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Homework Help: Multipoles physics problem

  1. Jan 17, 2010 #1
    1. The problem statement, all variables and given/known data
    find:

    [tex]\vec{p}[/tex] (dipole moment)
    [tex]\itshape V(r,\theta)[/tex] (dipole potential)

    of a spherical shell which has a charge area of [tex]\itshape \sigma= k cos \theta [/tex].

    2. Relevant equations
    [tex]\itshape \vec{p}= \int\int \vec{r} \sigma (r\stackrel{2}{} sin \theta d\theta d\phi)[/tex]
    [tex]\itshape V(r, \theta )= \frac{\hat{R} \bullet \vec{p}}{4 \pi \epsilon R^{2}}[/tex]

    3. The attempt at a solution
    I came up with 0 both times and i know that's wrong. plz help
     
  2. jcsd
  3. Jan 17, 2010 #2

    vela

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    Re: Multipoles

    Show us your work so we can see where you went wrong.
     
  4. Jan 17, 2010 #3
    Re: Multipoles

    I can tell you that either your integral is wrong or you are not accounting for [itex]\vec{r}[/itex] in the integral. Since I know the equation is correct, you are not accounting for [itex]\vec{r}[/itex] in your calculation.

    Do you know what [tex]\vec{r}[/tex] is?
     
    Last edited: Jan 17, 2010
  5. Jan 17, 2010 #4
    Re: Multipoles

    Work: OK i re-did the problem this time taking a different approach....

    [tex]\itshape\vec{p}=\int^{\pi}_{0}\int^{2\pi}_{0}\vec{r}\sigma\:d\vec{a}[/tex]

    [tex]\itshape \sigma=kcos\:\theta\:,\:d\vec{a}=r^{2}sin\:\theta\:d\theta\:d\phi\:\hat{r}\:,\:k=constant[/tex]

    so...
    [tex]\itshape \vec{p}=\int^{\pi}_{0}\int^{2\pi}_{0}r^{3}k(cos^{2}\:\theta\:sin\:\theta\:sin\:\phi\:\hat{x}\:+\:sin^{2}\:\theta\:cos\:\theta\:cos\:\phi\:\hat{y}\:+\:cos^{2}\:\theta\:sin\:\theta\:\hat{z})\:d\theta\:\:d\phi[/tex]
    so when...
    [tex]\itshape R\geq\:r\;\Rightarrow\;\vec{p}=\frac{4}{3}\pi r^{3}k\:\hat{z}[/tex]
    and when...
    [tex]\itshape r\geq\:R\;\Rightarrow\;\vec{p}=\frac{4}{3}\pi R^{3}k\:\hat{z}[/tex]

    and finally...
    [tex]\itshape \hat{r}\bullet\vec{p}=cos\:\theta\:\hat{z}\bullet\frac{4}{3}\pi r(or\:R)^{3}k\:\hat{z}=\frac{4}{3}\pi r(or\:R)^{3}k\:cos\:\theta\:[/tex]

    so when...
    [tex]\itshaper R\geq\:r\;\Rightarrow\;V_{dip}(r,\theta)=\frac{kR^{3}}{3\epsilon r^{2}}\:cos\:\theta[/tex]
    and when...
    [tex]\itshape r\geq\:R\;\Rightarrow\;V_{dip}(r,\theta)=\frac{kr}{3\epsilon}\:cos\:\theta[/tex]
    but what disturbs me is i thought i was supposed to be finding an approximate amount of V... and that was the exact answer, did i go wrong somewhere and just get really lucky giving me the exact amount, or am i right... and if im right what does using a higher multipole change?
     
    Last edited: Jan 17, 2010
  6. Jan 17, 2010 #5
    Re: Multipoles

    @jdwood983 the first time i didn't account for [tex]\itshape \vec{r}=r\hat{r}[/tex], and i always forget to because I'm being lazy, the second time i did though.
     
    Last edited: Jan 17, 2010
  7. Jan 17, 2010 #6
    Re: Multipoles

    Is it not possible that this charge distribution has exactly the electric field of a certain dipole? In which case higher order terms would not be necessary.
     
  8. Jan 17, 2010 #7
    Re: Multipoles

    I am not sure, nor do i know a way to check that. Is the a formula or and of the given i had above that would tell me this, I'm self teaching Griffiths' book, which has no answers and no teacher to ask, so I'm a bit lost.
     
  9. Jan 17, 2010 #8
    Re: Multipoles

    It is certainly possible, and your result seems like more than a coincidence given what he asks in part b.
     
  10. Jan 17, 2010 #9
    Re: Multipoles

    Because what really drove the integral above was the [tex]\itshape \sigma[/tex] which says that the charge density on the x and y axis are both k (when z=0) would it be safe to say the the dipole moment will be in the z direction in cases like this? Also because of circle symmetry the net charge is 0 at the origin (I assume is the center) is the V going to be defined as a dipole, which i think might be mentioned in the book?
     
  11. Jan 17, 2010 #10

    vela

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    Re: Multipoles

    It's been too long since I've done this stuff, so I might be totally off here. But I'd guess the result isn't too surprising because [itex]\sigma \propto Y^0_1[/itex], where [itex]Y^m_l[/itex] are the spherical harmonics.
     
  12. Jan 17, 2010 #11
    Re: Multipoles

    the whole Y thing went right past me, but what u said i think clarified my final troubles on this problem. Thank you everyone for the help on this problem :)
     
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