Chap 18 Solns-6E

# Chap 18 Solns-6E - = 0.05 V 2 m = 2.5 x 10-2 V/m 201...

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CHAPTER 18 ELECTRICAL PROPERTIES PROBLEM SOLUTIONS 18.5 (a) In order to compute the resistance of this copper wire it is necessary to employ Equations (18.2) and (18.4). Solving for the resistance in terms of the conductivity, R = ρ l A = l σ A From Table 18.1, the conductivity of copper is 6.0 x 10 7 ( -m) -1 , and R = l σ A = 2 m 6.0 x 10 7 ( Ω - m) - 1 [ ] ( π ) 3 x 10 - 3 m 2 2 = 4.7 x 10 -3 (b) If V = 0.05 V then, from Equation (18.1) I = V R = 0.05 V 4.7 x 10 - 3 = 10.6 A (c) The current density is just J = I A = I π d 2 2 = 10.6 A π 3 x 10 - 3 m 2 2 = 1.5 x 10 6 A/m 2 (d) The electric field is just E = V l

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Unformatted text preview: = 0.05 V 2 m = 2.5 x 10-2 V/m 201 18.28 (a) No hole is generated by an electron excitation involving a donor impurity atom because the excitation comes from a level within the band gap, and thus, no missing electron is created from the normally filled valence band. (b) No free electron is generated by an electron excitation involving an acceptor impurity atom because the electron is excited from the valence band into the impurity level within the band gap; no free electron is introduced into the conduction band. 202...
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Chap 18 Solns-6E - = 0.05 V 2 m = 2.5 x 10-2 V/m 201...

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