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Relative error

  1. Feb 11, 2014 #1
    Hello,
    If I have a measurement of 1cm, and have an absolute error of 0.1 cm, I know that I can write my measurement as (1±0.1)cm.
    If I want to write it with its relative error instead of an absolute error, can I still use the bracket?
    i.e) (1±10%)cm ???

    Thank you!!
     
  2. jcsd
  3. Feb 11, 2014 #2
    10% is not in cm. So it does not make sense to write it inside the parenthesis.
     
  4. Feb 11, 2014 #3

    jtbell

    User Avatar

    Staff: Mentor

    To put it another way, the relative error would be 10% no matter what units you're using. So I would write it as 1cm ± 10%.
     
  5. Feb 11, 2014 #4
    I, on the other hand, would write it as (1±10%)cm. % is not in cm so we must multiply it by cm in order to get the correct unit. Keep in mind that % is short hand for 1/100.
     
  6. Feb 11, 2014 #5
    It works here (the value is 1 cm) but what if it's 2 cm with an error of 10%?

    Then you will multiply 10% by cm to get 0.1 cm?
    But the error is actually 0.2 cm.
     
  7. Feb 11, 2014 #6

    jbriggs444

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    Science Advisor

    You multiply 2 cm by +/- 10% to get +/- 0.2 cm.
    Or you multiply cm by +/- 10% and take two of them to get +/- 0.2 cm.
    Or you multiply 2 by +/- 10% and use units of cm to get +/- 0.2 cm

    Yes. That's right. No matter which of the three ways you choose to interpret 10%. There may be motivations to choose one interpretation over the others, but an off-by-a-factor-of-two problem is not one of them.
     
  8. Feb 11, 2014 #7
    No.

    2cm±10% is 2cm±0.2cm
    (2±10%)cm is (2±0.2)cm

    It works either way.
    In most cases, I have seen the first one (2cm±10%) much more often.
    The second on is common in specifications. For example:
    Pin hole depth (cm): 2±10%

    Or in table form:
    Parameter ... Value ... units
    Pin hole depth ... 2±10% ... cm
     
  9. Feb 11, 2014 #8
    I have never seen that.
    I guess it would be 2(cm±10%) !?!
     
  10. Feb 11, 2014 #9
    Millikan oil drop

    hmm I guess I could use it either way then.

    I actually have another question at this point, it's about Millikan oil drop experiment.

    I've got
    mg=kvf, when the e-field is zero, (taking downwards direction as positive), k is some constant and vf is the terminal velocity of an oil drop.
    Then when the e-field is on, mg+kve=Eq, where Eq is the force from the electric field, and k is the same constant and ve is the drift velocity of an oil drop.
    When I isolated q (charge), i got

    q=[(ve+vf)/vf]*dmg/V

    and keep in mind that q=ne, where n is the number of charge and e is an elementary charge (q is of course the number of charge in an oil drop)

    I got something like 8*10^-18 for q, and I'm trying to find n, so that I can plot q vs. n to find the slope of the line (which is e)

    but in order for me to find the number of charge (n), don't I have to divide the q by e?
    I'm a little confused here because I thought the whole point of doing this experiment is to determine e. but by dividing q by e to obtain n, aren't I misinterpreting the whole the experiment?

    How can i find n without dividing q by e?
     
  11. Feb 12, 2014 #10
    You find n by looking at many drops and noticing that they form bunches. Each bunch correspond to a different value of n. So you just count the bunches starting from n=0 for those drops that were unaffected by the electric field.
     
  12. Feb 12, 2014 #11

    Dale

    Staff: Mentor

    Last edited: Feb 13, 2014
  13. Feb 13, 2014 #12
    That post was intended for a different thread (oops). Does anybody know how I can remove it from here?
     
  14. Feb 13, 2014 #13

    Dale

    Staff: Mentor

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