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Electric dipole moment

  1. Nov 6, 2009 #1
    Hi

    I want to calculate the electric dipole moment of point charges along the z-axis with distances a and with the charge distribution

    [tex] \varrho (\vec{x}) = q \delta (\vec{x}) - 2q \delta (\vec{x} - \vec{a}) + q \delta (\vec{x} - 2 \vec{a}) [/tex]

    and of course [tex]\vec{a} = a \vec{e}_{z} [/tex]

    I did the following:

    [tex] \vec{p} = \int \vec{x}' \varrho (\vec{x'}) d^{3}x' [/tex]

    [tex]= q \int \vec{x}' \delta (\vec{x}) d^{3}x' - 2q \int \vec{x}' \delta (\vec{x} - \vec{a}) d^{3}x' + q \int \vec{x}' \delta (\vec{x} - 2 \vec{a}) d^{3}x' [/tex]

    Now I have some questions:

    1.) I guess I should write a prime in the arguments of the delta functions. Is this true? (The definition of my electric dipole moment is with prime, the given distribution without but that makes no sense? I should write a prime to all x vectors or no primes. correct?

    2.) How to evaluate the integrals further? I know that the delta function is only one at the points of the charges and everywhere else zero but what to do with the x-vectors?

    If this would be a normal integral I would do integration by parts, but this makes no sense here.

    In general I know the relation that

    [tex]\int f(x) \delta (x-a) dx = f(a)[/tex]

    but here I have no Function f because x is a vector and I am in 3-d space.

    I am confused. Please help me.

    edit:

    I wanted to post it in Classical Physics and not here. Wrong forum. Sorry... maybe a nice mentor will move it? :)
     
    Last edited: Nov 6, 2009
  2. jcsd
  3. Nov 6, 2009 #2

    clem

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    Science Advisor

    1. All integration variables are x', including in the delta.
    2. The f(x) here is just f(x)=x. You are integrating only along the x axis.
     
  4. Nov 9, 2009 #3
    thanks, I got it
     
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