Three point charges Q, +2q and +q are placed | vedclass

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(B) 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 configuration, t

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$-\frac{\sqrt{2}}{3}q$

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(B) 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 configuration, the charges are $Q$, $+2q$, and $+q$. The sides of the triangle are $\sqrt{2}a$, $\sqrt{2}a$, and the hypotenuse is $\sqrt{(\sqrt{2}a)^2 + (\sqrt{2}a)^2} = \sqrt{2a^2 + 2a^2} = \sqrt{4a^2} = 2a$.The potential energy is $U = \frac{1}{4\pi\epsilon_0} [\frac{Q(2q)}{\sqrt{2}a} + \frac{Q(q)}{\sqrt{2}a} + \frac{(2q)(q)}{2a}] = 0$.Multiplying by $4\pi\epsilon_0$ and simplifying:$\frac{2Qq}{\sqrt{2}a} + \frac{Qq}{\sqrt{2}a} + \frac{2q^2}{2a} = 0$$\frac{3Qq}{\sqrt{2}a} + \frac{q^2}{a} = 0$$\frac{3Qq}{\sqrt{2}} = -q^2$$Q = -\frac{\sqrt{2}q}{3}$.

Three point charges $Q$, $+2q$ and $+q$ are placed at the vertices of a right-angled isosceles triangle of side length $\sqrt{2}a$. The net electrostatic potential energy of the configuration is zero, if $Q$ is equal to

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