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Capacitance with a dielectric or metal slab partly filling the gap

This type of question, asked 12 times: May 2021 S1, 4 May 2023 S1, 19 Apr 2025 S1, 21 Apr 2025 S2, 23 Apr 2025 S1, 11 Apr 2026 S1, 13 Apr 2026 S1, 16 Apr 2026 S2, 19 Apr 2026 S1, 12 May 2026 S1, 13 May 2026 S2, 15 May 2026 S2. All 12 questions below, oldest first.

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  1. Question 1

    May 2021 S1

    15
    Separation between the plates of parallel plate capacitor is dd and the area of each plate is AA. When a slab of material of dielectric constant kk and thickness tt (t<d)(t < d) is introduced between the plates, its capacitance becomes (ε0\varepsilon_0 = permittivity of free space)

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  2. Question 2

    4 May 2023 S1

    39
    A parallel plate capacitor has plate area 40 cm240\text{ cm}^2 and plate separation 2 mm. The space between the plates is filled with a dielectric medium of thickness 1 mm and dielectric constant 5. The capacitance of the system is (ε0=\varepsilon_0 = permittivity of vacuum)

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  3. Question 3

    19 Apr 2025 S1

    18
    The plates of a parallel plate capacitor are separated by a distance ' dd ' with air as the medium between them. A dielectric slab of dielectric constant 3 is introduced between the plates so as to increase the capacity by 50%50\%. The thickness of the dielectric slab is

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  4. Question 4

    21 Apr 2025 S2

    48
    A parallel plate capacitor has plate area 50 cm250{\text{ }cm}^{2} and plate separation 3 mm . The space between the plates is filled with a dielectric medium of thickness 1 mm and dielectric constant 4. The capacitance becomes ( ε0=\varepsilon_{0}= permittivity of free space)

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  5. Question 5

    23 Apr 2025 S1

    32
    A parallel plate air capacitor has capacity ' CC ' and distance of separation between plates is ' d '. If a conducting sheet of thickness 2d3\frac{2d}{3} is inserted in between the plates, the capacitance becomes C1C_{1}. The ratio of C1C\frac{C_{1}}{C} is

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  6. Question 6

    11 Apr 2026 S1

    127
    A parallel plate capacitor has plate area 40 cm240\ \mathrm{cm^2} and plates separation 22 mm. The space between the plates is filled with a dielectric medium of a thickness 11 mm and dielectric constant 55. The capacitance of the system is

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  7. Question 7

    13 Apr 2026 S1

    126
    The capacitance of a capacitor becomes 76\dfrac{7}{6} times the original value if dielectric slab of thickness t=2d3t = \dfrac{2d}{3} is introduced between the plates, where 'd' is the distance of separation between the plates. The dielectric constant of the slab is

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  8. Question 8

    16 Apr 2026 S2

    128
    A capacitor has capacity C when it's parallel plates are separated by air medium of thickness 'd'. A slab of material of dielectric constant K having area equal to that of plates but thickness d2\dfrac{d}{2} is inserted between the plates. Capacitance of the capacitor in the presence of slab will be

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  9. Question 9

    19 Apr 2026 S1

    126
    A slab of material of dielectric constant KK has the same area as the plates of a parallel plate capacitor but has a thickness (4/5)d(4/5)d, where dd is the separation of the plates. The capacitance in the presence and absence of dielectric are CC and C0C_0 respectively. The ratio (C/C0)(C/C_0) is

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  10. Question 10

    12 May 2026 S1

    127
    A parallel plate air capacitor consists of two metal plates of area AA and separation dd. A slab of thickness tt and dielectric constant 2k2k is inserted between the plates with its faces parallel to the plates. The effective capacitance of the system is (ϵ0\epsilon_0 = permittivity of free space)

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  11. Question 11

    13 May 2026 S2

    128
    A slab of material of dielectric constant kk has the same area as the plates of the parallel plate capacitor but has thickness 3d4\dfrac{3d}{4}, where dd is the separation of the plates. How is the capacitance changed, when the slab is inserted between the plates? (CC = original capacitance)

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  12. Question 12

    15 May 2026 S2

    128
    A parallel plate capacitor of plate separation 22 mm and plate area 40 cm240\ \mathrm{cm^2} is filled with dielectric whose area is same as plates of capacitor. Thickness is 11 mm and dielectric constant is 55. New capacitance is (ϵ0\epsilon_0 = permittivity of free space)

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