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Dipole moment question

  1. Jun 1, 2012 #1
    1. The problem statement, all variables and given/known data
    4 charges are placed each at a distance 'a' from origin. The dipole moment of the configuration is:
    28h27mf.jpg
    a)2qa[itex]\hat{j}[/itex]
    b)3qa[itex]\hat{j}[/itex]
    c)2qa[[itex]\hat{i}+\hat{j}[/itex]]
    d)none



    2. Relevant equations



    3. The attempt at a solution
    How do i determine dipole moment here when there is no dipole present?
    Dipole consists of two charges equal in magnitude and opposite in sign but i don't see such situation here.
     
  2. jcsd
  3. Jun 2, 2012 #2
    You can hypothetically assume that the charges are in dipole form. What I mean is, the -2q charge on negative x axis can be split(hypothetically) as -q and -q charges at the same point. This creates a dipole with the negative y axes +q charge. The 3q charge on y axis can be split as 2q and q, do you see the dipoles now?
     
  4. Jun 2, 2012 #3
    Thanks for the reply Infinitum! :smile:
    I do see the dipoles now, i will get back with a solution.
     
  5. Jun 2, 2012 #4

    ehild

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    Infinitum is right, you can split the charges so as they form dipoles. But there is a definition of the dipole momentum for a charge distribution. For a system of point charges it is P=∑qiri, where ri means the position vector of the i-th point charge. You can count the position vectors from the centre of the square. So the position vector of the -2q charge on the right is ai, that of the 3q charge is aj, and so on. The result is the same that you get with the splitting method.


    ehild
     
  6. Jun 2, 2012 #5
    Thank you both Infinitum and ehild! :smile:
    I get my answer to be 2qa[itex]\hat{j}[/itex]. ehild, thank you for the formula, that might come in handy because i have similar problems like this one.
     
  7. Jun 2, 2012 #6

    ehild

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    You will learn that formula sooner or later. :smile:

    ehild
     
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