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What is the total dipole moment

  1. Jun 17, 2012 #1
    Sometimes I am asked to compute the total dipole moment of a charge configuration. Normally you work with dipole moment per unit volume, so you can find the above by integrating over the entire volume, which is quite easy. I'm curius though, what is the physical interpretation of the total dipole moment of a charge configuration? What does this quantity tell us and is it used in any calculations? Like often you find the total dipole moment to be zero - what does that show us?
     
  2. jcsd
  3. Jun 17, 2012 #2

    gabbagabbahey

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    Total dipole moment (usually just called dipole moment) of any charge distribution [itex]\rho(\mathbf{r})[/itex] is defined as [itex]\mathbf{p}\equiv \int\mathbf{r}'\rho(\mathbf{r}')\text{d}^3 r'[/itex]. From its definition, it should be clear that it is position independent, and so an observer should measure the same value no matter where they are.

    The primary reasons for defining such a quantity are:

    (1) The dipole contribution to the electrostatic potential can be easily expressed in terms of [itex]\mathbf{p}[/itex] as [tex]V_{\text{dip}}(\textbf{r}) = \frac{1}{4\pi\epsilon_0}\frac{\mathbf{p} \cdot \mathbf{r}}{r^3}[/tex] and this contribution typically dominates at large distances from a distribution with little or no net charge (a very common scenario!)

    (2) The force, torque and electrostatic energy of as perfect dipole are easily expressed in terms of [itex]\mathbf{p}[/itex] and often make good approximations for calculating the dynamics of a neutral charge distribution, relatively far from it.
     
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