Three charges Q, (-2q) and (-2q) are placed at the vertices of an isosceles right angled triangle as shown in figure. The net electrostatic potential energy is zero if Q is equal to

The electrostatic potential energy of the system is the sum of pairwise interactions between the three charges. Let $Q_1 = Q$, $Q_2 = -2q$, and $Q_3 = -2q$ be placed at the vertices of an isosceles right triangle with legs of length $l$.

For charges separated by distance $r$, the potential energy is $U = k\frac{q_a q_b}{r}$, where $k = \frac{1}{4\pi\epsilon_0}$. The pairwise distances are:

$r_{12} = l$, $r_{13} = l$, and $r_{23} = l\sqrt{2}$.

The total potential energy is:

$U_{\text{total}} = k\left(\frac{Q(-2q)}{l} + \frac{Q(-2q)}{l} + \frac{(-2q)(-2q)}{l\sqrt{2}}\right)$

Simplifying:

$U_{\text{total}} = k\left(-\frac{4Qq}{l} + \frac{4q^2}{l\sqrt{2}}\right)$

Setting $U_{\text{total}} = 0$ and solving:

$-\frac{4Qq}{l} + \frac{4q^2}{l\sqrt{2}} = 0$

$\frac{4Qq}{l} = \frac{4q^2}{l\sqrt{2}}$

$Q = \frac{q}{\sqrt{2}}$

The net electrostatic potential energy is zero when $Q = \frac{q}{\sqrt{2}}$.