Physics - Capacitor Question with Solution | TestHub

The capacitance of a parallel plate capacitor is given by $C=\frac{{ϵ}_{0}A}{d}$. Given $A=200c{m}^2=200\times 10^{-4}{m}^2=2\times 10^{-2}{m}^2$. Given $d=1mm=1\times 10^{-3}m$. Given ${ϵ}_0=8.85\times 10^{-12}F/m$.$C=\frac{(8.85\times 10^{-12}F/m)\times (2\times 10^{-2}{m}^2)}{1\times 10^{-3}m}=8.85\times 2\times 10^{-12}\times 10^{-2}\times 10^{3}F=17.7\times 10^{-11}F=177\times 10^{-12}F$. Since $1pF=10^{-12}F$, the capacitance is $177pF$.