A parallel plate capacitor with air between the plates has a capacitance of 9 pF. What will be the capacitance if the distance between the plates is reduced to half and the space between them is filled by a substance of dielectric constant 7?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 126 pF
The capacitance of a parallel plate capacitor is inversely proportional to the plate separation and is multiplied by the dielectric constant \( K \) when the space is filled with a dielectric.
\( \displaystyle C_0 = \frac{\varepsilon_0 A}{d} = 9\,\mathrm{pF} \)
\( \displaystyle C = \frac{K\varepsilon_0 A}{d/2} = 2K\,\frac{\varepsilon_0 A}{d} = 2K\,C_0 \)
\( \displaystyle C = 2 \times 7 \times 9\,\mathrm{pF} = 126\,\mathrm{pF} \)
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
