A parallel plate capacitor with plate area \(100 \mathrm{~cm}^{2}\) and separation between the plates \(1…

A parallel plate capacitor with plate area \(100 \mathrm{~cm}^{2}\) and separation between the plates \(1.0\) \(\mathrm{cm}\) is connected across a battery of emf \(24 \mathrm{~V}\). The force of attraction between the plates is
  1. \(1.0 \times 10^{-7} \mathrm{~N}\)
  2. \(2.5 \times 10^{-7} \mathrm{~N}\)
  3. \(4 \times 10^{-5} \mathrm{~N}\)
  4. \(1.6 \times 10^{-5} \mathrm{~N}\)

Solution



Force between the two plates of capacitor \(\mathrm{C}=\mathrm{A} \varepsilon / \mathrm{d}\)
\(\begin{array}{l}
\mathrm{F}=\frac{\mathrm{Q}^{2}}{2 \varepsilon_{0} \mathrm{~A}}=\frac{\left(\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}} \cdot \mathrm{V}\right)^{2}}{2 \varepsilon_{0} \mathrm{~A}} \\
=\frac{\varepsilon_{0}^{2} \mathrm{~A}^{2} \mathrm{~V}^{2}}{\mathrm{~d}^{2} 2 \varepsilon_{0} \mathrm{~A}}=\frac{\varepsilon_{0} \mathrm{AV}^{2}}{2 \mathrm{~d}^{2}} \\
=\frac{8.85 \times 10^{-12} \times 100 \times 10^{-4} \times 24 \times 24}{2 \times 10^{-4}}=2.5 \times 10^{-7} \mathrm{~N}
\end{array}\)

Asked in: JEE Mains - Capacitance - Chapter Test

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