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

Expression for the energy stored in a capacitor : Consider an uncharged capacitor of capacitance 'c' and its initial will be zero. Now it is connected across a battery for charging then the final potential dfference across the capacitor be 'V' and final charge on the capacitor be 'Q'
`therefore` Average potential difference `V_(A)=(O+V)/(2)=(V)/(2)`
Hence workdone to move the charge `Q=W=V_(A)xxQ=(VQ)/(2)`
This is stored electrostatic potential energy 'U'.
`therefore U=(QV)/(2)`
We know `Q=CV` then 'U' can be written as given below.
`U=(QV)/(2)=(1)/(2)CV^2=(Q^2)/(2C)`
`therefore` Energy stored in a capacitor `U=(QV)/(2)=(1)/(2)CV^2=(1)/(2)(Q^2)/(C)`
Effect , of Dielectric on energy stored :
Case (a): When the charging battery is disconnected from the circuit :
Let the capacitor is charged by a battery and the disconnected from the circuit. Now the space between the plates in filled with a dielectric of dielectric constant 'K' then potential decreases by `(10/(K)` times and charge remains constant.
Capacity increases by 'K' times
New capacity `C' =(Q)/(V')=(Q)/(V)=K(Q)/(V)=KC [ because V'=(V)/(K),C=(Q)/(V)]`
`therefore C =KC`
Energy stored `U'=(1)/(2)QV'=(1)/(2)Q(V)/(K)=(U)/(K)`
`U'=(U)/(K)`
`therefore` Energy stored decreases by `(1)/(K)` times.
Case (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 potential decreases by `(1)/(K)` times and charge 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)Q'V=(1)/(2)(KQ)V=KU`
`U'=KU`
`therefore` Energy stored in the capacitorf increases by 'K' times.