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Three charges $Q,+y$ and $+q$ are placed at the vertices of a right-angle isosceles triangle as shown below. The net electrostatic energy of the configuration is zero, if the value of $\mathrm{Q}$ is
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The correct answer is:
$\frac{-\sqrt{2} q}{\sqrt{2}+1}$
Net electrostatic energy for the system
$\mathrm{U}=\mathrm{K}\left[\frac{\mathrm{q}^{2}}{\mathrm{a}}+\frac{\mathrm{Qq}}{\mathrm{a}}+\frac{\mathrm{Qq}}{\mathrm{a} \sqrt{2}}\right]=0$
$\begin{aligned} \Rightarrow q &=-Q\left[1+\frac{1}{\sqrt{2}}\right] \\ & \Rightarrow Q=\frac{-q \sqrt{2}}{\sqrt{2}+1} \end{aligned}$
$\mathrm{U}=\mathrm{K}\left[\frac{\mathrm{q}^{2}}{\mathrm{a}}+\frac{\mathrm{Qq}}{\mathrm{a}}+\frac{\mathrm{Qq}}{\mathrm{a} \sqrt{2}}\right]=0$
$\begin{aligned} \Rightarrow q &=-Q\left[1+\frac{1}{\sqrt{2}}\right] \\ & \Rightarrow Q=\frac{-q \sqrt{2}}{\sqrt{2}+1} \end{aligned}$
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