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Dipole moment

  1. Feb 17, 2008 #1

    tony873004

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    A needle suspended from a string hangs horizontally. The electric field at the needle’s location is horizontal with a magnitude 3.7*103 N/C and is at an angle of 30° with the needle. There is no net electrical force acting on the needle, but the string exerts a torque of 3.7*10-3 to hold the needle in equilibrium. What is the needle’s dipole moment?

    I know that [tex]
    \overrightarrow \tau = \overrightarrow {\rm{p}} \times \overrightarrow {\rm{E}} \,\,
    [/tex]

    But I can't simply re-write it as [tex]
    \frac{{\overrightarrow \tau }}{{\overrightarrow {\rm{E}} }} = \overrightarrow {\rm{p}} \,\,\,\,
    [/tex]
    , can I? I don't think you can do this to a cross-product. How can I solve for the dipole moment?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 17, 2008 #2
    You will only be able to that if p and E are perpendicular, in which case they are not. Do you remember how to find the magnitude of a cross product? Sin(x) maybe?mm?
     
  4. Feb 17, 2008 #3

    tony873004

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    I know its |A x B| = |A||B| sin(x). But in this problem, I don't know what A is. That's what I'm trying to solve for.

    |A x B|
    ------- = |A|. This looks like a dead end.
    |B|
     
  5. Feb 17, 2008 #4
    how about
    [tex] p=\frac{\tau}{sin(\theta) E}[/tex] ?
     
  6. Feb 17, 2008 #5

    tony873004

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    If I do that, then I'm dividing units of Nm by N/C, which gives me mC. The book doesn't talk about this, and my class notes are not very good, but I think that is the right unit for dipole moment, meters*coulombs?

    If so, I get 3.7e-4/sin(30) *3.7e3 = 2.738 mC for the answer.

    If this is correct, don't go away, because I'm still not confident I understand this.
     
  7. Feb 17, 2008 #6
    m*C are the correct units.
    I see some inconsistencies in your signs and values for your exponents to the ones you stated originally.
     
  8. Feb 17, 2008 #7

    tony873004

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    Yes, my original question should have been 3.7*10^3 N/C and 3.7*10^-4 for torque. (I imagine you knew that 103 meant 10^3 :) )
     
  9. Feb 17, 2008 #8

    tony873004

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    t is a vector, and so is E. But in the division that got me 2.738 mc, I divided only their magnitudes. Was this right? I'm guessing they want p to be a vector with direction. It will either be into or out of the board. How do I know which?
     
  10. Feb 18, 2008 #9

    tony873004

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    Ok, now I follow your chain of logic. Thanks for your help. I've got a scalar as an answer, and I still need to assign it a direction. Any thoughts?

    [tex]
    \begin{array}{l}
    \overrightarrow \tau = \overrightarrow {\rm{p}} \times \overrightarrow {\rm{E}} \, \\
    \\
    \left| {\overrightarrow {\rm{p}} \times \overrightarrow {\rm{E}} } \right| = \left| {\overrightarrow {\rm{p}} } \right|\left| {\overrightarrow {\rm{E}} } \right|\sin \theta \\
    \\
    \overrightarrow \tau = \left| {\overrightarrow {\rm{p}} } \right|\left| {\overrightarrow {\rm{E}} } \right|\sin \theta \,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left| {\overrightarrow {\rm{p}} } \right|\, = \frac{{\,\overrightarrow \tau }}{{\left| {\overrightarrow {\rm{E}} } \right|\sin \theta }}\,\,\, \\
    \\
    \left| {\overrightarrow {\rm{p}} } \right|\, = \frac{{3.7 \times 10^{ - 4} {\rm{N}} \cdot {\rm{m}}}}{{\sin \theta \cdot 3.7 \times 10^3 {\rm{N/C}}}} = 2.7{\rm{ C}} \cdot m \\
    \,\,\,\, \\
    \end{array}
    [/tex]
     
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