736_PartUniversity Physics Solution

736_PartUniversity Physics Solution - 24-26 Chapter 24...

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24-26 Chapter 24 24.77. IDENTIFY: The object is equivalent to two identical capacitors in parallel, where each has the same area A , plate separation d and dielectric with dielectric constant K. SET UP: For each capacitor in the parallel combination, 0 A C d = P . EXECUTE: (a) The charge distribution on the plates is shown in Figure 24.77. (b) 2 9 00 4 2(4.2) (0.120 m) 22 . 3 8 1 0 F 4.5 10 m A C d ⎛⎞ == = × ⎜⎟ × ⎝⎠ PP . EVALUATE: If two of the plates are separated by both sheets of paper to form a capacitor, 9 0 2.38 10 F 24 A C d × P , smaller by a factor of 4 compared to the capacitor in the problem. Figure 24.77 24.78. IDENTIFY: As in Problem 24.72, the system is equivalent to two capacitors in parallel. One of the capacitors has plate separation d , plate area ( ) wL h and air between the plates. The other has the same plate separation d , plate area wh and dielectric constant K . SET UP: Define eff K by eff 0 eq K A C d = P , where Aw L = . For two capacitors in parallel, eq 1 2 CC C =+ . EXECUTE: (a) The capacitors are in parallel, so 0 () 1 w L h K wh wL Kh h C dd d L L = + P . This gives eff 1 K hh K L L =+ − . (b) For gasoline, with 1.95: K = 1 4 full: eff 1.24 4 L Kh ; 1 2 full: eff 1.48 2 L ; 3 4 full: eff 3 1.71. 4 L (c) For methanol, with 33: K = 1 4 full: eff 9 4 L = = ; 1 2 full: eff 17 2 L = = ; 3 4 full: eff 3 25. 4 L (d) This kind of fuel tank sensor will work best for methanol since it has the greater range of eff K values. EVALUATE: When 0 h = , eff 1 K = . When hL = , eff K K = .
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25-1 C URRENT , R ESISTANCE , AND E LECTROMOTIVE F ORCE 25.1. IDENTIFY: / IQ t = . SET UP: 1.0 h 3600 s = EXECUTE: 4 (3.6 A)(3.0)(3600 s) 3.89 10 C. QI t == = × EVALUATE: Compared to typical charges of objects in electrostatics, this is a huge amount of charge. 25.2. IDENTIFY: / t = . Use d I nqvA = to calculate the drift velocity d . v SET UP: 28 3 5.8 10 m . n 19 1.60 10 C q . EXECUTE: (a) 2 420 C 8.75 10 A. 80(60 s) Q I t = × (b) d . = This gives 2 6 d 28 19 3 2 8.75 10 A 1.78 10 m s. (5.8 10 )(1.60 10 C)( (1.3 10 m) ) I v nqA π −− × = × ×× × EVALUATE: d v is smaller than in Example 25.1, because I is smaller in this problem. 25.3. IDENTIFY: / t = . / J IA = . d J nqv = SET UP: 2 (/ 4 ) A D = , with 3 2.05 10 m D . The charge of an electron has magnitude 19 1.60 10 C. e += × EXECUTE: (a) (5.00 A)(1.00 s) 5.00 C. t = The number of electrons is 19 3.12 10 . Q e (b) 62 23 2 5.00 A 1.51 10 A/m . 4 ) 4 ) ( 2 . 0 51 0 m ) I J D ππ = × × (c) 4 d 28 3 19 1.51 10 A/m 1.11 10 m/s 0.111 mm/s. (8.5 10 m )(1.60 10 C) J v nq × = EVALUATE: (a) If I is the same, / = would decrease and d v would decrease. The number of electrons passing through the light bulb in 1.00 s would not change. 25.4. (a) IDENTIFY: By definition, J = I/A and radius is one-half the diameter. SET UP: Solve for the current: I = JA = J π ( D /2) 2 EXECUTE: I = (1.50 × 10 6 A/m 2 )( π )[(0.00102 m)/2] 2 = 1.23 A EVALUATE: This is a realistic current.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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736_PartUniversity Physics Solution - 24-26 Chapter 24...

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