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

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Question

(A) The net electrostatic potential energy $U$ of a system of point charges is given by the sum of the potential energies of all pairs of charges: $U

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

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(A) The net electrostatic potential energy $U$ of a system of point charges is given by the sum of the potential energies of all pairs of charges: $U = \sum \frac{k q_i q_j}{r_{ij}}$.For the given configuration,the charges are $+Q$,$+2q$,and $+q$. The distances between them are $a$,$a$,and $\sqrt{a^2 + a^2} = a\sqrt{2}$.The total potential energy is:$U = \frac{1}{4 \pi \varepsilon_0} \left( \frac{Q(2q)}{a} + \frac{Q(q)}{a} + \frac{(2q)(q)}{a\sqrt{2}} ight) = 0$Since $\frac{1}{4 \pi \varepsilon_0} 
eq 0$,we have:$\frac{2Qq}{a} + \frac{Qq}{a} + \frac{2q^2}{a\sqrt{2}} = 0$$\frac{3Qq}{a} + \frac{\sqrt{2}q^2}{a} = 0$$3Qq + \sqrt{2}q^2 = 0$$3Q = -\sqrt{2}q$$Q = -\frac{\sqrt{2}}{3} q$

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

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