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Three point charges $+Q,+2 q$ and $+q$ are placed at the vertices of a right angled isosceles triangle. The net electrostatic potential energy of the configuration is zero, if $\mathrm{Q}$ is equal to
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Verified Answer
The correct answer is:
$-\frac{\sqrt{2}}{3} q$
Net electrostatic potential energy of the system is,
$$
\begin{aligned}
& \mathrm{U}=\frac{1}{4 \pi \varepsilon_0}\left(\frac{\mathrm{Qq}}{\mathrm{a}}+\frac{2 \mathrm{Qq}}{\mathrm{a}}+\frac{2 \mathrm{qq}}{\sqrt{2} \mathrm{a}}\right)=0 \\
\mathrm{Q} & +2 \mathrm{Q}+\frac{2 \mathrm{q}}{\sqrt{2}}=0 \\
3 \mathrm{Q} & +\sqrt{2} \mathrm{q}=0 \\
\therefore \quad & \mathrm{Q}=\frac{-\sqrt{2} \mathrm{q}}{3}
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{U}=\frac{1}{4 \pi \varepsilon_0}\left(\frac{\mathrm{Qq}}{\mathrm{a}}+\frac{2 \mathrm{Qq}}{\mathrm{a}}+\frac{2 \mathrm{qq}}{\sqrt{2} \mathrm{a}}\right)=0 \\
\mathrm{Q} & +2 \mathrm{Q}+\frac{2 \mathrm{q}}{\sqrt{2}}=0 \\
3 \mathrm{Q} & +\sqrt{2} \mathrm{q}=0 \\
\therefore \quad & \mathrm{Q}=\frac{-\sqrt{2} \mathrm{q}}{3}
\end{aligned}
$$
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