Feb 14, 2002
Notes for Lecture
Notes for Lecture #12
Dipole and quadrupole fields
The dipole moment is defined by
with the corresponding potential
and electrostatic field
|
E(r) = |
1
4pe0
|
|
ì
ï
í
ï
î
|
r3
|
- |
4p
3
|
p d3(r) |
ü
ï
ý
ï
þ
|
. |
| (3) |
The last term of the field expression follows from the following
derivation. We note that Eq. (3) is poorly defined as
r® 0, and consider the value of a small integral of
E(r) about zero. (For this purpose, we are supposing that
the dipole p is located at r = 0.) In this case we
will approximate
|
E(r » 0) » |
æ
è
|
|
ó
õ
|
sphere
|
E(r) d3r |
ö
ø
|
d3(r). |
| (4) |
First we note that
|
|
ó
õ
|
r £ R
|
E(r) d3r = - R2 |
ó
õ
|
r = R
|
F(r) |
^
r
|
dW. |
| (5) |
This result follows from the Divergence theorm:
|
|
ó
õ
|
vol
|
Ñ·V d3r = |
ó
õ
|
surface
|
V · dA. |
| (6) |
In our case, this theorem can be used to prove Eq. (5) for each cartesian coordinate if
we choose V º [^(x)] F(r)
for the x- component for example:
|
|
ó
õ
|
r £ R
|
ÑF(r) d3r = |
^
x
|
|
ó
õ
|
r £ R
|
Ñ·( |
^
x
|
F) d3r + |
^
y
|
|
ó
õ
|
r £ R
|
Ñ·( |
^
y
|
F) d3r + |
^
z
|
|
ó
õ
|
r £ R
|
Ñ·( |
^
z
|
F) d3r, |
| (7) |
which is equal to
|
|
ó
õ
|
r = R
|
F(r) R2 dW |
æ
è
|
( |
^
x
|
· |
^
r
|
) |
^
x
|
+ ( |
^
y
|
· |
^
r
|
) |
^
y
|
+ ( |
^
z
|
· |
^
r
|
) |
^
z
|
|
ö
ø
|
= |
ó
õ
|
r = R
|
F(r) R2 dW |
^
r
|
. |
| (8) |
Thus,
|
|
ó
õ
|
r £ R
|
E(r) d3r = - |
ó
õ
|
r £ R
|
ÑF(r) d3r = - R2 |
ó
õ
|
r = R
|
F(r) |
^
r
|
dW. |
| (9) |
Now, we notice that the electrostatic potential can be determined
from the charge density r(r) according to:
|
F(r) = |
1
4pe0
|
|
ó
õ
|
d3r¢ |
r(r¢)
|r-r¢|
|
= |
1
4pe0
|
|
å
lm
|
|
4p
2l+1
|
|
ó
õ
|
d3r¢ r(r¢) |
rl <
rl+1 >
|
Y*lm( |
^
r
|
) Ylm( |
^
r¢
|
). |
| (10) |
We also note that the unit vector can be written in terms of
spherical harmonic functions:
|
|
^
r
|
= |
ì
ï
ï
ï
ï
í
ï
ï
ï
ï
î
|
|
|
sin(q) cos(f) |
^
x
|
+ sin(q) sin(f) |
^
y
|
+ cos(q) |
^
z
|
|
|
|
|
æ
ú
Ö
|
|
|
æ
ç
è
|
Y1-1( |
^
r
|
) |
[^(x)]+[^(y)]
Ö2
|
+ Y11( |
^
r
|
) |
[^(x)]-[^(y)]
Ö2
|
+ Y10( |
^
r
|
) |
^
z
|
ö
÷
ø
|
|
|
|
|
| (11) |
Therefore, when we evaluate the integral over solid angle W
in Eq. (5), only the l = 1 term contributes and the effect
of the integration reduced to the expression:
|
- R2 |
ó
õ
|
r = R
|
F(r) |
^
r
|
dW = - |
1
4pe0
|
|
4 pR2
3
|
|
ó
õ
|
d3r¢ r(r¢) |
r <
r2 >
|
|
^
r¢
|
. |
| (12) |
The choice of r < and r > is a choice between the integration
variable r¢ and the sphere radius R. If the sphere
encloses the charge distribution r(r¢),
then r < = r¢ and r > = R so that Eq. (12)
becomes
|
- R2 |
ó
õ
|
r = R
|
F(r) |
^
r
|
dW = - |
1
4pe0
|
|
4 pR2
3
|
|
1
R2
|
|
ó
õ
|
d3r¢ r(r¢) r¢ |
^
r¢
|
º - |
p
3 e0
|
. |
| (13) |
If the charge distribution r(r¢) lies outside of
the sphere, then r > = r¢ and r > = R so that Eq. (12)
becomes
|
- R2 |
ó
õ
|
r = R
|
F(r) |
^
r
|
dW = - |
1
4pe0
|
|
4 pR2
3
|
R |
ó
õ
|
d3r¢ |
r(r¢)
r¢2
|
|
^
r¢
|
º |
4 pR3
3
|
E(0), |
| (14) |
which is consistent with the mean value theorem for the electrostatic
potential.
File translated from TEX by TTH, version 2.20.
On 14 Feb 2002, 21:24.