Two small charged spheres are placed in vacuum on the $x$ -axis:
$+3.0 \mu \mathrm{C}$ at $x=0$ and $-5.0 \mu \mathrm{C}$ at $x=40 \mathrm{~cm} .$ Where must a third charge $q$ be placed if the force it experiences is to be zero?
The situation is represented in Fig. 24-4. We know that $q$ must be placed somewhere on the $x$ -axis. (Why?) Suppose that $q$ is positive. When it is placed in interval $B C$, the two forces on it are in the same direction and cannot cancel. When it is placed to the right of $C$, the attractive force from the $-5 \mu \mathrm{C}$ charge is always larger than the repulsion of the $+3.0 \mu \mathrm{C}$ charge. Therefore, the force on $q$ cannot be zero in this region. Only in the region to the left of $B$ can cancellation occur. (Can you show that this is also true if $q$ is negative?)
For $q$ placed as shown, when the net force on it is zero, we have $F_{E 3}=F_{E 5}$ and so, for distances in meters,
$$
k_{0} \frac{q\left(3.0 \times 10^{-6} \mathrm{C}\right)}{d^{2}}=k_{0} \frac{q\left(5.0 \times 10^{-6} \mathrm{C}\right)}{(0.40 \mathrm{~m}+d)^{2}}
$$
After canceling $q, k_{0}$, and $10^{-6} \mathrm{C}$ from each side, cross-multiply to obtain
$$
5 d^{2}=3.0(0.40+d)^{2} \text { or } d^{2}-1.2 d-0.24=0
$$
Using the quadratic formula,
$$
d=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}=\frac{1.2 \pm \sqrt{1.44+0.96}}{2}=0.60 \pm 0.775 \mathrm{~m}
$$
Two values, $1.4 \mathrm{~m}$ and $-0.18 \mathrm{~m}$, are therefore found for $d$. The first is the correct one; the second gives the point in $B C$ where the two forces have the same magnitude but do not cancel.