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736_PartUniversity Physics Solution

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

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24-26 Chapter 24 24.77. I DENTIFY : 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. S ET U P : For each capacitor in the parallel combination, 0 A C d = P . E XECUTE : (a) The charge distribution on the plates is shown in Figure 24.77. (b) 2 9 0 0 4 2(4.2) (0.120 m) 2 2.38 10 F 4.5 10 m A C d = = = × × P P . E VALUATE : If two of the plates are separated by both sheets of paper to form a capacitor, 9 0 2.38 10 F 2 4 A C d × = = P , smaller by a factor of 4 compared to the capacitor in the problem. Figure 24.77 24.78. I DENTIFY : As in Problem 24.72, the system is equivalent to two capacitors in parallel. One of the capacitors has plate separation d , plate area ( ) w L h and air between the plates. The other has the same plate separation d , plate area wh and dielectric constant K . S ET U P : Define eff K by eff 0 eq K A C d = P , where A wL = . For two capacitors in parallel, eq 1 2 C C C = + . E XECUTE : (a) The capacitors are in parallel, so 0 0 0 ( ) 1 w L h K wh wL Kh h C d d d L L = + = + P P P . This gives eff 1 Kh h K L L = + . (b) For gasoline, with 1.95: K = 1 4 full: eff 1.24 4 L K h = = ; 1 2 full: eff 1.48 2 L K h = = ; 3 4 full: eff 3 1.71. 4 L K h = = (c) For methanol, with 33: K = 1 4 full: eff 9 4 L K h = = ; 1 2 full: eff 17 2 L K h = = ; 3 4 full: eff 3 25. 4 L K h = = (d) This kind of fuel tank sensor will work best for methanol since it has the greater range of eff K values. E VALUATE : When 0 h = , eff 1 K = . When h L = , eff K K = .
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