A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has a thickness \( \left(\displaystyle\frac{1}{4}\right)d \) and placed between plates of parallel plate capacitor, where d is the separation of the plates. The ratio of capacitances of capacitor with dielectric and with air as medium \( \left(\displaystyle\frac{c}{c_0}\right) \) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{4K}{3K+1} \)
With charge \( Q \) on plates of area \( A \), the field in the air is \( E_0 = Q/(\varepsilon_0 A) \), and inside the dielectric slab it is reduced to \( E_0/K \). The potential difference is the sum over the air gap \( (d - t) \) and the slab \( (t) \).
\( \displaystyle V = E_0\,(d - t) + \frac{E_0}{K}\,t = \frac{Q}{\varepsilon_0 A}\left(d - t + \frac{t}{K}\right) \)
\( \displaystyle C = \frac{Q}{V} = \frac{\varepsilon_0 A}{d - t + t/K} \)
Putting \( t = d/4 \):
\( \displaystyle C = \frac{\varepsilon_0 A}{\displaystyle\frac{3d}{4} + \frac{d}{4K}} = \frac{4K\,\varepsilon_0 A}{d\,(3K + 1)} \)
With air only, \( C_0 = \varepsilon_0 A/d \), so
\( \displaystyle \frac{C}{C_0} = \frac{4K}{3K + 1} \)
Shortcut: Use limiting cases. With \( K = 1 \) the slab is just air, so the ratio must be 1; this rules out \( K/(3K+1) \). With \( K \to \infty \) the slab behaves like a metal of thickness \( d/4 \), the gap becomes \( 3d/4 \), and the ratio must be \( 4/3 \). Of the remaining options, \( (3K+1)/4K \to 3/4 \) and \( 4K/(K+3) \to 4 \); only \( 4K/(3K+1) \to 4/3 \).
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Effect of Dielectric on Capacitance – …
