Two point charges \(\mathrm{q}_1=6 \mu \mathrm{C}\) and \(\mathrm{q}_2=4 \mu \mathrm{C}\) are kept at points A and B in air where \(\mathrm{AB}=10 \mathrm{~cm}\). What is the increase in potential energy of the system when \(\mathrm{q}_2\) is moved towards \(\mathrm{q}_1\) by 2 cm ?
\(\left(\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \text { SI units }\right)\)

The potential energy between two charges is given as \(U=\frac{K q_1 q_2}{r}\)
Initial potential energy is \(U_i=\frac{K_1 q_2}{r}\)
When charge \(\mathrm{q}_2\) moves towards the \(\mathrm{q}_1\) the separation between the charges becomes \(\mathrm{d}-\mathrm{x}\)
The final potential energy is \(U_f=\frac{K_1 q_2}{(d-x)}\)
The increase in potential energy is
\(\therefore \Delta U =U_f-U_i \)
\( \therefore \Delta U =\frac{K q_1 q_2}{(d-x)}-\frac{K q_1 q_2}{d}=K q_1 q_2\) \(\left(\frac{1}{d-x}-\frac{1}{d}\right)\)
\(\therefore \Delta U=\frac{\mathrm{Kq}_1 \mathrm{q}_2 \mathrm{x}}{\mathrm{~d}(\mathrm{~d}-\mathrm{x})}\)
Substituting given values,
\(\begin{array}{ll}
\therefore & \Delta \mathrm{U}=\frac{\left(9 \times 10^9\right) \times\left(6 \times 10^{-6}\right) \times\left(4 \times 10^{-6}\right) \times 0.02}{(0.1)(0.1-0.02)} \\
\therefore & \Delta U=\frac{4.32 \times 10^{-3}}{0.1 \times 0.08}=0.54 \mathrm{~J}
\end{array}\)