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Homework Help: Force between pt charge and polarized atom

  1. Nov 30, 2008 #1
    1. A point charge q is distance r from a neutral atom of polarizability [tex] \alpha [/tex]. What's the power of r of the force of attraction between them?

    2. Relevant equations

    Dipole moment induced by an external field is

    [tex] p = \alpha E_{ext} [/tex].

    The dipole moment (or its magnitude to be precise) is

    [tex] p = dq [/tex]

    where d is the distance between the charges +q and -q.

    Field from a dipole goes as

    [tex] E_{dip} \propto \frac{p}{r^3} [/tex].

    Field from a point charge goes as

    [tex] E_{ext} \propto \frac{q}{r^2} [/tex].

    3. The attempt at a solution

    Force of attraction is

    [tex] q E_{dip} \propto \frac{q p}{r^3} = \frac{q \alpha E_{ext}}{r^3} \propto q^2\alpha~\frac{1}{r^5} [/tex].

    Sanity check: force of attraction between a dipole [tex] p = dq [/tex] and a point charge q distance r away (on the axis of the dipole) using Coulomb's law.

    [tex] F = \frac{q^2}{4\pi\epsilon_0} \left ( \frac{1}{(r-d/2)^2} - \frac{1}{(r+d/2)^2} \right ) = \frac{q^2}{4\pi\epsilon_0}\frac{1}{r^2} \left ( \frac{1}{(1-d/2r)^2} - \frac{1}{(1+d/2r)^2} \right ) [/tex]

    For [tex] r >> d [/tex]

    [tex] F \approx \frac{q^2}{4\pi\epsilon_0}\frac{1}{r^2} \left ( 1 + 2\left(\frac{d}{2r}\right)^2 - 1 + 2\left(\frac{d}{2r}\right)^2 \right ) = \frac{1}{4\pi\epsilon_0}\frac{(dq)^2}{r^4} = \frac{1}{4\pi\epsilon_0}\frac{p^2}{r^4} [/tex].

    The dipole moment has r dependence

    [tex] p = \alpha E_{ext} \propto \frac{\alpha}{r^2} [/tex],

    which implies

    [tex] F \propto \frac{p^2}{r^4} \propto \frac{\alpha^2}{r^8} [/tex].

    Both calculations cannot be correct. So which one is right, and why is the other one wrong? Thanks!
    Last edited: Nov 30, 2008
  2. jcsd
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