A material slab of dielectric constant K has the same area (A) as the plates of a parallel plate capacitor, but its thickness is d/4, where d is the separation of the plates. The change in capacitance when the material slab is inserted between the plates of the parallel plate capacitor is given by:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → ((K − 1)/(3K + 1)) × (Aε0/d)
With a slab of thickness t and dielectric constant K between the plates, the field inside the slab is reduced by the factor K, so the slab acts like an air gap of thickness t/K. The effective air separation becomes \( d - t + t/K \).
\( \displaystyle C = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}} \)
Putting \( t = d/4 \):
\( \displaystyle C = \frac{\varepsilon_0 A}{\frac{3d}{4} + \frac{d}{4K}} = \frac{4K}{3K + 1}\,\frac{\varepsilon_0 A}{d} \)
The change in capacitance is this value minus the air capacitance \( C_0 = \varepsilon_0 A/d \):
\( \displaystyle \Delta C = \left(\frac{4K}{3K + 1} - 1\right)\frac{\varepsilon_0 A}{d} \)
\( \displaystyle \Delta C = \left(\frac{K - 1}{3K + 1}\right)\frac{A\varepsilon_0}{d} \)
Shortcut: Check with K = 1 (the slab is just air): the change must be zero. Only Options (1) and (4) give zero, and Option (4) has the wrong dimensions (\( d\varepsilon_0/A \) is not a capacitance), leaving Option (1).
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 – …
