Answer
Place $\mathrm{q}_{2} \ \ \ 0.12\mathrm{m}$ to the right of $\mathrm{q}_{1}.$
Work Step by Step
Apply $F =k\displaystyle \frac{|q_{1}||q_{2}|}{r^{2}},\qquad (19- 5)$
Take +x to be righward.
Let x be the distance between $\mathrm{q}_{1}$ and $\mathrm{q}_{2}.$
$\mathrm{q}_{1}$ acts with a leftward attracting force on $\mathrm{q}_{2}$,
$\mathrm{q}_{3}$ acts with a rightward attracting force on $\mathrm{q}_{2}.$
The two forces are in opposite directions, and we want them to cancel each other, so the net force is zero.
$k\displaystyle \frac{|q_{1}||q_{2}|}{x^{2}}=k\frac{|q_{2}||q_{3}|}{(0.32\mathrm{m}-x)^{2}}$
$|q_{1}|(0.32\mathrm{m}-x)^{2}=|q_{3}|\mathrm{x}^{2}$
$(q)(0.32\mathrm{m}-x)^{2}=(3.0q)x^{2}\qquad /\sqrt{...}$
$0.32\mathrm{m}-x=x(\pm\sqrt{3.0})$
$0.32\mathrm{m}=x(1.0\pm\sqrt{3.0})$
$\displaystyle \mathrm{x}=\frac{0.32\mathrm{m}}{1.0\pm\sqrt{3.0}}$
We obtain two values, $\mathrm{x}=-0.45\mathrm{m}$ and $\mathrm{x}=0.12\mathrm{m}.$
Since the negative value would place $\mathrm{q}_{2}$ left of $\mathrm{q}_{1}$, we discard it.
