# Calculating Relative Error of Measurement: (1±0.1)cm or (1±10%)cm?

• jenny777
In summary: Yes. 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.It works either way.In most cases, I have seen the first one (2cm±10%) much more often.
jenny777
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!

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

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%.

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.

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.

nasu said:
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.

nasu said:
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

jbriggs444 said:
Or you multiply cm by +/- 10% and take two of them to get +/- 0.2 cm.
I have never seen that.
I guess it would be 2(cm±10%) ?

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?

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.

dauto said:
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?

## What is the formula for calculating relative error of measurement?

The formula for calculating relative error of measurement is (|x-y|)/x, where x is the true value and y is the measured value.

## How do you interpret relative error of measurement?

Relative error of measurement is expressed as a percentage and represents the difference between the measured value and the true value, relative to the true value. A lower percentage indicates a more accurate measurement.

## What is the significance of using (1±0.1)cm or (1±10%)cm for relative error of measurement?

The values (1±0.1)cm and (1±10%)cm represent a range of possible measurements within a 10% margin of error. This is a common method for expressing relative error of measurement and allows for more flexibility in the measured value.

## Can relative error of measurement be negative?

Yes, relative error of measurement can be negative if the measured value is greater than the true value. This indicates an overestimation of the measurement.

## How can relative error of measurement be reduced?

Relative error of measurement can be reduced by improving the precision and accuracy of the measuring instrument, taking multiple measurements and averaging the results, and minimizing external factors that may affect the measurement.

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