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Dipole moment.

  1. Sep 25, 2006 #1
    From Griffiths, Problem 4.1
    A sphere of radius R carries a polarization

    where k is constant and r is the vector from the center.
    a. Calculate [tex]\sigma_b[/tex] and [tex]\rho_b[/tex].
    b. Find the field inside and outside the sphere.

    part a is handled simply by [tex]\sigma_b=\textbf{P}\cdot\hat{n}[/tex] and [tex]\rho_b=-\nabla\cdot\textbf{P}[/tex].

    part b is handled most easily by using the bound charges found and gauss's law, giving: [tex]\textbf{E}=\frac{-kr}{\epsilon_0}\hat{r}[/tex] and 0 outside.

    part b can also be handled by first getting the potential through:[tex]\frac{1}{4\pi \epsilon_0}\int_v \frac{\hat{r}\cdot\textbf{P}}{r^2}d\tau[/tex]. Then just get the negative gradient of V to yield E. This gives the same result as method 1 for both inside and outside (P=0 when dealing with outside case).

    Now, i tried to get the field, again by computing for V but this time with:
    [tex]V(r)=\frac{1}{4\pi \epsilon_0}\oint_s\frac{\sigma_b}{r}da + \frac{1}{4\pi \epsilon_0}\int_v\frac{\rho_b}{r}d\tau[/tex]. The answer i get using this integral is different. i would like to know if in the first place this integral is applicable? if i'm trying to get the potential inside, do i still consider the surface bound charge?

    the best i managed was an answer negative to the correct one. (but this had 1 dubioius step involved). also, i could only attempt to get the potential inside. can anyone make the last method work (for both outside and inside)? or share why it doesnt?

  2. jcsd
  3. Sep 25, 2006 #2

    Meir Achuz

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    \sigma_b is not spherically symmetric, so E iinside is not.
    You can't use Gauss as you do in that case.
    Your integrals should give the right answer if you mean r-r'
    when you write r.
    Last edited: Sep 25, 2006
  4. Sep 25, 2006 #3
    Gauss's law actually gives the right answers. [tex]\sigma_b=kR[/tex] everywhere on the surface if you apply the definition. likewise, [tex]\rho_b=-3k[/tex]. these are symmetric and so is E inside and out.

    as for the r-r' hint. i don't see it yet, but i will try it. thanks.

    i was wrong though, about my claim that i got E=0 outside using the second method. that's another issue - how to get that result.
  5. Sep 27, 2006 #4

    Meir Achuz

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    \sigma_b does not equal kR. It equals kR cos\theta because of the dot product. Go back and study Griffriths some more, then come back to the problem.
  6. Sep 30, 2006 #5


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    Meir Achuz, I don't quite understand this.

    If [itex]P(r)=kr \vec{e_r} [/itex]
    Then, won't [tex]\sigma_b[/tex] be

    [tex]\sigma_b = \vec{P}.\hat{n} = k R (\vec{e_r}.\vec{e_r}) = kR [/tex]
    Last edited: Sep 30, 2006
  7. Sep 30, 2006 #6

    Meir Achuz

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    Whoops. I apologize. I had read it as P=a fixed vector.
    Using Gauss as in the first post gives the right answer.
    To use the potential integrals still requires r-r'.
    Thanks for catching my carelessness.
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