Take uncharged conductors $1$ and $2$ as shown in figure.

Imagine a process of transferring charge from conductor $2$ to conductor $1$ bit by bit so that at the ends, conductor $1$ gets charge.

In transferring positive charge from conductor$2$ to conductor 1 work will be done externally, because at any stage conductor $1$ is at a higher potential than conductor $2$ .

To calculate the total work by in a small step involving transfer of an infinitesimally amount of charge.

Consider the situation when the conductors $1$ and $2$ have charges $Q'$ and $- Q'$ respectively.

The potential difference $V^{\prime}$ between conductors $1$ to $2$ is $V^{\prime}=\frac{Q^{\prime}}{C}$ where $C$ is the capacitance of the system.

A small charge $\delta Q^{\prime}$ is transferred from conductor $2$ to $1$ , then work done, $\delta \mathrm{W}=\mathrm{V}^{\prime} \delta Q^{\prime}$

$\therefore \delta \mathrm{W}=\frac{\mathrm{Q}^{\prime} \delta \mathrm{Q}^{\prime}}{\mathrm{C}}$

$\ldots$ $(1)$

Total work done to bringing charge $Q$ from conductor 2 to 1 is obtain by integration, $\mathrm{W}=\int d \mathrm{~W}$

$\therefore \mathrm{W}=\int_{0}^{\mathrm{Q}} \frac{\mathrm{Q}^{\prime}}{\mathrm{C}} \cdot \delta \mathrm{Q}^{\prime} \quad \therefore \mathrm{W}=\frac{1}{\mathrm{C}} \int_{0}^{\mathrm{Q}} \mathrm{Q}^{\prime} \delta \mathrm{Q}^{\prime}=\frac{1}{\mathrm{C}}\left[\frac{\left(\mathrm{Q}^{\prime}\right)^{2}}{2}\right]_{0}^{\mathrm{Q}}$

$\frac{1}{\mathrm{C}}\left[\frac{\mathrm{Q}^{2}}{2}\right] \therefore \mathrm{W}=\frac{\mathrm{Q}^{2}}{2 \mathrm{C}}$