7.2 Energy stored in a charged conductor

3.1 When equal and opposite charges placed on plates
3.2 When unequal charges are placed on the two plates

6.1 Working method to find capacitance of any capacitor

7.1 Force between the Plates of a Capacitor
7.2 Energy stored in a charged conductor

8.1 Loss of energy during redistribution of charge

9.1 Effect of Dielectric
9.2 Capacitance of a Capacitor Partially Filled with Dielectric
9.3 Quantities after inserting dielectric in a capacitor (fully)

12.1 Important points in $R$-$C$ circuits

15.1 Principle of Operation
15.2 Construction

Work has to be done in charging a conductor (capacitor) against the force of repulsion by the already existing charges on it. The work is stored as a potential energy in the electric field of the conductor. Suppose a conductor of capacity $C$ is charged to a potential ${V_ \circ }$ and let ${q_\circ}$ be the charge on the conductor at this instant. The potential of the conductor when (during charging) the charge on it was $q$ (< ${q_\circ}$) is,

$$V = \frac{q}{C}$$

Now, work done in bringing a small charge $dq$ at this potential is, $$dW = Vdq = \left( {\frac{q}{C}} \right)dq$$total work done in charging it from $0$ to $ (< {q_\circ})$ is,$$W = \int\limits_0^{{q_ \circ }} {dW = } \int\limits_0^{{q_ \circ }} {\frac{q}{C}dq = \frac{1}{2}\frac{{{q_ \circ }^2}}{C}} $$

This work is stored as the potential energy,$$U = \frac{1}{2}\frac{{{q_ \circ }^2}}{C}$$

Now, we can use ${q_ \circ } = C{V_ \circ }$ and write this expression also as,$$\begin{equation} \begin{aligned} {q_ \circ } = C{V_ \circ } \\ U = \frac{1}{2}C{V^2} = \frac{1}{2}\frac{{{q^2}}}{C} = \frac{1}{2}qV \\\end{aligned} \end{equation} $$