An electric point charge of Q1 = 8.91 nC is placed at the origin of the real axis. Another point charge of Q2 = 2.15 nC is placed at a position of p = 2.52 m on the real axis. At which position can a third point charge of q = -9.56 nC be placed so that the net electrostatic force on it is zero?
Let the sign of Q2 be changed from positive to negative. At which position can the point charge q be placed now so that the net electrostatic force on it is zero?
For the first part, I got 1.69m which is correct, I just cannot get a real number as an answer for the second part. Any help would be greatly appreciated!
You probably just made some calculation error.
For the second case where $Q_1 > 0$ and $Q_2 < 0$, intuitively the electric force is balanced on the right hand side where it's closer to the weaker charge at $p_2 = 2.52$ and further away from the stronger charge $p_1 = 0$ at the origin.
$$ \frac{Q_1}{x^2} - \frac{|Q_2|}{(x - 2.52)^2} = 0$$
The magnitude of the test charge $q$ doesn't matter. The solution is $x \approx 4.953$, necessarily unique just like the first case.
The top figure below is for the first case $Q_2 > 0$ you solved, and the lower two figures are for the $Q_2 = -2.15$, where the third (bottom) figure is a zoom-in.
The blue curves are the electric potential and the yellow curves are the electric force (positive meaning pointing towards $+\hat{x}$.
In all figures, the desired position is where the yellow curve intersect with the $x$-axis.