Answer
(a) $35cm$
(b) No. The forces would reverse the direction but would still balance.
Work Step by Step
(a) We know that
$x=\frac{d}{\sqrt{\frac{q_1}{q_2}-1}}$
We plug in the known values to obtain:
$x=\frac{10cm}{\sqrt{\frac{9.9\mu C}{5.1\mu C}}-1}$
$x=25.4cm$
Thus, the equilibrium distance from $q_2$ is $25.4cm$ and from $q_1$ is $10cm+25.4cm=35.4cm\approx 35cm$
(b) We know that the force will balance at this point whether $q_3$ is positive or negative. The sign of $q_3$ determines whether forces on it due to $q_1$ and $q_2$ are attractive or repulsive.
