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

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

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

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-2q

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Solution

The correct option is A $\frac{- 2 q}{2 + \sqrt{2}}$
Net electrostatic energy $= 0$
$U_{Q q} + U_{q q} + U_{Q q} = 0$

$∴ \frac{Q q}{a} + \frac{q^{2}}{a} + \frac{Q q}{a \sqrt{2}} = 0 o r Q \sqrt{2} + q \sqrt{2} + Q = 0$
$Q (\sqrt{2} + 1) = - q \sqrt{2} o r Q = \frac{- 2 q}{(2 + \sqrt{2})}$

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

The correct option is A −2q2+√2Net electrostatic energy=0UQq+Uqq+UQq=0∴Qqa+q2a+Qqa√2=0orQ√2+q√2+Q=0Q(√2+1)=−q√2orQ=−2q(2+√2)

The correct option is A $\frac{- 2 q}{2 + \sqrt{2}}$
Net electrostatic energy $= 0$
$U_{Q q} + U_{q q} + U_{Q q} = 0$

$∴ \frac{Q q}{a} + \frac{q^{2}}{a} + \frac{Q q}{a \sqrt{2}} = 0 o r Q \sqrt{2} + q \sqrt{2} + Q = 0$
$Q (\sqrt{2} + 1) = - q \sqrt{2} o r Q = \frac{- 2 q}{(2 + \sqrt{2})}$