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

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(A) The 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 = \s

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$(-q/2)$

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(A) The 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 system of three charges $Q, q, q$ at the vertices of an equilateral triangle of side $l$,the pairs are $(Q, q)$,$(Q, q)$,and $(q, q)$.The total potential energy is:$U = \frac{kQq}{l} + \frac{kQq}{l} + \frac{kq^2}{l} = 0$Factor out $\frac{kq}{l}$:$\frac{kq}{l} (Q + Q + q) = 0$Since $\frac{kq}{l} 
eq 0$,we have:$2Q + q = 0$$2Q = -q$$Q = -\frac{q}{2}$

Three charges $Q, (+q)$ and $(+q)$ are placed at the vertices of an equilateral triangle of side $l$ as shown in the figure. If the net electrostatic potential energy of the system is zero,then $Q$ is equal to

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