A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in
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(c) Battery in disconnected so $Q$ will be constant as $C \propto \,K$. So with introduction of dielectric slab capacitance will increase using $Q = CV$, $V$ will decrease and using $U = \frac{{{Q^2}}}{{2C}}$, energy will decrease.
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