The dielectric strength of air is about \(3\times10^{6}\ \mathrm{V\,m^{-1}}\). A parallel plate capacitor has air between its plates. The largest surface charge density that its plates can carry without breakdown of the air is about (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) → \(2.66\times10^{-5}\ \mathrm{C\,m^{-2}}\)
Dielectric strength is the largest field that air can bear before it stops insulating. Between the plates of a parallel plate capacitor the field depends only on the surface charge density, \( E=\sigma/\varepsilon_0 \). So the largest field fixes the largest \( \sigma \), whatever the plate area or the gap.
\( \begin{aligned} \sigma_{max} &= \varepsilon_0E_{max} \\ &= 8.85\times10^{-12}\times3\times10^{6} \\ &= 2.655\times10^{-5} \\ &\approx 2.66\times10^{-5}\ \mathrm{C\,m^{-2}} \end{aligned} \)
NCERT (2026-27):
Capacitors and Capacitance – “The maximum electric field that a dielectric medium can withstand without break-down (of its insulating property) is called its dielectric strength; for air it is about 3 × 106 Vm–1. …”
The Parallel Plate Capacitor – “In the inner region between the plates 1 and 2, the electric fields due to the two charged plates add up, giving”
\( E=\frac{\sigma}{2\varepsilon_0}+\frac{\sigma}{2\varepsilon_0}=\frac{\sigma}{\varepsilon_0} \)\( \qquad (2.41) \)