چون سوال نسبتی است برای راحتی کار: $x = 3cm$ و ${q_1} =  + 1\mu c$

$\left. \begin{gathered}
  {F_{21}} = \frac{{90 \times 2 \times 1}}{{3 \times 3}} = 20N \hfill \\
  {F_{31}} = \frac{{90 \times 4 \times 1}}{{9 \times 9}} = \frac{{40}}{9}N \hfill \\ 
\end{gathered}  \right\}{F_1} = 20 - \frac{{40}}{9} = \frac{{140}}{9}N$

$\left. \begin{gathered}
  {F_{23}} = \frac{{90 \times 2 \times 4}}{{6 \times 6}} = 20N \hfill \\
  {F_{13}} = {F_{31}} = \frac{{40}}{9}N \hfill \\ 
\end{gathered}  \right\}{F_3} = 20 + \frac{{40}}{9} = \frac{{220}}{9}N$

در نتیجه:

$\frac{{{F_1}}}{{{F_3}}} = \frac{{\frac{{140}}{9}}}{{\frac{{220}}{9}}} = \frac{{14}}{{22}} = \frac{7}{{11}}$