Three charges Q, +q and +q are placed at the vertices of the right...

The net electrostatic potential energy of a system of charges is the sum of the potential energies of all pairs of charges.Let the charges be placed at the vertices of the right-angled isosceles triangle. From the given figure, the charges are $Q$, $+q$, and $+q$.The distances between the pairs of charges are:1. Distance between $Q and +q$ (vertical side) $= a$2. Distance between $+q and +q$ (horizontal side) $= a$3. Distance between $Q and +q$ (hypotenuse) $= \sqrt{a^2 + a^2} = a\sqrt{2}$The total electrostatic potential energy $U$ of the system is given by:$U = \dfrac{1}{4\pi\epsilon_0} \left( \dfrac{Q \cdot q}{a} + \dfrac{q \cdot q}{a} + \dfrac{Q \cdot q}{a\sqrt{2}} \right)Given that the net electrostatic energy of the configuration is zero, we have$U = 0$:\dfrac{1}{4\pi\epsilon_0} \left( \dfrac{Qq}{a} + \dfrac{q^2}{a} + \dfrac{Qq}{a\sqrt{2}} \right) =0$Dividing the entire equation by $\dfrac{q}{4\pi\epsilon_0 a}$ (assuming $q ≠ 0$), we get:$Q + q + \dfrac{Q}{\sqrt{2}} =0$Grouping the terms with $Q$:$Q \left(1 + \dfrac{1}{\sqrt{2}} \right) +$q = 0 Q \left( \dfrac{\sqrt{2} + 1}{\sqrt{2}} \right) =$-$q Q = \dfrac{-\sqrt{2}q}{\sqrt{2} + 1}$Answer:$\dfrac{-\sqrt{2}q}{1+\sqrt{2}}