Charge Is Quantized — Question 1

Section 21-4 · Physics pages 3432–3433

Question 1 An exact elementary charge converts a net charge into an electron imbalance Use e = 1.602176634 × 10−19 C, exact. A small object has charge Q = −4.806529902 × 10−12 C, specified as an exact model value. Only electrons have been added to an initially neutral object; the positive constituents stay fixed. object Ne− Schematic; labels, not drawing lengths, specify the data. (a) Find the number of excess electrons and the charge change needed to return the object to neutrality. (b) If instead the displayed magnitude had only three significant figures, explain why the quotient by e would be an estimate rather than an exact electron count. Estimate the number of electron counts within a rounding half-width of .005 × 10−12 C.
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Question 1 - Solution An exact elementary charge converts a net charge into an electron imbalance object Ne− (a) Method and calculation Q = −Ne ⇒N = 4.806529902 × 10−12 1.602176634 × 10−19 = 3.0000000 × 107. ∆Q = +Ne = +4.806529902 × 10−12 C. (b) Interpretation and further calculation δN = δQ/e = .005 × 10−12 1.602176634 × 10−19 ≈3.12075 × 104. Nallowed count ≈2δN ≈6.24 × 104. Many adjacent integer electron counts fit a rounded macroscopic charge even though every individual object has a definite integer imbalance in this model. Physical check The exactness of e does not remove uncertainty or rounding in a measured charge.

Question 1

An exact elementary charge converts a net charge into an electron imbalance

Use e=1.602176634×10−19e=1.602176634\times10^{-19} C, exact. A small object has charge Q=−4.806529902×10−12Q=-4.806529902\times10^{-12} C, specified as an exact model value. Only electrons have been added to an initially neutral object; the positive constituents stay fixed.

Diagram shown in the original SVG worksheet.

  1. Find the number of excess electrons and the charge change needed to return the object to neutrality.

  2. If instead the displayed magnitude had only three significant figures, explain why the quotient by ee would be an estimate rather than an exact electron count. Estimate the number of electron counts within a rounding half-width of .005×10−12.005\times10^{-12} C.

Question 1 - Solution

An exact elementary charge converts a net charge into an electron imbalance

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Q=−Ne⇒N=4.806529902×10−121.602176634×10−19=3.0000000×107.Q=-Ne\Rightarrow N=\frac{4.806529902\times10^{-12}}{1.602176634\times10^{-19}}=3.0000000\times10^7.ΔQ=+Ne=+4.806529902×10−12C.\Delta Q=+Ne=+4.806529902\times10^{-12}\,\mathrm C.

(b) Interpretation and further calculation

δN=δQ/e=.005×10−121.602176634×10−19≈3.12075×104.\delta N=\delta Q/e=\frac{.005\times10^{-12}}{1.602176634\times10^{-19}}\approx3.12075\times10^4.Nallowedcount≈2δN≈6.24×104.N_{\mathrm{allowed\ count}}\approx2\delta N\approx6.24\times10^4.Many adjacent integer electron counts fit a rounded macroscopic charge even though every individual object has a definite integer imbalance in this model.

Physical check

The exactness of ee does not remove uncertainty or rounding in a measured charge.