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 energy of the system is zero, then $Q$ is equal to
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(a) Potential energy of the system
$U = k\frac{{Qq}}{l} + \frac{{k{q^2}}}{l} + \frac{{kqQ}}{l} = 0$
$==>$ $\frac{{kq}}{l}(Q + q + Q) = 0$ $==>$ $Q = - \frac{q}{2}$
$U = k\frac{{Qq}}{l} + \frac{{k{q^2}}}{l} + \frac{{kqQ}}{l} = 0$
$==>$ $\frac{{kq}}{l}(Q + q + Q) = 0$ $==>$ $Q = - \frac{q}{2}$
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