The introduction of dielectric slab between the capacitor plates increases its capacitance due to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → decrease in the electric field
A dielectric placed in the field of a charged capacitor is polarised. The induced surface charges on its faces produce a field opposite to the applied one, so the net field between the plates decreases. For the same charge \( Q \), the potential difference \( V = Ed \) falls, and so the capacitance rises:
\( \displaystyle C = \frac{Q}{V} = \frac{Q}{E d} \)
Option (1) increase in the electric field — Incorrect. The polarisation field opposes the applied field, so the field decreases, not increases.
Option (2) increase in the electric potential — Incorrect. The potential difference \( V = Ed \) falls with the field; a larger \( V \) for the same \( Q \) would lower \( C \).
Option (3) increase in the charge — Incorrect. For an isolated capacitor the charge on the plates does not change; capacitance is \( Q/V \), a property of the arrangement, and it rises because \( V \) falls.
Option (4) decrease in the electric field — Correct. The reduced field lowers \( V \) for the same \( Q \), so \( C = Q/V \) increases.
NCERT (2026-27):
Dielectrics and Polarisation – “… Clearly, the field produced by these surface charges opposes the external field. The total field in the dielectric is, thereby, reduced from the case when no dielectric is present.”
Effect of Dielectric on Capacitance – “… The capacitance C, with dielectric between the plates, is then
\( \displaystyle C = \frac{\varepsilon_0 K A}{d} \qquad (2.51) \)”