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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?

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1 Answer 1

up vote 5 down vote accepted

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 $r = -\tfrac{1}{3}$.

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I suspected we were searching for some squares, but I didn't think of square my equation because I thought I was lacking a more fundamental step –  mattecapu May 17 '14 at 7:54
We're all so used to writing $V \propto 1/r$ that it's very easy to forget it's actually $V \propto 1/|r|$ :-) –  John Rennie May 17 '14 at 7:59
I think it's because $r$ is considered geometrically (thus always positive), while now we talk about it as a signed number, a point in a Cartesian plane. Does it make sense? –  mattecapu May 17 '14 at 8:54
@mattecapu Yes, you are right. –  Awesome May 17 '14 at 9:03
However, thank you very much for your detailed answer! I really appreciated it –  mattecapu May 17 '14 at 11:42

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