Two point charges ' $q 1$ ' and 'q2' are separated by a distance ' $\mathrm{d}$ '. What is the increase in…
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
$\begin{array}{ll}
\therefore & \Delta U=U_f-U_f \\
\therefore & \Delta U=\frac{k q_1 q_2}{d}-\frac{k q_1 q_2}{(d-x)} \\
\therefore & \Delta U=k q_1 q_2\left(\frac{1}{d}-\frac{1}{d-x}\right) \\
\therefore & \Delta U=\frac{-k q_1 q_2 x}{d(d-x)}
\end{array}$