A parallel-plate capacitor has plates with an area of 383 cm^2 and an air-filled

Question

A parallel-plate capacitor has plates with an area of 383 cm^2 and an air-filled gap between the plates that is 2.55 mm thick. The capacitor is charged by a battery to 575 V and then is disconnected from the battery. How much energy is stored in the capacitor? The separation between the plates is now increased to 5.1 mm. How much energy is stored in the capacitor now? How much work is required to increase the separation of the plates from 2.55 mm to 5.1 mm? Explain your reasoning. This answer has not been graded yet.

Explanation / Answer

part a )

C = eo*A/d

C = 8.85 x 10^-12 x 383 x 10^-4 / 2.55 x 10^-3

C = 1.329 x 10^-10 F = 132.9 x 10^-12 F

Ui = 1/2*CV^2

Ui = 2.197 x 10^-5 J

part b )

C' = eo*A/d

C' = 6.646 x 10^-11 F = 66.46 x 10^-12 F

Q remain same when pull apart the plate

Q = CV = 7.64 x 10^-8 C

Uf = 1/2 * Q^2/C'

Uf = 4.39 x 10^-5 J

part c )

w = Uf - Ui

w = 2.196 x 10^-5 J

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