4): (2.215)
Therefore, if satis
es the Laplace equation, r2 = 0;
center'1
66X
i=1i; (2.216)
valid to order 3:
For 2-dimensional problems in which z-dependence is suppressed, we have
center'1
44X
i=1i: (2.217)
The equation can be applied to each sub-unit having a volume 3(3-D) or area 2(2-D). Resultant
simultaneous equations can be solved numerically.
Example 10 Potential in a Long Cylinder
Consider a conducting cylinder having a cross-section as shown in Fig. ??. The periodic upper
electrode is at a potential Vand the at lower electrode is grounded. In order to apply the
nite
element method, we divide the cross section into sub-sections and allocate 10 nodes points as shown.
47
Applying Eq. (2.217) to the potentials i(i= 1 10);we obtain
41= 100 + 2+ 2 3;
42= 1+ 2 4;
43= 100 + 1+ 4+ 5;
44= 2+ 3+ 6;
45= 100 + 4+ 6+ 9;
46= 4+ 5+ 10;
47= 300 + 8;
48= 200 + 7+ 9;
49= 2 5+ 8+ 10;
410= 2 6+ 9:
Solutions are: 1= 63:7;2= 31:0;3= 61:8;4= 30:2;5= 53:5;6= 27:9;7= 97:1;
8= 88:2;9= 55:7;10= 27:9all in Volts. A larger number of node points will improve
accuracy.
Cross-section of long cylinder with periodic anode structure.
48
Problems
2.1 A ring charge of total charge qand radius bis coaxial with a long grounded conducting
cylinder of radius a(<b):Determine the potential everywhere.
2.2 A ring charge of total charge qand radiusbis coaxial with a long uncharged dielectric cylinder
of permittivity "and radius a:Determine the potential everywhere.
2.3 A charge qis placed at an axial distance bfrom a conducting disk of radius a:Determine
the potential everywhere. Consider two cases, (a) the disk is grounded, and (b) the disk is
oating.
2.4 Show that a charge qat a distance dfrom the center of a oating conducting spherical shell
of radiusaraises the sphere potential to
s=q
4"0d;ifd>a (qoutside the sphere) ;
or
s=q
4"0a;ifd<a (qinside):
2.5 A large grounded conducting plate has a hemispherical bob of radius a:A chargeqis placed
at an axial distance dfrom the center of the bob. Find the force on the charge.
2.6 Show that the capacitance per unit length of a parallel wire transmission line with a common
wire radius aand separation distance dis
C
l="0
ln
d+p
d2 4a2
2a!:
2.7 A coaxial cable having inner and outer radii aandbis bent to form a thin toroidal capacitor
with a major radius R( a; b):Find the capacitance.
2.8 Show that the mutual capacitance between conducting spheres of radii aandbseparated by
a large distance d a;bis approximately given by
Cab'4"0ab
d;
and that the capacitance of the sphere of radius ais a¤ected by the sphere of radius bas
Caa'4"0a
1 +ab
d2
:
49
2.9 Rigorous analysis of potential problems involving two conducting spheres can be made by
using the bispherical coordinates de
ned by
x=asincos
cosh cos;
y=asinsin
cosh cos;
z=asinh
cosh cos:
=constant surface is a sphere described by
x2+y2+ (z acotanh)2=a
sinh2
;
and=constant surface is
( acotan)2+z2=a
sin2
;
which is spindle-like shape.
(a) Finding the metric coe¢ cients h;h;andh ;show that the Laplace equation in the
bispherical coordinates is
@
@1
cosh cos@
@
+1
sin@
@sin
cosh cos@
@
+1
sin2(cosh cos)@2
@ 2= 0:
(b) Show that the general solution to the Laplace equation is in the form
(;; ) =p
cosh cosX
l;m(Alme(l+1
2)+Blme (l+1
2))Pm
l(cosh)eim :
As in the oblate spheroidal coordinates, in this coordinate system too, the Laplace
equation is not separable.
2.10 Using the inversion method, show that the capacitance of a solid or closed conducting hemi-
sphere of radius ais given by
C= 8"0a
1 1p
3
:
2.11 Find a 2D Green s function for the interior of long cylinder having a semicircular cross-section
of radiusa:
50
Hint: Assume
G(r;r0) =(P
mAm(=0)msin(m ) sin(m 0); <0<a;
P
m[Bm(=0)m+Cm(0=)m] sin(m ) sin(m 0); 0<<a;
where r= (; );r0= (0; 0):
2.12 A conducting disk of radius ais placed parallel to an external electric
eld E0:The dominant
perturbation to the potential is dipole as shown in Example 6. What is the leading higher
order correction?
2.13 Cylindrical capacitors have cross-sections as shown. Estimate graphically the capacitance per
unit length for each. For (a), analytic expression for the capacitance is
C
l=2"0
cosh 12
1+2
2 d2
212;
where1= 4a;2= 2a;andd=a:For (b), one has to resort to numerical analysis for an
exact value.
2.14 Find numerically the capacitance of a conducting cube of side a:What do you estimate for
the lower and upper bounds of the capacitance?
2.15 Derive Eq. ( ??), the expression for the potential due to a point charge in the oblate spheroidal
coordinates (;; ):The charge is at
0;0; 0
:
2.16 Derive Eqs. (2.138) and (2.139), the Green s function for an oblate spheroid described by
=0(const.)
2.17 The prolate spheroidal coordinates (;; )is convenient to solve potential problems involving
prolate spheroids (sphere elongated along the zaxis). The coordinate transformation is
51
de
ned by
x=asinhsincos ;
y=asinhsinsin ;
z=acoshcos:
In the limit of !0; =const. surface describes a thin rod having a length 2a;and in the
limit! 1;it approaches a sphere with radius acosh'asinh. Show that the metric
coe¢ cients are:
h=aq
sinh2+ cos2=h;
h =asinhcos:
Then, show that general solution of Laplace s equation r2 = 0 in the prolate spheroidal
coordinates is in the form
(;; ) =X
l;m[AlmPm
l(cosh) +BlmQm
l(cosh)] [ClmPm
l(cos) +DlmQm
l(cos)]eim :
In the lowest order l= 0;possible one dimensional solutions are
() =Q0(cosh) = ln coth
2
;
() =Q0(cos) = ln cot
2
:
52