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

i) Consider an uncharged capacitor of capacitance ‘C’. The potential of the plates will be initially zero. If the condenser is connected across a battery with potential difference V across its terminals, the condenser starts charging. The final potential difference across its plates will also be V. If Q is the charge on the condenser Q = CV.
ii) In the process of charging, for any additional charge to be placed on the plates of the condenser, work need to be done against the potential v existing on the plates.
iii) The amount of work done to place an additional charge ‘dq’ on the condenser is given by
dW = Vdq.
The total work done in charging the condenser to ‘Q’ is given by
$\mathrm{W}=\int \mathrm{dW}={\int }_0^{\mathrm{Q}} V\mathrm{dq}$
But $\mathrm{V}=\frac{\mathrm{q}}{\mathrm{C}}$
$\mathrm{w}={\int }_0^{\mathrm{Q}} \frac{\mathrm{q}}{\mathrm{C}}\mathrm{dq}={\left[\frac{{\mathrm{q}}^2}{2\mathrm{C}}\right]}_0^{\mathrm{Q}}=\frac{{\mathrm{Q}}^2}{2\mathrm{C}}$
This work done is stored in the form of electric field then
$\therefore$ The energy stored in a condenser U is given by $\mathrm{U}=\frac{{\mathrm{Q}}^2}{2\mathrm{C}}$
It can also be written as $(\therefore \mathrm{Q}=\mathrm{CV})$
$\begin{array}{l}\mathrm{U}=\frac{1}{2}{\mathrm{CV}}^2\\ =\frac{1}{2}\mathrm{QV}\end{array}$
Effect of dielectric on energy stored :
i) When the charging battery is disconnected from the circuit. Let a capacitor of capacity Co be charged by using a battery. Let the charge be Q. The potential difference between the plates of the capacitor is V_o.
The charge on the capacitor Q = C_oV_o
Energy stored in the capacitor is given by ${\mathrm{U}}_0=\frac{{\mathrm{Q}}^2}{2{\mathrm{C}}_0}$
After charging, the battery is disconnected and a dielectric slab of constant K is introduced between the plates. Now the capacity of the capacitor = KC₀

The charge on the capacitor remains same=Q
The potential difference across the capacitor
$\mathrm{V}=\frac{\mathrm{Q}}{\mathrm{C}}=\frac{\mathrm{Q}}{{\mathrm{KC}}_0}=\frac{{\mathrm{V}}_0}{\mathrm{K}}$
Then energy stored in the capacitor becomes
$\begin{array}{l}\mathrm{U}=\frac{{\mathrm{Q}}^2}{2\mathrm{C}}=\frac{{\mathrm{Q}}^2}{2{\mathrm{KC}}_0}\\ =\frac{1}{\mathrm{K}}\left(\frac{{\mathrm{Q}}^2}{2{\mathrm{C}}_0}\right)=\frac{{\mathrm{U}}_0}{\mathrm{K}}\end{array}$
The energy stored is reduced by a factor K times of original value
ii) When the charging battery is not disconnected in the circuit. Let a capacitor is charged by connecting to a battery. The capacity of the capacitor is C_o and potential is V_o.
Energy stored in the capacitor is given by
${\mathrm{U}}_0=\frac{1}{2}{\mathrm{C}}_0{\mathrm{V}}_0^2$

Now without disconnecting the battery, a dielectric slab of constant K is introduced between the plates. The capacity increases to
C = K C₀.
The battery maintains the same potential V=V_o.
The energy stored bocomes
$\begin{array}{l}\mathrm{U}=\frac{1}{2}{\mathrm{CV}}^2\\ =\frac{1}{2}{\mathrm{KC}}_0{\mathrm{V}}_0^2={\mathrm{KU}}_0\end{array}$
The energy stored in the capacitor increases by K times of original value