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 (∈r = 6) is inserted between the plates. The change in its potential energy after inserting the dielectric slab is _______× 10–12 J.
JEE -2024
Solution:
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