Question

A parallel plate capacitor with plate area $100c{m}^2$ and sepration between the plate 5 mm is connected across a 24 V battery. The force of attraction between the plates is of the order of

• A. ${10}^{-6}N$
• B. ${10}^{-8}N$
• C. ${10}^{-4}N$
• D. ${10}^{-7}N$

Answer 1

Answer: (A)

${10}^{-6}N$

The capacitance of parallel plate capacitor is $C=\frac{A{ϵ}_0}{d}$
The force of attraction between the plates is $F=\frac{Q^2}{2A{ϵ}_0}=\frac{C^2{V}^2}{2A{ϵ}_0}$
$F=(\frac{A{ϵ}_0}{d}{)}^2.\frac{V^2}{2A{ϵ}_0}=\frac{A{ϵ}_0{V}^2}{d^2}=\frac{100\times {10}^{-2}\times 8.85\times {10}^{-12}\times {24}^2}{5\times {10}^{-3}}={10}^{-6}N$