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