"Pumping" Charges — Question 2

Section 27-1 · Physics pages 4314–4315

Question 2 Electron transfer through a source has the same energy requirement as conventional pumping Use e = 1.602176634 × 10−19 C, exact. An ideal 9.00 V source moves N = 2.00 × 1018 electrons internally from its positive terminal to its negative terminal in 5.00 s. + − electron motion conventional current Schematic; labels, not drawing lengths, specify the data. (a) Find the conventional current direction and magnitude. (b) Find each electron potential-energy increase and the total non-electrostatic work.
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Question 2 - Solution Electron transfer through a source has the same energy requirement as conventional pumping + − electron motion conventional current (a) Method and calculation |Q| = Ne = .320435 C, I = Ne/t = .0640871 A. Conventional current inside source: negative terminal →positive terminal. (b) Interpretation and further calculation ∆Ue = (−e)(V−−V+) = 9.00e = 1.44196 × 10−18 J. W = N∆Ue = (Ne)(9.00) = 2.88392 J. Physical check A negative charge increases its potential energy by moving to a lower electric potential.

Question 2

Electron transfer through a source has the same energy requirement as conventional pumping

Use e=1.602176634×10−19e=1.602176634\times10^{-19} C, exact. An ideal 9.00 V source moves N=2.00×1018N=2.00\times10^{18} electrons internally from its positive terminal to its negative terminal in 5.00 s.

Diagram shown in the original SVG worksheet.

  1. Find the conventional current direction and magnitude.

  2. Find each electron potential-energy increase and the total non-electrostatic work.

Question 2 - Solution

Electron transfer through a source has the same energy requirement as conventional pumping

Diagram shown in the original SVG worksheet.

(a) Method and calculation

|Q|=Ne=.320435C,I=Ne/t=.0640871A.|Q|=Ne=.320435\,\mathrm C,\quad I=Ne/t=.0640871\,\mathrm A.Conventional current inside source: negative terminal→positive terminal.\text{Conventional current inside source: negative terminal}\to\text{positive terminal}.

(b) Interpretation and further calculation

ΔUe=(−e)(V−−V+)=9.00e=1.44196×10−18J.\Delta U_e=(-e)(V_--V_+)=9.00e=1.44196\times10^{-18}\,\mathrm J.W=NΔUe=(Ne)(9.00)=2.88392J.W=N\Delta U_e=(Ne)(9.00)=2.88392\,\mathrm J.

Physical check

A negative charge increases its potential energy by moving to a lower electric potential.