Answer
$Q_5=-6.1\times10^{-6}C$; Stable equilibrium.
Work Step by Step
$|F_{12}|=|F_{32}|=k\frac{(6.4\times10^{-6}C)^2}{(0.092m)^2}=43.55N$
$|F_{42}|=k\frac{(6.4\times10^{-6}C)^2}{(0.13m)^2}=21.78N$
$F_{2x}=F_{2y}=|F_{32}|+|F_{42}|\sin(45^o)=58.95N$
$\sum F_2=\sqrt{(F_{2x})^2+(F_{2y})^2}=83.37N$
$|F_{52}|=k\frac{(6.4\times10^{-6}C)Q_5}{(0.0651m)^2}=83.37N$
$Q_5=-6.1\times10^{-6}C$
Because the forces on all charges are balanced, these 5 charges are in equilibrium. Assuming there are no external forces or charges, if this arrangement is slightly disturbed, the forces on the charges will act to return them back to this original position. Therefore, this is stable equilibrium.
