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Question

Three charges $$Q,+q$$ and $$+q$$ are placed at the vertices of a right angle isosceles triangle as shown. The net electrostatic energy of the configuration is zero, if $$Q$$ is equal to :

125039_a319cbdf683d4b6482b116bbaab249fa.png


A
q1+2
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B
2q2+2
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C
2q
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D
+q
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Solution

The correct option is B $$\dfrac { -2q }{ 2+\sqrt { 2 } } $$
As triangle is isosceles right angle, so $$ AB=BC=a$$ and $$AC=\sqrt{2}a$$
The net electrostatic energy of the system is $$U=U_{AB}+U_{BC}+U_{AC}=0$$
or $$\dfrac{1}{4\pi\epsilon_0}(\dfrac{Qq}{a}+\dfrac{q^2}{a}+\dfrac{Qq}{\sqrt{2}a})=0$$
or $$Q(1+\dfrac{1}{\sqrt{2}})=-q$$
or $$Q=\dfrac{-\sqrt{2}q}{\sqrt{2}+1}=\dfrac{-2q}{2+\sqrt 2}$$

189433_125039_ans.png

Physics

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