A capacitor of capacity \(C_0\) is connected to a battery of emf \(V_0\). When steady state is attained a dielectric slab of dielectric constant (K) is slowly introduced in the capacitor. Mark the Correct statement(s), in final steady state :
Magnitude of induced charge on the each surface of slab is \(C_0V_0(K - 1)\).
Net electric force due to induced charges on the plate is zero.
Force of attraction between plates of capacitor is \(\frac{K(C_0V_0)^2}{2 \varepsilon_0 A}\).
Net field due to induced charges in dielectric slab is \(\frac{8V_0(K-1)^2}{K \varepsilon_0 A}\)
Solution:
Final charge on plates \(Q = KC_0V_0\). Induced charge \(q_{\text{ind}} = Q(1 - 1/K) = KC_0V_0(1 - 1/K) = C_0V_0(K-1)\). This is correct. (C) Force of attraction between plates \(F = \frac{1}{2} C'V_0^2 / d = \frac{1}{2} (KC_0) V_0^2 / d\). Since \(C_0 = varepsilon_0 A / d\), \(F = \frac{1}{2} K (\varepsilon_0 A / d) V_0^2 / d = \frac{K \varepsilon_0 A V_0^2}{2d^2}\). This can be rewritten as \(\frac{K(C_0V_0)^2}{2 \varepsilon_0 A}\). Both A and C are correct statements. Option A is chosen as the primary answer.
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