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Text book solution

  1. Apr 18, 2012 #1
    1. The problem statement, all variables and given/known data
    please look at equations II.2.8 , II.2.9, II.2.10 in the attached scan,

    2. Relevant equations
    none


    3. The attempt at a solution
    im unable to get from II.2.8 to II.2.9, i wonder where they got the extra r in the denominator of the stuff in the brackets,
    here is what i get for II.2.10:
    [tex] \frac{1}{18} \left( \frac{ (n+2) n C} { r^{n+3}} - \frac{3 M e^2}{r^4} \right) [/tex]
    what im i doing wrong?
     

    Attached Files:

  2. jcsd
  3. Apr 18, 2012 #2

    rude man

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    Looks like sloppy work to me:

    1. in going from II.2.8 to II.2.9 somehow E changed to e.

    2. There should be a minus sign in front of II.2.9, otherwise it's correct.

    remember: d/dr(rn) = nrn-1 etc.

    No way I can determine the correctness of II.2.10 since a new parameter (r0) is introduced.
     
  4. Apr 19, 2012 #3
    how did u do that? in II.2.8 there is (for stuff after partial differentiation***{EDIT})
    [tex] \frac{1}{6r^2} \left( ME^2 \frac{1}{r} - \frac{nC}{r^n} \right) [/tex]
    how does this end up being
    [tex] \frac{1}{6} \left( ME^2 \frac{1}{r^4} - \frac{nC}{r^{n+3}} \right) [/tex]
    when the stuff just outside the bracket gets taken in? r is only to the inverse second power, why does r in II.2.9 end up to the inverse fourth power? 2+1 =4 ? remember the partial diff hasn't yet been done.

    ***EDIT
    i mean partial differentiation sign not after performing the operation.
     
    Last edited: Apr 19, 2012
  5. Apr 19, 2012 #4
    though i think the [itex] r_o [/itex] is the r corresponding to minimum P.E., looking at II.2.7 and the paragraph preceeding it.
     
  6. Apr 19, 2012 #5

    rude man

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    Oops - I hadn't noticed - in II.2.9 there should not be a partial differential anymore. The derivative is taken already in going from II.2.8 to II.2.9.

    Like I said - sloppy work! And there still needs to be a minus sign in front of the corrected II.2.9.
     
  7. Apr 19, 2012 #6
    Ok, i see that now, i had not noticed that the differentiation had already been done, so that's why you kept on insisting on the negative sign, ( i admit i was clueless why until now).
    thanks. Let me work on it and see whether i can better this guy's work.
    cheers.
     
  8. Apr 19, 2012 #7
    one quick last one, has anyone seen this question before? is n=8.1 really? is that the correct answer that i need to find when i correct this guy's unholy mess?
     
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