Answer
$F_{1}$ = $(1.8\times10^{-4} N)$ in positive x-direction.
Work Step by Step
$F_{3 on 1}$ = $9.0\times10^{9}\times\frac{5.0\times10^{-9}\times1.0\times10^{-9}}{(0.02)^{2}}$
$F_{3 on 1}$ = $1.125\times10^{-4} N$ in positive x-direction ($q_{1}$ is repelled by $q_{3}$)
$F_{2 on 1}$ = $9.0\times10^{9}\times\frac{3.0\times10^{-9}\times1.0\times10^{-9}}{(0.02)^{2}}$
$F_{2 on 1}$ = $6.75\times10^{-5} N$ in positive x-direction ($q_{1}$ is attracted by $q_{2}$)
$F_{1}$ = $F_{3 on 1}$ + $F_{2 on 1}$
$F_{1}$ = $(1.8\times10^{-4} N)$ in positive x-direction.
