Answer
-4.00
Work Step by Step
According to Coulomb's law,
$F_{31}=\frac{1}{4\pi\epsilon_{0}}\frac{q_{3}q_{1}}{(L_{12}+L_{23})^{2}}=\frac{1}{4\pi\epsilon_{0}}\frac{q_{3}q_{1}}{(2L_{23})^{2}}$ and
$F_{32}=\frac{1}{4\pi\epsilon_{0}}\frac{q_{3}q_{2}}{L_{23}^{2}}$
Given: $F_{31}+F_{32}=0$
Or $F_{31}=-F_{32}$
$\implies \frac{1}{4\pi\epsilon_{0}}\frac{q_{3}q_{1}}{4.00\times L_{23}^{2}}=-\frac{1}{4\pi\epsilon_{0}}\frac{q_{3}q_{2}}{L_{23}^{2}}$
which gives $\frac{q_{1}}{4.00}=-q_{2}$
Or $\frac{q_{1}}{q_{2}}=-4.00$
