Discussion
In Millikan's oil drop experiment the electronic charge e could be written as \(K\eta^{1.5}\), where K was a function of other experimental parameters with negligible error, while the viscosity of air \(\eta\) assumed was 0.4% too low. What was the error in e ?
Answer: C
The error in e was 0.6%.
For a power law \(e = K\eta^{n}\) with K exact, relative errors scale by the exponent:
\[ \begin{aligned} \frac{\delta e}{e} &= n\frac{\delta\eta}{\eta} \\ &= 1.5 \times 0.4\% = 0.6\% \end{aligned} \]
Since \(\eta\) was too low, e came out about 0.6% too low. 0.4% ignores the exponent, 1.5% confuses the exponent with the error, and 0.7% has no basis.
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