# Electric Field - Three Charges

## Homework Statement

Three charges, q1 = 8.59nC, q2 = -5.67nC and q3 = 2.98nC are at the corners of an equilateral triangle, as shown in the figure below.

The angle alpha is 60.0° and L = 0.537 m. We are interested in the point midway between the charges q1 and q2 on the x axis.
For starters, calculate the magnitude of the electric field due only to charge q3 at this point

## Homework Equations

F = k (qa)(qb)/r^2
E = k (q)/r^2
E = F/q0

## The Attempt at a Solution

Okay so Im at an utter loss... I thought at first you should find all the Forces from each point on 3. using the first equation. And so well I did that. Then I took those forces and used the angle of 60 degrees to get them all going in the same direction that way I could get the net force. After gaining the net force I used the x and the y to find the magnitude of the net force. Then I used the 3rd equation listed. so I divided the F by q3 and recieved an answer. But I feel like my thought process is wrong. I don't have a scanner or anything. Or I would upload my hand written work. But I typed each step that I took.

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AEM
A couple of things are not clear in your problem statement. What is it you are asked to find? Since you didn't provide a picture, you need to describe the arrangement of the charges a little better. Are q1 and q2 on the x axis while q3 is above them on the y axis? Is L the length of the sides of the equilateral triangle?

The question asked is to calculate the magnitude of the electric field due only to charge q3 at this point? This is the last statement of the problem. In the picture provided (I tried to copy and paste but it didn't work.) Q1 is touching at the origin and Q2 is a little to the right on the x-axis. Since the triangle is an equilateral triangle Q3 is half way between them to the right off the y-axis in quadrant I. The length of Side Q3 to Q2 is .537m. I made the assumption that all the sides were this long now because its an equilateral triangle. Also Each angle of the triangle is 60 degrees or at least should be since its an equilateral triangle. Sorry if I wasn't clear before.

Thanks for the help AFM but I was able to figure out this problem on my own after looking at it for a little while. I messed up on a negative sign and it threw all of my numbers off.

AEM
Thanks for the help AFM but I was able to figure out this problem on my own after looking at it for a little while. I messed up on a negative sign and it threw all of my numbers off.
Good for you for figuring it out!