# Electric Fields

1. Mar 11, 2008

### soul5

1. The problem statement, all variables and given/known data
A charge of 3.5*10^-6 is fixed on the x-axis at x= 0.55m, while a charge of -15*10^-6 is fixed at the origin. What is the net electric field on the x-axis at 0.8m?

2. Relevant equations
Fe = qE

3. The attempt at a solution
Used Fe = qE, but that's it.

2. Mar 11, 2008

### tiny-tim

Hi soul5!

How about using $$q/r^2$$?

3. Mar 11, 2008

### Mike Cookson

Apply the principal of super position. The field at any point is the sum of the two individual fields at that same point when acting independently...IE work out the individual fields at x=0.55m and then add them together.