# Net Electric Field of 2 charged particles fixed to X-axis

1. Sep 22, 2010

### Oaksmack

2 Particles are fixed to an x-axis:
q1=2.1*10^-8 C @ x=.2m
q2=-4.00q1 @ x=.7m

We are trying to find the coordinate on the x-axis where the net electric field produced by the particles = 0.
The equation I have is

E=((8.99*10^9)q)/r^2

I got what i thought was q2 first, by multiplying 2.1*10^-8 by -4.00 to get -8.4*10^-8. From there I equated E1 & E2:
((8.99*10^9)(2.1*10^-8))/r^2 = ((8.99*10^9)(-8.4*10^-8))/r^2
from there, I had no idea how to solve for r.
I simplified the constant 8.99*10^9 with the respective q's, but just couldn't figure out how to solve for r without it canceling into 1 and having two numbers equaling each other.

2. Sep 22, 2010

### Awatarn

Almost correct except $$r$$ in your equation
$$r$$ is a distance from the charge you are considering. So each $$r$$ on the left side and right side of your equation should not be equal. One should be include the distance between two point charges.

3. Sep 23, 2010

### Oaksmack

Wait, so one of the r's should be .5m, the distance between the two particles?