Today in Physics 217: electric dipoles and their
interactions
‰ Origin of coordinates for
multipole moments
‰ Force, torque, and
potential energy for
dipoles in uniform electric
fields
Felectron
‰ Force on dipoles in
nonuniform electric fields
‰ Example: dipole vs. dipole
‰ Permanent and induced
dipoles in atoms and
molecules
23 October 2002
Physics 217, Fall 2002
Fproton
Induced dipole
moment
E
1
Choice of coordinate origin matters a lot in
multipole expansion
Changing the origin doesn’t change the physics, but it can
radically change the terms in the series. Consider a point
charge at the origin:
q
V=
Monopole only, of course
r
and one not at the origin, say at ( r ,θ , φ ) = ( a ,θ 0 , 0 ) :
2
3

q
a
a
a
 
 
V = 1 + cosθ 0 +   P2 ( cosθ 0 ) +   P3 ( cosθ 0 ) + …
r 
r
r
r

How did a point charge get all these nonzero multipole
moments? It didn’t, really; the situation still amounts to one
point charge, but the accounting is more complicated, when
1 r ≠ 1 r.
23 October 2002
Physics 217, Fall 2002
2
Choice of coordinate origin matters a lot in
multipole expansion (continued)
‰ So when someone gives you a charge distribution and
asks what all the moments are, they also have to tell you
what coordinate system to use.
‰ One useful exception: if the total charge is zero, then the
monopole moment is zero, and the dipole moment is
independent of the choice of coordinate origin.
Consider such a situation, and suppose the dipole
moment were p originally and pa in a coordinate system
with its origin displaced by some vector a:
pa = ra′ ρ ( r ′ ) dτ ′ =
∫
∫ ( r ′ − a ) ρ ( r ′) dτ ′ 0
= ∫ r ′ρ ( r ′ ) dτ ′ − a ∫ ρ ( r ′ ) dτ ′ = p − qa = p
23 October 2002
Physics 217, Fall 2002
.
3
Force and torque on dipole in uniform E field
If the dipole moment is constant, the net force is zero, because
the charges get pulled equally and oppositely.
There is a torque, though, that tends to align the dipole
moment vector with the applied field:
x
N = r+ × F+ + r− × F−
+q
d
 d
= × qE +  −  × ( − qE )
2
 2
= qd × E = p × E
= − pEyˆ , in this case.
23 October 2002
d
-qE
Physics 217, Fall 2002
qE
z
E
-q
4
Potential energy of dipole in uniform E field
(Griffiths problem 4.7)
Consider a dipole initially perpendicular to the field (0). The
field tends to pull it into alignment (toward 1); we have to
push to make it move toward 2. Thus the work we do is
θ
W=
∫
p × E dθ ′
x
π 2
θ
= pE
∫
2
sin θ ′dθ ′
π 2
= − pE cos θ = − p ⋅ E
0
θ
1
z
E
U= .
(Work we do = potential energy.)
23 October 2002
Physics 217, Fall 2002
5
Force on dipole in nonuniform E field
If E changes with position, the forces on each charge in a
simple dipole will no longer in general be equal in
magnitude, leading to a net force F on the dipole. Suppose
the dipole is very short; infinitesimal, in fact:
F = q ( E+ − E− ) = q∆E
x
+q
Definition of gradient:
∆Ez
= ( —Ez )x ⇒ ∆Ez = d ⋅ —Ez
d cos α
∆Ex = d ⋅ —Ex (if we had chosen to
∆Ey = d ⋅ —Ey
give E a y or an x
component)
-qE
α
d
qE
z
E
-q
∴ F = q (d ⋅ —) E = ( p ⋅ —) E
23 October 2002
Physics 217, Fall 2002
6
Torque on dipole in nonuniform E field
x
The torque on a small dipole,
about its own center, is still
θ
N = p× E
since the same arguments
apply as before. But since
there’s a net force F on the
dipole, there’s an extra torque
of r×F about any other point:
N = p× E + r × F
F
ƒ
N
F
ƒ
z
E
x
r
N
E
z
23 October 2002
Physics 217, Fall 2002
7
Dipole vs. dipole: force and torque
Griffiths problems 4.5 and 4.29: Two perfect (infinitesimal)
dipoles p1 and p2 are perpendicular and lie a distance r apart.
What is the torque on p1 (about its center) due to p2? What is
the torque on p2 (about its center) due to p1? What are the
forces on each, due to each? Why are the torques not equal
and opposite?
z
p2
p1
r
y
x
23 October 2002
Physics 217, Fall 2002
8
Dipole vs. dipole: force and torque (continued)
The field of each, at each other’s position:
2 p1 cosθ
p1 sin θ ˆ p1 ˆ
p1
p1
E1 ( 2 ) =
rˆ +
θ = 3 θ = − 3 zˆ = − 3 zˆ
3
3
r
r
r
y
a
2 p2 cosθ ′
2 p2
2 p2
2 p2
p2 sin θ ′ ˆ
ˆ
ˆ
ˆ
ˆ
′+
′=−
′=
E2 ( 1 ) =
r
θ
r
y
y
=
r ′3
r ′3
r ′3
y3
a3
The torque on each:
2p p
ˆy = − 1 2 xˆ
a3
a3
p1 p2
 p1 ˆ 
ˆ
N 2 = p2 × E1 ( 2 ) = p2 y ×  − 3 z  = − 3 xˆ
a
 a 
N 1 = p1 × E2 ( 1 ) = p1 zˆ ×
23 October 2002
2 p2
Physics 217, Fall 2002
9
Dipole vs. dipole: force and torque (continued)
Force on 2 due to field of 1:
∂
F2 = ( p2 ⋅ — ) E1 ( 2 ) = p2
E1 ( 2 )
∂y
y=a
∂  p1
= p2
− 3
∂y  y

3 p1 p2
=
zˆ 
zˆ .
4

a
 y=a
By Newton’s third law, the force on 2 by 1 is
F1 = ( p1 ⋅ — ) E2 ( 1 ) = −
23 October 2002
3 p1 p2
Physics 217, Fall 2002
a
4
zˆ .
10
Dipole vs. dipole: force and torque (continued)
The forces the dipoles exert on each other are equal and
opposite. Why aren’t the torques? Because we calculated the
torques about different centers. If we refer both torques to the
coordinate origin (i.e. the position of dipole 1), then
N 1 ( 0 ) = p1 × E2 ( 1 ) = −
2 p1 p2
a
3
xˆ (still)
N 2 ( 0 ) = p2 × E1 ( 2 ) + r × F2 = −
p1 p2
= − 3 xˆ +
a
3 p1 p2
a3
xˆ =
p1 p2
3
a
2 p1 p2
a3
xˆ + ayˆ ×
3 p1 p2
a
4
zˆ
xˆ = − N 1 ( 0 ) ,
as you always thought it should be.
23 October 2002
Physics 217, Fall 2002
11
Permanent and induced dipoles in atoms and
molecules
Some naturally-occurring charge distributions, such as many
simple molecules, have permanent, built-in dipole moments:
H
+
H
+
O
--
p
H
+
H
+
N
--
H
+
p
and the electrical properties of materials made with these
components are decisively influenced by the behaviour of
these dipoles in each others’ fields.
Some materials are of course made up of neutral, non-polar
components, but even these can have dipole moments induced
by external fields. Consider even the lowly hydrogen atom:
23 October 2002
Physics 217, Fall 2002
12
Permanent and induced dipoles in atoms and
molecules (continued)
Electron
Fproton
Proton
Felectron
a
Induced dipole
moment
E
x
23 October 2002
Physics 217, Fall 2002
13
Permanent and induced dipoles in atoms and
molecules (continued)
The electron and proton move in opposite directions until the
force on the proton by the displaced electron balances the
force on the proton by the external field. Suppose (crudely)
that the electron has uniform charge density and stays
spherical through this process. If the equilibrium position of
the proton is a distance d from the center of the electron, then
4 3
πd
qd
pinduced
1 3
ˆ
ˆ
ˆ
E− q ( q ) = − x 2 q
= −x 3 = −x
= − Eexternal , or
3
4
d
a
a
π a3
3
pinduced = α Eexternal , α = a3
(
)
= 4πε 0 a3 in MKS .
α, called the polarizability, is thus proportional to atomic
volume.
23 October 2002
Physics 217, Fall 2002
14
Atomic polarizabilities
23 October 2002
(
Element
α 10 −24 cm -3
H
Li
Na
K
0.66
12.0
27
34
Physics 217, Fall 2002
)
15