Answer
r = - 0.28 [m]
Work Step by Step
Remember that Net Force Must Be Zero;
$F_{net} = F_{2on1} - F_{1on2}$
0 = $F_{2on1} - F_{1on2}$
Transpose to the other side;
$F_{1on2} = F_{2on1}$
Expand into Coulomb's Law:
$\frac{kq_{1}q}{r^{2}} = \frac{kq_{2}q}{r^{2}}$
Replace r with (x) and (0.6-x)
$\frac{kq_{1}q}{x^{2}} = \frac{kq_{2}q}{(0.6-x)^{2}}$
We can cancel k and q on both sides.
$\frac{q_{1}}{x^{2}} = \frac{q_{2}}{(0.6-x)^{2}}$
Substitute known values
$\frac{2.5\times10^{-6}}{x^{2}} = \frac{3.5\times10^{-6}}{(0.6-x)^{2}}$
Cancel out $10^{-6}$ on both sides
$\frac{2.5}{x^{2}} = \frac{3.5}{(0.6-x)^{2}}$
Isolate x
$\sqrt \frac{(0.6-x)^{2}}{x^{2}}=\sqrt \frac{3.5}{2.5}$
Simplify
0.6 - x = 1.18x
2.18x = -0.6
x =-0.28 [m]
