24-26Chapter 2424.77. 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,0ACd=P. EXECUTE:(a)The charge distribution on the plates is shown in Figure 24.77. (b) 290042(4.2)(0.120 m)22.3810F4.510mACd−−⎛⎞===×⎜⎟×⎝⎠PP. EVALUATE:If two of the plates are separated by both sheets of paper to form a capacitor, 902.3810F24ACd−×==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 ()w Lh−and air between the plates. The other has the same plate separation d, plate area whand dielectric constant K. SET UP:Define effKby eff0eqKACd=P, where AwL=. For two capacitors in parallel, eq12CCC=+. EXECUTE:(a)The capacitors are in parallel, so 000()1w LhKwhwLKhhCdddLL−⎛⎞=+=+−⎜⎟⎝⎠PPP. This gives eff1KhhKLL⎛⎞=+−⎜⎟⎝⎠. (b) For gasoline, with 1.95:K=14full: eff1.244LKh⎛⎞==⎜⎟⎝⎠; 12full: eff1.482LKh⎛⎞==⎜⎟⎝⎠; 34full:eff31.71.4LKh⎛⎞==⎜⎟⎝⎠(c) For methanol, with 33:K=14full: eff94LKh⎛⎞==⎜⎟⎝⎠; 12full: eff172LKh⎛⎞==⎜⎟⎝⎠; 34full: eff325.4LKh⎛⎞==⎜⎟⎝⎠(d) This kind of fuel tank sensor will work best for methanol since it has the greater range of effKvalues. EVALUATE:When 0h=, eff1K=. When hL=, effKK=.
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