A slab of material of dielectric constant k has the same area as the plates of a parallel plate capacitor, but has a thickness \( \displaystyle\frac{3d}{4} \). Where d is the distance between plates of capacitor. The ratio of the capacitance with dielectric inside it to its capacitance without the dielectric is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{4k}{(k+3)} \)
A dielectric slab of thickness \( t \) reduces the field inside it by the factor \( k \), so the potential difference falls and the capacitance becomes
\( \displaystyle C = \frac{\varepsilon_0 A}{d - t + \frac{t}{k}} \)
Without the slab, \( C_0 = \varepsilon_0 A/d \). Putting \( t = \frac{3d}{4} \):
\( \displaystyle d - \frac{3d}{4} + \frac{3d}{4k} = \frac{d}{4} + \frac{3d}{4k} = \frac{d\,(k + 3)}{4k} \)
\( \displaystyle \frac{C}{C_0} = \frac{d}{d\,(k+3)/(4k)} = \frac{4k}{k + 3} \)
Shortcut: Put \( k = 1 \): the slab is then just air and the ratio must be 1. Option (1) gives 3/5, Option (3) gives 5/3 and Option (4) gives 1/4; only Option (2) gives 1.
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
Effect of Dielectric on Capacitance – …
