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Potential of uniformly polarized sphere

  1. Feb 12, 2010 #1
    1. The problem statement, all variables and given/known data

    This is problem 4.12 in Griffiths: "Calculate the potential of a uniformly polarized sphere (Ex. 4.2) directly from Eq 4.9"

    2. Relevant equations

    Eq 4.9:
    V=\frac{1}{4\pi\epsilon_{o}}\int_{V}\frac{ \hat{\textbf{r}}\cdot \textbf{P}({\textbf{r'}})}{r^2}d{\tau}'[/tex]
    (Sorry, the formatting is a bit off and r is actually script r, which I don't know how to do here)

    3. The attempt at a solution
    The solution manual simply factors P out of the integral and then uses Gauss's law (by noticing that the integral left over is the same as the electric field of a uniformly charged sphere divided by the charge density). That's nice, but how can you factor out P?

    My conceptual understanding of the integral is that you are summing up a bunch of dipoles, and script r gives you the vector from a given dipole to the field point (which doesn't move)--so the dot product of script-r-hat and P changes and it should stay in the integral. Can someone justify to me how
    [tex]\int_{V}\frac{ \hat{\textbf{r}}\cdot \textbf{P}({\textbf{r'}})}{r^2}d{\tau}' = \textbf{P}\cdot\left\{\int_{V}\frac{ \hat{\textbf{r}}}{r^2}d{\tau}'\right\}

    (I guess something about the symmetry of the problem allows you to do this?)
  2. jcsd
  3. Feb 13, 2010 #2
    First off, as a student you shouldn't have a copy of the solution manual; it will only impede your learning of the material. You should throw it away immediately.

    Second, the question states that the polarization, [itex]P[/itex], is uniform. If [itex]P[/itex] is uniform, it does not depend on [itex]r[/itex] and can be pulled out of the integral.
  4. Feb 13, 2010 #3


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    P is a constant vector the direction and magnitude of which do not depend on the integration variable r'. By contrast, the vector



    does depend on r'. You can either first take all the little individual dot products of P with


    and then add them (P inside the integral) or you can first find the vector


    then take a single dot product with P (P outside the integral).
  5. Feb 13, 2010 #4
    Ah, now that I look at the problem it seems so silly. For some reason it seemed like you couldn't just take the constant out of the integration because one of the terms of the dot product wasn't constant. But since the dot product is distributive it makes perfect sense. Guess I just needed to sleep on it.
  6. Feb 13, 2010 #5

    I am glad you see how [itex]P[/itex] can be pulled out of the integral, but I hope you plan on getting rid of your ill-obtained solution manual (ie. students shouldn't have a copy, professors only--it actually says this on his website). You will not learn a thing about E&M while using it.
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