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Reproducing dirac-delta function in electric dipole field

  1. Nov 22, 2014 #1
    1. The problem statement, all variables and given/known data
    [problem 3.49 from Griffiths' Introduction to Electrodynamics 4th edition. The relevant equations from the book are reproduced in "relevant equations" below] In Ex. 3.9, we obtained the potential of a spherical shell with surface charge ##\sigma(\theta) = k\cos\theta##. In Prob. 3.30, you found that the field is pure dipole outside; it's uniform inside. Show that the limit ##R## → 0 reproduces the delta function term in Eq. 3.106.

    2. Relevant equations
    Dipole moment of this charge configuration:
    [tex] \vec{p} = \frac{4}{3}\pi kR^3\hat{z}[/tex]
    Potential inside shell:
    [tex] V_{in} = \frac{k}{3\epsilon_0}r\cos\theta[/tex]
    Potential outside shell:
    [tex] V_{out} = \frac{k}{3\epsilon_0}\frac{R^3}{r^2}\cos\theta[/tex]
    Eq. 3.106 from the book:
    [tex] \vec{E}_{dip}(\vec{r}) = \frac{1}{4\pi\epsilon_0}\frac{1}{r^3}[3(\vec{p}⋅\hat{r})\hat{r} - \vec{p}] - \frac{1}{3\epsilon_0}\vec{p}\delta^3(\vec{r})[/tex]
    3. The attempt at a solution
    First of all I'm not sure how am I supposed to get a delta function out of a limit.
    From the above potentials, I calculated the electric field inside and outside the shell (from the definition of the potential: ##\vec{E} = -\nabla V##).
    Electric field outside shell:
    [tex] \vec{E}_{out} = \frac{k}{3\epsilon_0}\frac{R^3}{r^3}(2\cos\theta\hat{r} + \sin\theta\hat{\theta})[/tex]
    It should be mentioned that the above expression is identical to the first term (the one without the delta function) in Eq. 3.106 cited above, the only difference being that in Eq. 3.106 it is written in coordinate-free form.
    Electric field inside shell:
    [tex] \vec{E}_{in} = -\frac{k}{3\epsilon_0}(\cos\theta\hat{r} - \sin\theta\hat{\theta})[/tex]
    I reasoned that in the limit ##R## → 0, the electric field everywhere but the origin will be zero (by taking the limit of ##\vec{E}_{out}## above) or at least an infinitesimal quantity depending on the infinitesimal ##R##. At the origin (or at an infinitesimal sphere of radius ##R##), the field will be uniform (in the sense that it doesn't depend on ##R##).
    Now I'm stuck - I can't figure out how to derive the delta function.
    Any help will be greatly appreciated - this problem really interests me now :p
     
  2. jcsd
  3. Nov 22, 2014 #2
    After some thinking, I realized that the field inside the shell is really uniform since ##\cos\theta\hat{r}-\sin\theta\hat{\theta} = \hat{z}##. Thus,
    [tex]\vec{E}_{inside}=-\frac{k}{3\epsilon_0}\hat{z}[/tex]
    We also know that ##\vec{p} = \frac{4}{3}\pi kR^3\hat{z}## and from this:
    [tex]\vec{E}_{inside}=-\frac{1}{4\pi\epsilon_0}\frac{\vec{p}}{R^3}[/tex]
    The last expression clearly blows up at ##R##→0 (although it was a bit of a cheat since I technically divided by zero in the previous step).
    I still can't figure out how to get to the delta function though, so any ideas will be appreciated!
     
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