What is the area of the plates of a $2\,\mu F$ parallel plate capacitor,given that the separation between the plates is $0.5\, cm$?

(N/A) The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$.
Given values are $C = 2\,\mu F = 2 \times 10^{-6}\, F$ and $d = 0.5\, cm = 0.5 \times 10^{-2}\, m$.
The permittivity of free space is $\epsilon_0 = 8.854 \times 10^{-12}\, F/m$.
Rearranging the formula for area $A$,we get $A = \frac{C d}{\epsilon_0}$.
Substituting the values: $A = \frac{(2 \times 10^{-6}) \times (0.5 \times 10^{-2})}{8.854 \times 10^{-12}}$.
$A = \frac{1 \times 10^{-8}}{8.854 \times 10^{-12}} \approx 1129.43\, m^2$.
Thus,the area of the plates is approximately $1130\, m^2$.