Feb 12, 2001

Notes for Lecture Notes for Lecture #12

Dipole and quadrupole fields

The dipole moment is defined by

p = ó
õ
d3r r(r) r,
(1)
with the corresponding potential
F(r) = 1
4pe0
p · ^
r
 

r2
,
(2)
and electrostatic field
E(r) = 1
4pe0
ì
ï
í
ï
î
3 ^
r
 
(p · ^
r
 
) - p

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

4p
3
 
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)


File translated from TEX by TTH, version 2.20.
On 12 Feb 2001, 10:47.