A capacitor is charged with a battery and then disconnected. A dielectric slab is then inserted between the plates. How are the
The stored energy related to the capacitor affected?

Explanation

Energy stored in the capacitor,
`U= (1)/(2)(Q^2)/(C ), " so, " U propto (1)/(C )` as Q = constant.
Due to the insertion of the dielectric slab, the amount of energy stored is reduced. The final energy stored in the capacitor will be `(1)/(k)` times of its previous value.