A large capacitor is constructed of square metal plates 2.0 cm on a side sandwic

Question

A large capacitor is constructed of square metal plates 2.0 cm on a side sandwiching a bakelite sheet 0.02 nun thick. Using the fact that the dielectric constant of bakelite is k=7.0 and that the permittivity of free space is epsilon 0 = 8.85 x 10-12 F/m, compute the capacitance of this capacitor in Farads. The capacitor above is placed next to a smaller capacitor and the two are wired together as shown below. The smaller capacitor is constructed from the same materials as the large one: however the bakelite sheet is 0.04 mm thick and the metal plates are 1.0 cm on a side. Compute the total capacitance of this device in Farads.

Explanation / Answer

c=A/d = orA/d = 8.85*10^-12*7(.02)2 /(2*10^-5) = 1.24nF

b) C of smaller /(4*10^-5) = .155nF

total capacitance = C of big + C of small = 1.395nF

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