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 $\left(N\frac{q_3{q}_2}{4\pi {\left(ϵ\right)}_0}\right).$ Find the value of $N$

Answer 1

Answer: 4

The potential energy when $q_3$ is at point $C$
$U_1=\frac{1}{4\pi {\epsilon }_0}\left[\frac{q_1{q}_3}{0.80}+\frac{q_2{q}_3}{\sqrt{(0.80+(0.60}}\right]$
The potential energy when $q_3$ is at point $D$ ,
$U_2=\frac{1}{4\pi {ϵ}_0}\left[\frac{q_1{q}_3}{0.80}+\frac{q_2{q}_3}{0.20}\right]$
Thus, change in potential energy is
$\Delta U=U_2-U_1$
$\therefore N\frac{q_2{q}_3}{4\pi {ϵ}_0}=\frac{1}{4\pi {ϵ}_0}\left[\frac{q_1{q}_3}{0.80}+\frac{q_2{q}_3}{0.20}-\frac{q_1{q}_3}{0.80}-\frac{q_2{q}_3}{1.00}\right]$
$\therefore N=\frac{1}{0.20}-1=4$