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The charges Q, +q and +q are placed at the vertices of an equilateral triangle of side $l$ . If the net electrostatic potential energy of the system is zero, then Q is equal to

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zero

$- \frac{q}{2}$

$- q$

$\frac{+ q}{2}$

answer is A.

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Detailed Solution

$k \frac{Q q + q^{2} + Q q}{a} = 0$

$2 Q q = - q^{2}$

$Q = - q / 2$

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