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Relative error 
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#1
Feb1114, 12:13 PM

P: 28

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
Feb1114, 01:05 PM

P: 1,994

10% is not in cm. So it does not make sense to write it inside the parenthesis.



#3
Feb1114, 02:01 PM

Mentor
P: 11,878

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%.



#4
Feb1114, 02:41 PM

Thanks
P: 1,948

Relative error
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.



#5
Feb1114, 03:12 PM

P: 1,994

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. 


#6
Feb1114, 03:30 PM

P: 999

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 


#7
Feb1114, 03:31 PM

P: 627

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 


#8
Feb1114, 03:33 PM

P: 627

I guess it would be 2(cm±10%) !?! 


#9
Feb1114, 07:26 PM

P: 28

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 efield is zero, (taking downwards direction as positive), k is some constant and vf is the terminal velocity of an oil drop. Then when the efield 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? 


#10
Feb1214, 11:05 AM

Thanks
P: 1,948

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.



#11
Feb1214, 07:02 PM

Mentor
P: 17,556



#12
Feb1314, 09:45 AM

Thanks
P: 1,948




#13
Feb1314, 10:27 AM

Mentor
P: 17,556

http://physics.nist.gov/Pubs/SP811/sec07.html They use % as simply the number 1/100. So they would express it as, e.g. 5 cm (1 ± 3 %). 


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