For a parallel plate capacitor having plate area \( 1.13 \times 10^{3}\,\mathrm{m^{2}} \) and separation between plates 0.5 cm, the capacitance is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 2.0\,\mu\mathrm{F} \)
The capacitance of a parallel-plate capacitor with air between the plates is proportional to the plate area:
\( \displaystyle C = \frac{\varepsilon_0 A}{d} \)
\( \varepsilon_0 = 8.85\times10^{-12}\,\mathrm{F\,m^{-1}} \)
\( d = 0.5\,\mathrm{cm} = 5\times10^{-3}\,\mathrm{m} \)
\( A = 1.13\times10^{3}\,\mathrm{m^{2}} \)
\( \displaystyle C = \frac{8.85\times10^{-12} \times 1.13\times10^{3}}{5\times10^{-3}} = \frac{1.0\times10^{-8}}{5\times10^{-3}} = 2.0\times10^{-6}\,\mathrm{F} = 2.0\,\mu\mathrm{F} \)