# Electric Fields and source charge

1. Feb 9, 2010

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

The source charge (the magenta circle at the origin) is a positive charge. A particle whose charge is -6e-09 C is placed at location D.

1. The electric field at location D has the value < -8000, 8000, 0 > N/C. What is the unit vector in the direction of E at this location?

2. What is the electric force on the -6e-09 C charge?

3. What is the unit vector in the direction of this electric force?

2. Relevant equations
E = kq/r^2

3. The attempt at a solution

1. E = sqrt ( -8000^2 + 8000^2 + 0^2) = 11300
tan^-1(-8000/8000) = 135 degrees

11300=(8.99E9)(6E-9) / r^2
r=0.07 m
r(x)=0.07cos(135) = -0.05 m
r(y)=0.07sin(135) = 0.05 m

This is wrong. Can anyone show me how to do it? I assume you can't do #2 without knowing the value of #1, and you can't do #3 without knowing the value of #2? What exactly is question #2 and #3 asking?

2. Feb 9, 2010

### cepheid

Staff Emeritus
Here's the problem. You were trying to use the definition of the electric field to calculate the distance D must be from the origin, but you used the value of the test charge instead of the value of the source charge.

Hint: The value of the source charge is not given because you don't need it. You can figure out what the unit vector is without knowing how far D is from the origin. All you have to do is calculate E/|E|.

Hint 2: For part three, there is a very simple relationship between the direction of the electric field and the direction of the force on the test charge. Their directions are either the same (in which case the unit vectors are the same), or they are exactly opposite (in which case the unit vectors are antiparallel.) Which one is it in this situation?