JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Electrostatics

Take the two equal legs of the right-angled isosceles triangle to be of length $a$, so the hypotenuse is $a\sqrt{2}$. Place $Q$ at one end of the hypotenuse: it then sits a distance $a$ from the $+q$ at the right-angle vertex and a distance $a\sqrt{2}$ from the $+q$ at the far vertex, while the two $+q$ charges are a distance $a$ apart. The total energy is $U=\frac{kQq}{a}+\frac{kQq}{a\sqrt{2}}+\frac{k{q}^2}{a}$. Setting $U=0$ and cancelling $kq/a$ gives $Q\left(1+\frac{1}{\sqrt{2}}\right)+q=0$, so $Q=-\frac{q}{1+\frac{1}{\sqrt{2}}}=-\frac{\sqrt{2}q}{\sqrt{2}+1}$.