Consider a plane parallel-plate capacitor made of two strips of aluminum foil se

Question

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 6.60 cm wide. The foil is 0.00400 mm thick, and the paper is 0.0200 mm thick and has a dielectric constant of 3.70. What length should the strips be if a capacitance of 4.50 108 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.)

Explanation / Answer

C = k*o*A / d
k = 3.70, d = 0.02*10^-3 m
So the required plate area is
A = C*d / (k*o)
A = 4.50*10^-8 * 0.02*10^-3 / (3.7*8.854*10^-12) = 0.02747 m^2

A = L * W
so the strip Length is
L = A / W = 0.02747m^2 / 6.60*10^-2m = 0.4162 m

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