Relative error (1 Viewer)

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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!!
 
3,706
402
10% is not in cm. So it does not make sense to write it inside the parenthesis.
 

jtbell

Mentor
15,193
2,787
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%.
 
1,948
200
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.
 
3,706
402
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.
 

jbriggs444

Science Advisor
6,881
2,238
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?
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

But the error is actually 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.
 

.Scott

Homework Helper
2,162
681
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.
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
 

.Scott

Homework Helper
2,162
681
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?
 
1,948
200
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.
 
1,948
200
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.
That post was intended for a different thread (oops). Does anybody know how I can remove it from here?
 

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