Question

A parallel-plate capacitor with the plate area 100 cm² and the separation between the plates 1⋅0 cm is connected across a battery of emf 24 volts. Find the force of attraction between the plates.

Answer 1

Area of the plates of the capacitor, A = 100 cm² = 10^$-$2 m²
Separation between the plates, d = 1 cm = 10^$-$2 m
Emf of battery, V = 24 V

Therefore,
Capacitance, $C=\frac{{\in }_{0}A}{d}=\frac{(8.85\times {10}^{-12})\times \left({10}^{-2}\right)}{{10}^{-2}}=8.85\times {10}^{-12}\mathrm{V}$

Energy stored in the capacitor, $$\begin{gathered} E=\frac{1}{2}C{V}^2=\frac{1}{2}\times (8.85\times {10}^{-12})\times (24{)}^2 \\ =2548.8\times {10}^{-12}\mathrm{J} \end{gathered}$$

Force of attraction between the plates, $F=\frac{E}{d}=\frac{2548.8\times {10}^{-12}}{{10}^{-2}}=2548.8\times {10}^{-10}\mathrm{N}$