Q108 / 125 · BITSAT 2009 · Physics
Calculate the area of the plates of a one farad parallel plate capacitor if separation between plates is $1 \mathrm{~mm}$ and plates are in vacuum
- A$18 \times 10^{8} \mathrm{~m}^{2}$
- B$0.3 \times 10^{8} \mathrm{~m}^{2}$
- C$1.3 \times 10^{8} \mathrm{~m}^{2}$
- Correct answerD$1.13 \times 10^{8} \mathrm{~m}^{2}$
Solution
For a parallel plate capacitor $\mathrm{C}=\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$
$$
\begin{aligned}
\therefore \quad & \mathrm{A}=\frac{\mathrm{Cd}}{\varepsilon_{0}}=\frac{1 \times 10^{-3}}{8.85 \times 10^{-12}} \\
&=1.13 \times 10^{8} \mathrm{~m}^{2}
\end{aligned}
$$
This corresponds to area of square of side $10.6 \mathrm{~km}$ which shows that one farad is very large unit of capacitance.
Hence, the correct answer is option: 'D'.