A capacitor of capacity $C_{0}$ is connected to a battery of voltage $V_{0}$ . 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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Force of attraction between plates of capacitor is $\frac{k {(C_{0} V_{0})}^{2}}{2 \in_{0} A}$

Magnitude of induced charge on the each surface of slab is $C_{0} V_{0} (k - 1)$

Electric force due to induced charges on any plate is zero

Field due to induced charges in dielectric slab is $\frac{(k - 1) C_{0} V_{0}}{\in_{0} A}$

answer is A, B, D.

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Detailed Solution

Final charge on the capacitor plates is $k C_{0} V_{0}$ that, induced charges on the dielectric slab surface are given as $q_{i} = q (1 - \frac{1}{k}) = C_{0} V_{0} (k - 1)$
As induced charge on the two surface of dielectric slab are equal so, force of attraction, $F = \frac{q^{2}}{2 ε_{0} A} = \frac{k^{2} C_{0}^{2} V_{0}^{2}}{2 ε_{0} A}$
Field inside the dielectric slab $E_{i} = \frac{q_{i}}{\in_{0} A} = \frac{(k - 1) C_{0} V_{0}}{\in_{0} A}$