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Dielectric slab inserted into a charged capacitor, battery connected or removed: change in energy or charge

This type of question, asked 4 times: 2021 Paper 23, 29 Jun 2022, 4 Apr 2024, 22 Jan 2025. All 4 questions below, oldest first.

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  1. Question 1

    2021 Paper 23

    22
    A parallel plate capacitor of capacitance 200μF200\mu F is connected to a battery of 200 V200\text{ }V. A dielectric slab of dielectric constant 2 is now inserted into the space between plates of capacitor while the battery remain connected. The change in the electrostatic energy in the capacitor will be ____\_\_\_\_ J.

  2. Question 2

    29 Jun 2022

    17
    A parallel plate capacitor filled with a medium of dielectric constant 10, is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant 15. Then the energy of capacitor will:

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  3. Question 3

    4 Apr 2024

    113
    A parallel plate capacitor of capacitance 12.5pF12.5pF is charged by a battery connected between its plates to potential difference of 12.0 V12.0\text{ }V. The battery is now disconnected and a dielectric slab (ϵr=6)\left( \epsilon_{r}= 6 \right) is inserted between the plates. The change in its potential energy after inserting the dielectric slab is ________ ×10−12 J\times10^{- 12}\text{ }J

  4. Question 4

    22 Jan 2025

    19
    A parallel-plate capacitor of capacitance 40μ F40\mu\text{ }F is connected to a 100 V power supply. Now the intermediate space between the plates is filled with a dielectric material of dielectric constant K=2K = 2. Due to the introduction of dielectric material, the extra charge and the change in the electrostatic energy in the capacitor, respectively, are -

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