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If a capacitor having capacitance 2 F and plate separation of 0.5 cm will have area
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$1130 k m^2$
$\because$ Capacitance of a parallel plate capacitor,
$\begin{aligned} \mathrm{C} & =\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}} \quad \mathrm{~A}=\frac{\mathrm{Cd}}{\varepsilon_0}=\frac{2 \times 0.5 \times 10^{-2}}{8.85 \times 10^{-12}} \\ & =1.130 \times 10^9 \mathrm{~m}^2=1130 \mathrm{~km}^2\end{aligned}$
$\begin{aligned} \mathrm{C} & =\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}} \quad \mathrm{~A}=\frac{\mathrm{Cd}}{\varepsilon_0}=\frac{2 \times 0.5 \times 10^{-2}}{8.85 \times 10^{-12}} \\ & =1.130 \times 10^9 \mathrm{~m}^2=1130 \mathrm{~km}^2\end{aligned}$
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