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 4.80 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 6.50 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.)

Explanation / Answer

capacitance = dielctric constant of medium * permittivity of free space * area / thickness

C = k e0 A/d

6.5E-8 = 3.7*8.85E-12*(4.8E-2*L)/(0.02E-3)

L=0.827 m

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