Charge Is Conserved — Question 6

Section 21-5 · Physics pages 3462–3463

Question 6 An emission experiment must count returned and escaped electrons A neutral emitter S sends 1000 electrons to an initially neutral collector C. Later, 200 of those electrons return from C to S and 300 leave C for a distant initially neutral enclosure E. No other particles move. Treat S, C and E as the complete isolated charge system. S C E 1000 200 300 Schematic; labels, not drawing lengths, specify the data. (a) Find the final charges of S, C and E in units of e. (b) Find the final total charge if an observer counts only S and C. Explain the apparent imbalance using the charge crossing that smaller system boundary.
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Question 6 - Solution An emission experiment must count returned and escaped electrons S C E 1000 200 300 (a) Method and calculation QS = (1000 −200)e = +800e. QC = (−1000 + 200 + 300)e = −500e, QE = −300e. QS + QC + QE = 0. (b) Interpretation and further calculation QS+C = 800e −500e = +300e. ∆QS+C = −Qoutward = −(−300e) = +300e. The two-component subsystem becomes positive because electrons leave it. The complete system remains neutral. Physical check All three arrows count electrons, so the signed charge carried along each arrow is negative.

Question 6

An emission experiment must count returned and escaped electrons

A neutral emitter S sends 1000 electrons to an initially neutral collector C. Later, 200 of those electrons return from C to S and 300 leave C for a distant initially neutral enclosure E. No other particles move. Treat S, C and E as the complete isolated charge system.

Diagram shown in the original SVG worksheet.

  1. Find the final charges of S, C and E in units of ee.

  2. Find the final total charge if an observer counts only S and C. Explain the apparent imbalance using the charge crossing that smaller system boundary.

Question 6 - Solution

An emission experiment must count returned and escaped electrons

Diagram shown in the original SVG worksheet.

(a) Method and calculation

QS=(1000−200)e=+800e.Q_S=(1000-200)e=+800e.QC=(−1000+200+300)e=−500e,QE=−300e.Q_C=(-1000+200+300)e=-500e,\quad Q_E=-300e.QS+QC+QE=0.Q_S+Q_C+Q_E=0.

(b) Interpretation and further calculation

QS+C=800e−500e=+300e.Q_{S+C}=800e-500e=+300e.ΔQS+C=−Qoutward=−(−300e)=+300e.\Delta Q_{S+C}=-Q_{\mathrm{outward}}=-(-300e)=+300e.The two-component subsystem becomes positive because electrons leave it. The complete system remains neutral.

Physical check

All three arrows count electrons, so the signed charge carried along each arrow is negative.