The electric field between the plates of a parallel plate air capacitor is \(4\times10^{5}\ \mathrm{V\,m^{-1}}\). The plates are 1.5 mm apart, and the charge on the capacitor is \(1.2\ \mu\mathrm{C}\). The capacitance of the capacitor is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(2\ \mathrm{nF}\)
The field between the plates of a parallel plate capacitor is uniform. So the potential difference between the plates is the field times the gap, \( V=Ed \). The capacitance then comes from \( C=Q/V \).
Given:
\( Q=1.2\ \mu\mathrm{C}=1.2\times10^{-6}\ \mathrm{C} \)
\( E=4\times10^{5}\ \mathrm{V\,m^{-1}} \)
\( d=1.5\ \mathrm{mm}=1.5\times10^{-3}\ \mathrm{m} \)
\( \begin{aligned} V &= Ed \\ &= 4\times10^{5}\times1.5\times10^{-3} \\ &= 600\ \mathrm{V} \end{aligned} \)
\( \begin{aligned} C &= \frac{Q}{V} \\ &= \frac{1.2\times10^{-6}}{600} \\ &= 2\times10^{-9}\ \mathrm{F} \\ &= 2\ \mathrm{nF} \end{aligned} \)
NCERT (2026-27):
The Parallel Plate Capacitor – “Now for uniform electric field, potential difference is simply the electric field times the distance between the plates, that is,”
\( V=E\,d \)\( \qquad (2.42) \)
Capacitors and Capacitance – “… Consequently, V is also proportional to Q, and the ratio Q/V is a constant:”
\( C=\frac{Q}{V} \)\( \qquad (2.38) \)