Answer
$0.254m$
Work Step by Step
We can find the required distance $d$ as follows:
$\Sigma F_x=-Kx+\frac{KQq}{(d-x)^2}=0$
This simplifies to:
$d-x=\pm\sqrt{\frac{KQq}{Kx}}$
$\implies d=x\pm\sqrt{\frac{KQq}{Kx}}$
We plug in the known values to obtain:
$d=0.124\pm\sqrt{\frac{(8.99\times 10^9)(8.55\times 10^{-6})(2.44\times 10^{-6})}{89.2}(0.124)}$
$d=0.254m$ or $-0.006m$
Given that $d\gt0$, $-0.006m$ is not possible.
Thus $d=0.254m$
