# A charge Q is placed at each of the two opposite corners of a square . A charge q is placed at each of the other two corners . If the resultant force on each charge q is zero , then

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

Answer : (d) $\;q=-2 \sqrt{2}\;Q$
Explanation :
$F_{net} \;on\;q=\overrightarrow{F}_{1} +\overrightarrow{F}_{2} + \overrightarrow{F}_{3}$
$0=F \hat{i} + F \hat{j} + \overrightarrow{F}_{3}$
$0=\large\frac{k\;q\;Q}{a^2}\;(\hat{i} + \hat{j})+ \large\frac{k\;q^2}{(\sqrt{2} a)^2}\;\large\frac{\hat{i} + \hat{j}}{\sqrt{2}}$
$0=(\large\frac{k\;q\;Q}{a^2}+\large\frac{k\;q\;Q}{2 \sqrt{2}\;a^2})\;(\hat {i}+ \hat{j})$
$q\;Q+\large\frac{q^2}{2\;\sqrt{2}}=0$
$q=-2 \sqrt{2}\;Q\;.$
in 3rd step ,value of f3 = kq^2/(√2a^2)×(i+j)/√2  ; where does it come from
plz explain