Question
$1 F\ ($farad$)$ is a very large unit for practical purposes. Explain with example.
✓
Answer
$\rightarrow$ Suppose,
for a parallel plate capacitor having $1 F$ capacitance, distance between two plates is $1 \ cm$ and area of each plate is $A.$
$\rightarrow$ Capacitance of parallel plate capacitor,
$C =\frac{\varepsilon_0 A}{d}$
$\therefore A =\frac{ C d}{\varepsilon_0}=\frac{1 \times 1 \times 10^{-2}}{8.85 \times 10^{-12}}$
$\therefore A \approx 10^9 m^2$
$\therefore A =10 \times 10^8 m^2$
$\rightarrow$ Considering the shape of plate to be a square,
$\therefore l^2=10 \times 10^8 m^2$
$\therefore l \approx 3 \times 10^4 m$
$\therefore l=30 \ km$
$\rightarrow$ Which is practically impossible.
So $,1 F$ is a unit too large for practice.
for a parallel plate capacitor having $1 F$ capacitance, distance between two plates is $1 \ cm$ and area of each plate is $A.$
$\rightarrow$ Capacitance of parallel plate capacitor,
$C =\frac{\varepsilon_0 A}{d}$
$\therefore A =\frac{ C d}{\varepsilon_0}=\frac{1 \times 1 \times 10^{-2}}{8.85 \times 10^{-12}}$
$\therefore A \approx 10^9 m^2$
$\therefore A =10 \times 10^8 m^2$
$\rightarrow$ Considering the shape of plate to be a square,
$\therefore l^2=10 \times 10^8 m^2$
$\therefore l \approx 3 \times 10^4 m$
$\therefore l=30 \ km$
$\rightarrow$ Which is practically impossible.
So $,1 F$ is a unit too large for practice.
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