Let's say we have two charges called $q_1$ and $q_2$, respectively $20 \, C$ and $-40\,C$, at a distance $d=1\,m$
We want to find all the points where electric potential is null.
I solved the equation
$$\frac{q_1}{4\pi\epsilon_0r_1} + \frac{q_2}{4\pi\epsilon_0(d-r_1)}=0$$
For $r_1$ (distance from $q_1$), and found $r_1=\frac13\,m$
However this is not the only solution: there is another point not in-between the charges, but $1\,m$ left from $q_1$ ($r_1=-1\,m$)
How can I set up an equation giving me both the solutions?
Best Answer
Let's draw the setup:
The expression for $V(r)$ is simply (I'll set $\kappa$ to 1 for convenience):
$$ V(r) = \frac{20}{r} - \frac{40}{r+1} $$
so the potential is zero when:
$$ \frac{20}{r} = \frac{40}{r+1} $$
Only this isn't quite right because the potential for each charge is symmetric so the potential due to charge $A$ obeys $V_A(r) = V_A(-r)$ and likewise for the other charge. It's because you're ignoring this that your equation gives you only one null point. The equation really should be:
$$ \frac{20}{|r|} = \frac{40}{|r+1|} $$
The easy way to deal with those modulus operators is to square both sides:
$$ \frac{400}{r^2} = \frac{1600}{(r+1)^2} $$
and if we rearrange this we get the quadratic equation:
$$ 3r^2 - 2r - 1 = 0 $$
Quadratic equations have two roots, and the two roots are going to give you the two null points. If we use the usual expression for the roots of a quadratic equation we get $r = 1$ and $r = -\tfrac{1}{3}$.