What is the energy stored when the space between the plates is filled with dielectric.
(a) With charging battery disconnected ?
(b) With charging battery connected in the circuit ?

Explanation

(A) When the charging battery is disconnected from the circuit
Let the capacitor is charged by a battery and disconnected from the circuit. Now the space between the plates is filled with a dielectric of dielectric constant K then potential decreases by `(1)/(k)` times and charge remains constant.
Capacity increases by K times.
New capacity `C' = (Q)/(V) = (Q)/((V)/(K)) = K (Q)/(V) = KC` `[ :' V' = (V)/(K) , C = (Q)/(V)]`
`:. C = KC`
Energy stored `U' = (1)/(2) QV' = (1)/(2) Q (V)/(K) = (U)/(K)`
`U' = (U)/(K)`
`:.` energy stored decreases by `(1)/(K)` times.
(b) when the charging battery is connected in the circuit
Let the charging battery is continue the supply of charge. when the dielectric is introduced then petential decreases by `(1)/(K)` times and charges on the plates increases until the potential difference attains the original value = V
new charge on the plates Q = KQ
Hence new capacity `C' = (Q)/(V) = (KQ)/(V) = KC`
Energy stored in the capacitor `U' = (1)/(2) QV = (1)/(2) (KQ) V = KU`
U' = KU
`:.` Energy stored in the capacitor increases by K times.