A parallel-plate capacitor has capacitance of 1.0 F. If the plates are 1.0 mm apart, what is the area of the plates?

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Solution

$C = \frac{ε_{o} A}{d}$

$∴ A = \frac{C d}{ε_{o}} = \frac{(1) (10^{- 3})}{8.86 \times 10^{- 12}}$

$= 1.1 \times 10^{8} m^{2}$

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A parallel-plate capacitor has capacitance of 1.0 F. If the plates are 1.0 mm apart, what is the area of the plates?

C=εoAd∴A=Cdεo=(1)(10−3)8.86×10−12=1.1×108m2

$C = \frac{ε_{o} A}{d}$

$∴ A = \frac{C d}{ε_{o}} = \frac{(1) (10^{- 3})}{8.86 \times 10^{- 12}}$

$= 1.1 \times 10^{8} m^{2}$