**1. The problem statement, all variables and given/known data**
A +2.0-nC charge is at the origin and a -4.0-nC charge is at x = 1.0-cm.

At what x-coordinate could you place a proton so that is would experience no net force?

[NOTE by me] I put q

_{1} as the positive charge (+2.0nC) and q

_{2} as the negative charge (-4.0nC). q

_{0}, of course, is the proton.

**2. Relevant equations**
- quadratic formula

- F

_{0} = F

_{01} - F

_{02} which becomes F

_{01} = F

_{02} when I replace F

_{0} with zero (no net force).

**3. The attempt at a solution**
So I know that the proton must be either to the left of the positive charge or the right of the negative charge. It can't be in between these two charges for it to have zero net force.

I plugged in numbers into this equation (F

_{01} = F

_{02}) which became:

(2 x 10

^{-9})/(r

^{2}) = (4 x 10

^{-9}) / ((1.0 x 10

^{-2}) + r)

^{2}
I simplified this and got:

0 = r

^{2} + (2 x 10

^{-2})r - (1 x 10

^{-4})

Then I solved for r using quadratic formula and got:

r = (+ 4.1421 x 10

^{-3}) or (- 2.4142 x 10

^{-2}).

I thought the negative number would automatically cancel out because r is distance but the actual answer goes with the negative number. I'm confused at how cramster.com explained it.

The book is called Physics for Scientists and Engineers A Strategic Approach (2nd) by Randall Dewey Knight) and its chapter 26 number 47. Here's a direct link (have to be a member though; it's free to look at odd numbers):

http://physics.cramster.com/physics-...-9-718769.aspx
Thanks.