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Homework Help: Quick errors help

  1. Jan 22, 2005 #1
    Hi guys..i always seem to forget how to calculate errors and end up confusing myself in a giant mess...so any help much appreciated.

    I have 24.98 +/- 1.05 = e/(nkT)

    And i need to work out n, i know all the other numbers (which are constants and have no errors). So firstly how do i calculate the error on n if all of the others are constant, and secondly how would i calculate the error on n if T also had an error? Cheers :)
     
  2. jcsd
  3. Jan 22, 2005 #2

    dextercioby

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    1.St question:do u know calculus??Your function mai be written (via logaritmation)
    [tex] nkT=\ln(24.98\pm1.05) [/tex]

    Express "n" and "[itex] \Delta n [/itex]" in terms "T" and "[itex] \Delta T[/itex]" respectively.

    Daniel.
     
  4. Jan 22, 2005 #3
    oops, sorry, i didnt explain the equation clearly enough. "e" in this case doesnt stand for exponential, its the charge of an electron (k is the boltzmann constant and T is temperature).
     
  5. Jan 22, 2005 #4

    dextercioby

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    AAAA...That changes thing a little bit,not too much.You didn't answer to my question,though...Do u know calculus??
    Then the function would be:
    [tex] n=\frac{e}{(24.98 \pm 1.05) k} \frac{1}{T} [/tex]

    Can u differentiate a function??

    Daniel.
     
    Last edited: Jan 22, 2005
  6. Jan 22, 2005 #5
    Hi,

    Yeah i can differentiate..but what bit would i need to differentiate and why?
     
  7. Jan 23, 2005 #6
    Has this thread been forgotton about? :( Sorry i just dont even know where to start with errors when it gets more complicated than adding/subtracting...

    Do you want me to differentiate every letter? Would that require using the product rule twice? Am i going in completly the wrong direction with this? Why do you need to differentiate to find errors? Thanks for any help so far :)
     
  8. Jan 23, 2005 #7

    dextercioby

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    Because that's the theory of errors...Involves partial (in this case one variable,namely T) differentiation.

    [tex] \Delta n=|\frac{dn(T)}{dT}|\Delta T [/tex]

    ,where [itex] \Delta T [/itex] is the error in temperature,and the corresponding [itex] \Delta n [/itex] is the error in concentration...

    Daniel.
     
  9. Jan 23, 2005 #8
    So if ive understood you correctly:

    [tex]\Delta n={-T}^{-2}\Delta T[/tex]

    ? :)
     
  10. Jan 23, 2005 #9

    dextercioby

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    U didn't.

    1.What happened to the constants???
    2.The minus does not exist.It's "erased" by the modulus/absolute value.There's never minus in error theory...All errors must add...

    Daniel.
     
  11. Jan 23, 2005 #10
    Whoops! Lemme try again...

    [tex]\Delta n=\frac{e}{(24.98 \pm 1.05) k}\ {T}^{-2}\Delta T[/tex]

    Is that right? Cheers for the patience :biggrin:
     
  12. Jan 23, 2005 #11

    dextercioby

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    It looks okay,though that [itex] \pm [/itex] in the denominator looks kinda weird...

    Daniel.
     
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