Why is a $1$ $F$ unit so big in practice?
(N/A) The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$.
For a capacitance of $C = 1$ $F$,with a plate separation of $d = 1$ $cm = 10^{-2}$ $m$,the required area $A$ is:
$A = \frac{C d}{\epsilon_0} = \frac{1 \times 10^{-2}}{8.854 \times 10^{-12}} \approx 1.13 \times 10^9$ $m^2$.
If the plates are square,the side length $L = \sqrt{A} \approx \sqrt{1.13 \times 10^9} \approx 33.6$ $km$.
Since it is physically impossible to construct a capacitor with such massive plates,$1$ $F$ is considered an extremely large unit for practical purposes.
For a capacitance of $C = 1$ $F$,with a plate separation of $d = 1$ $cm = 10^{-2}$ $m$,the required area $A$ is:
$A = \frac{C d}{\epsilon_0} = \frac{1 \times 10^{-2}}{8.854 \times 10^{-12}} \approx 1.13 \times 10^9$ $m^2$.
If the plates are square,the side length $L = \sqrt{A} \approx \sqrt{1.13 \times 10^9} \approx 33.6$ $km$.
Since it is physically impossible to construct a capacitor with such massive plates,$1$ $F$ is considered an extremely large unit for practical purposes.
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