Three point charges $+\mathrm{q},+2 \mathrm{q}$ and $+\mathrm{Q}$ are placed at the three vertices of an…
Three point charges $+\mathrm{q},+2 \mathrm{q}$ and $+\mathrm{Q}$ are placed at the three vertices of an equilateral triangle. If the potential energy of the system of three charges is zero, the value of $Q$ in terms of $q$ is
$\mathrm{Q}=-\frac{2 \mathrm{q}}{3}$
$\mathrm{Q}=-\frac{1}{3} \mathrm{q}$
$\mathrm{Q}=\frac{3 \mathrm{q}}{2}$
$\mathrm{Q}=\frac{\mathrm{q}}{2}$
Solution
Potential energy due to $+\mathrm{q}$ and $+2 \mathrm{q}$,
$U_1=\frac{K q(2 q)}{r}$
$\therefore \quad$ Potential energy due to $+\mathrm{q}$ and $+\mathrm{Q}$,
$\mathrm{U}_2=\frac{\mathrm{KqQ}}{\mathrm{r}}$
$\therefore \quad$ Potential energy due to $+\mathrm{Q}$ and $+2 \mathrm{q}$,
$\mathrm{U}_3=\frac{\mathrm{KQ}(2 \mathrm{q})}{\mathrm{r}}$
Given: Potential energy of system $=0$
$\begin{aligned}
\therefore \quad & \mathrm{U}_1+\mathrm{U}_2+\mathrm{U}_3=0 \\
& \frac{\mathrm{Kq}(2 \mathrm{q})}{\mathrm{r}}+\frac{\mathrm{KqQ}}{\mathrm{r}}+\frac{\mathrm{KQ}(2 \mathrm{q})}{\mathrm{r}}=0 \\
& 2 \mathrm{q}+\mathrm{Q}+2 \mathrm{Q}=0 \\
& 3 \mathrm{Q}=-2 \mathrm{q} \\
& \mathrm{Q}=\frac{-2}{3} \mathrm{q}
\end{aligned}$