# Electric field problem

1. Feb 13, 2006

### Punchlinegirl

An equilateral triangle with sides 66 m has charges +5 $$\mu C$$, -5 $$\mu C$$, and -9.3 $$\mu C$$ at its vertices and charge + 18.6 $$\mu C$$ at the midpoint on its side between the vertices with charges $$+5 and -5 \mu C$$. What is the magnitude of the x-component of the electric field at point P, the point forming a second equilateral triangle with +5 $$\mu C and -5 \mu C$$? Answer in units of N/C.
Sorry I don't have a picture but it's 2 equilateral triangles that look like a mirror image of each other.
First I drew a picture of what the electric fields would look like. The -9.3 and 18.6 $$\mu C$$ would not have a y-component, so I ignored those for now.
so I found that
k * 5 x 10^-6/ 66^2 * 33/66 = 5.17
and k * -5 x 10^-6 / 66^2 *57.2/66 = 8.95 (since it's absolute value of the charge)
So E_x= 14.12... which isn't right.. help please?

2. Feb 14, 2006

### Punchlinegirl

Ok I figured out what I was doing wrong... the second part of the problem asks for the magnitude of the y-component of the electric field at point P.
E_1 = k * 5 x 10^-6 / 66^2 * 57.2/66 = 8.93
E_2 = 8.93 (same set up as E_1)
E_3 = k * 18.6 x 10^-6 / 57.2^2 = -51.2, since the field would be downward
E_4 = k * 9.3 x 10^-6 / 114.4^2 = 6.40
8.93+8.93-51.2+6.40= -26.94
This isn't right.. can someone help me?