Answer
$R_{13}$ = - 0.144 m
Work Step by Step
First, calculate the net force $F_{1}$ exerted by $F_{2on1}$ and $F_{3on1}$, so
$F_{1}$=$F_{2on1}$-$F_{3on1}$
Transpose to find $F_{3on1}$
$F_{3on1}$=$F_{2on1}$-$F_{1}$
Remember $F_{1}$ is given, = 7 [N]
Hence,
$\frac{kq_{1}q_{3}}{r_{1on3}^{2}}$ = $\frac{kq_{2}q_{1}}{r_{2on1}^{2}}$ + 7
Add the right-hand side,
$\frac{q_{1}q_{3}}{r_{1on3}^{2}}$ = $\frac{(9x10^{9})(5x10^{-6})(3x10^{-6})}{0.200^{2}}$ + 7
$\frac{kq_{1}q_{3}}{r_{1on3}^{2}}$ = 10.375
Isolate $r_{1on3}$
$r_{1on3}$ = $\sqrt \frac{kq_{1}q_{3}}{10.375}$
Substitute known terms
$r_{1on3}$ = $\sqrt \frac{(9x10^{9})(3x10^{-6})(8x10^{-6})}{10.375}$
Hence,
$R_{13}$ = - 0.144 m
Note that the distance is negative due to its reference with regards to the x-axis.
