A slab of material of dielectric constant K has the same area as the plates of a parallel-plate capacitor but has a thickness (1/4)d, where d is the separation of the plates. If the slab is inserted between the plates of the capacitor, the potential difference between the plates will be
(Given: Potential difference across the capacitor without dielectric is \( V_0 \))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( V=V_0\left(\frac{3K+1}{4K}\right) \)
The capacitor is charged and then the slab is inserted, so the charge \( Q \) on the plates stays the same (the battery is taken as disconnected). The potential difference is the sum of the drops across the air gap and the slab; inside the slab the field is reduced by the factor \( K \).
\( \displaystyle E_0 = \frac{Q}{\varepsilon_0 A} \;\text{in air}, \qquad \frac{E_0}{K} \;\text{in the slab} \)
With slab thickness \( t = d/4 \), the air gap is \( d - t = 3d/4 \):
\( \displaystyle V = E_0\left(\frac{3d}{4}\right) + \frac{E_0}{K}\left(\frac{d}{4}\right) = E_0 d\left(\frac{3K + 1}{4K}\right) \)
Without the slab, \( V_0 = E_0 d \), so
\( \displaystyle V = V_0\left(\frac{3K+1}{4K}\right) \)
Shortcut: Limiting case: put \( K = 1 \) (a slab of air). Nothing changes, so \( V \) must equal \( V_0 \). Option (2) gives \( V_0 \), while Options (1) and (4) give \( 7V_0/4 \) and Option (3) gives \( V_0/2 \).
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 – …
