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 (ε r =6) is inserted between the plates. The change in its potential energy after inserting the dielectric slab is _ _ _ _ ×10⁻¹² ~J.

Electrostatics: JEE Main 2024 Physics question

JEE Main 2024 · 4 Apr, Evening shift · Q57

JEE Main 20244 Apr, Evening shiftPhysicsHardElectrostatics167× since 2002

A parallel plate capacitor of capacitance

12.5 \mathrm{~pF}

is charged by a battery connected between its plates to potential difference of

12.0 \mathrm{~V}

. The battery is now disconnected and a dielectric slab

(\epsilon_{\mathrm{r}}=6)

is inserted between the plates. The change in its potential energy after inserting the dielectric slab is ________

\times10^{-12} \mathrm{~J}

Your answerType your answer. Integers and decimals are accepted.

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