Problem A parallel-plate capacitor has plates of area A=100 cm2A = 100\ \text{cm}^2A=100 cm2, separation d=1 mmd = 1\ \text{mm}d=1 mm, and an air dielectric. Calculate the capacitance using the formula C=ε0A/dC = \varepsilon_0 A/dC=ε0A/d. Solution Unit conversion. Area in square metres: A=100 cm2=100⋅10−4 m2=1⋅10−2 m2,d=1 mm=1⋅10−3 mA = 100\ \text{cm}^2 = 100\cdot 10^{-4}\ \text{m}^2 = 1\cdot 10^{-2}\ \text{m}^2, \qquad d = 1\ \text{mm} = 1\cdot 10^{-3}\ \text{m}A=100 cm2=100⋅10−4 m2=1⋅10−2 m2,d=1 mm=1⋅10−3 m Capacitance of the parallel-plate capacitor. With ε0=8,85⋅10−12 F/m\varepsilon_0 = 8{,}85\cdot 10^{-12}\ \text{F/m}ε0=8,85⋅10−12 F/m: C=ε0 Ad=8,85⋅10−12⋅1⋅10−21⋅10−3=8,85⋅10−141⋅10−3=8,85⋅10−11 FC = \varepsilon_0\,\frac{A}{d} = 8{,}85\cdot 10^{-12}\cdot \frac{1\cdot 10^{-2}}{1\cdot 10^{-3}} = \frac{8{,}85\cdot 10^{-14}}{1\cdot 10^{-3}} = 8{,}85\cdot 10^{-11}\ \text{F}C=ε0dA=8,85⋅10−12⋅1⋅10−31⋅10−2=1⋅10−38,85⋅10−14=8,85⋅10−11 F In picofarads: C≈88,5 pF\ev{C \approx 88{,}5\ \text{pF}}C≈88,5 pF Links Topics: Electric field and potential Concepts: Capacitance and capacitor Objects: Parallel-plate capacitor