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In the figure, the value of $Q$ so that the electrostatic potential energy of the system becomes zero is
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Verified Answer
The correct answer is:
$\frac{2 q}{2-\sqrt{2}}$
We have
$\mathrm{U}_{\text {System }}=\mathrm{K}\left[\frac{-\mathrm{q} \cdot \mathrm{q}}{\mathrm{x}}+\frac{\mathrm{qQ}}{\mathrm{x}}+\frac{-\mathrm{qQ}}{\sqrt{2} \mathrm{x}}\right]$
$\begin{aligned} & \Rightarrow 0=-q^2+q Q-\frac{q Q}{\sqrt{2}} \\ & \Rightarrow 0=-q+Q-\frac{Q}{\sqrt{2}} \\ & \Rightarrow Q\left(1-\frac{1}{\sqrt{2}}\right)=q \\ & \Rightarrow Q=\frac{\sqrt{2} q}{\sqrt{2}-1} \\ & \Rightarrow Q=\frac{2 q}{2-\sqrt{2}}\end{aligned}$
$\mathrm{U}_{\text {System }}=\mathrm{K}\left[\frac{-\mathrm{q} \cdot \mathrm{q}}{\mathrm{x}}+\frac{\mathrm{qQ}}{\mathrm{x}}+\frac{-\mathrm{qQ}}{\sqrt{2} \mathrm{x}}\right]$
$\begin{aligned} & \Rightarrow 0=-q^2+q Q-\frac{q Q}{\sqrt{2}} \\ & \Rightarrow 0=-q+Q-\frac{Q}{\sqrt{2}} \\ & \Rightarrow Q\left(1-\frac{1}{\sqrt{2}}\right)=q \\ & \Rightarrow Q=\frac{\sqrt{2} q}{\sqrt{2}-1} \\ & \Rightarrow Q=\frac{2 q}{2-\sqrt{2}}\end{aligned}$
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