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Electric field of a filament

  1. Feb 10, 2008 #1

    tony873004

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    1. The problem statement, all variables and given/known data
    A long glass filament carries a charge density [tex]\lambda = - 2.7{\rm{ nC/m}}[/tex] . What is the magnitude of the electric field 0.67 mm from the filament?
    3. The attempt at a solution

    Did I do this problem correctly? I'm guessing not because my answer has an extra /m . Shouldn't it simply be N/C?
    [tex]
    E = \left| {\frac{{k\,d\lambda }}{{r^2 }}} \right| = \left| {\frac{{8.99 \times 10^9 {\rm{N}} \cdot \frac{{{\rm{m}}^{\rm{2}} }}{{{\rm{C}}^{\rm{2}} }} \cdot - 2.7{\rm{ nC/m}} \cdot 10^{ - 9} {\rm{C/nC}}}}{{{\rm{0}}{\rm{.67mm}} \cdot 0.001{\rm{m/mm}}}}} \right| = 54000000{\rm{ N/Cm}}
    [/tex]
     
    Last edited: Feb 10, 2008
  2. jcsd
  3. Feb 10, 2008 #2
    Isn't the filament a long line of charge? So, the field strength is proportional to 1/r.
     
  4. Feb 10, 2008 #3

    Dick

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    Basically, except isn't E=2*k*lambda/r? I think you missed a '2'. Your equation is for the infinitesimal element of E before it is integrated.
     
  5. Feb 10, 2008 #4

    tony873004

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    Thanks. Ok, then would it be:
    [tex]E = \left| {\frac{{2k\lambda }}{r}} \right| = \left| {\frac{{2 \cdot 8.99 \times 10^9 {\rm{N}} \cdot \frac{{{\rm{m}}^{\rm{2}} }}{{{\rm{C}}^{\rm{2}} }}\left( { - 2.7{\rm{nC/m}} \cdot 10^{ - 9} {\rm{C/nC}}} \right)}}{{{\rm{0}}{\rm{.67mm}} \cdot 0.001{\rm{m/mm}}}}} \right| = 72000\,{\rm{N/C}}[/tex]
    ?
    Where did E=2k*lambda/r come from? Is this just a standard formula I can use as a starting point, or would I be expected to integrate and come up with this formula?
     
  6. Feb 10, 2008 #5
  7. Feb 10, 2008 #6

    Dick

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    I know it's k*lambda/r. I had to look it up to get the numerical factor of 2. Just looking at the problem statement, it doesn't look like a 'derive the formula' type problem, though the derivation isn't that hard. I would suspect the formula is in your book or notes and you can just use it. Check though.
     
  8. Feb 10, 2008 #7

    tony873004

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    Thanks. I'd like to understand how to do it even if we can start with the formula. Awvvu's link shows how. Stay tuned... I might have questions about that link.
     
  9. Feb 10, 2008 #8

    tony873004

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    I though of another way. From our class notes, we did the integration of the same problem, except the filament had a finite length. The formula was [tex]
    {\overrightarrow {E_y } = \frac{{k\lambda L}}{{y^2 \sqrt {1 + \left( {\frac{L}{{2y}}} \right)^2 } }}}
    [/tex]

    To convert this to use for an infinite length:

    As L approaches infinity, the 1 under the radical becomes insignificant, allowing me to eliminate it and replace [tex]
    \sqrt {\left( {\frac{L}{{2y}}} \right)^2 }
    [/tex] with [tex]{\frac{L}{{2y}}}[/tex].
    . Then my L's cancel leaving me with [tex]\frac{{2k\lambda }}{y}[/tex]

    Thanks for checking my work. I probably would have submitted my original attempt if the two of you didn't correct me.
     
  10. Feb 10, 2008 #9

    Dick

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    That works fine. Glad you got it.
     
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