February 5, 2006
Notes for Lecture #11
Dipole fields
The dipole moment is defined by
p=
Z
d3 rρ(r)r,
(1)
1 p · r̂
,
4πε0 r2
(2)
with the corresponding potential
Φ(r) =
and electrostatic field
(
1
E(r) =
4πε0
)
3r̂(p · r̂) − p 4π
−
p δ 3 (r) .
3
r
3
(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
Z
E(r)d3 r δ 3 (r).
E(r ≈ 0) ≈
(4)
sphere
First we note that
Z
E(r)d3 r = −R2
Z
r≤R
Φ(r)r̂dΩ.
(5)
V·dA.
(6)
r=R
This result follows from the Divergence theorm:
Z
Z
3
∇ · Vd r =
vol
surface
In our case, this theorem can be used to prove Eq. (5) for each cartesian coordinate if we
choose V ≡ x̂Φ(r) for the x− component for example:
Z
3
∇Φ(r)d r = x̂
r≤R
Z
3
∇ · (x̂Φ)d r + ŷ
r≤R
Z
3
∇ · (ŷΦ)d r + ẑ
r≤R
Z
∇ · (ẑΦ)d3 r,
(7)
r≤R
which is equal to
Z
Φ(r)R2 dΩ ((x̂ · r̂)x̂ + (ŷ · r̂)ŷ + (ẑ · r̂)ẑ) =
r=R
Z
Φ(r)R2 dΩr̂.
(8)
r=R
Thus,
Z
r≤R
3
E(r)d r = −
Z
r≤R
3
∇Φ(r)d r = −R
2
Z
r=R
Φ(r)r̂dΩ.
(9)
Now, we notice that the electrostatic potential can be determined from the charge density
ρ(r) according to:
l
1 X 4π Z 3 0 0 r<
1 Z 3 0 ρ(r0 )
∗
0
dr
=
d
r
ρ(r
)
Φ(r) =
l+1 Ylm (r̂)Ylm (r̂ ).
4π0
|r − r0 |
4π0 lm 2l + 1
r>
(10)
We also note that the unit vector can be written in terms of spherical harmonic functions:
(
r̂ =
sin(θ) cos(φ)x̂ + sin(θ) sin(φ)ŷ + cos(θ)ẑ 4π
√
√
Y1−1 (r̂) x̂+iŷ
+ Y11 (r̂) −x̂+iŷ
+ Y10 (r̂)ẑ
3
2
2
q
(11)
Therefore, when we evaluate the integral over solid angle Ω in Eq. (5), only the l = 1 term
contributes and the effect of the integration reduced to the expression:
−R
2
1 4πR2 Z 3 0 0 r< 0
d r ρ(r ) 2 r̂ .
Φ(r)r̂dΩ = −
4π0 3
r>
r=R
Z
(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 ρ(r0 ), then r< = r0 and r> = R so that Eq.
(12) becomes
Z
−R2
r=R
Φ(r)r̂dΩ = −
p
1 4πR2 1 Z 3 0 0 0 0
d r ρ(r )r r̂ ≡ −
.
2
4π0 3 R
3ε0
(13)
If the charge distribution ρ(r0 ) lies outside of the sphere, then r> = r0 and r> = R so that
Eq. (12) becomes
−R
2
1 4πR2 Z 3 0 ρ(r0 ) 0 4πR3
R d r 0 2 r̂ ≡
E(0).
Φ(r)r̂dΩ = −
4π0 3
3
r
r=R
Z
(14)