Physics · Electrostatic Potential
हिन्दीA parallel plate capacitor with vacuum between its plates has a plate separation of \( 0.885\text{ mm} \). What plate area \( A \) is required to produce a capacitance of \( 2\text{ nF} \)? (Take \( \varepsilon_0 = 8.85 \times 10^{-12}\text{ C}^2\cdot\text{N}^{-1}\cdot\text{m}^{-2} \))
Correct answer · Option 2
\( 0.2\text{ m}^2 \)
Explanation
\( A = \frac{C d}{\varepsilon_0} = \frac{(2 \times 10^{-9}\text{ F}) \times (0.885 \times 10^{-3}\text{ m})}{8.85 \times 10^{-12}\text{ F/m}} = 0.2\text{ m}^2 \).