Three identical charges each of $2 \mu \mathrm{C}$ are placed at the vertices of a triangle ABC as s

Question

Three identical charges each of $2 \mu \mathrm{C}$ are placed at the vertices of a triangle ABC as shown in the figure. If $\mathrm{AB}+\mathrm{AC}=12 \mathrm{~cm}$ and $\mathrm{AB} \cdot \mathrm{AC}=32 \mathrm{~cm}^{2}$, the potential energy of the charge at A is\

Professor PC Thomas Classes · Answer

Given each charge $\mathrm{q}=2 \mu \mathrm{C}, \mathrm{q}=2 	imes 10^{-6} \mathrm{C}$
$$
\begin{equation*}
\mathrm{AB}+\mathrm{AC}=12 	ag{i}
\end{equation*}
$$
$$
\begin{equation*}
\mathrm{AB}.\mathrm{AC}=32	ag{ii}
\end{equation*}
$$

From equation (i), $\mathrm{AB}=12-\mathrm{AC}$
Put the value of $A B$ in equation (ii)
$(12-\mathrm{AC}) \mathrm{AC}=32$
$A C^{2}-12 A C+32=0$
On solving, we get $A C=8$ or 4
$	herefore$ If $A C=8 \mathrm{~cm}, A B=4 \mathrm{~cm}$
Or $\mathrm{AB}=0.04 \mathrm{~m}, \mathrm{AC}=0.08 \mathrm{~m}$
The PE of the charge at $A$ is $U$
$$
\mathrm{U}=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{\mathrm{q}^{2}}{\mathrm{AB}}+\frac{\mathrm{q}^{2}}{\mathrm{AC}}ight]
$$
$=9 	imes 10^{9}\left[\frac{4 	imes 10^{-12}}{0.04}+\frac{4 	imes 10^{-12}}{0.08}ight]$