The separation between a parallel plate capacitor having capacitance \(2 \mathrm{~F}\) is \(0.5…

The separation between a parallel plate capacitor having capacitance \(2 \mathrm{~F}\) is \(0.5 \mathrm{~cm}\). The area of the plates of capacitor is
  1. \(3.03 \times 10^{9} \mathrm{~m}^{2}\)
  2. \(1.13 \times 10^{9} \mathrm{~m}^{2}\)
  3. \(3.13 \times 10^{9} \mathrm{~m}^{2}\)
  4. \(0.13 \times 10^{9} \mathrm{~m}^{2}\)

Solution

\(\mathrm{C}=2 \mathrm{~F} ; \mathrm{d}=0.5 \times 10^{-2} \mathrm{~m} ; \varepsilon_{0}=8.85 \times 10^{-12} \mathrm{~F} / \mathrm{m}\)
\(\mathrm{C}=\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}\)
\(\quad \mathrm{Or}, \quad \mathrm{A}=\frac{\mathrm{Cd}}{\varepsilon_{0}}\)
\(=\frac{2 \times 0.5 \times 10^{-2}}{8.85 \times 10^{-12}} \mathrm{~m}^{2}\)
\(=1.13 \times 10^{9} \mathrm{~m}^{2}\)

Asked in: JEE Mains - Capacitance - Chapter Test

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