The electric field between the plates of a capacitor having area \( 2\,\mathrm{m^{2}} \) is \( 100\,\mathrm{V/m} \). The separation between the plates of capacitor is \( 5\,\mathrm{mm} \). The energy stored in the capacitor is :
(Permittivity of free space \( = 8.854\times10^{-12}\,\mathrm{C^{2}N^{-1}m^{-2}} \))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 4.4\times10^{-10}\,\mathrm{J} \)
The energy of a charged capacitor is stored in the field between its plates, with energy density \( u = \frac{1}{2}\varepsilon_0E^{2} \) spread uniformly over the volume \( Ad \).
\( A = 2\,\mathrm{m^{2}} \)
\( d = 5\,\mathrm{mm} = 5\times10^{-3}\,\mathrm{m} \)
\( E = 100\,\mathrm{V/m} \)
\( \displaystyle U = \frac{1}{2}\varepsilon_0E^{2}\,(Ad) \)
\( \displaystyle U = \frac{1}{2}\,(8.854\times10^{-12})\,(100)^{2}\,(2\times5\times10^{-3})\ \mathrm{J} \)
\( \displaystyle U = \frac{1}{2}\times8.854\times10^{-12}\times10^{4}\times10^{-2} = 4.427\times10^{-10}\,\mathrm{J} \)
\( U \approx 4.4\times10^{-10}\,\mathrm{J} \)
The same result follows from \( U = \frac{1}{2}CV^{2} \) with \( C = \varepsilon_0A/d \) and \( V = Ed \).
Shortcut: All options have the digits 4.4, so only the power of ten matters: \( \frac{1}{2}\times8.854 \approx 4.4 \) and \( 10^{-12}\times10^{4}\times10^{-2} = 10^{-10} \).