An air-filled parallel-plate capacitor is connected to a battery and allowed to charge up. Now a slab of dielectric material is placed betwe

Question

An air-filled parallel-plate capacitor is connected to a battery and allowed to charge up. Now a slab of dielectric material is placed between the plates of the capacitor while the capacitor is still connected to the battery. After this is done, we find that

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Delwyn 4 years 2021-08-12T19:54:47+00:00 1 Answers 353 views 0

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  1. Euphemia
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    2021-08-12T19:55:47+00:00
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    Answer:

    The voltage across the capacitor will remain constant

    The capacitance of the capacitor will increase

    The electric field between the plates will remain constant

    The charge on the plates will increase

    The energy stored in the capacitor will increase

    Explanation:

    First of all, if a capacitor is connected to a voltage source, the voltage or potential difference across the capacitor will remain constant. The electric field across the capacitor will stay constant since the voltage is constant, because the electric field is proportional to the voltage applied. Inserting a dielectric material into the capacitor increases the charge on the capacitor.

    The charge on the capacitor is equal to

    Q = CV

    Since the voltage is constant, and the charge increases, the capacitance will also increase.

    The energy in a capacitor is given as

    E = \frac{1}{2}CV^{2}

    since the capacitance has increased, the energy stored will also increase.

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