Coulomb's Law — Question 8

Question 8

A force-ratio experiment can test an assumed distance exponent

A fixed pair of charges is modeled by $F=A/r^p$, with unknown positive $A$ and exponent $p$. Ideal reference measurements give $F(.100\,\mathrm m)=90.0\,\mu$N and $F(.200\,\mathrm m)=22.5\,\mu$N. A third reading at .300 m has a hard allowed interval $[10.6,11.4]\,\mu$N; distances and reference values are treated as exact for this model test.

1. Infer $p$ from the two reference values and predict the third force.

2. Determine whether all three measurements fit one unchanged inverse-power model. Explain why the discrepancy alone does not prove that Coulomb law fails.

Question 8 - Solution

A force-ratio experiment can test an assumed distance exponent

(a) Method and calculation

$$\frac{90.0}{22.5}=\left(\frac{.200}{.100}\right)^p\Rightarrow4=2^p.$$$$p=\frac{\ln4}{\ln2}=2,\quad F(.300)=90.0(.100/.300)^2=10.0\,\mu\mathrm N.$$

(b) Interpretation and further calculation

$$10.0\notin[10.6,11.4]\quad\Rightarrow\quad\text{no common fixed-parameter fit}.$$$$F_{\mathrm{measured,min}}-F_{\mathrm{predicted}}=.6\,\mu\mathrm N.$$Charge leakage, changed geometry, background force or invalid uncertainty bounds could break the experimental model. The calculation rejects that full set of assumptions, not a fundamental law in isolation.

Physical check

This consistency test uses hard bounds, so no probability or confidence level follows from the interval alone.