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Two Electric Dipoles-Net Charge of Zero

  1. Jan 18, 2014 #1
    Hi there,
    Can someone help out with this problem?
    Thank you.:redface:
    1. The problem statement, all variables and given/known data
    Consider two electric dipoles in empty space. The net charge of each dipole is zero, so does an electric force exist between them?



    2. Relevant equations
    F=kq1q2/r2



    3. The attempt at a solution
    If I solve for F with Coulomb's equation it yields zero, but that just seems too simple.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 18, 2014 #2

    CAF123

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    Gold Member

    Can you show your work? It is correct to treat each dipole to be formed from two point charges of opposite polarity. It can be shown that, for particular orientations of the two dipoles, Newton's third law is not obeyed in the strong form.
     
  4. Jan 19, 2014 #3
    With regard to showing my work, I used Coulomb's Law and inserted a net charge of zero which yields: Force = zero.

    Would it be correct to state that the net force between electric dipoles is zero, but since they are polarized, (the positive and negative charges redistributing themselves to a certain degree, slightly more positive on one end, slightly more negative on the other.) the slightly separated internal charges create their own field resulting in the dipoles aligning themselves positive to negative.

    Cheers
     
  5. Jan 19, 2014 #4

    CAF123

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    Gold Member

    Place two dipoles on the x-y plane. Label the four charges by 1,2,3 and 4, where 1 and 2 constitute one dipole and 3 and 4 the other. Charge 1 experiences a force from 2,3 and 4. Charge 2 experiences a force from 1,3 and 4. Similarly for 3 and 4. Write explicit expressions for the force on each charge using Coulomb's law.

    The resulting analysis will show that, as you pointed out, the force on, say 1 from 2 is internal and does not contribute to the net force on the dipole.
     
  6. Jan 19, 2014 #5
    Thank you!
     
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