A parallel plate capacitor has air between its plates and a capacitance of 25 pF. The plates are 2 mm apart. The dielectric strength of air is \(3\times10^{6}\ \mathrm{V\,m^{-1}}\). The largest charge that the capacitor can hold without breakdown of the air is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(1.5\times10^{-7}\ \mathrm{C}\)
The dielectric strength is the largest electric field that the air can bear without breakdown. Between the plates the field is \( E=V/d \), so this sets the largest potential difference. The charge then follows from \( Q=CV \).
Given:
\( C=25\ \mathrm{pF}=25\times10^{-12}\ \mathrm{F} \)
\( d=2\ \mathrm{mm}=2\times10^{-3}\ \mathrm{m} \)
\( \begin{aligned} V_{max} &= E_{max}\,d \\ &= 3\times10^{6}\times2\times10^{-3} \\ &= 6\times10^{3}\ \mathrm{V} \end{aligned} \)
\( \begin{aligned} Q_{max} &= CV_{max} \\ &= 25\times10^{-12}\times6\times10^{3} \\ &= 1.5\times10^{-7}\ \mathrm{C} \end{aligned} \)
Therefore, the capacitor can hold at most \( 1.5\times10^{-7}\ \mathrm{C} \) before the air between the plates breaks down.
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. …”
Summary – “9. A capacitor is a system of two conductors separated by an insulator. Its capacitance is defined by C = Q/V, where Q and –Q are the charges on the two conductors and V is the potential difference between them. …”