Case: Read the passage and answer the question.
A parallel plate capacitor is first charged using a battery and then isolated. A dielectric slab is then introduced between the plates.
How does the potential difference across the capacitor change?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → It decreases
The battery is removed, so the charge Q on the plates stays fixed. A dielectric slab increases the capacitance, so the potential difference \( V=Q/C \) changes with it.
Option (1) It increases — Incorrect. With Q fixed, V rises only when the capacitance falls, as when the plates are pulled apart. A dielectric slab raises the capacitance.
Option (2) It decreases — Correct. The polarised dielectric carries surface charges that oppose the field between the plates, so the field and the potential difference fall. For a slab filling the gap, the potential difference drops from \( V_0 \) to \( V_0/K \).
Option (3) It remains constant — Incorrect. The potential difference stays fixed only while the battery remains connected. Here the capacitor is isolated, so the charge is fixed instead.
Option (4) Cannot be said — Incorrect. The charge is fixed and the capacitance increases. These two facts are enough to decide the change in \( V=Q/C \). So the potential difference falls.
NCERT (2026-27):
Effect of Dielectric on Capacitance – “… When there is vacuum between the plates,”
\( \displaystyle E_0=\frac{\sigma}{\varepsilon_0} \)
“and the potential difference V0 is”
\( V_0=E_0d \)
Effect of Dielectric on Capacitance – “… where K is a constant characteristic of the dielectric. Clearly, K > 1. We then have”
\( \displaystyle V=\frac{\sigma d}{\varepsilon_0K}=\frac{Qd}{A\varepsilon_0K} \)\( \qquad (2.50) \)