NOTE 12 11/06/2007 3.2 The Method of Images
Suppose a point charge q is held a distance d above an infinite grounded conducting plane.
What is the potential in the region above the plane?
Boundary Conditions: (The first uniqueness theorem guarantees one solution only.) (p. 121-122) (1) V 0 at z 0 (since the conducting plane is grounded.)
(2) V 0 far from the charge (i.e. r(x2y2z2)1/2 >> d )
,( ), { 2 2 2 2/1 2 2 2 2/1 }
]) ( [ ]) ( 4 [
1
dz y x
q dz
y x
q
zy o
xV
(3.9)Above solution satisfies both boundary conditions.
3.2.2 Induced Surface Charge
(2.35) VaboveVbelow o nˆ Vbelow 0 (for conductor)
( )ˆ ( )ˆ z0
z V o above
o above
o V n V z above
0
]) ( [
) ( ])
( [
) ( 4
1 { }
),
( 2 2 2 2/3 2 2 2 2/3
z
dz y x
dz q dz
y x
dz o q
yx o
( , ) 21 [ 2 2 2]3/2 d y x y qd
x
(3.10)
2
0 0 0
) ( )
(
2 2 2 3/2 rdrd 2 2 1/2 q da
Q r d
qd d
r induced qd
total (3.11)
3.2.3 Force and Energy z
F d
q
o 22 ˆ
) 2 41 (
One could determine the energy by calculating the work required to
bring q in from infinity. The force required (to oppose the electric force) is z
d q
o 22 ˆ
) 2 41 (
,
d qz d qd
z q
o o
o z dz
W 4 4
1 4
4 1 )
2 4 (
1 2 2
2
2 ˆ ( )
3.2.4 Other Image Problems:
Example 3.2 A point charge q is situated a distance a from the center of a grounded conducting sphere of radius R. Find the potential outside the sphere.
HM#7 Problems: 3.7, 3.10, 3.12 (Due: 11/22)
期中考:11 月 15 日(四),10:10-12:00 am, LH108, 範圍: p.58-126
q
V=0 z
d
V (x,y,z)
-q d
q
q' Ra , b Ra2
( , ) [ ]
cos ) / ( 2 ) / (
) / ( cos
41 2
2 2 2 2 2
2
a R r a R r
q a R ra
a r
q
r
oV
0 ]
[
( / ) 2 cos) / ( cos
41 2
2 2 2
2
ra R R ra qa R ra
a r
q
o 2 2 2 2 2
) 4 (
1 )
( ' 4
1
R a
Ra q b
a qq
o
F o
(3.18)
3.3 Separation of Variations
Example 3.3: Boundary conditions:
(i) V 0 at y 0 (ii) V 0 at ya (iii) V Vo( y) at x 0 (iv) V 0 as
x
2 0
2 2
2
y V x
V , and V(x,y) X(x)Y(y) ( ) 2 2( ) ( ) 2 2( ) 0
dy y Y d dx
x X
d X x
y Y
2 0
2 2
2 ( )
) (1 ) ( ) (
1
dy y Y d y dx Y
x X d x
X 1( ) 2 2( ) k2
dx x X d x
X and Y(1y)d2dyY(2y) k2
0 ) 2 ( ) (
2
2 k X x
dx x X
d X(x) Aekx Bekx, Y(y)CsinkyDcosky
V(x,y)(Aekx Bekx)(CsinkyDcosky)
(iv) condition V(x,y)ekx(CsinkyDcosky) (i) condition V(x,y)Cekxsinky
(ii) condition sinka0 kan, i.e. k n/a (n=1,2,3,….)
( , ) / sin( / )
1C e n y a
y x
V n x a
n n
(iii) condition (0, ) sin( / ) ( )
1
y V a y n C
y
V o
n
n
a a
n Cn n y a m y a dy Vo y m y a dy
0 0
1 sin( / )sin( / ) ( )sin( / )
a n m
m a n
dy a y m a y n 0
) (
) ( 0/2 )
/ sin(
) /
sin(
Cn aaVo y n y a dy
0
2 ( )sin( / ) , if Vo(y) is a constant potential, then
( )
) ( 0 / 4 2
0
2 sin( / ) (1 cos ) even
n odd n V
a V a
n Vo n y a dy o n o
C
( , ) / sin( / )
...
5 , 3 , 1 4 1
a y n e
y x
V n x a
n n
Vo
(3.36)
Example 3.4: Boundary conditions (i) V 0 at y 0
(ii) V 0 at ya (iii) V Vo at x b (iv) V Vo at x b
2 0
2 2
2
y V x
V
V(x,y)(Aekx Bekx)(CsinkyDcosky) )
, ( ) ,
( x y V x y
V
A B
V(x,y)coshkx(CsinkyDcosky)
(i) and (ii) conditions D0 and k n/a
V(x,y)Ccosh(nx/a)sin(ny/a)
1 cosh( / )sin( / ) )
, (
n Cn n x a n y a
y x
V
(iii) condition o
n Cn n b a n y a V
y b
V
1 cosh( / )sin( / ) )
,
(
(( ))
0 /
) 4 /
cosh( even
n odd
n n b a Vo
C
( , ) cosh(cosh( // ))sin( / )
...
5 , 3 , 1 4 1
a y n y
x
V nn bx aa
n n
Vo
3.3.2 Spherical Coordinates
0 )
(sin )
(
22 2 2 2
2 sin
1 sin
2 1
2
1
V
r rr
Vr r
V r V (3.53) azimuthal symmetry: r(r2Vr)sin1 (sin V)0 (3.54)) ( ) ( ) ,
(r R r
V (3.55)
[See Example 3.5 and do it by yourself.]
(3.56) (3.65)
0
) (cos ) (
) ,
( 1
l
r l B l
lr P
A r
V ll (3.65)
1 )
0(cos
P , P1(cos)cos , 2
/ ) 1 cos 3 ( )
(cos 2
2
P , P3(cos)(5cos33cos)/2
Example 3.6: The potential Vo()is specified on the surface of a hollow sphere of radius R.
Find the potential inside the sphere.
0
) (cos )
, (
l
l l lr P A r
V , ( , ) (cos ) ( )
0
o
l
l l
lR P V
A R
V
1 2 2
0 '
{ 0 sin
) (cos ) (cos
l l
l P d
P
, ll''ll or ll' 2l21
AR P Pl d Vo Pl d
l
l l
l (cos ) (cos )sin ( ) '(cos )sin
' 0
0 0
1 ' 22 ' ' 0
1 22 '
0 0 (cos ) '(cos )sin
l l ll
l ll l l l
l l
l
lR P P d AR A R
A
d P
V
Al lRl o( ) l(cos )sin
2 0 1 2
Let Vo()ksin2(/2), sin2(/2) 12(1cos) 21[Po(cos)P1(cos)], then
d P
P P
Al lRl k [ o(cos ) (cos )] l(cos )sin
0 1
2 2 1
2
] sin ) (cos ) (cos sin
) (cos ) (cos
[ 0 0 1
2 2 1
2lRl k
Po Pl d
P Pl d
] sin ) (cos ) (cos sin
) (cos ) (cos
[ 0 0 1
2 2 1
2lRl k
Po Pl d
P Pl d
2 2 2 1 k2 k
Ao ,A1 23R k2(32)2kR, Al1 0
) 1
( cos ) ( )
(cos )
,
( 0 1 1 1 2 2 2 cos
0 k r R
R k o k
o l
l l
lr P A r P Ar P r
A r
V
(3.71)
Strongly suggested examples: Examples: 3.7, 3.8, 3.9