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Question 23.1: The plates of a parallel-plate capacitor are separated in ai......

The plates of a parallel-plate capacitor are separated in air by a distance d = 1 mm. (a) Find the capacitance of this capacitor if its area is A = 1 cm². (b) What must be the plate area if its capacitance is to be 1 F?

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(a) From Eq. 23.5, we have:

C={\frac{\epsilon_{o}A}{d}}        (Parallel-plate capacitor)        (23.5)

={\frac{(8.85\times10^{-12}\,{\mathrm{F/m}})(1\times10^{-4}\,{\mathrm{m}}^{2})}{(1\times10^{-3}\,{\mathrm{m}})}}=\;8.85\times10^{-13}\,{\mathrm{F}}=\;0.885\,{\mathrm{pF}}

(b) From Eq. 23.5, we have:

A={\frac{C d}{\epsilon_{o}}}={\frac{(1\,{\mathrm{F}})(1\times10^{-3}\,{\mathrm{m}})}{(8.85\times10^{-12}\,{\mathrm{F}}/{\mathrm{m}})}}=1.13\times10^{8}\,{\mathrm{m}}^{2}

This is an area of a square that has a side of more than 10.6 km. Therefore, the farad is indeed a large unit. However, modern technology has permitted the construction of a 1 F capacitor of a very modest size. This capacitor is used as a backup power supply (up to many months) for computer memory chips in case of a power failure.

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