An air filled parallel plate capacitor has capacitance $C_{0}$ . Now the air column in the capacitor is completely replaced by a dielectric slab having a dielectric constant $K$ and the capacitor is connected to a battery of emf $E$ and then the slab is taken out. In the process of taking out the dielectric, it is observed that

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an energy $C_{0} E^{2} (K - 1)$ is absorbed by the cell

a work $\frac{1}{2} C_{0} E^{2} (K - 1)$ has to be done by the external agent to take the slab out

a charge $C_{0} E (K - 1)$ flows through the cell

the energy stored in the capacitor reduces to $C_{0} E^{2} (K - 1)$

answer is B, C, D.

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Detailed Solution

$Q_{i} = C_{0} E a n d Q_{f} = k C_{0} E △ Q = (k - 1) C_{0} E U_{i} = \frac{1}{2} C_{0} E^{2} a n d U_{f} = \frac{1}{2} k C_{0} E^{2} W_{b} = W o r k d o n e b y t h e b a t t e r y = △ Q \times E = (k - 1) C_{0} E^{2} = E n e r g y a b s o r b e d b y t h e c e l l . W o r k d o n e b y e x t e r n a l a g e n t = W_{e} N o w, W_{b} + W_{e} = U_{f} - U_{i} + L o s t e n e r g y . S i n c e t h e r e i s n o r e s i s t i v e e l e m e n t i n t h e c i r c u i t a n d t h e d i e l e c t r i c s l a b i s r e m o v e d f r o m t h e c a p a c i t o r v e r y s l o w l y, l o s t e n e r g y c a n b e i g n o r e d . S o W_{e} = \frac{1}{2} C_{0} (k - 1) E^{2}$