# Finding electric potential where electric field is zero

1. Feb 24, 2017

### cmkluza

1. The problem statement, all variables and given/known data
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?

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

3. 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?

2. Feb 24, 2017

### Staff: Mentor

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?

3. Feb 26, 2017

### rude man

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 +.

Last edited: Feb 26, 2017
4. Feb 26, 2017

### haruspex

For the excellent reason cmkluza gave.
That is indeed the error.
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.