Two point charges ' \(q 1\) ' and 'q2' are separated by a distance ' \(\mathrm{d}\) '. What is the increase in potential energy of the system when ' \(q 2\) ' is moved towards ' \(q 1\) ' by a distance ' \(x\) '?
\((\mathrm{x} < \mathrm{d})\left(\frac{1}{4 \pi \varepsilon_0}=\mathrm{K}, \text { constant }\right)\)

The potential energy between two charges is given as \(U=\frac{k q_1 q_2}{r}\)
Initial potential energy is \(\mathrm{U}_{\mathrm{f}}=\frac{\mathrm{kq}_1 \mathrm{q}_2}{\mathrm{r}}\)
When charge \(q_2\) moves towards the \(q_1\) the separation between the charges becomes \(\mathrm{d}-\mathrm{x}\)
The final potential energy is \(\mathrm{U}_{\mathrm{f}}=\frac{\mathrm{kq}_1 \mathrm{q}_2}{(\mathrm{~d}-\mathrm{x})}\)
The increase in potential energy is
\(
\begin{array}{ll}
\therefore & \Delta U=U_f-U_f \\
\therefore & \Delta U=\frac{k q_1 q_2}{d}-\frac{k q_1 q_2}{(d-x)} \\
\therefore & \Delta U=k q_1 q_2\left(\frac{1}{d}-\frac{1}{d-x}\right) \\
\therefore & \Delta U=\frac{-k q_1 q_2 x}{d(d-x)}
\end{array}
\)