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A parallel plate capacitor is maintained at a certain potential difference. When a 3 mm thick slab is introduced b/w plates, in order to maintain the same Potential difference, the distance b/w the plates is increased by 2.4 m. The dielectric constant of the slab is __________.

    Correct answer is '5'. Can you explain this answer?
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    A parallel plate capacitor is maintained at a certain potential differ...
    If d is the separation between the plates (each of area A0) of a parallel plate condenser in air, then its capacitance in air, then its capacitance




    If a slab of thickness t is introduced between the plates with new separation d’, then its new capacitance,




    As Q = CV, the charge on the capacitor is same in both cases therefore to maintain the same potential difference the capacitance C and C’ must be same i.e, from (1) and (2)


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    A parallel plate capacitor is maintained at a certain potential differ...
    Given:
    - A parallel plate capacitor maintained at a certain potential difference.
    - Introduction of a 3 mm thick slab between the plates.
    - In order to maintain the same potential difference, the distance between the plates is increased by 2.4 m.

    To find:
    The dielectric constant of the slab.

    Explanation:
    When a dielectric material is introduced between the plates of a capacitor, it affects the capacitance of the capacitor. The capacitance of a parallel plate capacitor with a dielectric material is given by the equation:

    C = (ε₀ * εᵣ * A) / d,

    where:
    C is the capacitance,
    ε₀ is the permittivity of free space (8.854 × 10⁻¹² F/m),
    εᵣ is the relative permittivity or dielectric constant of the slab,
    A is the area of the plates, and
    d is the distance between the plates.

    Step 1: Find the initial capacitance without the slab.
    Since the potential difference across the capacitor remains the same, the initial capacitance (C₀) can be represented as:

    C₀ = (ε₀ * A) / d₀,

    where d₀ is the initial distance between the plates.

    Step 2: Find the final capacitance with the slab.
    After introducing the slab, the distance between the plates increases to (d₀ + 2.4 m). Therefore, the final capacitance (C) can be represented as:

    C = (ε₀ * εᵣ * A) / (d₀ + 2.4 m).

    Step 3: Equate the initial and final capacitance.
    Since the potential difference remains the same, we can equate C₀ and C:

    (ε₀ * A) / d₀ = (ε₀ * εᵣ * A) / (d₀ + 2.4 m).

    Step 4: Solve the equation for εᵣ.
    By cross-multiplying and rearranging the equation, we get:

    (ε₀ * A) * (d₀ + 2.4 m) = (ε₀ * εᵣ * A) * d₀.

    Simplifying further:

    d₀ + 2.4 m = εᵣ * d₀.

    Dividing both sides by d₀:

    1 + (2.4 m / d₀) = εᵣ.

    Step 5: Substitute the given values and calculate εᵣ.
    The given thickness of the slab is 3 mm, which is equivalent to 0.003 m. The increase in distance between the plates is 2.4 m. Substituting these values into the equation:

    1 + (2.4 m / d₀) = εᵣ,
    1 + (2.4 m / d₀) = 5.

    Therefore, the dielectric constant of the slab is 5.
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    Community Answer
    A parallel plate capacitor is maintained at a certain potential differ...
    As here with introduction of slab or with change in separation, potential difference between the plates is changed, the charged capacitor is isolated, i.e., q = q'
    So, C0V0 = CV [as q = CV]
    or, C0 = C [as V = V0 given]
    Now as with introduction of dielectric slab the capacity increases from

    by increasing d to (d + 0.24 cm) the capacity C1, becomes



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