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Q.

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

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a

$\frac{- q}{1 + \sqrt{2}}$

b

$\frac{- 2 q}{2 + \sqrt{2}}$

c

$- 2 q$

d

$+ q$

answer is B.

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Detailed Solution

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Net electrostatic energy should be zero.

$U = \frac{k Q q}{a} + \frac{k q^{2}}{a} + \frac{k Q q}{a \sqrt{2}} = 0$

$\Rightarrow \frac{k q}{a} (Q + q + \frac{Q}{\sqrt{2}}) = 0 \Rightarrow (Q + q + \frac{Q}{\sqrt{2}}) = 0 \Rightarrow Q (1 + \frac{1}{\sqrt{2}}) = -q \Rightarrow Q (\frac{\sqrt{2} + 1}{\sqrt{2}}) = -q⇒ Q = - q (\frac{\sqrt{2}}{1 + \sqrt{2}}) = - q (\frac{\sqrt{2}}{1 + \sqrt{2}}) \frac{\sqrt{2}}{\sqrt{2}} \Rightarrow Q = - \frac{2 q}{2 + \sqrt{2}}$

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