jackson3.7sol
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Problem 3.7
Classical Electrodynamics(third edition)
J. D. Jackson
Three point charges (q, -2q, q) are located in a straight line with separation a and with themiddle charge (-2q) at the origin of a grounded conducting spherical shell of radius b.
3.7a:
Write down the potantial of the three charges in the absence of the grounded sphere. Find thelimiting form of the potantial as a
2 = Q remains finite. Write your
answer in spherical coordinates.
The potential for an arbitrary point is:
( )0
1 2 1
4
q x
x x a x aπε + −
Φ = − + − −
G. Nootz
y
a
aq
q
-2q
x
z
b
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The same equation in spherical coordinates:
( )2 2 2 2
0
2 2 2 20
1 2 1, ,
4 2 cos 2 cos( )
1 2 1
4 2 cos 2 cos( )
qr
r r a ra r a ra
q
r r a ra r a ra
θ ϕ πε θ π θ
πε θ θ
Φ = − + + − + − −
= − + + − + +
This becomes a
( )2 2 2 2
0
0
1 2 1,0,
4 2 2
1 2 1
4
qr
r r a ra r a ra
q
a r r a r
ϕ πε
πε
Φ = − +
+ − + + = − + − +
For r > a we take r commen, since we wont to expand in a/r << 1
( )
( ) ( )
1 1
0
1
10 0
1
10 0
1 1 2,0, 1 1
4
1 11
1 11 1 1
l l
ll l
l l
l l
ll l
q a ar
r r r r r
a a a
r r r r r
a a ar r r r r
ϕ πε
− −
− ∞ ∞
+= =
− ∞ ∞
+= =
Φ = − + + −
− = =
+ = − = −
∑ ∑
∑ ∑
Comparing this to the solution of the Laplace equation in spherical coordinates:
( ) ( )( 1)
0
, , cosl
l l
l
r A r Pθ ϕ θ ∞
− +
=
Φ = ∑
We find:
( ) ( ) ( ) ( )
( )( ) ( ) ( )
( )
1 1
0 00
2
21 2 10 00 0
2
22 110
2, , cos 1 cos
4
2 2 11 1 cos cos
4 2
cos2
l ll
l ll l
l l
l ll
l ll ll l
l
lll
q a ar P P
r r r
q a q aP P
r r r r
q aP
r
θ ϕ θ θ
πε
θ θ πε πε
θ πε
∞ ∞
+ += =
∞ ∞
+ += =
∞
+=
Φ = + − −
= + − − = −
=
∑ ∑∑ ∑
∑
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In the limit of a qa2 = Q:
( ) ( )23
0
1, , cos
2
Qr P
r θ ϕ θ
πε Φ =
3.7b:
The presence of the grounded sphere alters the potential for r < b. The added potential can beviewed as caused by image charges. Use linear superposition to satisfy the boundary
conditions and find the potential everywhere inside the sphere for r < a and
r > a. Show that in the limit a
( ) ( )5
23 5
0
1, , 1 cos
2
Q r r P
r bθ ϕ θ
πε
Φ = −
( )*' '
' '0
1 2 2
4
q q q qq q x
x x a x a x a x abπε + + − −
+ + − −
− Φ = + + + + + − − − −
The 2q
b term is compensating for the potential of -2q at the surface of the grounded sphere.
a−
a+ -2q
x
z
b
y
q+
'q+
'a+
'a−
'q−
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In spherical coordinates this becomes:
( )
( ) ( )
'
2 2 2 '2 '0
'
2 2 2 '2 '
'
2 2 2 '2 '0
'
2 2 2 '2
1 2, ,
4 2 cos 2 cos
2
2 cos 2 cos
1 2
4 2 cos 2 cos
2
2 cos
q q qr
r r a ra r a ra
q qq
b r a ra r a ra
q q q
r r a ra r a ra
q qq
b r a ra r a
θ ϕ πε θ θ
θ π θ π
πε θ θ
θ
+ +
+ + + +
− −
− − − −
+ +
+ + + +
− −
− − −
−Φ = + ++ − + −
+ + + + − + + − +
−= + ++ − + −
+ + ++ + + + '2 cosra θ −
( )
( ) ( ) ( ) ( )
' '
2 2 2 '2 ' 2 2 2 '2 '0
' '
' '0
1 2 2, ,
4 2 2 2 2
1 2 2
4
q q q qq qr
r br a ra r a ra r a ra r a ra
q q q qq q
r br a r a r a r a
θ ϕ πε
πε
+ + − −
+ + + + − − − −
+ + − −
+ + − −
− Φ = + + + + + + − + − + + + +
= ± ± − + ± ± − − + +
Keeping only the solutions, which lead to the right answer and substituting what we know
from problem 2.2.
2' ' and
b ba q q
a a= = −
( )( ) ( )
( ) ( )
( ) ( )
2 20
2 2
0
2 20
1 2 2, ,
4
1 2 2 1
4
1 2 2 1
4
q qb q q q qbr
r a r b r ab ba r a r
a a
q b b
r a r b r ab ba r a r a a
q b b
r a r b r ab ba r a r
a a
θ ϕ πε
πε
πε
Φ = + + − + + − − + − + = + + − + + −
− + − + = + + − + + − − + − +
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( )( ) ( )
( ) ( )
02 2
1 11 1
2 2
0
1 1 2 2 1 1, ,
41 1
1 2 2 11 1
4
qr
ar ar r a r b r ab b
b b
q ar ar r a r a
b b r b b b
θ ϕ πε
πε
− −− −
Φ = + + − + + − − + − + = + − − − − + + + − +
For r > a we expand in terms of a/r and ar/b 2:
( ) ( ) ( )
( ) ( )
( )( ) ( )
2 20 0 0 00
1 1 2 1 2 1
0 0 0 00
100
1 1 2 2 1 1, , 1 1
4
2 21 1
42
1 1 1 14
l l l l l ll l
l l l ll l l l
l l l l l ll l
l l l l
l l l l
ll
ll
q a a r a a r r
r r b b r b r r b b
q a a a r a r
r r r b b bq a
r r
θ ϕ πε
πε
πε
∞ ∞ ∞ ∞
= = = =
∞ ∞ ∞ ∞
+ + + +
= = = =∞
+=
Φ = − − + + − − −
= + − − − + − −
= + − − − + −
∑ ∑ ∑ ∑
∑ ∑ ∑ ∑
∑ ( ) 2 10
2 2 2
2 1 4 11 10
2 2 2 22
2 1 4 1 2 1 4 11 10 0
2
2
4
2 2 1 1
4 2
l ll
ll
l l l
l ll l
l l l ll
l l l ll l
a r
b b
q a a r
r b
q a a r r a
r b r b
πε
πε πε
∞
+=
∞ ∞
+ += =
∞ ∞
+ + + += =
−
= −
= − = −
∑
∑ ∑
∑ ∑
Comparing this to the General solution of Laplace equation in spherical coordinates:
( ) ( )( 1)
0
, , cosl
l l
l
r A r Pθ ϕ θ ∞
− +
=
Φ = ∑
we find:
( ) ( )2
2
22 1 4 110
1, , cos
2
ll
ll ll
q r r a P
r bθ ϕ θ
πε
∞
+ +=
Φ = − ∑
With qa2 = Q as a
( ) ( )
( )
22
22 1 4 110
23 5
0
lim 1 1, , cos0 2
1 51 cos
2
ll
ll ll
r r qa Pa r b
Q r P
r b
θ ϕ θ πε
θ πε
∞
+ +=Φ = − →
= −
∑
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For r < a we expand in terms of r/a and ar/b 2:
( ) ( ) ( )
( )
1 11 1
2 2
0
1 1 1 1
2 2
0
0 00
1 2 2 1, , 1 1
4
1 1 2 2 1 11 1 1 14
1 1 2 11
4
l l ll
l ll l
q ar ar r r a r a
b b r b b b
q r ar r ar a a b b r b a a b b
q r r a r
a a a a r b
θ ϕ πε
πε
πε
− −− −
− − − −
∞ ∞
= =
Φ = ± − ± − ± + ± + ± +
= − − − − + + − − +
= + − − −∑ ∑ ( )2 20 0
2 2 2
2 1 4 10 00
22
2 1 4 110
1 21
1 1
2
1 1 1
2
l l ll
l ll l
l l l
l ll l
ll
l ll
a r
b b b b
q r a r
a r b b
q ar
a b r a
πε
πε
∞ ∞
= =
∞ ∞
+ += =
∞
+ +=
+ − − = − − − = − − +
∑ ∑
∑ ∑
∑
-1
0
1
-1
0
1
-2
-1
0
1
-1
0
1
-1
0
1