Two point charges $\mathrm{q}_1=6 \mu \mathrm{C}$ and $\mathrm{q}_2=4 \mu \mathrm{C}$ are kept at points A…
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 ?
$\left(\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \text { SI units }\right)$
0.27 J
$\quad 0.54 \mathrm{~J}$
0.81 J
54 J
Solution
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
$\begin{aligned}
& \therefore & \Delta U & =U_f-U_i \\
& \therefore & \Delta U & =\frac{K q_1 q_2}{(d-x)}-\frac{K q_1 q_2}{d}=K q_1 q_2\left(\frac{1}{d-x}-\frac{1}{d}\right)
\end{aligned}$
$\therefore \quad \Delta U=\frac{\mathrm{Kq}_1 \mathrm{q}_2 \mathrm{x}}{\mathrm{~d}(\mathrm{~d}-\mathrm{x})}$ Substituting given values,
$\begin{array}{ll}
\therefore & \Delta \mathrm{U}=\frac{\left(9 \times 10^9\right) \times\left(6 \times 10^{-6}\right) \times\left(4 \times 10^{-6}\right) \times 0.02}{(0.1)(0.1-0.02)} \\
\therefore & \Delta U=\frac{4.32 \times 10^{-3}}{0.1 \times 0.08}=0.54 \mathrm{~J}
\end{array}$