The Electric Field Due to a Point Charge — Question 3

Question 3

Uncertain source size and distance produce a worst-case field range

Use $k=8.99\times10^9\,\mathrm{N\,m^2/C^2}$; treat the specified charges as points in vacuum. A positive point charge has $Q\in[1.90,2.10]$ nC and its distance from P lies in $[.0980,.1020]$ m. All combinations are allowed hard bounds.

1. Find the allowed field-magnitude interval.

2. Compare with the nominal value at 2.00 nC and .100 m, and derive the first-order relative sensitivity to small changes in Q and r.

Question 3 - Solution

Uncertain source size and distance produce a worst-case field range

(a) Method and calculation

$$E_{\min}=k\frac{1.90\times10^{-9}}{.1020^2}\approx1641.77\,\mathrm{N/C}.$$$$E_{\max}=k\frac{2.10\times10^{-9}}{.0980^2}\approx1965.74\,\mathrm{N/C}.$$

(b) Interpretation and further calculation

$$E_{\mathrm{nom}}=1798\,\mathrm{N/C},\quad\frac{dE}{E}=\frac{dQ}{Q}-2\frac{dr}{r}.$$$$|\delta E|/E\lesssim.050+2(.020)=.090\quad\text{to first order}.$$The exact interval need not be symmetric around the nominal field; linear sensitivity is only an approximation.

Physical check

Minimum charge with maximum distance gives the minimum field, not minimum charge with minimum distance.