$(a)\;q=2\;Q\qquad(b)\;q=-\sqrt{2}\;Q\qquad(c)\;q=2 \sqrt{2}\;Q\qquad(d)\;q=-2 \sqrt{2}\;Q$

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Answer : (d) $\;q=-2 \sqrt{2}\;Q$

Explanation :

$F_{net} \;on\;q=\overrightarrow{F}_{1} +\overrightarrow{F}_{2} + \overrightarrow{F}_{3}$

$0=F \hat{i} + F \hat{j} + \overrightarrow{F}_{3}$

$0=\large\frac{k\;q\;Q}{a^2}\;(\hat{i} + \hat{j})+ \large\frac{k\;q^2}{(\sqrt{2} a)^2}\;\large\frac{\hat{i} + \hat{j}}{\sqrt{2}}$

$0=(\large\frac{k\;q\;Q}{a^2}+\large\frac{k\;q\;Q}{2 \sqrt{2}\;a^2})\;(\hat {i}+ \hat{j})$

$q\;Q+\large\frac{q^2}{2\;\sqrt{2}}=0$

$q=-2 \sqrt{2}\;Q\;.$

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