A parallel plate capacitor with air between the plates has a capacitance of 6 pF. What will be the capacitance if the distance between the plates is reduced to half and the space between them is filled with a substance of dielectric constant 5?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 60 pF
The capacitance of a parallel plate capacitor is proportional to the dielectric constant and inversely proportional to the plate separation.
\( \displaystyle C_0 = \frac{\varepsilon_0 A}{d} = 6\,\mathrm{pF} \)
With the separation \( d/2 \) and a dielectric of constant \( K = 5 \):
\( \displaystyle C = \frac{K\,\varepsilon_0 A}{d/2} = 2K\,\frac{\varepsilon_0 A}{d} = 2 \times 5 \times 6\,\mathrm{pF} = 60\,\mathrm{pF} \)
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 – …
