Three charges Q,(-2q) and (-2q) are placed at | vedclass

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(C) The electrostatic potential energy $U$ of a system of point charges is given by $U = \sum \frac{k q_i q_j}{r_{ij}}$.For the given system,the charg

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

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(C) The electrostatic potential energy $U$ of a system of point charges is given by $U = \sum \frac{k q_i q_j}{r_{ij}}$.For the given system,the charges are $q_1 = Q$,$q_2 = -2q$,and $q_3 = -2q$. The distances between them are $r_{12} = l$,$r_{13} = l$,and $r_{23} = \sqrt{l^2 + l^2} = l\sqrt{2}$.The total potential energy is:$U = \frac{k Q(-2q)}{l} + \frac{k Q(-2q)}{l} + \frac{k (-2q)(-2q)}{l\sqrt{2}} = 0$Dividing by $k$ and simplifying:$-\frac{2Qq}{l} - \frac{2Qq}{l} + \frac{4q^2}{l\sqrt{2}} = 0$$-\frac{4Qq}{l} + \frac{4q^2}{l\sqrt{2}} = 0$$\frac{4Qq}{l} = \frac{4q^2}{l\sqrt{2}}$$Q = \frac{q}{\sqrt{2}}$Thus,the correct option is $C$.

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

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