How should I calculate the error in my e/m ratio calculation for my homework?

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SUMMARY

The discussion focuses on calculating the error in the e/m ratio of an electron using the formula \(\frac{e}{m} = (32/D^2)*(nA^2\pi)^2*(Va/I_s^2)*(I_s/V_s)^2\). The participant is uncertain about how to incorporate measurement errors for the variables Va and I_s, especially after obtaining a high percentage error from traditional methods. They explore the possibility of using gradient errors from their graphs instead, which yield a significantly lower error estimate of approximately 10%. The key conclusion is that utilizing gradient errors may provide a more accurate representation of uncertainty in this context.

PREREQUISITES
  • Understanding of the e/m ratio and its significance in physics.
  • Familiarity with error analysis and propagation techniques.
  • Proficiency in using graphing software to analyze experimental data.
  • Knowledge of basic measurement principles and equipment increments.
NEXT STEPS
  • Research error propagation methods in experimental physics.
  • Learn about using gradient errors in data analysis.
  • Explore statistical methods for combining percentage errors.
  • Investigate best practices for measuring voltage and current in laboratory settings.
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Students in physics, particularly those working on experimental projects involving the e/m ratio, as well as educators and researchers focused on error analysis in measurements.

Phystudent91
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Homework Statement


I'm in the middle of a project finding the e/m ratio of the electron by the use of the formula below. I know all required values and can find a result, I'm unsure how to calculate an error for the calculation.
In the formula below, D, n, A and f are all given quantities (therefore no error/the error can be taken as 0, right?)

Homework Equations



[itex]\frac{e}{m}[/itex] = (32/D2)*(nA2∏)2*(Va/Is2)*(Is/Vs)2

The Attempt at a Solution



The last 2 parts of the equation are where the trouble arises. I have had to measure and record results in two different ways to find values and to plot the 2 graphs - Va/Is2 and Is/Vs - and the software used gives a percentage error in the line.
Do I need to find the error in each measurement for Va and Is etc?

The current way I know to calculate error would mean using the smallest value measured and half of the smallest possible measurement (half the equipment increment). This would result in a stupidly high error, eg:
smallest V measurement was 5V, this is also the smallest increment on the equipment, therefore error is ±2.5V meaning an error in that set of measurements of 50%??
I'm then taught to add the percentage errors... this leads to 350% error.
However, because I know the gradients of the graph, is it acceptable to only use these errors? This would give a total are of ≈10%. Much better!

Which is the correct way of going about this? And if the answer is neither of these, what is the correct way of going about this!?

Thanks in advance!
 
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