The separation between the plates of a parallel plate capacitor of capacitance 2 pF is 0.5 mm. The area of the plates of the capacitor is (Given: \( \varepsilon_0 = 8.85 \times 10^{-12}\,\mathrm{C^2\,N^{-1}\,m^{-2}} \))
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 1.13\,\mathrm{cm^2} \)
The capacitance of a parallel plate capacitor with air between the plates is \( C = \varepsilon_0 A/d \), so the area is
\( \displaystyle A = \frac{C\,d}{\varepsilon_0} \)
\( C = 2\,\mathrm{pF} = 2\times10^{-12}\,\mathrm{F} \)
\( d = 0.5\,\mathrm{mm} = 5\times10^{-4}\,\mathrm{m} \)
\( \displaystyle A = \frac{2\times10^{-12} \times 5\times10^{-4}}{8.85\times10^{-12}} = 1.13\times10^{-4}\,\mathrm{m^2} \)
\( 1\,\mathrm{m^2} = 10^{4}\,\mathrm{cm^2} \)
\( A = 1.13\times10^{-4} \times 10^{4} = 1.13\,\mathrm{cm^2} \)