The Electric Field — Question 8

Question 8

Force and charge bounds produce asymmetric field bounds

A positive probe charge lies in $[1.90,2.10]$ nC. Its positive x-directed electrical force lies in $[.570,.630]$ microN. Treat these as independent simultaneous hard bounds and assume the force is entirely electric.

1. Find the exact permitted interval for $E_x=F_x/q$.

2. Compare with the central-value estimate and compute the largest relative deviation from that estimate. Explain why dividing matching lower endpoints is not the minimum field.

Question 8 - Solution

Force and charge bounds produce asymmetric field bounds

(a) Method and calculation

$$E_{\min}=\frac{.570\times10^{-6}}{2.10\times10^{-9}}\approx271.429\,\mathrm{N/C}.$$$$E_{\max}=\frac{.630\times10^{-6}}{1.90\times10^{-9}}\approx331.579\,\mathrm{N/C}.$$

(b) Interpretation and further calculation

$$E_{\mathrm{central}}=.600\times10^{-6}/(2.00\times10^{-9})=300\,\mathrm{N/C}.$$$$\max|E-E_{\mathrm{central}}|/E_{\mathrm{central}}=\frac{331.579-300}{300}\approx.105263.$$For positive quantities, a quotient decreases when its numerator decreases or its denominator increases.

Physical check

The extrema occur at opposite corners of the allowed force-charge rectangle.