Finding electric potential where electric field is zero

[cmkluza]

Homework Statement

A -10.0 nC point charge and a +20.0 nC point charge are 15.0 cm apart on the x-axis.
What is the electric potential at the point on the x-axis where the electric field is zero?

Homework Equations

$E = k\frac{Q}{r^2}$
$V = k\frac{Q}{r}$

The Attempt at a Solution

I've set up my problem as follows:

I'm looking for the $r$ such that $E = E_1 + E_2 = 0$. Setting up my equations gets me:
$$k\frac{(-10\times10^{-9})}{r^2} + k\frac{(20\times10^{-9})}{(0.15 - r)^2} = 0$$
$$\frac{20}{(0.15 - r)^2} = \frac{10}{r^2}$$
I used Wolfram to skip some algebra and got $r = -\frac{3}{20}(1 + \sqrt{2})$ and $r = \frac{3}{20}(\sqrt{2}-1)$. Given that there has to be an electric field between the two point charges (unless I really need to brush up on my physics), I use the negative value for $r$.
I set up the potential and try to solve as follows:
$$V = k\frac{(-10\times10^{-9})}{-\frac{3}{20}(1 + \sqrt{2})} + k\frac{(20\times10^{-9})}{(0.15 + \frac{3}{20}(1 + \sqrt{2}))}$$
I've done this a few times, even with some different numbers, and I keep getting the same answer of 599.2, but this is incorrect. What am I missing/doing wrong?

 

[gneill]

Consider whether the math takes into account the change in field direction when it "passes through" one of the charges. If not, what can you do to avoid the problem?

 

[rude man]

I used Wolfram to skip some algebra and got $r = -\frac{3}{20}(1 + \sqrt{2})$ and $r = \frac{3}{20}(\sqrt{2}-1)$. Given that there has to be an electric field between the two point charges (unless I really need to brush up on my physics), I use the negative value for $r$.

EDIT: sorry, I didn't note your valid argument about r having to be <0. So you picked the right r < 0. But what I said below holds.
Also, when computing potentials, distance is always +.

 

[haruspex]

why did you pick a negative value for r?

For the excellent reason cmkluza gave.

when computing potentials, distance is always +.

That is indeed the error.

Consider whether the math takes into account the change in field direction when it "passes through" one of the charges.

Despite cmkluza's first equation, the second equation has equated the magnitudes of the fields, which handles that issue at the expense of introducing the spurious positive root.