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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!
     
  4. Sep 14, 2017 #3
    I know this thread is near three years old, and OP is likely long gone. But I've spent some significant time on the problem so thought I ought to share my thoughts. Having spent so much time, I would like to know the cleanest solution. I am not totally confident in the soundness of my own argument, but here goes. As problem states, we are looking to reproduce the delta function term in the dipole field, [itex] -\frac{\vec{p}}{3 \epsilon_0} \delta^3(\vec{r})[/itex], from the uniformly polarized sphere.

    As OP notes, the field inside the sphere is uniform. So, we can take a integral of the field (which we take as zero outside the sphere of radius [itex]R[/itex]) over all space, divide through by the volume of the sphere and arrive at the original field. That is,

    [tex]\vec{E_{avg}} = \frac{1}{\frac{4}{3} \pi R^3} \int_{all space}d^3\vec{r}\vec{E_{in}}=-\frac{k}{3\epsilon_0}\hat{z}.[/tex]

    We define [itex] \vec{G_{in}} = \frac{\vec{E_{in}}}{\frac{4}{3} \pi R^3}[/itex]. So,

    [tex] \int_{all space}d^3\vec{r}\vec{G_{in}}=-\frac{k}{3\epsilon_0}\hat{z}.[/tex]

    We note that since [itex]\vec{E_{in}}[/itex] is uniform (for [itex] r < R [/itex] and 0 otherwise), [itex]\vec{G_{in}}[/itex] essentially has the form [itex]1/R^3[/itex] for [itex] r < R [/itex] (and 0 otherwise). It is easy to show, by integrating it against a test function, that in the limit [itex]R \rightarrow 0[/itex], [itex]\vec{G_{in}}[/itex] is a (multiple of) delta function.

    What is that multiple [itex]C[/itex]? We take the limit [itex]R \rightarrow 0 [/itex] of the above equation, replacing [itex]\vec{G_{in}}[/itex] with [itex] C \hat{z} \delta^3(\vec{r}) [/itex], and find that [itex]C[/itex] must be [itex] -\frac{k}{3 \epsilon_0} [/itex]. Solving for the original field, we have

    [tex] \lim_{R \rightarrow 0} \vec{E_{in}} = -\frac{\frac{4}{3} \pi R^3 k}{3 \epsilon_0} \hat{z} \delta^3(\vec{r}), [/tex]

    or recognizing [itex] \frac{4}{3} \pi R^3 k \hat{z} = \vec{p} [/itex] (dipole aligned to z-axis),

    [tex] \lim_{R \rightarrow 0} \vec{E_{in}} = -\frac{\vec{p}}{3 \epsilon_0} \delta^3(\vec{r}). [/tex]

    Thoughts? Maybe it's simpler (re: necessary?) if you allow [itex]k[/itex] to increase with shrinking [itex]R[/itex] such that [itex] \vec{p} [/itex] is constant.

    Edit: more I think about it, less convinced I am. I think there's a problem in multiplying [itex] \lim_{R \rightarrow 0}\vec{G_{in}} = \lim_{R \rightarrow 0}\frac{1}{\frac{4}{3} \pi R^3}\vec{E_{in}} [/itex] by [itex] \frac{4}{3} \pi R^3 [/itex] and just cancelling [itex] R^3[/itex] on inside and out of limit. Damn. BUT, if you let [itex] k [/itex] be proportional to [itex] \frac{1}{R^3} [/itex], then there is no need to do the [itex] \vec{G_{in}} [/itex] substitution, and the same essential argument works fine.

    Again, though, if there's a simpler way, someone do tell!
     
    Last edited: Sep 14, 2017
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