Question

Two charges $q_{1}$ and $q_{2}$ are placed $A$ and $B$ (as shown in figure). A third charge $q_{3}$ placed at point $C$ is moved along the arc to $D$ . The change in the potential energy of the system is $\left(N \frac{q_{3} q_{2}}{4 \pi \left(\epsilon \right)_{0}}\right).$ Find the value of $N$

Answer 1

The potential energy when $q_3$ is at point $C$
$ U _1=\frac{1}{4 \pi \varepsilon_0}\left[\frac{ q _1 q _3}{0.80}+\frac{ q _2 q _3}{\sqrt{(0.80+(0.60}}\right] $
The potential energy when $q_{3}$ is at point $D$ ,
$U_{2}=\frac{1}{4 \pi \epsilon _{0}}\left[\frac{q_{1} q_{3}}{0 . 80} + \frac{q_{2} q_{3}}{0 . 20}\right]$
Thus, change in potential energy is
$\Delta U=U_{2}-U_{1}$
$\therefore N\frac{q_{2} q_{3}}{4 \pi \epsilon _{0}}=\frac{1}{4 \pi \epsilon _{0}}\left[\frac{q_{1} q_{3}}{0 . 80} + \frac{q_{2} q_{3}}{0 . 20} - \frac{q_{1} q_{3}}{0 . 80} - \frac{q_{2} q_{3}}{1 . 00}\right]$
$\therefore N=\frac{1}{0 . 20}-1=4$