JEE Main 2024 Physics question
A parallel plate capacitor of capacitance 12.5 pF is charged by a battery connected between its plates to potential difference of 12.0 V. The battery is now disconnected and a dielectric slab ( = 6) is inserted between the plates. The change in its potential energy after inserting the dielectric slab is × 10 J.
Answer: 750
Solution
With Q fixed, U = Q²/2C falls by a factor 6: = ½ × 12.5 pF × 144 = 900 pJ; = 150 pJ; the change is 750 × 10⁻¹² J (a decrease).