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# An air capacitor of capacity $C=10 \mu F$ is connected to a constant voltage battery of 12 V. Now the space between the plates is filled with a liquid of dielectric constant 5. The additional charge that flows now, from the battery to the capacitor is:(1) $120 \mu C$(2) $600 \mu C$(3) $480 \mu C$(4) $24 \mu C$

Last updated date: 16th Sep 2024
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Hint: We know that dielectric constant, also called relative permittivity or specific inductive capacity, property of an electrical insulating material (a dielectric) equal to the ratio of the capacitance of a capacitor filled with the given material to the capacitance of an identical capacitor in a vacuum without the dielectric material. Vacuum, Solids, Liquids and Gases can be a dielectric material. Some of the examples of solid dielectric materials are ceramics, paper, mica, glass etc. Liquid dielectric materials are distilled water, transformer oil etc. Gas dielectrics are nitrogen, dry air, helium, oxides of various metals etc.

We know that dielectric, insulating material or a very poor conductor of electric current. When dielectrics are placed in an electric field, practically no current flows in them because, unlike metals, they have no loosely bound, or free, electrons that may drift through the material. Instead, electric polarization occurs. Pure water is a non-polar dielectric. But they are not at rest and can't induce charges to produce electric fields like a solid dielectric. The motion of water molecules varies the capacity of a capacitance constantly. Therefore, water can't be used as a dielectric in a capacitor.
Dielectric materials are used in many applications such as: Electronic components such as capacitors (responsible for energy storage properties of the device) High-K / low-K materials widely used in Semiconductors to enhance performance and reduce device size (where K refers to permittivity of dielectric constant).
Initial charge ${{\text{q}}_{\text{i}}}=\text{CV}=10\times 12=120\mu \text{C}$
After filled with dielectric liquid the charge ${{\text{q}}^{\prime }}={{\text{C}}^{\prime }}\text{V}=\text{KCV}$
After we put the values we get: $5\times 10\times 12=600\mu \text{C}$
The (additional) charge that flows now from battery to the
capacitor $=\mathrm{q}_{\mathrm{f}}-\mathrm{q}_{\mathrm{i}}=600-120=480 \mu \mathrm{C}$

So, the correct option is option C.

Note: We know that a capacitor (originally known as a condenser) is a passive two-terminal electrical component used to store energy electrostatically in an electric field. The forms of practical capacitors vary widely, but all contain at least two electrical conductors (plates) separated by a dielectric ( insulator). The potential energy in a capacitor is stored in an electric field, where a battery stores its potential energy in a chemical form. However, in general batteries provide higher energy density for storage, while capacitors have more rapid charge and discharge capabilities (greater Power density). Ceramic capacitors are well-suited for high frequencies and high current pulse loads. Because the thickness of the ceramic dielectric layer can be easily controlled and produced by the desired application voltage, ceramic capacitors are available with rated voltages up to the 30 kV range.