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Physics / Chapter 21: Electric Charge / Section 21-4

Charge Is Quantized — Question 3

Section 21-4 · Physics pages 3436–3437

Question 3 A measurement interval can isolate one allowed integer charge Use e = 1.602176634 × 10−19 C, exact. An isolated object built from ordinary atoms is assumed to have Q = ne with integer n. A calibrated hard measurement bound gives −1.33 × 10−18 ≤Q ≤−1.27 × 10−18 C. Treat the endpoints as exact bounds. Q/e -10 -9 -8 -7 -6 Schematic; labels, not drawing lengths, specify the data. (a) Find all integer charge states compatible with the interval. (b) The same instrument later reports −1.45 × 10−18 ≤Q ≤−1.10 × 10−18 C for another object. Find the allowed integer states and explain why rounding the midpoint alone is insufficient.
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Question 3 - Solution A measurement interval can isolate one allowed integer charge Q/e -10 -9 -8 -7 -6 (a) Method and calculation −1.33 × 10−18 e ≤n ≤−1.27 × 10−18 e ⇒−8.30121 ≲n ≲−7.92672. nmin = ⌈−8.30121 · · · ⌉= −8, nmax = ⌊−7.92672 · · · ⌋= −8. (b) Interpretation and further calculation −9.05019 ≲n ≲−6.86566 ⇒n ∈{−9, −8, −7}. Q ∈{−9e, −8e, −7e}. The whole interval must be intersected with the discrete lattice of allowed charges; a single rounded central estimate would discard compatible possibilities. Physical check The negative endpoints require care: the ceiling of a negative lower bound moves toward zero.

Question 3

A measurement interval can isolate one allowed integer charge

Use e=1.602176634×10−19e=1.602176634\times10^{-19} C, exact. An isolated object built from ordinary atoms is assumed to have Q=neQ=ne with integer nn. A calibrated hard measurement bound gives −1.33×10−18≤Q≤−1.27×10−18-1.33\times10^{-18}\le Q\le-1.27\times10^{-18} C. Treat the endpoints as exact bounds.

Diagram shown in the original SVG worksheet.

  1. Find all integer charge states compatible with the interval.

  2. The same instrument later reports −1.45×10−18≤Q≤−1.10×10−18-1.45\times10^{-18}\le Q\le-1.10\times10^{-18} C for another object. Find the allowed integer states and explain why rounding the midpoint alone is insufficient.

Question 3 - Solution

A measurement interval can isolate one allowed integer charge

Diagram shown in the original SVG worksheet.

(a) Method and calculation

−1.33×10−18e≤n≤−1.27×10−18e⇒−8.30121≲n≲−7.92672.\frac{-1.33\times10^{-18}}e\le n\le\frac{-1.27\times10^{-18}}e\Rightarrow -8.30121\lesssim n\lesssim-7.92672.nmin=⌈−8.30121⋯⌉=−8,nmax=⌊−7.92672⋯⌋=−8.n_{\min}=\lceil-8.30121\cdots\rceil=-8,\quad n_{\max}=\lfloor-7.92672\cdots\rfloor=-8.

(b) Interpretation and further calculation

−9.05019≲n≲−6.86566⇒n∈{−9,−8,−7}.-9.05019\lesssim n\lesssim-6.86566\Rightarrow n\in\{-9,-8,-7\}.Q∈{−9e,−8e,−7e}.Q\in\{-9e,-8e,-7e\}.The whole interval must be intersected with the discrete lattice of allowed charges; a single rounded central estimate would discard compatible possibilities.

Physical check

The negative endpoints require care: the ceiling of a negative lower bound moves toward zero.

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