In Figure a, a particle of elementary charge +e is initially at coordinate z = 20 nm on the dipole...

Question

In Figure a, a particle of elementary charge +e is initially at coordinate z = 20 nm on the dipole axis (here a z axis) through an electric dipole, on the positive side of the dipole. (The origin of z is at the center of the dipole.) The particle is then moved along a circular path around the dipole center until it is at coordinate z = –20 nm, on the negative side of the dipole axis. Figure b gives the work Wa done by the force moving the particle versus the angle ? that locates the particle relative to the positive direction of the z axis. The scale of the vertical axis is set by Was = 4.0 A? 10–30 J. What is the magnitude of the dipolemoment?

David · Accepted Answer

The work done  by the force due to an electric dipole 
2 2
2
cos0 cos180
4 4
2
4
p p
W q V q
r r
pq
r
 

 
      
  
  

 (1) 
And from the table, we have  
304 10
20 0.2
W J
r cm m
  
 
 
9 2 21 9 10 .

In Figure a, a particle of elementary charge +e is initially at coordinate z = 20 nm on the dipole axis (here a z axis) through an electric dipole, on the positive side of the dipole. (The origin of z...

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