The electric field between the plates of a parallel plate air capacitor is \(2\times10^{5}\ \mathrm{V\,m^{-1}}\). Each plate has an area of \(0.01\ \mathrm{m^{2}}\), and the plates are 1 mm apart. The energy stored in the capacitor is (Take \(\varepsilon_0=8.85\times10^{-12}\ \mathrm{C^{2}\,N^{-1}\,m^{-2}}\).)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(1.77\times10^{-6}\ \mathrm{J}\)
The energy of a charged capacitor is stored in the electric field between its plates. The energy per unit volume, called the energy density, is \( u=\frac{1}{2}\varepsilon_0E^{2} \). The total energy is \( u \) times the volume \( Ad \) between the plates.
\( \begin{aligned} u &= \frac{1}{2}\varepsilon_0E^{2} \\ &= \frac{1}{2}\times8.85\times10^{-12}\times(2\times10^{5})^{2} \\ &= 0.177\ \mathrm{J\,m^{-3}} \end{aligned} \)
\( \begin{aligned} Ad &= 0.01\times10^{-3} \\ &= 10^{-5}\ \mathrm{m^{3}} \end{aligned} \)
\( \begin{aligned} U &= u\times Ad \\ &= 0.177\times10^{-5} \\ &= 1.77\times10^{-6}\ \mathrm{J} \end{aligned} \)
NCERT (2026-27):
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