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 (εr = 6) is inserted between the plates. The change in its potential energy after inserting the dielectric slab is ________ × 10⁻¹² J. Numerical answer type. Work it out, then open the answer.

Answer: 750. Battery removed, Q constant: Uf=(Ui)/(εr). Ui=1/2(12.5×10⁻¹²)(12)²=900×10⁻¹² J Uf=900/6=150×10⁻¹² J |Δ U|=900-150=750×10⁻¹² J (a decrease). Answer: 750

A parallel plate capacitor of capacitance 12.5 pF is charged by a… | JEE Main 2024 (4 Apr, Shift 2)

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Dielectrics

A parallel plate capacitor of capacitance $12.5\text{ pF}$ is charged by a battery connected between its plates to potential difference of $12.0\text{ V}$ . The battery is now disconnected and a dielectric slab $(\epsilon_r = 6)$ is inserted between the plates. The change in its potential energy after inserting the dielectric slab is ________ $\times 10^{-12}\text{ J}$ .

Numerical answer type. Work it out, then open the answer.

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Answer: 750

Battery removed, $Q$ constant: $U_f=\frac{U_i}{\epsilon_r}$ .
$U_i=\frac{1}{2}(12.5\times10^{-12})(12)^2=900\times10^{-12}\text{ J}$
$U_f=\frac{900}{6}=150\times10^{-12}\text{ J}$
$|\Delta U|=900-150=750\times10^{-12}\text{ J}$ (a decrease). Answer: 750

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