Little confused here, Consider a plane parallel-plate capacitor made of two stri

Question

Little confused here,

Consider a plane parallel-plate capacitor made of two strips of aluminum foil separated by a layer of paraffin-coated paper. Each strip of foil and paper is 6.80 cm wide. The foil is 0.00400 mm thick, and the paper is 0.0350 mm thick and has a dielectric constant of 3.70. What length should the strips be if a capacitance of 5.00 times 10-8 F is desired? (If, after this plane capacitor is formed, a second paper strip can be added below the foil-paper-foil stack and the resulting assembly rolled into a cylindrical form-similar to that shown in the figure below-the capacitance can be doubled because both surfaces of each foil strip would then store charge. Without the second strip of paper, however, rolling the layers would result in a short circuit.) m

Explanation / Answer

since given, width,w = 6.8 cm = 0.068 m
d = thickness of paper = 3.5*10^-2 mm = 3.5*10^-5 m
k = 3.7
let length = l
area = width*length = wl = 0.068l
capacitance is given by, C = k0A/d = 3.7*8.854*10^-12*0.068l/3.5*10^-5 = 5*10^-8 F

this gives l = 0.7856 m Ans

if the capacitor is rolled, the required length would get halved

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