# Electric forces and fields

1. Nov 5, 2004

### nemzy

Two 2.06e-6 C point charges are located on the x axis. One is at 1.06 M and the other is at -1.06 M

(a) Determine the electric field on the y axis at .48 M

This is how i did it:

Etotal=sum of E vectors...

E=ke*q/r^2

r^2=1.06^2+.48^2

q=2.06e-6

E1 = Ke*q/(1.06^2+.48^2)
E2= same thing

E1+E2 = total..
answer is wrong..what am i doing wrong

2. Nov 5, 2004

Both charges are positive. The electric field equation is given by

Kq/r^2 * r-hat, which is a unit vector pointing radially away from the charge.

If you draw a picture of the two e-fields, you'll see that the components don't add exactly (in fact, the x components cancel out), so you need to take that into account.

3. Nov 5, 2004

### nemzy

yeah when i was trying to do this i just realized that too..but i still dont nkow how to find the electric field

4. Nov 5, 2004

You should ask yourself: what is the unit vector, r-hat, that points from the point charge to the coordinate in consideration?

IE, for one point charge, what is the unit vector that points from (1.06, 0) to (0, 0.48)? Knowing this, you can express the unit vector r, in terms of cartesian coordinates, and you should see, that when you add the E-fields together, the x-hat components cancel.