ch26_p53 - e = 1.60 10 19 C) 1.9 10 18 J for the work done...

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53. (a) Referring to Fig. 26-34, the electric field would point down (towards the bottom of the page) in the strip, which means the current density vector would point down, too (by Eq. 26-11). This implies (since electrons are negatively charged) that the conduction- electrons would be “drifting” upward in the strip. (b) Eq. 24-6 immediately gives 12 eV, or (using
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Unformatted text preview: e = 1.60 10 19 C) 1.9 10 18 J for the work done by the field (which equals, in magnitude, the potential energy change of the electron). (c) Since the electrons dont (on average) gain kinetic energy as a result of this work done, it is generally dissipated as heat. The answer is as in part (b): 12 eV or 1.9 10 18 J....
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This note was uploaded on 01/25/2011 for the course PHYSICS 17029 taught by Professor Rebello,nobels during the Spring '10 term at Kansas State University.

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