Electric Charge — Question 8

Section 21-1 · Physics pages 3386–3387

Question 8 A charge measurement can establish magnitude without establishing sign Two independent charge estimates for the same sample give hard allowed intervals [−.20, +.40] nC and [+.10, +.50] nC. The intervals are bounds, not standard deviations. Assume both bounds are valid. Q 0 I1 I2 Schematic; labels, not drawing lengths, specify the data. (a) Find the joint allowed interval and determine whether the sample sign is established. (b) A calibration correction b is subsequently discovered, with −.15 ≤b ≤+.05 nC; the true charge is the intersected estimate plus b. Find the full corrected interval and reassess the sign.
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Question 8 - Solution A charge measurement can establish magnitude without establishing sign Q 0 I1 I2 (a) Method and calculation I = [−.20, .40] ∩[.10, .50] = [.10, .40] nC. min I = .10 > 0 ⇒Q > 0 before the correction. (b) Interpretation and further calculation Itrue = [.10 −.15, .40 + .05] = [−.05, .45] nC. Q < 0, Q = 0, Q > 0 are all compatible with the corrected bounds. A common unknown bias cannot be reduced by intersecting repeated measurements as if it were independent random noise. Physical check Hard interval endpoints combine by extremization, without a root-sum-square rule.

Question 8

A charge measurement can establish magnitude without establishing sign

Two independent charge estimates for the same sample give hard allowed intervals [−.20,+.40][-.20,+.40] nC and [+.10,+.50][+.10,+.50] nC. The intervals are bounds, not standard deviations. Assume both bounds are valid.

Diagram shown in the original SVG worksheet.

  1. Find the joint allowed interval and determine whether the sample sign is established.

  2. A calibration correction bb is subsequently discovered, with −.15≤b≤+.05-.15\le b\le+.05 nC; the true charge is the intersected estimate plus bb. Find the full corrected interval and reassess the sign.

Question 8 - Solution

A charge measurement can establish magnitude without establishing sign

Diagram shown in the original SVG worksheet.

(a) Method and calculation

I=[−.20,.40]∩[.10,.50]=[.10,.40]nC.I=[-.20,.40]\cap[.10,.50]=[.10,.40]\,\mathrm{nC}.min⁡I=.10>0⇒Q>0before the correction.\min I=.10>0\Rightarrow Q>0\quad\text{before the correction}.

(b) Interpretation and further calculation

Itrue=[.10−.15,.40+.05]=[−.05,.45]nC.I_{\mathrm{true}}=[.10-.15,.40+.05]=[-.05,.45]\,\mathrm{nC}.Q<0,Q=0,Q>0are all compatible with the corrected bounds.Q<0,\ Q=0,\ Q>0\quad\text{are all compatible with the corrected bounds}.A common unknown bias cannot be reduced by intersecting repeated measurements as if it were independent random noise.

Physical check

Hard interval endpoints combine by extremization, without a root-sum-square rule.