February 27, 2000
Notes for Lecture #13
Dipole and quadrupole elds
The dipole moment is dened by
p=
with the corresponding potential
Z
( )r;
d3 r r
(1)
1 p ^r ;
(r) = 4"
r2
(2)
0
and electrostatic eld
(
1 3^r(p ^r) p
E(r) = 4"
r3
0
)
4 pÆ3(r)
3
(3)
:
The last term of the eld expression follows from the following derivation. We note that Eq.
(3) is poorly dened 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
Z
3
E(r 0) E(r)d r Æ3(r):
(4)
sphere
First we note that
Z
r R
E(r)d r = R
3
2
Z
r=R
(r)^rd
:
(5)
VdA:
(6)
This result follows from the Divergence theorm:
Z
vol
r Vd r =
3
Z
surface
In our case, this theorem can be used to prove Eq. (5) for each cartesian coordinate if we
choose V x^(r) for the x component for example:
Z
r(r) = x^
rR
Z
r (x^)d3r + y^
rR
Z
r (y^)d3r + ^z
rR
which is equal to
Z
Thus,
r=R
(r)R2d
((x^ ^r)x^ + (y^ ^r)y^ + (^z ^r)^z) =
Z
E
(r)d3 r =
rR
Z
r
(r)d3r = R2
rR
Z
r=R
Z
r=R
Z
rR
r (^z)d3r;
(r)R2d
^r:
(r)^rd
:
(7)
(8)
(9)
Now, we notice that the electrostatic potential can be determined from the charge density
(r) according to:
Z
1
(r) = 4
0
d3 r 0
(r0) = 1 X 4 Z
jr r0j 40 lm 2l + 1
d3 r 0 (r0 )
l
r<
l+1 Ylm
r>
(^r)Ylm(r^0):
(10)
We also note that the unit vector can be written in terms of spherical harmonic functions:
(
^ + sin() sin()y^ + cos()^z
q ) cos()x
^r = sin(
4 Y (^r) x^p+y^ + Y11 (^r) x^p y^ + Y10 (^r)^z
1 1
3
2
2
(11)
Therefore, when we evaluate the integral over solid angle in Eq. (5), only the l = 1 term
contributes and the eect of the integration reduced to the expression:
R2
Z
1 4R
(r)^rd
= 4
r =R
0 3
2
Z
d3r0(r0) rr<2 r^0:
>
(12)
The choice of r< and r> is a choice between the integration variable r0 and the sphere radius
R. If the sphere encloses the charge distribution (r0 ), then r< = r 0 and r> = R so that Eq.
(12) becomes
R2
Z
4
R2 1
1
(r)^rd
= 4 3 R2 d3r0(r0)r0r^0 3p" :
r=R
0
0
Z
(13)