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 energy stored in the capacitor change?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → It decreases
Once the capacitor is isolated, the charge Q on its plates cannot change. The energy stored is \( U=Q^{2}/2C \), and a dielectric slab increases the capacitance (to \( KC \), with \( K\gt1 \), when it fills the gap).
Option (1) It increases — Incorrect. The energy increases when the capacitor stays connected to the battery. Then V is fixed and \( U=\frac{1}{2}CV^{2} \) grows with C. Here the battery is removed, so Q is fixed instead.
Option (2) It decreases — Correct. With Q fixed and C increased by the slab, \( U=Q^{2}/2C \) becomes smaller. For a slab filling the gap, the energy falls to \( U_0/K \).
Option (3) It remains constant — Incorrect. The energy would stay the same only if the capacitance did not change, but the dielectric slab raises it.
Option (4) Cannot be said — Incorrect. The charge is fixed and the capacitance increases. These two facts are enough to decide the change in \( Q^{2}/2C \). So the energy falls.
NCERT (2026-27):
Energy Stored in a Capacitor – “We can write the final result, in different ways
\( \displaystyle W=\frac{Q^{2}}{2C}=\frac{1}{2}CV^{2}=\frac{1}{2}QV \)\( \qquad (2.69) \)”
Effect of Dielectric on Capacitance – “…
\( \displaystyle K=\frac{C}{C_0} \qquad (2.54) \)
Thus, the dielectric constant of a substance is the factor (>1) by which the capacitance increases from its vacuum value, when the dielectric is inserted fully between the plates of a capacitor.”