DIPOLES II
We now consider the effect of fields on dipoles. The situation is as shown in the
following drawing.
In general, the field produces both a torque and a force.
Torque




       
   
   
  R  F   q E  r        q  E  r  
2

2  2 

2


 
 
  p
 
p
  E r     E r  
2
2 2
2



Now let   0 . Then
   
  pE r
Force



   
   
F  q E r    q E r  

2

2
Hence if


       
E r    E r  
2
2



Then F = 0. In other words, a dipole feels no net force in a uniform field. In a non-uniform field
there will be a force. To find it we consider a special case in which the dipole is parallel to the
field
Then
 

 


E  q E  r    q E  r    xˆ
2
2 




If E is due to another dipole,

p 2  p 2 xˆ
located at the origin then
 2kp 2
E
xˆ
R3
and

1
1
2kqp 2 
1
1



F  q 2kp 2 

xˆ 



3
3
3
3
r3  
 
 
  


  r   
1
r   
1   
 
  2r 
2

2 
 2r  
Now (1 + x)n =1 + nx + …..
 2kqp 2 
  
   2kqp 2  3 
6kp1p2
F 
 
1  3   1  3   
xˆ



2r  
2r  
r 3 
r3  r 
r4
Since τ ~1/r3 and F ~ 1/r4 it is normally torque that dominates.
Potential Energy of Dipole in Electric Field
Consider assembling the dipole from infinity.
PE = PEi + work I do to move charges from infinity to their final positions
First move both to the center of the dipole. Since the charges are equal and opposite, this takes
no work. Now move them at right angles to the field to a distance ℓ apart. This also takes no


work because F  qE is perpendicular to the distance moved. At this point we have
Now rotate by an angle  to
This requires work

 

w  qE sin    q  E   sin 
2
 2

 q E sin   q E cos 
 
 pE cos   p  E
 
 PE  p  E
Torque on One Dipole Due to Another
 
k
k




 
 
  p2  E  p 2  3  p1  rˆ  rˆ  p1   3  p1  rˆ  p 2  rˆ  p 2  p1 
r3
r3


Suppose p1  p1xˆ and p 2  p 2 yˆ . Then
2kp1p 2
 k
  3p1  p2zˆ   p1p 2zˆ   
zˆ
r3
r3
Hence dipole 2 will rotate counter clockwise, as seen from above page, until it lines up with
dipole 1.