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1
Chapter 2 Electrostatics
2.1 Poisson’s Eq. and Green’s Theorem
(i) Maxwell eqs (cgs)
.
.
= . = ~ ﾞ
= . ﾞ
0
4
t c
B E
E
ﾝ
ﾝ
ﾎﾏ v
v
v
= .ﾞ
v
E V
.
Poisson’s eq. ﾞ = . 2
4 V ﾎﾏ
(2.1)
Since ﾞ
.
= . . 2 1 4 v v
v v
r r
r r
'
( ') ﾎﾂ ,
it is easy to verify that
V r r
r r
d r ( ) ( ' )
| ' |
' v
v
v v
v =
.
ﾏ 3
(2.2)
satisfies Poisson’s eq.
2
If the electrostatics problems involved charge with no
boundary surfaces, the above general solution would be the
most convenient and straightforward solution to any problem.
Actually, of course, many involve finite regions of space, with
or without charge insides, and with prescribed boundary
conditions on the bounding surfaces. These BCs may be
simulated by an appropriate distribution of charges outside the
region of interest (perhaps at infinity), but the above solution
becomes inconvenient, except in simple cases(e.g., method of
images).
(ii) To handle the BC, it is necessary to develop some new
mathematical tools, theorems due to George Green.
The divergence theorem for any vector field
v F
ﾞ . = .
v v v v Fd r F nd
V
3 ﾐ
ﾐ
Let
v F
= ﾞ ﾓ ﾕ, where
ﾓ and ﾕ
are arbitrary scale fields.
Now, ﾞ . = ﾞ . ﾞ = ﾞ + ﾞ .ﾞ
v F
( ) ﾓ ﾕ ﾓ ﾕ ﾓ ﾕ 2
and ﾓ ﾕ ﾓ
ﾝﾕ
ﾝ
ﾞ . = vn
n
ﾝ
ﾝ n
is the normal derivative at the surface S.
Green’s first identity
3
( )
V S
d r
n
d ﾞ +ﾞ .ﾞ = ﾓ ﾕ ﾓ ﾕ ﾓ
ﾝﾕ
ﾝ
ﾐ 2 3v
If we write down the above eq. again with ﾓ and ﾕ
interchanged ( ﾓ ﾕ . ), and then subtract it from the eq.
ﾞ .ﾞ ﾓ ﾕ terms
cancel, and we obtain Greens second identity or
Green’s theorem
. = ﾞ . ﾞ
S V
d
n n
r d ﾐ
ﾝ
ﾝﾓ
ﾕ
ﾝ
ﾝﾕ
ﾓ ﾓ ﾕ ﾕ ﾓ v 3 2 2 ) (
Poisson’s diff. eq. ｨ integral
eq. if we choose
| ' |
1 1
r r R v v .
= = ﾕ
v r -- observation
point, vr' -- integration variable. Put
ﾓ = V , ﾞ = . 2 4 V ﾎﾏ
. From
ﾞ
.
= . . 2 1 4 v v
v v
r r
r r
'
( ') ﾎﾂ ,
we have
[ ( ' ) ( ' ) ( ' )] ' [
'
( )
'
] ' . . + = . 4 4 11 3 ﾎ ﾂ
ﾎ
ﾏ
ﾝ ﾝ
ﾝ ﾝ
ﾐ V r r r
R
r d r V
n R R n
V d
V S
v v v v v
If vr ｸ vol,
4
V r r
r r
d r
R n
V V
n R
d
vol S
( ) ( ' )
| ' |
' [
' '
( )] ' v
v
v v
v =
.
+ .
ﾏ
ﾎ
ﾝ ﾝ
ﾝ ﾝ
ﾐ 3 1
4
1 1
(2.3)
If . rv vol, L.H.S. is zero)
Two remarks
(1) If S ｨ , and
v E
on S falls off faster than R.1, then
ｨ 0
S
, and the eq. reduces to the familiar result.
(2) for a charge free volume, V anywhere inside vol ( a sol. to
Laplaces’s eq. ﾞ = 2 0 V )
is expressed in the above eq. in terms
of potential and its normal derivative only on the surface. This
rather surprising result is not a solution to a boundary value
problem, but only an integral eq., since the specification of both
V and
ﾝ ﾝ V n
(Cauchy BC) is an over specification of the
problem.
2.2 Uniqueness of the Solution with Dirichlet or
Neumann Boundary Conditions
The question arises as to what are BCs appropriate for
Poisson’s (or Laplace’s) eq. in order that a unique and wellbehaved
(i.e., physically reasonable) solution exist inside the
bounded region. Physical experience leads us believe that
5
(1) specification of the potential on a closed S (e.g., a system of
conductors held at different potentials) define a unique potential
problem-Dirichlet BC.
(2) Similarly, specification of the
v E
(normal derivative of
the potential) everywhere on the S (corresponding to surface
charge density) also define a unique problem- -Neumann BC.
We now prove these expectations.
ﾞ = . 2 4 V ﾎﾏ
(2.4)
with either Dirichlet or Neumann.
We suppose, to the contrary, that there exist V1 and V2 ,
let U V V = . 2 1,
then ﾞ = 2 0 U inside Vol, U = 0 or
ﾝ ﾝ U n
= 0 on S
From Green’s first identity, with ﾓ ﾕ = =U
U U U Ud r U U
n
d
S vol
ﾞ +ﾞ .ﾞ =
2 3v ﾝ
ﾝ
ﾐ
ﾋ ﾞ = | | U
d r
vol
2 3 0 v
6
which implies ﾞ = U
0.
Consequently inside vol, U is constant.
U=0 ﾋ V V 1 2 = (DBC)
V is unique, apart from an unimportant arbitrary additive
const. (NBC)
From R. H. S. of the eq., there is also a unique sol. to a
problem with mixed BC (i.e., DBC over part of S, NBC over
the remaining part).
However, it should be clear that a sol. with both V and
ﾝ ﾝ V n
specified on a closed boundary does not exist, since there
are unique solutions for D & N condition separately and these
will in general not be consistent.
2.3 Formal Solution with Green’s
Function
The sol. of Poisson’s or Laplaces eq. in a finite vol. with
either DBC or NBC can be obtained by means of Green’s
theorem and so--called "Green’s function"
In obtaining result
V r r
r r
d r
R n
V V
n R
d
vol S
( ) ( ' )
| ' |
' [
' '
( )] ' v
v
v v
v =
.
+ .
ﾏ
ﾎ
ﾝ ﾝ
ﾝ ﾝ
ﾐ 3 1
4
1 1
7
--- not a solution -- we chose ﾕ=
.
1
v v r r'
, it being the potential
of a unit point charge,
ﾞ
.
= . . '
'
( ') 2 1 4 v v
v v
r r
r r ﾎﾂ
The function 1
v v r r . '
is only one of a class of functions
depending on the variable vr
and rr' , and called Green’s
function G r r (
, ' ) v v , which satisfy ﾞ =. . ' ( , ') ( ') 2 4 G r r r r v
v v v ﾎﾂ
In general, G r r
r r
F r r ( , ' )
'
( , ' ) v v
v v
v v =
.
+ 1
where F satisfies ﾞ = ' ( ,
') 2 0 F r r v v
In facing the problem of satisfying the prescribed BC on
V or
ﾝ ﾝ V n
, we can find the key by considering the integral eq.
As pointed out, this is not sol. satisfying the correct BC because
both V &
ﾝ ﾝ V n
appear. It is at best an integral relation for V.
With the generalized concept of a GF and its additional freedom
[via F r r ( , '
) v v , these arise the possibility that we can use
Green’s
theorem with ﾕ= G r r ( , ' ) v v and choose F r r ( , '
) v v to eliminate
one or the other of the two surface integral.
8
Thus we obtain a result which involves any DBC or NBC.
Of course, if the necessary G r r ( , ' ) v v depended in detail on the
exact form of the BC, the method would have little generality.
As will be seen, this is not required, and G r r ( , ' ) v v satisfies
rather simple BC on S.
With Green’s Theorem , ﾓ ﾕ = = V Grr , (,') v v ,
. = ﾞ . ﾞ
S V
d
n n
r d ﾐ
ﾝ
ﾝﾓ
ﾕ
ﾝ
ﾝﾕ
ﾓ ﾓ ﾕ ﾕ ﾓ ] [ ) ( 3 2 2 v
ﾞ = . 2 4 ﾓ ﾎﾏ, ﾞ
= . . 2 4 ﾕ
ﾎﾂ( ') v v r r
V r r G r r d r G r r V
n
V G r r
n
d
vol S
( ) ( ') ( , ') ' [ ( , ')
'
( , ')
'
] ' v v v v v v v
v v
= + . ﾏ
ﾎ
ﾝ ﾝ
ﾝ
ﾝ
ﾐ 3 1
4
(2.5)
The freedom available in the definition of G means that we can
make the surface integral depend only on the chosen type of
BCs.
For DBCs, we demand
G r r D(
, ' ) v v = 0 for
vr'
on S
V r r G r r d r V r
n
G d D
vol
D
S
( ) ( ' ) ( , ' ) ' [ ( ' )
'
' v v v v v v = . ﾏ
ﾎ
ﾝ ﾝ
ﾐ 3 1
4
(2.6)
9
For Neumann Bcs, we must be careful, the obvious choice of
BC on G r r ( ,
' ) v v seems to be
ﾝ
ﾝ
G r r
n
N ( , ' )
'
v v
= 0 for vr' on S
But an application of Divergence theorem to
ﾞ .ﾞ = ﾞ . ' '
( , ') ' ' ( , ') ' ' G r r d r G r r n d
vol S
v v v v v v 3 ﾐ
.
=
ﾝ
ﾝ
ﾐ G r r
n
d
S
( , ' )
'
'
v v
. . =. 4 4 3 ﾎﾂ ﾎ ( ') ' v v v r r d r
vol
Consequently, the simplest allowable BC on GN is
ﾝ
ﾝ
ﾎ G r r
n S
N ( , ' )
'
v v
= . 4 for vr' on S
where S is total area of B.S.. Then the sol. is
V r V r G r r d r V
n
G d S N
vol
N
S
( ) ( ') ( , ') ' [
'
' v v vv v =< > + + ﾏ
ﾎ
ﾝ
ﾝ
ﾐ 3 1
4
(2.7)
10
< > = V
S
V r d S
S
1 ( ' ) ' v ﾐ --- the average area S.
The customary Neumann problem is the so-called " exterior
problem" in which the volume is bounded by two surfaces, one
closed and finite, the other at infinite,
ｨ < > ｨ S VS ; 0.
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