# Calculating the electric field

1. Jan 27, 2007

### symmet

Here's the problem:

Three charges are at the corners of an equilateral triangle as in the figure below:

.............(+)q_1
............/...\
.........../.....\ L
........../.......\
...q_0(+)____(-)q_2

L = .06 m
q_0 = 1.8*10^-6 C
q_1 = 6.6*10^-6 C
q_2 = -4.6*10^-6 C

Calculate the electric field at the position of q_0 (the origin) due to q_1 and q_2.
Answer is a vector in terms of i and j.

My approach was to break down q_1 and q_2 into vector components. For example, to find the horizontal components, I did:

E_x = k(q/x^2)

where x is the horizontal distance from position q_0 and q is the charge. (k = 8.99*10^9)

So, plugging this in for both q_1 and q_2, I got:

q_1: E_x = -659266.67 N/C
q_2: E_x = 114872.22 N/C

I put a - on q_1 because of the direction of the vector at q_0. Adding these together I get -544394.45 for the total horizontal component. But this isn't right. Can anyone tell me where I've gone wrong in my thought process? Thanks.

Last edited: Jan 27, 2007
2. Jan 28, 2007

### mace2

Do you know what the answer is?

Umm... well, it's a vector, so you would need a cosine for your horizontal q_1 field at point q_0.

3. Jan 28, 2007

### symmet

It's online homework, so no I don't know what it is.

Well since it is an equilateral triangle, all sides are the same length L and all angles are 60 degrees. I know that the q_1 charge is L/2 away from q_0 horizontally because it is equilateral (and Lcos60 = L/2). That is what would be substituted for x in the equation.