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Q.

A capacitor of capacity C0 is connected to a battery of voltage V0. When steady state is attained a dielectric slab of dielectric constant k is slowly inserted in the capacitor to fill the space between the plates completely. In final steady state which of the following statements is/are correct

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a

Magnitude of induced charge on the each surface of slab is C0V0k−1

b

Electric force due to induced charges on any plate is zero

c

Force of attraction between plates of capacitor is kC0V022∈0A

d

Field due to induced charges in dielectric slab is k−1C0V0∈0A

answer is A.

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Detailed Solution

Final charge on the capacitor plates is kC0V0 that, induced charges on the dielectric slab surface are given as qi=q1−1k=C0V0k−1 As induced charge on the two surface of dielectric slab are equal so, force of attraction, F=q22ε0A=k2C02V022ε0A Field inside the dielectric slab Ei=qi∈0A=k−1C0V0∈0A

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